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# Algebra Symbols
A comprehensive collection of 225+ symbols used in algebra, categorized by subject and type into tables along with each symbol's name, usage and example.
Algebra is a subfield of mathematics pertaining to the manipulation of symbols and their governing rules. The following is a compilation of symbols from the different branches of algebra, which include basic algebra, number theory, linear algebra and abstract algebra.
For readability purpose, these symbols are categorized by their function and topic into charts and tables. Other comprehensive lists of symbols — as categorized by subject and type — can be also found in the relevant pages below (or in the navigational panel).
Get the master summary of mathematical symbols in eBook form — along with each symbol’s usage and LaTeX code.
## Constants
In algebra, constants are symbols used to denote key mathematical elements and sets. The following tables document the most common of these — along with each symbol’s name, usage and example.
(For common constants in general, see common math constants.)
### Key Mathematical Sets
In algebra, certain sets of numbers (or other more elaborated objects) tend to occur more frequently than others. These sets are often denoted by some variants of alphabetical letters — many of which are of the blackboard bold typeface.
## Variables
Since algebra is concerned with the manipulation of mathematical symbols, it often draws upon a wide range of variables as placeholders for varying objects and quantities. The following table documents the most common of these — along with their respective usage and example.
## Delimiters
In mathematics, delimiters are symbols used to denote the separation between independent mathematical entities. The following table features some of the most common delimiters in algebra. For common delimiters in general, see common delimiters.
## Function-related Symbols
As a foundational component of algebra, function plays a key role in establishing the rules pertaining to the manipulation of symbols. The following table documents some of the most common function-related operators and notational symbols — along with their meaning and example.
## Operators
In algebra, operators can be thought of as a special type of function mapping one or multiple mathematical entities to another, and are often given special names or notations due to their repeated occurrences.
In particular, these operators are often related to numbers, key functions, linear algebra and abstract algebra — the vast majority of which are found in the tables below. For common operators in general, see common operators.
## Relational Symbols
In algebra, relational symbols are used to express the relationship between two mathematical entities, and are often related to concepts such as equality, comparison, divisibility and other higher-order relationships. The following tables document the most common of these — along with their usage and meaning.
### Relational Symbols in Abstract Algebra
For the master list of symbols, see mathematical symbols. For lists of symbols categorized by type and subject, refer to the relevant pages below for more. | 3.265625 | CC-MAIN-2021-04 |
## Friday, March 4, 2011
### ## Spatial weighting and Voronoi tessellation.
I've been spending far too much time lately on triangular meshes and Voronoi tessellation. The idea for combining temperature measurements is to weight stations according to the area they could be said to "represent". This is important in Antarctica, for example, where the distribution is very uneven, and the Peninsula is likely to dominate just because of the number of stations.
For the global indices, this is handled indirectly through gridding. A cell average is calculated, and cells combined, usually weighted by their easily calculated area.. There's a problem, usually unresolved, with empty cells. And this is one reason why I'm attracted to Voronoi tessellation. Every cell has a reading, provided you are prepared to create a new tessellation every month.
I experimented with a simpler variable cell scheme last year. But there were still empty cells.
A Voronoi tessellation just divides the space up according to the nearest nodes. The patch around node A consists of points nearer to A than any other node. There are still some practical problems in getting it into TempLS, because the tessellation would have to be done for every month. So it has to be reasonably efficient, but more importantly, done without any intervention to fix things. In this post I'll describe my experience so far.
Voronoi tessellation follows from triangular gridding. If you can form a triangular mesh connecting the nodes, then the bisector of each line is likely to be a cell boundary. It's not certain, because one of the nodes in the adjoining cells could be closer.
I found mesh and Voronoi routines in the R packages geometry, alphahull and tripack. The latter seemed the most used, and is called by alphahull. The key routine is called delvor, and does both. However, I found it advisable to run the mesher tri.mesh() first, because it had a flag for removing duplicate nodes.
Here is a Delaunay mesh of Antarctic stations. I've marked a quadrilateral to explain what Delaunay means. It is usually expressed as a requirement that the circumcircles of each triangle include no other nodes. I find that hard to visualize - a simpler version is that in each quad like that marked, the angles opposite the central line add up to less than 180. That gives the simple algorithm for converting any mesh to Delaunay. Since the sum of angles in the quad is 360, if the angles added up to more than 180, you could flip that central line to cross the other way, and the sum would then be less than 180. Of course this could disturb the property for some other quad, so the process can go on for a while. There are more efficient algorithms, but for small meshes like this, flipping is OK. In fact, you can make it quite efficient by doing an occasional sort.
The mesh includes quite a lot of ocean. That's because tri.mesh, and everything in R that I found, insists on meshing the convex hull of the region. The convex hull is what you would get by running a taut string around the points. I did my own experimenting to try to generate a mesh where I can chip away to make the mesh better follow the land. I'll probably be able to do this with the tri.mesh structure. But there's the problem of doing it mechanically so it can be done every month as stations reporting change.
As I said above, the Voronoi tessellation is formed by drawing the perpendicular bisectors of the lines, and surrounding each node by the inner envelope. The Delaunay property means that you can be fairly sure that this means just connecting up the circumcentres. But there is a remaining problem, as you'll see from this result:
Although the circumcircles can't include other nodes, near the boundary, this isn't much of a restriction, and the Voronoi lines go a long way. I want to use the areas for weights, so something has to be done. Oddly, I couldn't find anything in R that confines the cells to the meshed region. So I added my own code to cut off at the boundary, which gave this:
So here, finally, is a map showing the weightings produced (by area of circle). A problem is that it is rather uneven. On the peninsula, say, just one or two stations get connected to a large area. I think something will need to be done to fix this by averaging - not to upweight the region, but to avoid suppressing some stations completely.
This effect is somewhat exaggerated because I've included all stations that have ever reported; in any given month it will be sparser.
I have posted the code here (Antvor.zip) on the doc depository. Antvor.r is the executable; idx.r is the set of station data which comes from Ryan's code.
1. Sorry to report, I can't access the code.
But, then, some weird seems to happen when I try to log in and post as well.
2. Sorry, Ron, I tried to give a direct like to the file, but maybe that doesn't work if you aren't the owner. I've replaced it with a link to the docs store (same as in the resources list, top right).
I see I've also messed up the figs - the last two are the same. I'll fix that when I get back home.
3. Looks interesting. But haven't you gone from one extreme to another??
eg. the coastal cells have gone from radiating out to infinity, to being restricted to the coast. Don't you need an algorithm to make sure the points/positions on the perimeter are in the centre of their cell?
(BTW I know nothing about the methods you are using)
4. Waemcast,
It's just for weighting. It doesn't in the end matter too much whether a node is a good representative of the exact area that's nominated. It's just accounting to ensure that regions aren't overrepresented. In that sense, if one node gets just a bit of coast, and another nearby gets a bigger chunk of interior, that's "fair". The region is weighted about right.
However, there is some loss of information. If that situation does arise (as it does) then there should be some local averaging. My thinking is that this analysis should be done to allocate the area, and then some spatial smoothing of the weightsso that the information in "unlucky" nodes can still count.
Quantitatively it could go like this. One could say that there is good correlation up to 250 km (say), but not so good beyond 500. So you want to have the 500 km scale properly allocated, and we've done that. The price is loss of effective dof, or increased variance, which goes with the variability of the weighting. If we can smooth the weighting locally, without spreading it beyond the 250 scale, then we ameliorate the variance issue without much loss.
5. So the bigger the area, the bigger the weighting??
BTW, if you have already explained this, just point me towards the relevant post(s).
:-)
6. I see some stations, surrounded by others, that have near zero weighting.
We know from previous studies that stations a few km from each other will have high correlation and the correlation decays with increasing distance.
I propose using that correlation vs distance function, combined with the distance to the grid cell being considered as the method of weighting stations.
That avoids the underweighting of stations just because there is a nearby station located between a station and a nearby grid cell.
In one sense, the traditional gridding of data is doing this same algorithm, but with only the discrete weighting coefficients of 0 or 1.
I really just suggesting traditional gridding, but using a fuzzy correlation function rather than a discrete yes/no, or station in or not in the cell.
Does this make any sense?
My other suggestions were roughly the same, but were described using each station as a reference rather than looking at it from the point of view of the grid cell.
7. Charlie,
The problem is that we mostly need a geometric criterion, because we have to grid monthly with the data varying. Now it's true, I expect, that the correlation function does not have to use the latest data etc.
I'm keen to develop my diffusion idea. It's a bit like multigrid. We've got the weighting approx right on a large scale, but as you say, unsatisfactory locally. So we can then smooth locally weights without ddisturbing the wider balance.
Specifically, I'm thinking this method. Usingthe mesh, we do a few diffusion steps (relaxation) in which, along lines of <100 km say, weight is exchanged. Each node might send out up to a third of what it has, and of course will receive back. The amount exchanged could taper with distance. Ten or so steps of that should even things up, and little weight would go beyond, say, 200 km.
8. The smoothly weighted grid method I propose has no trouble with missing months.
I'm assuming that the weighting vs. distance function would stay constant over the run. I only mention the station to station correlation as a way of justifying/selecting a particular weighting function. One could also use some arbitrary weighting such as inverse distance squared.
The handling of missing data would be no different than for traditional grid methods. In those, you just count the number of stations in the grid with data for that month and average them together.
In terms of my proposed solution I would describe that averaging as combining all stations inside the grid cell with weighting coefficient of 1. Then of course, you divide the summed station values by the number of stations to get the average.
In the modified method, all that changes is the weighting would not be limited to either 1 or 0. The result for each grid is simply the summation of each station value times its weight, divided by the total of all weights.
The more I look at that method, the more I'm convinced that it is not merely a quick and dirty method, but that it is also perhaps the most accurate way of estimating values for each grid cells.
The handling of a variety of combinations does what I would want an algorithm to do.
For nearly co-located stations, it removes the influence of the direction between the stations. For what would be empty grid cells in a normal gridding method, the algorithm "reaches out" to include far away stations. In grids with cells in it and nearby, the far away stations are included in the calculations, but have small contributions compared to the nearby stations.
The only other tweaking the algorithm might need is a way to truncate and ignore far away stations once a sufficient number of closer stations have been included.
========================
The approach I'm suggesting is so obvious that undoubtably somebody has already used it for spatial interpolation to infill missing data or something similar.
I recognize that all this is very much a divergence away from the direction you are going and will stop cluttering up the thread at this point.
9. Charlie,
I think you are on to something there. It's more than just spatial weighting though - it's an inadequacy in least squares. The weighting function really should incorporatee a correlation matrix. Here is Joe Triscari making that point.
My defence is that much work is still done with OLS and the matrix kernel complicates the numerical algebra a lot. But yes, it should be done, one day.
It's related to a controversy about S09 originally not allowing for first order correlation. That led to a corrigendum in which a Quenouille correction was used. I haven't found where O10 did that, but I presume they did. But this correction is a small step, and doesn't actually change the expected values - just the CI's.
10. Nick,
"That led to a corrigendum in which a Quenouille correction was used. I haven't found where O10 did that, but I presume they did."
Yes, indeed, I think we stated as much in the paper. And, unlike S09, Ryan posted full code so people can check that we actually did what we said we did. It is in lmFn, which is called by getRecon to do the trend and trend CI calculations:
### Apply DoF correction
Q = sqrt((1 - r) / (1 + r))
SE = SE / Q
I think you will find that the RLS method we used in effect incorporated full spatial correlation information, as derived from the AVHRR satellite data. Although this data suffers form temporal drift and inhomogeneities, that doesn't degrade the spatial correlation information nearly as much as it affects the accuracy of its temporal data.
11. Thanks, Nic
Yes, it wasn't hard to see. | 2.6875 | CC-MAIN-2021-21 |
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Posted on
Daily Challenge #237 - Delete Extra Occurrences
Alice and Bob were on a holiday. Both of them took many pictures of the places they've been, and now they want to show Charlie their entire collection. However, Charlie doesn't like these gallery sessions, since the setting usually repeats. He isn't fond of seeing the Eiffel tower 40 times in a row. He tells them that he will only sit during the session if they only show the same setting at most N times. Luckily, Alice and Bob are able to encode each setting as a number. Can you help them to remove numbers such that their list contains each number only up to `N` times, without changing the order?
Given a list `lst` and a number `N`, create a new list that contains each number of `lst` at most `N` times without reordering. For example if `N = 2`, and the input is `[1,2,3,1,2,1,2,3]`, you take `[1,2,3,1,2]`, drop the next `[1,2]` since this would lead to 1 and 2 being in the result 3 times, and then take 3, which leads to `[1,2,3,1,2,3]`.
Examples
`delete_nth([1,1,1,1],2)` # return `[1,1]`
`delete_nth([20,37,20,21],1)` # return `[20,37,21]`
Tests
`delete_nth([1,1,3,3,7,2,2,2,2])`, `N= 3`
`delete_nth([20,37,20,21])`, `N = 1`
Good luck!
This challenge comes from JustyFY on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!
Want to propose a challenge idea for a future post? Email [email protected] with your suggestions!
Discussion (8)
Vidit Sarkar
Here is C++ solution,
``````vector<int> delete_nth(vector<int> arr, int occ){
// stores the numbers and its corresponding count in answer vector
unordered_map<int, int> numCount;
vector<int> ans;
// if count of any number is less than occ,
// only then add it in ans
for(int num : arr){
if(numCount[num] < occ){
ans.push_back(num);
numCount[num]++;
}
}
return ans;
}
``````
sai kiran jv
What if given numbers are not positive ?
Vidit Sarkar
`ans` vector contains each number only up to `occ` times. So, `occ` can never be negative. If user passed a negative number as `occ`, then `ans` vector will be empty.
If given numbers are negative, still there should not be any problem. Example,
``````delete_nth({-1,-1,-1,-2,-3,-2,-4,-3}, 1) => {-1,-2,-3,-4}
delete_nth({1,-2,3,-7,-2,3,-8},1) => {1,-2,3,-7,-8}
``````
Javscript
With the assumption that in a long array, most elements will be seen enough times, and we need to quickly drop them:
``````const delete_nth = (lst, n) => {
const tired = new Set()
const seen = new Map()
return lst.filter(x => {
if (tired.has(x)) return false
const new_count = (seen.get(x) || 0) + 1
else seen.set(x, new_count)
return true
})
}
``````
Without that assumption:
``````const delete_nth = (lst, n) => {
const seen = new Map()
return lst.filter(x => {
const count = seen.get(x) || 0
if (count >= n) return false
seen.set(x, count + 1)
return true
})
}
``````
peter279k
Here is the Python solution:
``````def delete_nth(order,max_e):
freq_counts = {}
for number in order:
if number not in freq_counts:
freq_counts[number] = 0
ans_order = []
for number in order:
freq_counts[number] += 1
if freq_counts[number] > max_e:
continue
ans_order.append(number)
return ans_order
``````
Matt Ellen • Edited on
Trying to figure out a version that dosn't use a list or map to remember what's been, I've created this mess in python, that is limited to a maximum count of 999:
``````def nomorethancount(arr, count):
if count < 1:
return []
if count > len(arr):
return arr
if count >= 1000:
raise ValueError('count must be less than 1000')
bigposnum = 0
bignegnum = 0
result = []
for item in arr:
if item < 0:
absitem = -1 * item
bignegnum = checkitem(absitem, item, bignegnum, count, result)
else:
bigposnum = checkitem(item, item, bigposnum, count, result)
return result
def checkitem(absitem, item, bignum, count, result):
if 1000**absitem > bignum:
bignum += 1000**absitem
result.append(item)
else:
currentcount = (bignum // 1000**absitem) % 1000
if currentcount < count:
result.append(item)
bignum += 1000**absitem
return bignum
`````` | 2.78125 | CC-MAIN-2022-27 |
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A certain amount was deposited at bank A at a rate of x
Author Message
Current Student
Joined: 11 May 2008
Posts: 551
A certain amount was deposited at bank A at a rate of x [#permalink]
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30 Aug 2008, 03:36
A certain amount was deposited at bank A at a rate of x percent compounded annually and half that amount was deposited at bank B at a rate of y percent compounded annually. In three years, will the total amount on deposit in bank A be bigger than that on deposit in bank B?
1. y = 0.8x
2. The difference between the amounts initially deposited was \$100
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Director
Joined: 23 Sep 2007
Posts: 761
Show Tags
30 Aug 2008, 13:01
arjtryarjtry wrote:
A certain amount was deposited at bank A at a rate of x percent compounded annually and half that amount was deposited at bank B at a rate of y percent compounded annually. In three years, will the total amount on deposit in bank A be bigger than that on deposit in bank B?
1. y = 0.8x
2. The difference between the amounts initially deposited was \$100
A
let A = 200, so that B = 100
statement1: if you start with a bigger amount, and earn a greater interest percentage on that amount, then it is obvious that you will have more. Suff
statement 2: Let A = 200 and B = 100, if x = 1% and y = 100%, then B will be bigger. But if x = 1% and y=1%, then A will be bigger. Insuff
Intern
Joined: 16 Jun 2008
Posts: 17
Location: Kiev, Ukraine
Show Tags
30 Aug 2008, 13:04
Although the topic is "simple interest", the task is for compound interests
The amount of money received after three years depends on two variables: interest rate and amount of deposited money.
As far as there is a direct correlation between these two variables, the more money we deposit or the higher the interest rate is, the more money we will get in 3 years.
We know that second amount is twice less. The 1-st condition tells us that interest rate in bank B is less than in bank A, so we deposited less money at less interest rate in bank B, and after 3 years we will get less money, of course, than from bank A. This condition is enough for us to answer the question.
2nd condition only tells us about the difference in amounts but nothing about the interest rates, so it is not sufficient to answer the question.
SVP
Joined: 07 Nov 2007
Posts: 1738
Location: New York
Show Tags
30 Aug 2008, 13:30
arjtryarjtry wrote:
A certain amount was deposited at bank A at a rate of x percent compounded annually and half that amount was deposited at bank B at a rate of y percent compounded annually. In three years, will the total amount on deposit in bank A be bigger than that on deposit in bank B?
1. y = 0.8x
2. The difference between the amounts initially deposited was \$100
Clear A.
1)
Bank A has.. more deposit and more interset rate
2)
we don't know about interst rates..
insuffcient
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This is not a quality discussion. It has been retired.
If you would like to discuss this question please re-post it in the respective forum. Thank you!
To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
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Re: simple interest [#permalink] 30 Aug 2008, 13:30
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 3.921875 | CC-MAIN-2018-26 |
# How to model a recognition experiment with two measurement time points?
Hello! In the course of my master’s degree in psychology we currently conducting a word recognition experiment online. The conditions are varied within-subjects. All participants have to pass both conditions, control and experimental condition on the first date (T1). On the first date the participants are receiving a treatment which is expected to improve the recognition performance in the experimental condition associated recognition task.
One week later the participants have to undergo control and experimental condition again (T2). We want to evaluate whether the treatment effect continues from T1 to T2, specifically that the recognition performance is still larger in the experimental condition.
I’m gonna use the Brms package to specifiy and calculate an ordinal probit model with heteroscedastic error.
The Variables are defined as follows:
Variables of the Model
item = factor variable with 2 levels [new;old]
condition = factor variable with 2 levels [K;E]
time2 = numeric with [0 = T1; 1= T2]
old = numeric with [0 = new; 1 = old]
1. The Interaction of condition x item is the difference (change in the probability of the categories likewise shift of the thresholds) in detection from old vs new items in the experimental group vs the control group for T1, isn’t it? This Interaction takes just the data from T1 and not the whole from T1 + T2(?).
2. I used the discrimination parameter to allow the standart deviation of the old (previous learned items) items differ from 1. Do I have to include more discrimination parameters for condition and time?
3. The Interaction (condition x time2 x item) is the difference of the detection from T1 to T2. Does a non significant coefficient of this interaction term imply an enduring effect of the treatment after 1 week for the case that the (condition x item) interaction was significant?
4. What is modeled with | i | in this case? Is the benefit of this a more accurate estimation of the coefficients?
The Model I used:
``````uvsdt <- brm(
bf(Response ~ 1 + item * condition * time2 + (1 + item * condition * time2 | i | ID_T1T2 ),
disc ~ 0 + old + (0 + old | i | ID_T1T2 )),
data = data_hypnomemory, family = cumulative("probit"),
iter = 2500, inits = 0
)
``````
1 Like
Hi,
sorry for taking quite long to respond:
One way to understand what interactions as `condition * item` mean is to expand the dummy coding that will be used in the model. Assuming the reference levels are `new` and `K`, we get four cases:
``````Case Intercept item condition item:condition
[new, K] 1 0 0 0
[new, E] 1 0 1 0
[old, K] 1 1 0 0
[old, E] 1 1 1 1
``````
I.e. ignoring all other elements in the formula the linear predictor for `[new, E]` will equal `1 * b_Intercept + 0 * b_itemold + 1* b_conditionE + 0 * b_itemconditionoldE = b_conditionE`. Similarly for `old,E` the linear predictor will equal `1 * b_Intercept + 1 * b_itemold + 1* b_conditionE + 1 * b_itemconditionoldE`. With that in mind we can see that:
``````([old,E] - [new,E]) - ([old,K] - [new, K]) = 1 * b_itemconditionoldE
``````
or equivalently
``````([old,E] - [old,K]) - ([new, K] - [new, E]) = 1 * b_itemconditionoldE
``````
Thinking about how the actual linear predictors are built is a very general way to understand what the interactions means even in more complex scenarios (e.g. factors with more than two levels).
The `condition * item` interaction would involve both time points, `condition * item * time2` would let you estimate it separately for each time point.
Hope that answers this part of you inquiry :-)
No. First I am not sure you are correctly describing what the interaction term actually does. Second you need to interpret the uncertainty in the coefficient. If by “non-significant” you mean something like “the 95% posterior interval includes 0” it is also important, how wide the interval is. If the interval is narrow around 0 then you can be somewhat confident that the interaction really is negligible. If it is wide it means your data aren’t enough to learn much useful about the interaction. Also note that “significant” is a frequentist term and does not directly translate to a Bayesian context.
Finally, you can’t ignore that you also have the ` (1 + item * condition * time2 | i | ID_T1T2 )` and for some inference tasks it might be important to consider this term as well. I discussed this in a slightly different context at: Inferences marginal of random effects in multi-level models
It means that all the terms that use `| i |` share a correlation matrix, e.g. if the `1 + item * condition * time2` coefficients tend to be similar/different for some values of `ID_T1T2` then the coefficient for `0 + old` will also be more likely to be similar/different for those values of `ID_T1T2`. The benefit depends on whether this assumption is correct. I find it slightly weird in your case, because I would not necessarily expect the `disc` and overall response to behave similarly.
Overall, this seems to be a very ambitious model that would require A LOT of data to learn anything useful about the parameters.
Best of luck with your project!
(Responding belatedly after a period of not having time to check here)
(Also, I suspect you may have a better understanding of this than I do, but I’ll summarize for others coming later)
As I think you have discerned, data like this (participants saying “old”-or-“new” to stimuli that are either old or new) are often modelled using a “Signal Detection Theory” framework that actually ends up being well-captured by standard hierarchical GLM. To do it properly, you model the participant’s response (literally what they said, not response accuracy) as a function of the response-associated stimulus identity (here, whether the stim was old or new) plus any other observed or manipulated variables. As I understand it, in your data, `old` reflects the response-associated stimulus identity, `condition` reflects the experimental-vs-control manipulation, `time2` encodes the two different session times. Now, you also have a variable `item` that you also indicate has two levels, “new” and “old”, but if this is not perfectly redundant to the `old` variable I am confused as to what it actually is. Is it perchance the actual word token, in which case shouldn’t it have many levels, not merely two?
Leaving my uncertainty wrt the `item` aside and continuing with my summary (again, likely for others later) and ignoring for now the multiple participants, the model :"
``````Response ~ old * condition * time2
``````
has (interpretations assume intercept-independent contrasts like sum contrasts or half-sum contrasts):
• intercept: overall response bias
• main effect of `old`: overall discrimination (literally d’ if using a probit link)
• main effect of `condition`: effect of condition on response bias
• main effect of `time2`: effect of time on response bias
• `condition:time2`: interaction between condition and time on response bias
• `old:condition`: effect of condition on discrimination
• `old:time2`: effect of time on discrimination
• `old:condition:time2`: interaction between condition and time on discrimination
When there are multiple participants, the full hierarchical model (which you should definitely employ, following Barr’s “keep it maximal” dictate) would be:
``````Response ~ old*condition*time2 + (1+old*condition*time2 | participant)
``````
where potential across-participant corrrelations among all the above denoted effects are modeled.
If I’m right that `item` is intended to encode the actual tokens used for the old/new judgement (ex. the actual words), and if tokens were randomly assigned to old/new identities across participants, then you may want to model systematic differences among items in the same way as the above models systematic differences among participants. This could be done via the model:
``````Response ~ (
old*condition*time2
+ (1+old*condition*time2 | participant)
+ (1+old*condition*time2 | item)
)
``````
Finally, you mention employing an ordered probit response link. Is my understanding correct that this is to avoid the assumption of equal variance in the latent “psychological newness” distributions of old and new items? I haven’t seen this done this way, but I think that makes sense and cool if this is how it’s supposed to be done. Possibly this also explains why you are modelling a seemingly separate outcome, `disc`? If not, and `disc` is truly a data parameter that you’ve computed from the recognition data separately in R, then hopefully you’ll see from the above that this is unnecessary as inference on discrimination performance comes from the GLM automatically.
Feel free to correct me if I’m misunderstanding anything!
1 Like
Quick note: `disc` is AFAIK a parameter of the cumulative probit distribution.
Ah, thanks to @matti’s awesome tutorial series on SDT (particularly part 3), I now see that the `disc` parameter is for enabling unequal variance and effects of variables thereon. Cool!
1 Like
The discrimination parameter: Is a way to let the standart deviation differ from 1 for the old items (unequal variance signal detection theory model). If I am right, the discrimination parameter is the natural logarithm from the standart deviation.
1 Like
@martinmodrak
If I wanna calculate the expected value of the latent variable familiarity for the first measurement time point doesn’t that mean, that the estimates for itemOld and conditionE are estimated for the participants measured at the first time point?
``````E( Y' | condition = 1, item = 1, time2 = 0) =
1 * intercept + 1 * b_itemold +1 * b_conditionE + 1 * b_itemOld:conditionE +
0 * b_time2 + 0 * b_time2:conditionE+ 0 * b_itemold:time2 +
0 * b_itemold:time2:conditionE
``````
The time dimension confuses me. I want to know if the treatment on the first time point is effective to enhance the recognition performance (detection). To evaluate this question I would consider the Interaction b_itemold:conditionE, wouldn’t I?
I am not sure I completely understand what you are asking. There might be some additional misunderstanding. We have `itemOld` and `conditionE` as population level effects - those are estimated once. Then we have the varying intercept from `(1+old*condition*time2 | participant)` which creates a separate set of parameters for each value `participant` can take. So in that sense, yes, if you have participants (call it `p1`) that were only measured at one time point, they will still have their own `time2[p1]` parameter, but this coefficient will never interact with the likelihood and your inferences should be the same as if this parameter didn’t exist.
If the condition represents the treatment on the first time point, then yes. (if the values of condition for the same participant can differ between time points than you would need to introduce new value “condition on first time point” and use it instead).
Your description of the experiment is slightly confusing to me. You say there is “condition” (which is control or experimental), but then that there is “treatment” which is something else? Or how is the treatment represented in the model? And what does “old” actually mean? I think clarifying what exactly happened and what do the data mean could help us get faster to a good solution.
Best of luck!
1 Like
First of all: I appreciate all your efforts :).
Let’s try to clarify the experiment:
The general research question: Can we enhance the memory performance in a visual recognition word task through hypnosis. (For the case we have an effect) Is the improvement still measurable after one week?
We have 2 measurement time points. The second is one week after the initial measurement (= T2). All Participants attend both: experimental and control condition (within-subject).
The treatment is the hypnosis which is intended to improve the performance in the recognition task. During the first session all participants receive a hypnosis which is then bound to a post-hypnotic trigger (a note with an “E”) - which is intended to enhance the memory performance later on at the recognition task. There is another note as a control cue with a “K” on it.
Let’s demonstrate the whole experiment for one Participant Xi:
T1 (Initial measurement/session):
1. The Participant Xi gets the controll note (‘K’). Then he receive the hypnosis during which he will get the post-hypnotic trigger (note with the “E”).
2. Xi learns Words. After the learning phase he has to take one of the notes (random “K” or “E”) and wear it near his body. Then he is asked to evaluate presented words regarding their previous occurrence in the learning phase (from 1 to 6: 1 - surely a new word,2,3,4,5, 6 - surely seen before). During this phase all the previous learned words occur and the same amount of “new” items (not learned during the learning phase).
3. Xi learns words again. He has to take the other note (if he had first “K”, he is now asked to wear the “E” note). He is again asked to rate the occurring words with regard to their previous occurrance.
-> So Xi did both, control and experimental condition
T2(after one week):
1. All Participants redo 2. and 3. from the initial measurement/session.
The idea is to model the data as an unequal variance signal detection model (varying standart deviation for the previous learned items). And this with the Brms package defined as an ordered probit model with heteroscedastic error. I hope this is a more comprehensible description.
The Variables:
item = previous learned (= old), not learned during the experiment (= new)
time2 = data measured at T1 (= 0), data measured at T2 (=1)
Response = the Rating for the presented word (1 to 6)
condition = control (=0), experimental (=1)
@martinmodrak The OP and Matti’s tutorial both use treatment contrasts, and I wonder if (1) that’s a source of confusion here (bc with interactions treatment contrasts are hard to think about IMO) and (2) if sum contrasts can still “work” for the unequal variances model (in the sense that the `disc` parameter is now not `1` for the `new` stimuli and `x` for the `old` stimuli but instead `disc_intercept-x/2` for the `new` and `disc_intercept+x/2` for the `old`). Unless I’m mistaken, fixing the `disc` value for the `new` stimuli at `1` is a means of ensuring identifiability amidst the cumulative probit, yes? In which case would a sufficiently narrow (or even fixed-at-1) prior on the `disc_intercept` achieve the same ends?
1 Like
I know @paul.buerkner wrote an inspiring paper on ordinal regression models. Maybe he has an idea if it’s better to model this via two models (one for T1 and one for T2) or in one and what discrimination parameters should be allowed to vary and how.
OK, this is clearer.
I think one way to make the data easier to analyze would be to recode the `new` words in reverse and ignore the `item` term in the model. This is not completely safe, but I think would help thinking about the model much easier, as all coefficients are now positive if they are associated with increase in performance and negative when performance decreases. Eventually, you will probably want to move back to the original model, but by then you IMHO might have a better sense of what is going on.
I am not completely sure your full model is justifiable. I would start with a much simpler model, i.e.
``````ResponseRecoded ~ time2 * condition + item + (1 | participant)
``````
where `ResponseRecoded` is the response with `new` words reversed. This should be much easier to interpret… Or
``````Response ~ item*time2*condition + (1 + item | participant)
``````
For the non-recoded version.
Then I would do a posterior predictve check (`pp_check`) for the `sd` of the response by various groups (`stat_grouped`) and only if I would see discrepancies would I model the `disc` as differing between subgroups.
Similarly you could do oher `pp_check`s to see if additional terms need to be added to the model.
Also unless you have a ton of participants (my guess would be at least a few hundreds), I don’t think you can reliably learn about all the parameters in the full model, so you IMHO need to simplify anyway. Even the simpler models I proposed might be too rich if you have just a few dozen participants…
Does that make sense?
Yeah it seems, that the sd for the previous learned words is indeed higher.
Is it possible to model different discriminations for the old (previous learned words) items only? My intend is to model discrimination parameters for the old items in the controll group (old), the experimental group (old x conditionE), for the old words to time point 2 (old x time2 = 1) and for the old items in the experimental condition to T2. If I try to model this I always get discrimination parameters for the new items as well.
My Idea is to compare the above model with the model that just have one parameter for the differing sd for the old items. OR. I do two models one for T1 and one for T2 for comparing the conditions, but then the estimated standard error is higher then.
Maybe you have a few ideas :) | 3.140625 | CC-MAIN-2022-21 |
# Inverse transform sampling
(Redirected from Inversion sampling)
Inverse transform sampling (also known as inversion sampling, the inverse probability integral transform, the inverse transformation method, Smirnov transform, golden rule[1]) is a basic method for pseudo-random number sampling, i.e. for generating sample numbers at random from any probability distribution given its cumulative distribution function.
Inverse transformation sampling takes uniform samples of a number ${\displaystyle u}$ between 0 and 1, interpreted as a probability, and then returns the largest number ${\displaystyle x}$ from the domain of the distribution ${\displaystyle P(X)}$ such that ${\displaystyle P(-\infty . For example, imagine that ${\displaystyle P(X)}$ is the standard normal distribution with mean zero and standard deviation one. The table below shows samples taken from the uniform distribution and their representation on the standard normal distribution.
Transformation from uniform sample to normal
${\displaystyle u}$ ${\displaystyle F^{-1}(u)}$
.5 0
.975 1.95996
0.995 2.5758
.999999 4.75342
1-2^{-52} 8.12589
Inverse transform sampling for normal distribution
We are randomly choosing a proportion of the area under the curve and returning the number in the domain such that exactly this proportion of the area occurs to the left of that number. Intuitively, we are unlikely to choose a number in the far end of tails because there is very little area in them which would require choosing a number very close to zero or one.
Computationally, this method involves computing the quantile function of the distribution — in other words, computing the cumulative distribution function (CDF) of the distribution (which maps a number in the domain to a probability between 0 and 1) and then inverting that function. This is the source of the term "inverse" or "inversion" in most of the names for this method. Note that for a discrete distribution, computing the CDF is not in general too difficult: we simply add up the individual probabilities for the various points of the distribution. For a continuous distribution, however, we need to integrate the probability density function (PDF) of the distribution, which is impossible to do analytically for most distributions (including the normal distribution). As a result, this method may be computationally inefficient for many distributions and other methods are preferred; however, it is a useful method for building more generally applicable samplers such as those based on rejection sampling.
For the normal distribution, the lack of an analytical expression for the corresponding quantile function means that other methods (e.g. the Box–Muller transform) may be preferred computationally. It is often the case that, even for simple distributions, the inverse transform sampling method can be improved on:[2] see, for example, the ziggurat algorithm and rejection sampling. On the other hand, it is possible to approximate the quantile function of the normal distribution extremely accurately using moderate-degree polynomials, and in fact the method of doing this is fast enough that inversion sampling is now the default method for sampling from a normal distribution in the statistical package R.[3]
## Definition
The probability integral transform states that if ${\displaystyle X}$ is a continuous random variable with cumulative distribution function ${\displaystyle F_{X}}$, then the random variable ${\displaystyle Y=F_{X}(X)}$ has a uniform distribution on [0, 1]. The inverse probability integral transform is just the inverse of this: specifically, if ${\displaystyle Y}$ has a uniform distribution on [0, 1] and if ${\displaystyle X}$ has a cumulative distribution ${\displaystyle F_{X}}$, then the random variable ${\displaystyle F_{X}^{-1}(Y)}$ has the same distribution as ${\displaystyle X}$ .
Graph of the inversion technique from ${\displaystyle x}$ to ${\displaystyle F(x)}$. On the bottom right we see the regular function and in the top left its inversion.
## The method
The problem that the inverse transform sampling method solves is as follows:
• Let ${\displaystyle X}$ be a random variable whose distribution can be described by the cumulative distribution function ${\displaystyle F_{X}}$.
• We want to generate values of ${\displaystyle X}$ which are distributed according to this distribution.
The inverse transform sampling method works as follows:
1. Generate a random number ${\displaystyle U}$ from the standard uniform distribution in the interval ${\displaystyle [0,1]}$.
2. Compute the value ${\displaystyle x}$ such that ${\displaystyle F_{X}(x)=u}$.
3. Take ${\displaystyle x}$ to be the random number drawn from the distribution described by ${\displaystyle F_{X}}$.
Expressed differently, given a continuous uniform variable ${\displaystyle U}$ in ${\displaystyle [0,1]}$ and an invertible cumulative distribution function ${\displaystyle F_{X}}$, the random variable ${\displaystyle X=F_{X}^{-1}(U)}$ has distribution ${\displaystyle F_{X}}$ (or, ${\displaystyle X}$ is distributed ${\displaystyle F_{X}}$).
A treatment of such inverse functions as objects satisfying differential equations can be given.[4] Some such differential equations admit explicit power series solutions, despite their non-linearity.[citation needed]
## Examples
• As an example, suppose we have a random variable ${\displaystyle U\sim Unif(0,1)}$ and a cumulative distribution function
{\displaystyle {\begin{aligned}F(x)=1-\exp(-{\sqrt {x}})\end{aligned}}}
In order to perform an inversion we want to solve for ${\displaystyle F(F^{-1}(u))=u}$
{\displaystyle {\begin{aligned}F(F^{-1}(u))&=u\\1-\exp \left(-{\sqrt {F^{-1}(u)}}\right)&=u\\F^{-1}(u)&=(-\log(1-u))^{2}\\&=(\log(1-u))^{2}\end{aligned}}}
From here we would perform steps one, two and three.
• As another example, we use the exponential distribution with ${\displaystyle F(x)=1-e^{-\lambda x}}$ for x ≥ 0 (and 0 otherwise). By solving y=F(x) we obtain the inverse function
${\displaystyle x=F^{-1}(y)=-{\frac {1}{\lambda }}\ln(1-y).}$
The idea is illustrated in the following graph:
Random numbers yi are generated from a uniform distribution between 0 and 1, i.e. Y ~ U(0, 1). They are sketched as colored points on the y-axis. Each of the points is mapped according to x=F−1(y), which is shown with gray arrows for two example points. In this example, we have used an exponential distribution. Hence, for x ≥ 0, the probability density is ${\displaystyle \varrho _{X}(x)=\lambda e^{-\lambda \,x}}$ and the cumulated distribution function is ${\displaystyle F(x)=1-e^{-\lambda \,x}}$. Therefore, ${\displaystyle x=F^{-1}(y)=-{\frac {\ln(1-y)}{\lambda }}}$. We can see that using this method, many points end up close to 0 and only few points end up having high x-values - just as it is expected for an exponential distribution.
Note that the distribution does not change if we start with 1-y instead of y. For computational purposes, it therefore suffices to generate random numbers y in [0, 1] and then simply calculate
${\displaystyle x=F^{-1}(y)=-{\frac {1}{\lambda }}\ln(y).}$
## Proof of correctness
Let F be a continuous cumulative distribution function, and let F−1 be its inverse function (using the infimum because CDFs are weakly monotonic and right-continuous):[5]
${\displaystyle F^{-1}(u)=\inf \;\{x\mid F(x)\geq u\}\qquad (0
Claim: If U is a uniform random variable on (0, 1) then ${\displaystyle F^{-1}(U)}$ has F as its CDF.
Proof:
{\displaystyle {\begin{aligned}&\Pr(F^{-1}(U)\leq x)\\&{}=\Pr(U\leq F(x))\quad &({\text{applying }}F,{\text{ to both sides}})\\&{}=F(x)\quad &({\text{because }}\Pr(U\leq y)=y,{\text{ when U is uniform on}}(0,1))\\\end{aligned}}}
## Reduction of the number of inversions
In order to obtain a large number (lets say M) of samples one needs to perform the same number of inversions ${\displaystyle F_{X}^{-1}(u)}$ of the distribution ${\displaystyle F_{X}}$. One possible way to reduce the number of inversions to only a few while obtaining a large number of samples is the application of the so-called the Stochastic Collocation Monte Carlo sampler (SCMC sampler), within a polynomial chaos expansion framework, allows us the generation of any number of Monte Carlo samples based on only a few inversions of the original distribution and independent samples from a variable for which the inversions are analytically available, like for example the standard normal variable.[6] | 3.9375 | CC-MAIN-2017-47 |
CLIC Targets wb 30/11/15
Here are our CLIC targets for this week:
C– Counting: Reading Numbers; I can read 4 digit numbers (Remember the ‘,’ is thousand!) e.g. 8,905 etc…
L– Learn Its: x9 tables; Be able to jumble the table and answer randomly e.g. 3×9, 8×9, 5×9, etc…
I– It’s Nothing New: Smile Multiplication; Step 4: I can do smile multiplication for tenths e.g.
3 x 0.7= 2.1
C– Calculations: Multiplication; Step 14: (6, 7, 8, 9 times tables) I can solve an 1d x 2d e.g. 6×23=
Source: Primary 6 | 3.21875 | CC-MAIN-2022-33 |
Sei sulla pagina 1di 37
BRANCH NAME: - AERONAUTICAL ENGINEERING
AE 1301- FLIGHT DYNAMICS
PART-A
Short questions and answers (Module-I & II)
1)
What is the need to define ISA and give its values at standard sea level condition?
Ans: Since atmospheric conditions like pressure, temperature and density are constantly
varying with altitude and location of the earth surface, there is a need for a normal
atmosphere to use it as standard for the design and flight evaluation of aircraft all over the
5 N/m2, temperature =288K,
and density =1.225 kg/m3.
world. Its standard sea level values are, pressure = 1.013 x10
2) Distinguish between Troposphere & Stratosphere.
Troposphere
Stratosphere
(a) It extends from earth surface to It extends from troposphere to about around 11km altitude. 50 km altitude. (b) It is a turbulent region and its Its temperature remain constant upto temperature decrease linearly about 25km and then increases upto
at the rate of 6.5 C/km
(288K to 216K at 11km)
o
47km at the rate of 3
to 53km, temperature remains constant
at 282K.
o C/km. At 47km
3) Define Geometric and Geopotential altitudes. Give the relation between ‘g’ at the
Absolute altitude (h ) in terms of earth’s surface gravity (g ) and geometric
a
o
altitude (h g ) and radius of earth (r).
Ans: Geometric altitude (hg) is the physical altitude above the sea level .Geopotential
altitude (h) is the fictious altitude and differs from geometric altitude (hg) and is
physically compatible with the assumption of constant acceleration due to gravity (go),
ie; h = (g/go) 0.5 hg.
The ratio of acceleration due to gravity at an altitude ha, to that on the earth surface (go)
is :
g/go = (r/ha) 2 , ie, g = go (r / (r+hg )) 2
4) Derive from 1 st principle the variation of pressure (dp) in ISA with respect to the variation of Geometric altitude (dh g ) . Ans: Assume an air column of height dhg and unit base area as shown in fig. The 3 forces acting on the tiny air column are,
(p+dp) 1 +ρgdhg = p.1 , ie , dp = -ρgdhg
5) Define the terms (a) Pressure altitude (b) Temperature altitude (c) Density altitude.
Ans: If an aircraft senses the actual outside air pressure which corresponds to some
standard altitude ,then that altitude is called pressure altitude .Similarly if an aircraft
senses the actual outside temperature and density which corresponds to some
standard altitude , those are called temperature altitude and density altitude
respectively.
6) What are the factors which decide the flying path of an airplane as a rigid body?
Ans: Airplane as a rigid body its flying path is decided by ,
a) Its inertia characteristic.
b) Earth‘s acceleration due to gravity
c) Propulsive force generated by power plant
d) Aerodynamic forces (L&D) and moments created on it due to the
reaction between airplane and air.
7) Why the airplane is considered as a dynamic system in six degrees of freedom?
What are the conditions to be satisfied for equilibrium along a straight
unaccelerated flight path? Ans: Airplane motion in air can be completely defined only if six velocity components (linear velocities u,v,w along X,Y,Z axes and angular velocities p,q,r about X,Y,Z ) are given . Hence airplane is considered as a dynamic system in six degrees of freedom. For equilibrium along a straight unaccelerated flight path, the
equation of static applied to each degree of freedom must be satisfied:
ie; Fx =0 , L = 0 ,
Fy = 0 , Fz =
0
M = 0 ,
N = 0
8) What is plane of symmetry of airplane and define symmetric and asymmetric degrees of freedom? Ans: Plane of symmetry of airplane is the X-Z plane that divides it into two symmetric halves. Symmetric degrees of freedom are those components of
9)
motion related to airplane’s longitudinal motion .They are motions along X
and Z axes and about Y-axis . Asymmetric degrees of freedom are those
components of motion related to airplane’s lateral motion. They are motions
along Y- axis and about X and Z axes .
What will decide the lateral (asymmetric) degrees of freedom and define
directional stability and dihedral effect.
Ans: Lateral degrees of freedom of an airplane is decided by the direction of relative
wind to the plane of symmetry . This angle is called sideslip (β) and the airplane
is designed to resist the development of sideslip during all normal flight
maneuvers .The ability of the airplane to create yawing moments that tend to
eliminate any sideslip is called directional stability (sometimes called
weathercock stability). The rolling moment created because of sideslip is
referred to as dihedral effect . The second type of static lateral stability is
associated with the dihedral effect .
10) Define compressibility of fluids and what decides the speed of propagation of small pressure waves in fluids ? Ans: Compressibility of fluids (K) is defined as the reciprocal of bulk modulus of 11) fluids ;or it is the volumetric strain (dv/v) developed per unit change in pressure. ( K = -(dv/v)/dp ) .Speed of propagation of small pressure waves (sound waves) is dependent on the elastic property (bulk modulus) of the fluids . Speed of propagation of sound wave is a = √(dp/dρ), where dp &dρ are change in pressure and density . How the sound waves travel in gases and give the expression for the speed of
sound in gases?
Ans: Sound waves travel in gases as a series of adiabatic compressions and rare fractions and hence the expression for the speed of sound waves is a = (γ(dp/dρ)) 0.5 = (γp/ρ) 0.5 = (γgRT) 0.5 . 12) What are the different power plants used in airplanes? Which power plant is
most efficient for subsonic airplanes . Ans: Latest power plants used in airplanes include; propeller driven by recipro- cating engines as well as turbines , turbojets ,ramjet and rockets .Here other
than rockets all other power plants are using air-breathing engines. Turbojet
power plants are most efficient for subsonic airplanes compared to Turboprop,
ramjet and rockets. Ramjet and rockets are efficient for short
supersonic flights whereas turbo propellers are less efficient and can be used
only at low speed.
duration
13) Write down the major differences between the turboprop and turbojet engines in
generating the propulsive power
Ans:
propel the airplane , 80-95% is derived from the propeller and the reminder only from
the exhaust jet reaction . In typical turboprop system, power(P= TV) is nearly
constant and hence thrust (T) decreases as speed (V) increases. In turbojet propulsion
system, the total useful work done to propel the airplane is produced by the expansion
of the combustion gases through exhaust nozzle. Thrust is made nearly constant by
increasing the mass flow rate (kg/s) at high speed (V) .
In airplanes using turbo propeller system, out of total useful work done to
14) What are the conditions for minimum drag and minimum power required for an
airplane? Mention them in drag coefficients also.
Ans: Minimum drag (D) occurs when L/D or CL/CD is a maximum .It occurs when parasite drag = induced drag or when CDf =CL**2/пAe . Minimum power (P min ) occurs when the parasite power is 1/3 of Induced power .In the drag coefficient form C Df = 1/3(CL**2/пAe) . 15) What causes induced drag? Ans: Induced drag is an unavoidable by-product of lift and it increases as the angle of
attack () increases . Up to a critical angle of attack (c at which CLmax occurs)
both lift and induced drag increases . Since there are two types of lift (dynamic & induced lift) , there are two types of drag also ; dynamic drag and induced drag . Above cr ,lift will decrease and drag will overcome the thrust causing reduction of speed and altitude. 16) Define skin friction drag and pressure drag. Ans: Drag caused by the shear force (viscous flow) in the boundary layer is called skin friction drag . The drag caused by the pressure variation over and below the
surface of the wing is called pressure drag. It is caused by flow of high pressure
air from under the wing to over the wing .Sum of the above two drags are called
profile drag. All drags other than induced drag is called parasite drag.
17) What are the conditions required for maximum drag and minimum power?
Ans: Maximum drag occurs when the angle of attack exceeds the critical angle of
18)
attack and the speed approaches the stalling speed . The left hand limit of PR
versus Vcurve shows the stalling speed corresponding to maximum lift (CLmax)
and maximum drag.
Minimum power occurs at the speed at which total power(P
occurs when parasite drag is 1/3 of induced drag .In drag coefficient form,
C
R ) is minimum .This
Df = 1/3(CL**2/пAe) .
Explain the significance of load factor.
Ans: Load factor (n) is defined as the lift (L) divided by the weight (W) at the stalling
19)
speed corresponding to C
increasing the angle of attack will result in the reduction of C
C
speed . Hence a load factor of 1, at C
Lmax
. Any attempt to increase the lift by further
L and increase of
D . This will result in increase of thrust required to maintain the flight at lower
Lmax
, only feasible provided enough
power is available. Plot the variation of power available (P A ) with flight speed (V) for a propeller
powered airplane and indicate the effect of altitude on the curve. Ans: Firm line (PAO) on the P A versus V graph shows the power available at sea level for a propeller powered airplane .The dotted line P A σ shows the power available at an altitude where the density ratio, σ = ρ of air at altitude/ ρ of air at sea level ,
P A σ = P A0 * √σ.
20) How load factor (L/W) is related to bank angle?
Ans: During turning, the bank angle (φ) is related to Lift (L) and weight (W) as
follows:
W v
g
W
R .
=
2
L Sinφ = C.F =
i.e, Sin φ =
v
2
gR L
v
2
1
gR
Hence bank angle (φ) decreases when load factor (L/W) increases.
21) Define service and absolute ceiling.
Ans: The system of PR and PA versus V curves, both at sea level (o) and altitude (σ)
are fundamental in determining climb and decent performance of airplane.
Minimum and maximum speeds (absolute ceilings) are determined by the
intersection of the two curves (PR and PA) taken for the same altitude . Any speed
between the maximum and minimum ceiling speeds can be the service speed .
Sometimes in altitude flight, the minimum speed will be higher than the stalling
speed and hence the stalling speed cannot be achieved in level continuous flight.
22) What is the compressibility speed correction of airplane and how Vcorr is related
to true speed V?
Ans: At or near the critical mach number (Mcr) of the air plane , the drag increases
at a greater rate than the parabolic variation . Speed correction curves (thrust
constant ) at different altitudes are available . A correction speed Vcorr is
obtained from the chart for any incompressible speed ,V. By the assumption
of constant thrust for turbojet between V and Vcorr ,the V can be related to
Vcorr as follows:
T = D
ie; CD. q. S = CD corr .q corr .S. Hence V = Vcorr ( CDcorr/CD) .
23) How the speed correction (Vcorr) due to compressibility effect is related to
incompressible speed (V) when speed, altitude, t/c ratio and CL increases.
Ans: For a specific V, Vcorr will be lesser than V and the deviation increases
with the speed , increase in t/c ratio and CL . For each case Mcr is the
upper limit. Above specified deviations between V and Vcorr are further
enhanced in higher altitude . From constant power curves , Vcorr is related to V as follows :
V = Vcorr(CDcorr/CD) 0.333 .
24) Define range and endurance of an airplane.
Ans: For an airplane-engine combination, the range (R- km ) may be determined By multiplying the total fuel (F-kg) available by the average km traveled per kg of fuel consumed .
ie; R (km) = F(kg) * (km/kg )
Similarly for an airplane –engine combination, the endurance (E-hrs) may be
computed by dividing the total fuel (kg) available by the average consumption
of fuel in kg/hr .
ie ; E (hrs) = F (kg) /(kg/hr)
25) For turbojet driven airplane at what V/VL/Dmax β
min or Ψ max occurs . Also for
the turbo-propeller driven airplane at what V/VL/Dmax, R/C
max or R/S min occurs.
Ans: For turbojet driven airplane , from the TA/TL/Dmax versus V/VL/Dmax curve , it is
found that at β min or Ψ max , the value of V/VL/D MAX =1 .
Similarly for the turbo-propeller driven airplane from the PA/PL/Dmax versus
V/VL/Dmax curve, it is found that at R/C
is equal to 0.76 .
max
or R/S
min
, the value of V/VL/Dmax
26) Draw TR versus V graph of turbojet airplane and indicate VmaxE and VmaxR.init.
Ans: VmaxE is the minimum TR point on the graph.
V maxR
is the tangent point on the graph.
27) Draw PR versus V graph of turbo propeller airplane and indicate VmaxE and VmaxR.
in it .
Ans: VmaxE is the minimum PR point on the graph . VmaxR is the tangent point on the graph. 28) What are the conditions for maximum endurance of a jet powered airplane? Ans: Equation for maximum endurance (Emax) for jet powered airplane is:
Emax = (1/c)*(L/D)max * ln (Wo/W1) For maximum endurance: 1) c(1/hr) should be small
2) (L/D) max should be high .
3) (Wo/W1) mass ratio should be high. 29) What is the condition for R min for turning an airplane in level flight? Also give the time (t) for 360 0 turn in terms of V and tanø (ø-turning or yawing angle). Ans: Considering the turning banking angle as ø, tan ø= V 2 /R.g , ie ; R = V 2 /g tan ø . For Rmin ; 1) V should be small, approaching to V stall . 2) tan ø should be high . Time for 360 0 turn (t) = 2пR/V =2ПV/g tan ø.
30) What is the impact of acceleration in R/C as compared to non-accelerating climb?
Ans: Angle of climb (γ ) is related to R/C as Sin γ = (R/C ) /V.
Also Sin γ =(T-D-F1)/W (where T is thrust , D- drag and F1 –inertia for
acceleration), then R/C (m/s) = V(T-D-F1)/W
--(1)
And R/C without acceleration =V(T-D)/W --(2)
If acceleration is not considered (2), the R/C will be higher than R/C with
acceleration (1). That is the time to climb to a particular altitude will be higher
by 25 – 30% than that of without acceleration .
31) Show by graphical solution of range (R), what is the additional fuel(dF) required
to get the same range when empty weight is increased by (W).
Ans: For keeping the range (R) same, the additional fuel (dF) required when empty
weight of airplane is increased by W , can be find out by equating area
A1= A2 , ie; C1.dF =C2. W
(where C1 and C2 are the range per kg at the start and end of flight .C2 will be
higher than C1.
Then dF = W* C2/C1 , ie; dF is greater than W .
32) What is meant by control of an airplane, how longitudinal, roll and directional controls are provided in airplane. Ans: When an airplane is stable, it is necessary for the pilot to be able to control it, so that he can maneuver it into any desired position.
Longitudinal control is provided by the elevator, ie; flaps hinged behind the tail plane. Roll control is provided by the ailerons, ie; flaps hinged at the rear of the Aero foils near each wing tip. Directional control is provided by the rudder, ie; a vertical tail hinged to the stern post . 33) What is meant by stability of an airplane and what way it is different from Balance? Ans: The stability of an airplane means its ability to return to some particular
condition of flight , after having disturbed from that condition ,without any
effort by the pilot. The stability is often confused with the balance or equilibrium
of an aircraft .For eg: when an airplane flies with one wing lower than the other
may often ,when disturbed from that attitude ,return to it . Such an airplane is
out of its balance or trim, but it is stable . Stability is sometimes called inherent
stability .
34) Define the three conditions of stability.
Ans: There is a half-way condition between stability and instability. If the airplane
on disturbance tends to move father away from its original position, it is
unstable. If it comes back to its original position, then it is in stable condition.
But sometimes the airplane may tend to do neither of the two and prefer to
remain in its new position. This condition of airplane is called neutral stability.
35) With the help of C
m vs C
unstable conditions of it.
L curve of an airplane, state the stable, neutral and
Ans: Draw a graph with Cm on Y axis and CL on X axis . In Y axis, represent
Cm =0 above origin to indicate both positive and negative Cm values.
Line (1)&(2) in which dCm/dCL 0 shows unstability or neutral stability
Line (3) in which dCm/dCL 0 stability . When Cm = 0, the airplane is in equilibrium. 36) What are the two methods for predicting fuselage
contribution
to
longitudinal stability of airplane? Write down the formulae for simpler method and explain the terms in it. Ans: Methods for predicting fuselage contribution to longitudinal stability are
(a) Computational method by dividing the fuselage ito many sections (b) A
Shorter method for which the following formulae is used. ( dCm/d CL ) fus = (Kf Wf**2 Lf)/Sw c aw (where Lf is the over all fuselage length ,Wf is its maximum width and Kf is an empirical factor developed from experimental evidence . It depends entirely on the wing root chord’s position on the fuselage. 37) Define neutral point.
Ans: It is the limit of the centre of gravity of the airplane at which the static
longitudinal stability becomes neutral (dCm/dCL) =0. This stability criterion of
the airplane (ie; neutral point) will be different at different operating conditions
like, stick-fixed with and without power ,stick free with power. Also there is aft
c.g limit as well as forward c.g limit . Between these limits of neutral points
only the airplane c.g can be allowed to shift during all maneuvering conditions
of airplane in order to maintain longitudinal stability.
38) What is the criterion for static longitudinal stability?
Ans: For static longitudinal stability of the airplane (AP) the over all value of
(dCm/dCL)AP 0 . This rate of change of coefficient of moment about Yaxis
(dCm) with respect to the change in lift coefficient (dCL) is the summation of
similar values contributed by different parts of airplane , like ,wing ,fuselage ,
engine power, tail/elevator etc. Some of them are positive and others are
negative. The cumulative effect should be such that over all value should be
negative (ie; (dCm/dCL)total
0 ) .
39) State two contributions for static longitudinal stability and indicate them with a
plot .
Ans: In Cm versus CL plot indicate the longitudinal stability contributions of wing
And fuselage, tail alone and the total for airplane. If the total longitudinal stability which is the sum of individual contributions is negative (dCm/dCL )0 then the airplane will be stable . 40) Briefly define wing`s vortex system with figure. Ans: Wing`s vortex system can be represented as shown in figure, consisting of `bound vortex ` located at the wing quarter chord (0.25 c) and a vortex sheet streaming from the wing trailing edge, which will roll up to form the familiar
two trailing vortices . The wing wake center line which is in the vortex sheet is displaced downward and deformed by the influence of the bound vortex and the powerful trailing vortices . The strength of the vortex system is prop-
ortional to the CL
therefore the downwash () at any particular point will be
proportional to CL . ie; = f( CL) . 41) What is Neutral point (N 0 ) of an airplane at stick fixed and power-off condition. Show the new position of ‘N 0 ’ at power-on condition relative to the earlier N 0 .
Ans: The aft position of c.g of the airplane obtained by equating the static longitu-
dinal stability equation (dCm/dCL)
stability at stick-fixed and power –off condition added to it. This c.g limit at
aft end with stick-fixed and power-off gives the limit for the stability . The
new position of the airplane neutral point No` at power-on condition will be
towards the forward side of the No, since the contribution to stability by the
power effect is slightly destabilizing .
total
to zero , with the contribution for
42) What are the two major effects of the running propeller that contribute to the
Longitudinal stability and define them .
Ans: Two major effects are (1) Direct propeller contribution arising as a result of
the forces created by the propeller itself . (2) Indirect effects which arises as
the result of the slip stream from the propeller and its interaction with the
wing and tail surfaces .Propeller forces ,both thrust (T) and normal force(Np)
create moments about c.g of the airplane and hence they have stability contri-
bution (dCmp/dCL) which influence the over all static longitudinal stability.
43) What are the major contributions of the indirect effects of the running propellers
on the static longitudinal stability ?
Ans: There are 4 major contributions making up the indirect effects of the running
propellers on static longitudinal stability . They are the effects of slip stream created by propeller on the following :
a. effect of slipstream on wing-fuselage moments
b. effect of slipstream on wing lift coefficient (CL)
c. effect of slipstream downwash at the horizontal tail
44) Define elevator power and write down the elevator power criterion equation Ans: The magnitude of pitching moment coefficient, Cm, obtained per degree deflection of the elevator is termed as the elevator power .It is the derivative of Cm w.r.t elevator deflection (dCm/dδe =Cmδ )
The elevator power criterion is mentioned below dCm/dδe =Cmδ 45) What are the direct propeller contributions to S.L.S arising due to the forces created by propeller and write their simple forms in terms of (h/c) and (lp/c).
Ans: Direct propeller contribution to S.L.S arising due to forces (T and Np)
created by propeller are thrust (T) contribution (dCmp/dCL)
Force (Np) contribution (dCmp/dCL)NP .
(dCmp/dCL)T
= 0.25 h/c
T and the normal
if c.g of airplane is above thrust line (h/c is +ve ), (dCmp/dCL) is positive
and hence destabilizing . If thrust line is above c.g (h/c is –ve),(dCmp/dCL)
is negative and hence stabilizing .Simlarly (dCmp/dCL)
NP
= 0.02lp/c .
46) Briefly explain the c.g limits with the help of a figure and what decides the
anticipated c.g travel of the airplane.
Ans: Between stick-fixed power-on condition neutral point (No) and forward c.g
limit at C
range required , the more powerful will be the elevator. It can be seen that
anticipated c.g travel of the airplane is decided by the elevator and the
horizontal tail .
L max landing condition ,the c.g of airplane can travel. More c.g
47) How the most forward c.g limit of the airplane is fixed?
Ans: An indication of the most forward c.g permissible comes from the elevator
theory. That is from the requirement that the elevator must always be capable
of bringing the airplane into equilibrium at C L max attainable by it. As the c.g moves forward ,the airplane becomes more stable and more up-elevator is required to trimout C Lmax . Obviously some forward c.g , the elevator will be just powerful enough to attain equilibrium at C Lmax . 48) Write down the expression for the maximum stability attainable by the elevator using its maximum up-elevation . Ans: The expression for the limiting stability can be obtained by substituting δ emax
and C Lmax in the eqation given below :
Elevator up- deflection δ e = δ eo – (dCm/dC L ) C L /C mδ and solving for
(dCm/dC L ) max = (δ eo - δ emax ) C mδ / C Lmax .
(where δ eo is the elevator angle at zero lift and C mδ is the elevator power .) 49) How the forward c.g limit is restricted by the ground effect or for landing maneuvers . Ans: The forward c.g location is limited by the influence of the ground on the
downwash during landing. The variables that will be altered by the ground
effect are wing`s angle of attack (α
in a t = dC Lt /dα t and dЄ/ dα
w ) ,elevator power (C
mδ
) due to increase
t = τ , the elevator effectiveness .
50) Write the total elevator hinge moment coefficient equation and define the
components it .
Ans: Elevator total hinge moment coefficient (Ch) equation can be written as:
Ch =
Ch
Ch
+Ch α α + Ch δ δ
o –is the hinge moment coefficient term at zero angle of attack.
α – the partial derivative of hinge moment coefficient w.r.t angle of attack
of the tail = Ch/∂α
Ch
δ - the partial derivative of hinge moment coefficient w.r.t control surface
(elevator ) deflection = Ch
δ /∂δ
Since most control surfaces (elevator) are symmetric airfoil sections the
Term Ch
o will not be included unless specifically asked for.
51) What is meant by aerodynamic balancing control surface and how it is most
frequently done?
Ans: Methods for controlling the parameters Chα and Chδ (partial derivatives of
hinge moment coefficients with respect to α and δ ) are called aerodynamic balancing control surfaces . One of the most frequently used methods is the set-back hinge. In this ,by suitable mechanism the hinge line is moved aft such that the sum of moments about the hinge line will become smaller . 52) What is a tab and how it is effective? Ans: Tab is a control surface and is an auxillary flap usually built into the trailing
edge of the main control surface . Because of its location, it can create very
powerful moments about the control surface hinge line. It is used as a trimming device,a balancing device and in some cases as a primary control
surface. 53) Get a relation for the control surface floating angle of an elevator fitted with tab also . Ans: Assuming linearity for the hinge moment coefficients, the total control surface hinge moment coefficient (Ch) can be written :
Ch = Ch α α +Ch δe δe +Ch δt δt
At the floating balanced condition, Ch = 0, and δe = δ
Ie; δ float = -
(Ch α / Ch δe ) α- (Ch δt / Ch δe ) δt
float
which indicates that deflecting the tab can change the control surface floating
angle .
54) How the effect of freeing the elevator changes the tail contribution to the
longitudinal stability ?
Ans: The effect of freeing the elevator enters the tail term of the longitudinal
stability as a multiplying factor ( 1- τ (Chα/ Chδe)). If an elevator has a large
floating angle , (Chα/ Chδe) will be large and positive .Then the stability
contribution of the horizontal tail can be reduced sufficiently . For eg: if
(Chα/ Chδe) = 2 and τ = 0.5, the floating of the elevator can make the whole
tail contribution of stability zero. This shows the importance of careful
elevator balance design to ensure proper hinge moment characteristics and
thereby good stick-free stability can be readily appreciated .
55) What is the effect of gradient of stick-force with velocity on the stability of the
airplane ?
Ans: The gradient of stick-force (Fs) with velocity is extremely important, as it plays a major role in determining the pilot`s feel of airplane stability . A large gradient (dFs/dV) will tend to keep the airplane flying at constant V and will resist the influence of disturbance towards changing V. It also enable the pilot to bring the airplane to trim easily and will not require a lot of pilot attention to hold the given Vtrim .
56) What are the two methods in determining the stick-free neutral point (No`) and
define them . Ans: Both methods are through flight test data evaluation .In 1 st method , the non- dimensional stick force equation (Fs/q) is differentiated w.r.t to C L and equate to zero. The c.g position for d(Fs/q)/dC L = 0 will give the neutral point. It can be seen that (No`) will decrease with increase of C L .In the second for getting neutral point (No`) is to obtain flight curves of tab angle (δe) to trim (Fs=0) versus speed for different c.g positions. The slope of α t vs C L is a
function of stick-free longitudinal stability criterion and the neutral point (No`)
is the c.g position when dδ
t /dC L =0 .
57) What are the commonly used gadgetries for improving stick force gradient?
Ans: There are a few new devices now available for giving constant pull force
on the stick . The most common of them are the down spring and bob weight,
vee tab or spring tab. The down spring and bob weight are not very effective
in ground operation where as the spring tabs are more efficient in giving a
constant torque about the hinge line of the elevator.
58) What is the restriction on the aft c.g imposed by the stick-free neutral point
(No`) ?
Ans: The concept of stick-free neutral point (No`) is that c.g location where
(dCm/dCL)free = 0 or where dFs/dV = 0 through trim speed where Fs=0.
This brings a new restriction on the aft limit of the allowable c.g range .
At present the Army and Navy call only for the most aft c.g to be ahead
Of the stick-free neutral point (No`).
(Show the new aft limit of c.g (No`) and usable c.g range in a figure usually
used .)
59) What are the two important maneuvering flights and their essential requirements?
Ans: Flight in curved paths are called maneuvering flight . Two important maneuvering flights (a) that taking place in vertical plane passing through
the plane symmetry of air plane called pull-up maneuvering
(b) that taking
place in horizontal plane called normal turn. In (a) , the net upward force (L-W ) act as the pull-up force perpendicular to the curved path .In (b) the
resultant of lift vector in the horizontal plane perpendicular to the flight path act as the centripetal force . 60) Define stick-fixed maneuvering point (Nm) and stick- free maneuvering point (Nm`) . Ans: If the c.g is moved aft behind (No) and (No`), the stability in accelerated flight will be reduced until at some c.g position the change in elevator angle (dδe ) and the stick force (Fs) required to accelerate flight will be zero. These
c.g`s are termed as the maneuvering points .The c.g where the elevator angle
(δe) required to accelerate the airplane vanishes is the stick-fixed maneuvering
point (Nm) and the c.g where Fs required to accelerate the airplane vanishes is
the stick- free maneuvering point (Nm`) .
61) What is the criterion for longitudinal stability and control in maneuvering
flight?
Ans: The increment of elevator angle (∆δe) and stick force (Fs) to produce an
increment in normal acceleration equal to 1g at constant speed ,both in pull-
ups and in steady turn is the criterion for longitudinal stability and control in
maneuvering flight . That is (dδe /dn ) and (dFs/dn) are criterion .
62) Define stick- fixed maneuvering point (Nm) and where it is located with
reference to neutral point at stick-fixed (No) .
Ans:As the airplane c.g is moved aft of (No) , the gradient (dδe /dn ) continues to
reduce until at some c.g position it vanishes. This c.g position at which
dδe/dn =0 is termed as the stick- fixed maneuver point (Nm). Since Nm – No=
(dCm/dCL)fix , the Nm is aft of No . The difference between Nm and No is the
greatest at sea level and for light wing loads (W/S)min .
63) Define stick- free maneuver point (Nm`) and where it is located with reference to neutral point at stick –free (No`) . Ans: The gradient of the stick force (dFs /dn) do not vanish at stick –free neutral point (at (dCm/dC L ) free =0 ) but shows that , if c.g is moved sufficiently aft of No` , a position will reach where dFs/dn =0 . This c.g position is termed as stick-free maneuver point (Nm`) . 64) What are the restrictions on the c.g travel imposed by the gradient of stick force
per g in modern airplane ? Ans: The upper limit or lower limit of stick-force per g (dFs/dn) required for modern airplanes imposed further restriction on the c.g travel of the airplane for maintaining S,L,S .The forward c.g is limited to max: gradient dFs/dn and the aft c.g is limited by the min: gradient of the same. However it is found that the stick-free neutral point (No`) comes between these two .
65) What is the final c.g travel limit of the airplane for S.L.S considering the flight
upto stick-free conditions ?
Ans: The critical limits on airplane`s c.g are :
1)
On the forward c.g,
(a)
(b)
Maximum gradient ,stick force per g (dFs/dn)
Elevator required to land at C
Lmax
max
2)On the aft c,g :
(a) Power –on stick-free neutral point (No`)
(b) Minimum gradient, stick force per g (dFs/dn)
min
The usable c.g range is between (dFs/dn)
max
point and power-on stick-free
Neutral point (No`)
Short questions and answers(M4 & M5)
66) What is meant by dihedral effect?
Ans: The angle that the relative wind makes with longitudinal axis of airplane is
called sideslip . This sideslip (β) alters the wing’s span wise lift distribution
to create a net rolling moment . This rolling moment due to sideslip is termed
as dihedral effect . It is the measure of change in rolling moment coefficient (dCl) per degree change in sideslip (β = -ψ for straight flight path ). ie; dihedral effect = dCl/dψ or Clψ ( +ve for stable ) 67) Define power of lateral or aileron control. Ans: The power of lateral or aileron control will be expressed as the change in rolling moment coefficient per degree deflection of the ailerons . It is
expressed as dCl /dδ a and it acts in such a way that to counter balance the
dihedral effect so that the wings can be held level from straight flight or maintained at some equilibrium angle of bank (Ф) during turn . 68) What are the basic requirements that are to be fulfilled by the lateral control system? Ans: The basic requirements that determine the size of the control and amount of aerodynamic balance are (a) it should be large enough to provide sufficient rolling moment at low speeds to counteract the unbalance tending to roll the
airplane ;(b) the second requirement is that it roll the airplane at a sufficiently
high rate at high-speed for a given stick force .
69) What is meant by aileron reversal speed?
Ans: The deflection of the aileron will create a pitching moment tending to twist
the wing . When the wing twists it rotates in a direction tending to reduce the
rolling moment created by the aileron. When the speed is high enough, a point
can be reached where the wing twist will just counter the aileron rolling
moment and lateral control will be lost. This speed is known as the aileron
reversal speed. Hence designer should ensure that wings are sufficiently rigid
in torsion so that the aileron reversal speed is higher than the maximum speed
anticipated by the airplane.
70) What are the advantages of sideslip?
Ans: Sideslip can be used (a) to increase the airplane drag and thereby its flight path
angle during an approach for landing, (b) useful in getting smooth aerobatics
such as slow rolls and (c ) finally it can help during flight with asymmetric
power.
71) How the total directional stability contributions of parts of airplane is made
more stabilizing ? Ans: Sum of the directional stability contributions of wings, fuselage and propeller and that of their interference effects will usually give an unstable effect . Hence an additional stabilizing surface must be incorporated not only to overcome the instability but also to give the desired level of directional stability. This stabilizing surface is the normal vertical tail placed as far aft of
the c.g of airplane as practicable.
72) What is the requirement of directional control – rudder? Ans: Although the airplane will normally be in equilibrium at zero sideslip (β=0), there are many maneuvers that introduce yawing moments which are to be opposed by some yawing moment control (directional control) to achieve zero sideslip . This yawing moment control is supplied by pilot by means of a rudder, normally a plane flap attached to the aft portion of the vertical tail. 73) What are the flight conditions or maneuvers that produce unbalance yawing
moments those are to be overcome by rudder?
Ans: One is (a) the adverse yaw moment, which happens due to turn during normal
flight (b) slipstream rotation ; the slip stream behind the propeller has rotational
components which changes the angle of attack on the vertical tail (α
create sideslip .This is critical at high power – low speed flight. © Cross wind
during take-off and landing. (d) spinning recovery by rudder (e) antisymmetric
power flight ,when one of the multi-engines of airplane fails.
v ) and
74) Define rudder power and how it is related to directional stability of airplane.
Ans: The rate of change of yawing moment coefficient (dCn) per degree change in
rudder angle (dδ ) is called the rudder power. That is equal to dCn/ dδ
C nδr . The directional stability of airplane is dCn/dψ = C nψ . If C nδr = -0.001 and
C nψ = - 0.001, it can be seen that 1 0 of rudder will produce 1 0 of yaw.
r
r or
75) What is adverse yaw effects and how it is controlled by rudder?
Ans: Rudder power required to overcome the adverse yaw during rolling maneuver
is usually not very high and not usually used for rudder design .Adverse yaw
is created by the normal action of the aileron together with the yawing
moment created by the twisting of wing itself. These adverse moments are
always critical at low speeds and rudder must be capable of overcoming them at speeds very close to stall . 76) How rudder power is estimated? Ans: Rudder power to overcome the adverse yawing moment (Cn) tail is estimated using wind-tunnel model test data .The adverse yawing moment coefficient due to rolling of the wing (Cn) roll can be estimated theoretically . So the
rudder power required to overcome adverse yaw can be expressed as follows:
dCn/dδ r =Cn δr = ((Cn) roll +(Cn) tail )/ rudder throw (=30 0 ) . 77) Why the rudder is designed to suit one-engine inoperative condition?
Ans: The yawing moment coefficient due to the asymmetric thrust (Cn th ) is proportional to 1/V 3 . The rudder at full deflection gives a constant corrective yaw moment coefficient (Cnr) . The intersection of these two Cn th and Cnr gives the critical speed of airplane below which Cn th is more than Cnr . Hence below the critical speed (near stalling at full power ) the full rudder will not
balance out the moment due to antisymmetric power . Since other types of
yawing moments are not critical , most rudder are designed only to fulfill the
antisymmetric power flight condition.
78) How the floating rudder (stick-free) affects the directional stability?
Ans: When rudder is left free to float in response to its hinge moment, it can have
large effects on the directional stability. It is in the similar way, the floating
elevator affects the longitudinal stability. When the airplane sideslips, the
restoring moment due to the tail will be decreased if the rudder floats with the
wind and will be increased if it floats against the wind . The floating rudder
changes the effective angle of attack of the total vertical tail .
79) What is the criterion to keep the directional stability with stick-free above
certain limit or not to lose much?
Ans: For high speed airplanes which require close aerodynamic balance of rudder
and hence the rudder pedal force to be applied by the pilot should be within
practicable limit. So it is essential that the ratio of derivative of H.M
coefficient with tail angle of attack (α
kept low so as not to lose too much directional stability with stick-free .
v )and rudder power (Ch
δr
) should be
80) What is the relation for the greatest of pedal force (PF) with respect to sideslip (ψ) and give its accepted value . Ans: The greatest of pedal force (PF) w.r.t to sideslip (ψ) can be derived as dPF/dψ = (G q η v S r C r Ch δr /Cn δr ) *(Cn ψ ) free ,where (Cn ψ ) free = (Cn ψ ) fix – ( Ch αv .Cn δr ) / Ch δr . It shows that dPF/dψ varies with V 2 for normal aerodynamic balance . A criterion of 2.25 kgf/deg of sideslip at 240 km/hr
81) What is meant by rudder lock? Ans: A typical curve of floating angle (δ float ) vs sideslip (ψ) can be made for a closely balanced rudder . The rudder angle (δ r ) required to produce the sideslip varies some what linearly upto high (ψ). The pedal force required is
a function of the difference between δ rreqd and δ float . At high ψ, the δ float may catch up to δ rreqd, at which point the PF becomes zero .This point of intersection is called rudder lock. Considerable force is required afterwards,
to break the lock and restore the airplane to zero sideslip .
82) How to avoid rudder lock?
Ans: Rudder lock can be avoided by two ways; 1) To cut down the rudder
effectiveness (dα
v /dδ
r ) , thereby increasing the δ
rreqd
at a given sideslip (ψ).
(2)Another way is to provide a dorsal fin which increases the fuselage stability
at high (ψ) as well as reduces the tendency of the vertical tail to stall .
83) What is the concern of the designer to keep the pedal force required within
suitable practicable values?
Ans: A fighter airplane climbing at full throttle (power) at 288 km/hr may push over
to a dive up to 720 km/hr speed . If the airplane is trimmed out directionally at
climb, the pilot may have to exert a PF as high as 92 kgf to maintain zero
sideslip in the dive. Hence it is essential for the designer to keep the pedal
force changes with speed as low as possible by suitable aerodynamic balance.
84) Why the study on dynamic characteristics of the airplane is necessary ?
Ans: In order to understand the requirements for static stability and control, it is
necessary
investigating the types of motions of the airplane in response to a disturbance
to study the dynamic characteristics of the airplane . It is done by
from some equilibrium flight condition and the nature of transient motions of the airplane in response to the movement of its controls . 85) What way the dynamic stability analysis of the airplane help the design of control systems and the pilot who operates it ? Ans: If the motion of the airplane in response to some disturbance is very slow divergence, the control requirements are different from those needed if the divergence is extremely rapid. The ability of the pilot to react and apply the
controls is a factor which must be kept in mind for all studies of airplane dynamics. The design of controls and the ability of the pilot to apply controls in time , requires some knowledge of the transient response of the airplane to a disturbance or to controls . 86) What are the 4- different modes of motion of a dynamic system when responding to a disturbance from an equilibrium position? Ans: Dynamic system in general have 4- different modes of motion when
responding to a disturbance from its equilibrium position. They are aperiodic
and oscillatory modes with and without damping . That is aperiodic can be
pure convergence if the motion is damped (stable) and divergence if
undamped (unstable) .Similarly oscillatory motions (periodic ) can be damped
oscillation (stable) or undamped oscillation (unstable) .
87) What are the six degrees of motion of a dynamic system and how it is formed
for the airplane ?
Ans: According to the Newtonian laws of motion which states that (a) the sum of
all external forces in any direction must be equal to the time rate of change of
momentum, and (b) the sum of all external moments of forces must be equal to
the time rate of change of moment of momentum, all measured w,r,t axes
fixed in space . The six equations of motion are: Fx = m a
Fz = m a z , L = dHx /dt , M = dHy/dt , N = dHz /dt
the mass , a
and L, M, N are external moments of forces and Hx ,Hy , Hz are moments of
momentum about X, Y, Z axes respectively .)
x , Fy = m a y ,
. (where `m` is
x ,a y , a z
are linear accelerations and Fx , Fy , Fz are external forces
88) How the true acceleration and moment of momentum of airplane w. r. t fixed
axes in space are generated ? Ans: Accelerations w. r. t moving axes (X, Y, Z ) as well as moments of momentum are generated 1 st w. r. t moving axes . But they will not be the true values because as per Newtonian laws they are to be worked with reference to fixed axes in space . To overcome this difficulty use is made of moving axes which coincide in some particular manner from instant to instant with a definite set of
89) What are the different ways the moving airplane axis system can be fixed with
reference to the airplane?
Ans: The moving airplane axis system can be fixed with reference to the airplane in
two different ways : (1) One is to consider the axes fixed to airplane under all
conditions ,and are called body axes in which X -axis is along the thrust axis.
(2) Other possibility is to consider the X- axis always pointing to the direction
of motion or into the relative wind , called wind axes .The wind axes are
usually convenient .
90) What are the equations of longitudinal motion with free control ?
Ans: It involves 3 symmetric equations in the plane of symmetry of airplane (X-Z
plane) and equation of motion of the elevator control about its own hinge.
They are: Fx = m ύ , Fz = -mVѓ , M = I
variables to be considered for dynamic analysis are the change in forward
speed V, ∆α ,change in altitude(∆θ) and change in elevator angle ∆δ
91) What are the different modes and stability criterion of dynamic longitudinal
Y q` and HM e =Ie δ e``. Major
e .
motion whose governing equation is a 4
th
degree quartic ?
Ans: If all the 4 roots (λ`s) come out as real numbers , the dynamic motion is
aperiodic , damped or convergent if the real root is negative and undamped or
divergent if the real root is positive .
If any of the roots (λ`s)
form a complex pair , the motion is oscillatory ;
it will be damped oscillation if the real part of the complex root is negative and
undamped oscillation if the real part is positive .
92) How the stability of dynamic motion can be judged using the coefficients of the
4 th
order quartic which govern the motion ?
Ans: If all coefficients (A to E ) are positive , there can be no positive real root and
no possibility of pure divergence . If the Routh`s discriminant(RD)= (BCD-
AD 2 –B 2 E) is positive , then there is no possibility of complex root with positive
real part and hence no undamped oscillation. If RD is zero, there will be
neutrally damped oscillation and if RD is negative, one root will be complex
pair with positive real part, representing undamped oscillation etc.
Ans: The derivatives
Cl p and Cn r are the damping derivatives in roll and yaw
respectively; where as the derivatives Clr and Cnp are usually referred to as
the rotary or cross derivative . They are the rolling moment (Cl) due to yawing velocity (r) and yawing moment (Cn) due to rolling velocity (p) .
94) What are the characteristic modes of stick-fixed airplane?
Ans: Characteristic modes of stick-fixed longitudinal motion for nearly all airplanes
longitudinal motion of
are two oscillations (1) Long period with poor damping called phugoid mode
(2) Short period with heavy damping is referred to as the short period mode or
second mode (Note: give figures also.) The damping of the phugoid motion is
therefore a direct function of C
the more difficult will be to ensure damping of phugoid mode .
D . The cleaner the designer makes the airplane,
95) What is proposing mode of dynamic motion?
Ans: The damping of long period mode (phugoid mode) can become very weak
under certain design conditions and can become neutrally damped or even
unstable. This mode when neutrally damped is usually called proposing.
Such modes have occurred in high speed airplanes, in which the oscillations
of normal acceleration become so severe that the pilot was injured seriously
and airplane damaged before the pilot could stop the oscillation by slowing
down.
96) How many degrees of freedom are there for lateral dynamic motion and what
are they ?
Ans: There are 5- degrees of freedom for the lateral dynamic motion, because there
are two lateral controls ; the rudder and aileron. The five degrees of freedom
are : a) Velocity along the Y –axis (b) rotation about X- axis (c) rotation about Z-axis (d) rotation of rudder about its hinge (e) rotation of aileron about its hinge. 97) Define spiral divergence in dynamic stability? Ans: This is a characteristics associated with lateral dynamic stability of modern airplane .The divergent mode (undamped) motion is known as spiral
divergence .It can be easily demonstrated in equilibrium flight by giving a kick
to rudder or aileron and watching the response . If it is uncontrolled, steep high- speed spiral divergence develops. Inspection of lateral dynamic stability quartic equation gives an idea about spiral divergence. If E in quartic is zero, it is spiral boundary and when E is negative it is unstable and positive, stable. Usually with C L more than 0.213, the airplane will be spirally unstable. 98) Define Dutch roll and its effects. Ans: Lateral short period oscillation with weak damping are objectionable and
such oscillations are called Dutch roll .In most airplane designs the short
period oscillation with control locked is not objectionable as the damping is
normally heavy .But with control free the Dutch roll can have very weak
damping which is quite objectionable. Neutrally damped Dutch roll can be
investigated by equating the Routh`s discriminant to zero ,ie;(BCD-AD 2 -B 2 E)
= 0, which is oscillation boundary between stable and unstable.
99) What is meant by snaking?
Ans: In most cases with inadequate damping of lateral oscillation, it is tied in with
the floating characteristic of the rudder (δ
certain conditions of aerodynamic balance can introduce a poorly damped
rfloat
) . This characteristic under
lateral oscillation known as snaking.
100) Briefly explain autorotation. (Note: give figure, α vs C
autorotation).
L and show the cause of
Ans: When the wing’s are stalled; ie; beyond the critical angle of attack (α
rising wing will have higher C
unbalanced couple about X-axis to rotate (roll) further which is called
L than the falling wing .This creates an
cr
) , the
autorotation. Rolling occurring in the almost horizontal plane is also a form of
autorotation . 101) Briefly explain the spinning of an aircraft. Ans: Spinning is caused by the autorotation developed by the unequal lift in the wings .In a spin the airplane follows a steep spiral path which is composed of varying degrees of yaw, pitch and roll. A flat spin is chiefly yaw and spinning nose-dive is chiefly roll. The amount of pitch depends on how much the wings are banked from the horizontal . Area and disposition of the fin , rudder and
tail plane exert considerable influence on the airplane to spinning . 102)How to get out of the spin smoothly ? Ans: In order to get out of a spin we must get out of the stalled (C Lmax ) state by putting the nose down (reduce α) and also we must stop rotation by applying `opposite rudder `. In practice the roll is stopped first, because it is found the elevators are not effective (to change α ) until the rotation is stopped . 103) State two basic requirements of aircraft control surface.
Ans: AP control surfaces are mainly ailerons, canard(forward tail), elevator, rudder
and spoilers(special ailerons). Most of the control surfaces are used to produce
additional forces when it is required at different locations of AP. Also they
produce counter-balancing about three major axes of the AP to improve stability.
It aids the pilot to produce control forces to come out of maneuring flights.
104) Distinguish between stability and controllability.
Ans: Stability and control characteristics of an AP are called flying qualities. Stability
is the property of an equilibrium state and it is divided into static and dynamic
stability. Static stability is the tendency of the AP to return to its equilibrium
after a disturbance. In dynamic stability, the vehicle’s motion with the time
after it is disturbed from its equilibrium point is considered. Controllability of a
system is related to its ability to transfer control requirements. A system is
controllable if it transfers any initial state Xi(t) to any final state Xf(t) in finite
time.
105) What is the need for aerodynamic balancing?
Ans: Control surfaces like elevators, ailerons and rudders are used for generating
forces and moments in different directions for controlling the AP. When using,
the methods of controlling parameters(H.M coefficients) Chα and Chδ are of prime interest and are referred to as methods of aerodynamic balancing. If these coefficients are not suitably controlled, the stability and control will be in danger and it will do harm. (eg, In (dC m / dC L ) free eqn, the effect of elevator freeing comes as 1-τ Chα / Chδ and if τ Chα / Chδ = 1, the whole tail contribution to S.L.S will become zero.
106) What is meant by weather cocking effect?
Ans: See Ans to the short question-9 above.
107) Explain the term stability derivatives.
Ans:- For the analysis of dynamic characteristics(Modes of motion) of AP either in
longitudinal or lateral direction, respective simultaneous homogenous
differential equations are to be developed. The constant coefficients of these
equations are made up of AP mass and inert parameters and so called stability
derivatives, like CLα, CDα and Cmα. CLα = dCL / dα which is the slope of the
lift curve with respect to angle of attack. Cmα = (dCm / dCL)(dCL / dα) =
(dCL / dα) (S.L.S) criteria.
PART –B (QUESTIONS& HINTS /ANS)
1) (i) Briefly explain with the temperature-altitude plot , the structure of the
standard atmosphere upto a 105 km altitude.
(10)
Hint: Draw the graph, temperature (K) VS altitude (km) up to 105 km and explain
different temp: gradient as well as isothermal layers upto 105 km.(see p2&3
of ISA notes in module-1)
(ii) Find out the geopotential altitude corresponding to 25km geometric
altitude (Take radius of earth = 6300km) .
(6)
Hint: (See p-5 of ISA Note given in Module-1). Relation between geopotential
altitude(h) and geometric altitude (hg) is ; h = (r/(r+ hg))hg ,where `r` is
ie;
h = (6300/6325) *25 = …… km.
2) (i) Derive the relation for the pressure variation in the gradient layers of
the ISA .
(10)
Hint: Use the relation dp/p = -(go/Ra) (dT/T) and integrate between the base of the gradient layer (p1) and to some point in it (p&T) and get the result , p/p1= (T/T1)**(- go/Ra) (where `a` is the lapse rate.)
(ii) Find the density at 9km altitude from earth’s surface which is on a gradient layer having lapse rate equal -6.5K/km. Take standard sea level as the base and R of air = 287m 2 /s 2 K.
(6)
Hint: Density variation in the gradient layer (ρ/ρ1) is obtained from the relation
ρ/ρ1 = (p/p1)*(T1/T) = (T/T1)**((-go/Ra) -1) ρ/ρ1 = (229.5/288)**((-9.81/2 87*0. 65)-1) = rp , hence ρ at 9 km = ρ1* rp (ρ1 at std.sea level = 1.226 kg/m**3)
3) (i) Derive the relation for pressure variation in the isothermal layers of the ISA. Find out the pressure in the isothermal layer of 165K and at a height of 5km from the base, given the base pressure as 30N/m 2 . (R of
air = 287m 2 /s 2 K)
(10)
Hint: For derivation use dp/p = - (go/RT)dh . Give the base value at h1 as p1, T1&ρ1 and at altitude (h) above base as p, T& ρ . Then integrate dp/p
between p1&p and right-hand-side between h1& h , the derivation is
obtained as p/p1 =e **(-go/RT)(h-h1)
and ρ/ρ1 = p/p1 , since T=T1 (for isothermal layer)
Problem :T=T1 = 165K , (h-h1) = 5 km , p1 = 30 N/m**2, R= 287 m**2/s*2 K
(-go/RT) (h-h1) = (-9.81/287*165)*5000 = -rp
p = p1(e**(-rp)) = 30 e**(-rp))
(ii) Find out the density at the isothermal layers of 165K at a height 10km
from the base, given the base density as 1.2x10
(R of air = 287m /s K)
2
2
-4
Kg/m
3
(6)
Hint: ρ/ρ1 = p/p1 = e**(-rp ), then ρ = ρ1 e**(-rp ) ,
where ρ1 =1.2*10
-4
kg/m**3
4) (i) Derive the relations for maximum L/D as well as V
L/Dmax
and VminPR .
(8)
Hint: For max: L/D , CD /CL should be minimum .
get CD = CDf + CL
equate to zero. Then (CD /CL)min or (CL /CD)max at CDf = CDi = CL
Then get (CL /CD)max = (L/D)max = 0.886 (Ae/ CDf)
From L= W , get V(L/D)max = (2W /ρ S CL)
For V(L/D)max at altitude use ρ at altitude or = Vo/(σ)
2 /пAe . Differentiate (CD /CL) function w.r.t CL and
2 /пAe
0.5
.
0.5 = Vo .
0.5
.
Vmin PR occurs when CDf = CDi /3 or CD= 4 CDi /3 .
This can be obtained by differentiating (C
to zero. Then repeating the above process ( from L=W ) get Vmin PR = 0.76*V(L/D)max at sea level . Vmin PR at altitude = 0.76 Vo/(σ) 0.5
D /C L 3/2 ) function w.r.t C L and equate
(ii) Find out (L/D) max , V L/D max, T L/D max and V min PR of an airplane with the following features:
W = 1200kgf, S = 20m 2 , A = 5, e = 0.8, f = 4m 2 and σ = 0.5kg/m 3
Hint: CDf = f/S
= 4m 2 /20 m 2 = 0.2
(8)
(L/D)max at sea level = 0.886(Ae/ CDf) 0.5 =0.886*20
V(L/D)max =(W/σS CL) ,at (L/D)max , CDf = CL**2/пAe , CL =(0.8 п)**0.5. T(L/D)max = D (L/D)max = W/(L/D)max =1200*9.81/((L/D)max
V min PR
= 0.76 * V(L/D)max
5) (i) Explain with the help of a simple pendulum, the characteristics of static and dynamic equilibrium and how its stability can be evaluated.
Hint: Draw the figure given in the note and at position A (top ) it is statically unstable and at position B (bottom) it is statically stable . Dynamic stability at bottom (position B ) can be analyzed by response analysis . In dynamic
sense it will be neutrally stable at B .
(ii) Derive the equilibrium equations for the longitudinal degrees of freedom
along a straight path and get the expressions for uniform flight velocity
and velocity of climb.
Hint: For longitudinal degrees of freedom along a straight line
L = W cos r
Uniform velocity V =(2Wcos r /ρ C
Rate of climb =Vc = R/C = (T-D)V/W
= 0.5 ρ C L V
2 S
L S )
0.5
6) (i) Derive from the fundamental equation for R/C, the non dimensional form
of thrust available (T
A ) vs velocity relation and plot T A /T
L/Dmax
vs
V/V L/Dmax curves. Also prove that angle of climb is a maximum (γ max ) at
V/V L/Dmax = 1.
(8)
(8)
(8)
Hint: R/C = (T A – T R )V/W . Multiply & divide R/C equation by T L/D max V L/Dmax
and put 2 T A /T L/Dmax =y1 , 2T R /T L/Dmax = Y and V/V L/Dmax = x
also we know that T
R = AV 2 + B/V 2
and T
L/Dmax
=A V 2
L/Dmax
+ B /V 2
L/Dmax
2 T R /T L/Dmax = ( V/V L/Dmax ) 2 + 1/( V/V L/Dmax ) 2 ie; y = x 2 + 1/x 2
The max: R/C or R/S occurs at maximum T .V or maximum
x y = x (y1 – y) = x(y1- x 2 -1/x 2 )
d(x y)/dx =0 = y1-3x 2 + 1/x 2
= 3x 4 -y1 x 2 -1 =0
x 2 =(y1+/- (y1 2 +12))/6 or x = ((y1 +/- (y1 2 +12))/6) 0.5
Also Sin r =(R/C)/V =(T-D)/W
ie; r = Sin -1 (T-D)/W =
For max: r , dr/dx =0 = -2x + 2/x 3 , -2x 4 + 2=0 ,
ie; x = 1 = V/V L/Dmax
(ii) Find out the velocity ratio (x) for max: R/C, when y 1 = (2T A /T L/Dmax ) is 2 and
4and also obtain the flight velocity of air plane if its V L/Dmax = 500km/hr.
(8)
Hint: x= ?, 2 T A / T L/Dmax =Y1 = 2 & 4 ,
(a) x = √(2+/-(4+12) (b) x= ((4+/-(16+12)
0.5
)/6) =1 , ie; V = x .V L/Dmax =500 km/hr
0.5 )/6) 0.5 = a , ie; V = a. V L/Dmax
7) (i) Derive the expression for drag polar and explain it with a neat plot .
(6)
Hint: See page 12-13 of Note of Module-1 (after ISA note)
P
(ii) Draw the thrust / power required (T
A ) curve for a jet engine and piston engine and state your observations.
R / P
R ) and thrust / power available (T A /
Hint: Page- 21 for jet engine & p-23 for turbo-propeller engine .Show both sea
level and altitude curves .
(10)
8) (i) Derive the two equations given below and represent it graphically .
D = A V 2 + B/V 2
; P = A V 3 + B / V ,
(6)
Hint : D= D Pf + D i or C D = C Df + C Di (=C L 2 /πAe)
= σ f V 2
+ (W/b) 2 /(σ e V 2 ) ie ; D= A V2+B /V 2
(1)
P = D.V =A V 3 + B/ V
……(2)
(ii) An aircraft weighing 25 k N has a wing area 100m
2 and its drag coefficient is
C D = 0,016 + 0.04 C L
2 , calculate the minimum thrust required for straight and
level flight ,and the corresponding true air speed at sea level and at 10km
(√σ =0.58).Calculate also the minimum power required and the corresponding
true air speeds at the above conditions .
(10)
Hint: Given C D =0.016 + 0.04C L 2 ie; C Df = 0.016, For min: drag C Df =C Di
ie; 0.016 =C L 2 /πAe ,or 1/ πAe =0.04 or Ae = 25/π ,
(L/D) max = (C L /C D ) max = (0.4)/2C Df = 20.8
V L/Dmax = (W/σ S C L ) =(2500/3.33) 0.5 = V O , Then Vσ =Vo/ √σ =V O /0.58
T min = W/(L/D) max = 25000/20.8 , P min = T min V minPR ; But V
L/Dmax
V minPR =0.76
V minPR = 0.76*V O , And V minPR at altitude = V minPR / √σ
9) (i) Assuming parabolic polar variation for C D vs C L , write down the thrust required
(T R ) and power required (P R ) equations of an airplane. Give the conditions
speed for minimum P
R and minimum T
R in terms of drag coefficients also.
of
(8)
Hint : Use T R = T RP + T Ri
= σ f V 2 + (W/b) 2 /(σ e V 2 )
Also P R = P RP + P Ri =
Speed at minimum T
Form C D = C Df
C D / C L , w.r.t
Then , the condition C
Then (L/D) max =(C L /C D ) max = C L /2 C Df = 0.886 (Ae/C Df )
Next using relation , L = W , V = (W/σ S C
σ f V 3 + (W/b) 2 /(σ e V )
R
2
L
is when (L/D) max or (C D /C L ) min
/πAe ; To get the (C D /C L ) min
+ C
, differentiate the equation
C
L and equate to zero .
Df
= C
2
L
/πAe is obtained .
L ) 0.5 = V L/Dmax
For P Rmin = T min V minPR , But V minPR = 0,76 V L/Dmax , T min = W/(L/D) max
(ii) For the maximum L/D as well as minimum P
R conditions of an airplane, find out
(L/D) max , V L/Dmax , T L/Dmax and V min PR
with the following features:
2
W= 1500kgf, S= 20m , A= 5, e= 0.8, f = 4m2 , σ = 0.5kg/m
3 (8)
Hint: Use the above method given for (i) and solve problem.
10) (i) Derive from the fundamental equation for R/C, the generalized form of thrust
vs velocity curves and show that T R /T L/Dmax =1 when V/V L/Dmax = 1.Also
indicate in the T A /T L/Dmax vs V/V L/Dmax graph, the regions of max R/C and min
R/S and speed for β min or γ max
(8)
(ii) Find out the velocity ratio (x) for max: R/C, when y 1 = (2T A /T L/Dmax ) is 2 and 5 and also obtain the flight velocity of air plane if its V L/Dmax = 600km/hr.
Hint: x for maximum
R/C = ((y 1 +/-(y 1 2 +12 ) 0.5 )/6) 0.5
(8)
When y1 = 2 , x = ((2+/-(4+ 12 ) 0.5 )/6) 0.5 = 1 ,hence V = V L/Dmax
When y1 = 5 , find out x by the above equation , let it be equal to `a` Then V = a. V L/Dmax
11) (i) Derive the relation for the maximum range of a turbojet airplane.
Hint: (See Page 23 of Note of Module – 2 )
(10)
R max = (2/c`)(CL 1/2 /C D ) max (W O /σ S) 0.5 (1- (W 1 /W O ) 0.5 )
(ii) Calculate the maximum endurance of a turbojet airplane having the
following flying characteristic.
c’=2.5hr -1 , W 0 /W 1 =1.2,
S= 40m 2 ,
e=0.8
and
A= 10,
f=2m 2
Hint: For maximum endurance, flight V =V
L/Dmax
E max = (1/c`)(L/D) max ln (W O /W I ) ; where L/D max =0.886 b(e/f)
and b = (A S ) 0.5 =20 m, then get L/D max and E max by substitution
0.5
12) (i) Derive the relation for the maximum range of a turbo-propeller airplane.
Hint: See page 16-17 of Note of Module -2, R
max =k1(L/D) max
(η/c) ln(Wo/W1)
(6)
10)
(ii) Calculate the maximum endurance of a turbo-propeller airplane
having the following flying characteristic.
Hint: E
max
η= 0.85, C’=0.45hr
C L =1.2 ,
-1
, W 0 /W 1 =1.2,
k=1.5m/s,
0.5 -1)
C
1
L
D = 0.083 , Wo/σS =100m 2 /s 2
5
/C
D ) (σS /W
O )
0.5
((W /W )
0
1
= k (η/c) (C
(6)
13) (i)Derive the relation for the take-off ground run distance (S) for an airplane in
terms of weight (W), take-off velocity (V TO ), effective thrust (Te) @0.7 V TO (8)
Hint: See page 31-32 of Note of Module -2
Sg =
W V 2 TO /(2gTe @0.7 V TO )
(ii ) Find out the take-off ground run distance for an airplane having W=1500kgf,
V TO =360km/hr and T E at 0.7V TO =3kN and then find the horizontal distance
to climb an vertical obstacle of 15m, if the ground run distance 0.8 times of total distance. Hint: Sg = 1500*9.81*10 4 /(2*9.81*3000) = 2500 m
Sg = 0.8* S tot , ie;
S tot = Sg / 0.8 = 3125 m ; S c =625 m
is
(8)
14)(i) Derive the relation for the total landing distance from over a 15m height
obstacle consideringV 15 =1.3V S and V L =1.15V S ,whereV S is stalling speed
Hint: S A = (W/F)((V 2 15 – V 2 L )/2g
+15
Sg = V 2 L / 2(-a)
, S tot =S A + Sg
; V S =100 m/s
(10)
(ii) Find out the total landing distance required for an airplane having W/F
(average resistance coefficient) 10, and stalling speed 360km/hr. Take
average deceleration during ground run as 2.3m/s
(6)
Hint: Use the above method given in Hint 14(i) and solve .
15) Derive from fundamentals that the relation for the maximum rate of climb,
(
R
/
C
)
max
=
η
P
W
K
W
C
σ
s
Df
0.5
and give discussion on it.
1.155
(
L
/
D
)
max
Ans: Refer page 12a of Module-2 Note.
16) For a straight and level flight, velocity for minimum power is 0.76 times velocity
for minimum drag.
Ans: Give the derivations for the velocity (V
and minimum power required(V
(Details given in page 4 & 4b of Module-2 Note)
(L/D)max
minPR
).
) for minimum drag(D=TR)
Descriptive Questions and hints(M3)
17)(a)With the help of a figure, write down the moment equation about the c.g of an airplane flying in power-off and stick fixed condition. Then show the conditions for equilibrium and longitudinal stability. Hint: See page 4-6 of Note of Module -3
(16)
18)(i) Derive the relation for Neutral point (N 0 ) of an airplane at stick-fixed Power-off condition. Hint: See the page 19-20 of Note of Module -3
(ii) Find out the type of stability of an airplane at the stick-fixed, power- off condition which is having neutral point at 55% of chord and c.g
located at 45 cm aft of a.c . Also state the stability condition if c.g is at
65cm aft of a.c. (Take ‘c’ of airfoil = 1m).
Hint : Neutral point at the stick-fixed power-off condition (No)= 55% of `c`
(c =1m ), which is aft of aero-dynamic centre (a.c) of the wing .
(1) Stability when c.g is at 45 cm aft of a.c = 45% of `c`,
Ie; dC m /Dc L = x
c.g
– No = - 10% of `c` , (stable)
(2) Stability when c.g is at 65 cm aft of a.c = 65% of `c`,
Ie; dC m /Dc L = x
c.g
– No = + 10% of `c` , (unstable)
19) (a) From fundamental, write down the equation for the Wing contribution
to Static Longitudinal Stability (S.L.S) and from the simplified form
explain how the location of c.g of airplane with respect to a.c of wing
decides the S.L.S of the airplane.
Hint:See page 7-9 of Note of Module -3
20)(i) Derive the equilibrium and S.L.S equation of a turbo-propeller airplane
taking the direct thrust alone of the propeller.
Hint: See page 22-23 of Note of Module- 3
(ii) Using the simplified terms of (dC
mp /dC L ) T show graphically (in C m vs C L
graph), the positions of,
a)
b)
c) if dC m /dC L = -0.1 for h/c= 0
dC m /dC L for h/c=0.2, dC m /dC L for h/c= -0.2,
Hint: dCm/dC L = -0.1 for h/c = 0 And (dCmp/ dC L ) T = 0.25 h/c
dCm/dC L
for h/c = 0.2 , (dCmp/ dC L ) T = 0.05, dCm/dC L = -0.1+0.05
= - 0.05 (stable )
dCm/dC L
for h/c = -0.2 , (dCmp/ dC L ) T = -0.05, dCm/dC L
(10)
(6)
(16)
(8)
= -0.1 - 0.05 = -0.15 (more stable)
21)(a)Discuss in detail the power effects on static longitudinal stability for a jet
powered airplane . Hint: Details given in pages 32- 34 of Note of Module – 3
(16)
22 (a) (i) How most forward c.g. for free flight of an AP is fixed and write the relation for (dC m /dC L ) max in terms of elevator deflection, its power(Cmδ) and
C Lmax .
(6)
Hint: Details given in page 41 of Note of Module – 3
(ii) Derive the relation for x
cg forward from fundamental C mcg equation at
power-off equilibrium with elevator deflection.
Hint:- Details given in page 44 of Note of Module – 3
22 (b) Write short notes on:
(i) One engine in-operative condition
(ii) Aileron reversal and adverse yaw.
Hint:- Refer note-Module-IV.
(10)
23(a) (i) Show that elevator angle for trim is given by
δe trim = - (Cmo C Lo + Cm α CL trim ) / (Cm δe CLα – Cm α CL δe )
Ans:- An AP is said to be trimmed if the forces and moments acting on the AP are
in equilibrium. Setting the pitching moment equation equal to zero, we can solve
for the elevator angle required to trim as follows.
Cm = 0 = Cmo + Cm α α + Cm δe
Or, δe trim = - (Cmo + Cm α α trim ) / Cm δe
The lift coefficient to trim is CL trim = CLα α trim + CL δe δe trim -----eqn(3)
We can use this equation to obtain the trim angle of attack(α trim ).
δe------------------------eqn(1)
------------------eqn(2)
α trim = CL trim - CL δe δe trim
/ CLα---------------eqn(4)
If we substitute this eqn(4) back into eqn(2) we get the following eqn for elevator angle to trim:
δe trim = - (Cmo C Lo + Cm α CL trim ) / (Cm δe CLα – Cm α CL δe ) (ii) Discuss the advantages and disadvantages of CANARD configuration. Ans:- A canard is a tail surface located ahead of the wing. It has several attractive
features. The canard, if properly positioned, can be relatively free from wing or
propulsive flow interference. Canard control is more attractive for trimming the large nose-down moments produced by high lift devices. To counteract the nose- down pitching moment, canard must produce lift that will add to the lift produced by the wing. An aft tail must produce a download to counteract the pitching
moment and thus reduce the airplane’s overall lift. The major disadvantages of the canard is destabilizing contribution to the aircraft’s static stability. However, it is not a severe limitation. By the proper location of the C.g, one can ensure the AP is
statically stable.
24(a) Using the floating angle of elevator (δe) and its H.M coefficient, derive the
relation for the stability criteria at stick-free condition. Also get the relation
for No-No’(difference between stick-fixed(No) and stick free(No’) neutral
points).
(16)
Hint:- Details given in pages 62-65 of Note of Module – 3.
25 (a) (i) Define stick-fixed maneuver point (Nm) and stick-free maneuver
point(Nm’). Write down the expression for Nm and Nm’ in terms of No
and No’ respectively. (8) Hint:- Details given in page 81 and pages 83 & 85 of Note of Module – 3. (ii) Briefly explain the limits on AP’s c.g with the help of a figure. (8)
Hint:- Details given in page 87 of Note of Module – 3.
26) (a) Discuss in detail the contribution of various components of the airplane
on its static directional stability .
(16)
Hint: See pages 10-13 -briefly of Note of Module -4)
27) (a) Discuss briefly the following :
(1) Directional control and the basic requirements of the rudder .
Hint: See page 16-17 of Note of Module – 4
(2) Rudder lock and how to avoid it.
(8)
(8)
Hint: See page 24- 25 of Note of Module -4
28) Discuss briefly the following :
(1) Basic requirements of the rudder (6) Hint: See page 16-17 of Note of Module -4 (2) Aileron reversal ; (Hint: See page 7 of Note of Module -4 ) (5) (3) Adverse yaw ; (Hint: See page 18-19 of Note of Module-4) (5)
29) (a)(i) A statically stable aircraft can be dynamically stable or unstable .
Give the dynamic conditions under which it happens
(6)
Hint:A statically stable aircraft can be dynamically stable if its dynamic
modes of motion are damped (Both aperiodic and periodic motions. Total
4 types of modes can be possible). Similarly a statically stable aircraft
can be dynamically unstable if its dynamic modes of motion are undamped
(Both aperiodic and periodic motions. Total 4 types of
possible) .
modes can be
(ii)Discuss various stability derivatives relevant to longitudinal dynamics .
Hint: See page 16-17 of Note of Module -5
(Explain aerodynamic coefficients like , C
above pages).
30)(a) Write short notes on the following :
D ,C L , C Dα , C Lα etc; given in
(1)Autorotation (Hint: page 48-49 of Module- 5 )
(2) Dutch roll
(3) Spiral divergence (Hint: page 43 & 45 of Module-5 )
(Hint: page 43
of Module- 5 )
31) Discuss briefly the following:
(i) Phugoid motion
(Hint: Page 23-24 of Note of Module- 5 )
(ii) Stability derivatives in longitudinal dynamic
Hint: See page 17 of Note of Module – 5.
(10)
(6)
(5)
(5)
(8)
(8)
------------------END------------------- | 3.578125 | CC-MAIN-2020-50 |
And that might be conservative. I’ve seen people projecting him to steal 60, 70, or even 80 bases since that’s what recent league leaders have done. However, I see no reason to try to project Hamilton’s steals by comparing him to recent league leaders. He is not any of those other players and has demonstrated that he is clearly an anomaly in terms of speed, which means that it makes sense to treat him as an individual case. Anyway, here’s my thinking, assumptions, and the Billy Hamilton steal calculation:
500 at bats x .295 OBP x running 80% of time x successful 85% of time = 100 SBs
Well, it actually equals 100.3, but who’s counting? Let’s go through the assumptions one at a time. Could he get less than 500 at bats? Sure, but it’s also possible that he gets more than 500. It’s important to remember that even if he starts fewer games, he could still make up for it by entering the game as a pinch runner. I think a .295 OBP is a reasonable expectation, despite concerns about his hitting. He is fast enough where he should be able to beat out routine ground outs, which, in turn, should raise his batting average on balls in play significantly. I think 80% of the time he’s on first base he’ll be running because, while he’d like to attempt every time, sometimes he may have a runner blocking him at second base. This does not include the potential for him stealing third base, which could add double digit steals next season. Hamilton being successful 85% of the time also seems reasonable to me based on his numbers in the minors and majors.
Now, back to what is always in the back of my mind: OPS leagues. In all likelihood, he will not have an OPS that will help your team. Although, he will have the great benefit of allowing you to carry fewer players who provide steals at the expense of your on-base and slugging categories. Additionally, since he may enter some games as a pinch runner, he will occasionally be able to provide you with steals without impacting your OPS. His OBP could surprise since he was able to walk at a decent rate in the minors, though it remains to be seen how well that will translate to the majors. Regardless, I’m definitely interested and plan to pick him a couple rounds earlier then I think he will normally be drafted, since I anticipate that questions about his hitting will suppress his price.
Tom Jacks is sporadically on Twitter @votetomjacks
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## 35 Responses (Jump straight to the comment form)
1. apoxonbothyourhouses says:
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if you think he’ll have a .295 OBP, then he’ll steal 100.
However, that .306 OBP in AAA and CAREER 451:215 K:BB ratio in the minors is VERY alarming. I hope the kid can put it together, as his speed is otherworldly and can single-handedly win people fantasy leagues. Yet, I’m not confident he’ll be anything other than Dee Gordon with more hype.
• Tom Jacks says:
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@apoxonbothyourhouses: True, his AAA wasn’t what you’d like to see. However, I don’t completely discount his minor league numbers in 2012. I believe he has a real chance of producing a slightly less than .300 OBP, which may be a little too optimistic, but also isn’t something completely unrealistic.
2. Matt L says:
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Your assumption that he’ll attempt 80% of times on first seems wrong. If the guy in front of him has a .300 obp, wouldn’t we expect he’ll be blocked at 2nd 30% of the time? And the guy in front of that might still be on, too. I’m sure there are statistics on what percent of time average hitter (or better yet, reds 8 hitter) batted with no ducks on pond. Or what percent of time Hamilton did in minors. Either of those would be better starting assumptions.
• Tom Jacks says:
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@Matt L: My thoughts are what Jeff said below about a 300 obp not meaning the guy is on second when you’re on first 30% of the time. I agree that it would be helpful to use the percentage based on where he’s batting but, at least when I wrote this, it wasn’t clear where Hamilton would be batting. I also imagine that that number would vary depending on the obp of the hitters in front of you.
3. Count de monetball says:
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Tom, I have to agree with Matt L. While 100 SB’s are clearly possible, running 80% of the time seems way high to the reasons matt L. noted. 60-70 seems like the ceiling due to runners in front of him. If you added in 10-20 steals of third base beyond the formula you presented, that seems a much more likely scenerio – 500 at bats x .295 OBP x running 60% of time x successful 85% of time = 75 SBs +15 steals of 3rd base =90 SB’s. I think it would take an OBP higher than .295 for him to reach 100 SB’s. It would probably be easier to project this if you knew how many times he steals third base per year, as I just guessed 10-20 times and concluded at 15… Just a thought, but nonetheless, he will be exciting this year.
• Tom Jacks says:
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@Count de monetball: I think it will be less than the 30% being blocked as Jeff noted below, but I agree that 80% may be the best case scenario. I like your third base steals projection, 10-15 seems possible for Mr. Hamilton. And yes, it should be a very exciting year. I’m just crossing my fingers that he reaches that number of at bats…
4. Jeff says:
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.300 obp doesn’t mean 30% singles/walks. Sometimes guys get extra base hits. Plus he’ll lead off innings sometimes. I don’t know the exact number, but I don’t think a guy on first is blocked by a guy on 2nd anywhere close to 30% of the time.
• Tom Jacks says:
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@Jeff: Thanks, you answered that perfectly for me.
5. Big AL says:
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So what do you think his average ADP will be ? what round do you think you’ll have to draft him ? In a 12 team H2H Points league what does that translate in fanasty points if he gets the 100 steals. Can he hit .275 ? If he can he’ll probably score 120 runs. If-if-if . Is he worth the risk of a mid pick for his upside ? I guess we will see.
• Tom Jacks says:
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@Big AL: That’s a good question. I think he can hit around .275. I also think his value will be significantly higher in standard leagues compared to points leagues. I’d definitely use an early mid-round pick for him in standard leagues, but likely wait til a late mid-round pick for him in points leagues. I’ll admit that I’m less familiar with points leagues though.
6. tgk says:
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The question is where Billy will be drafted, and strategically is it better to own one guy who will take care of all your speed needs or is it safer to spread it around to 3-4 players who can steal 20+? If he’s projected to be the unquestioned league leader in steals, that would probably put him in the first 3-4 rounds?
• Tom Jacks says:
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@tgk: It could be safer to spread the steals around, but I think he’s worth the gamble depending on the price. I wouldn’t use a top 5 pick, but I could see using a 6-10 pick, as the season draws closer and we have a better idea on the team’s plans for him.
• jvetter says:
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He went round 5 pick 74 in one of the first NFBC draft champions league this year (real \$\$\$ on the line – 15 teams – standard roto).
• Tom Jacks says:
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@jvetter: Interesting. It would be tough for me to use more than my 6th pick in a 10 teamer, so 51-60th overall.
7. Carnac says:
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Billy Hamilton excites me. And Mrs. Carnac is ok with that because she knows I won’t stray too far because I’ve been burned before. See: Grey’s 2012 Dee Gordon pre-season profile; https://razzball.com/national-dee-keeper-society/
Also, see the similarly yawnstipating BB rates, lousy K%, playing time issues, the fact that they both weigh less than me, etc. Put it together and it’s a risky play for single (maybe double) category success. Might I miss out on a lot of excitement by not considering Hamilton in the first several rounds? Sure. And I’m ok with that.
• Tom Jacks says:
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@Carnac: Yeah, the Dee Gordon comparison is tempting. I think he’ll be better than Gordon, but the question is how much better and will it be enough to maintain ML success.
• fri-guy says:
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@Tom Jacks: i’m better than dee gordon. almost.
• Tom Jacks says:
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@fri-guy: Ha! I couldn’t argue with you there…
• Carnac says:
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hey now, don’t pick on people smaller than you! *opens tiny locker and a sheepish Dee Gordon emerges*
• Tom Jacks says:
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@Carnac: Leprechaun X: In the big leagues?
• P Swayz on a Horse says:
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@Tom Jacks: no, no, X is the one where he went to Vegas and killed Grey and his groomsmen. You’re thinking of a XII.
• Tom Jacks says:
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@P Swayz on a Horse: Ha! That would be a terrible way to go…
8. Swagger Jackers says:
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Good to see you again Tom Jacks.
In my OBP and QS roto auction league, pick 5:
Cano (\$21)
Scherzer (\$15)
Gomez (\$11)
Bautista (\$10)
Minor (\$6)
Myers (\$4)
Donaldson (\$4)
Segura (\$4)
Billy Hamilton (\$4)
Wacha (\$4)
• Tom Jacks says:
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@Swagger Jackers: Good to see you too. I’ll pick Cano, Scherzer, Myers, Segura, and Wacha.
9. Wake Up says:
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Per usual, I will be looking to grab HRs and OPS at the point in the draft where Hamilton will probably go…30 ribbies and .240 avg are both significant losses, as well…
• Tom Jacks says:
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@Wake Up: Yeah, I agree. In OPS leagues, his value drops quite a bit compared to standard leagues.
10. danny almonte says:
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Hey Tom, in a new dynasty league in a vacuum, is it crazy to take machado over c.Davis?
• Tom Jacks says:
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@danny almonte: No, I don’t think that’s crazy in a dynasty league.
• robdouth says:
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@danny almonte: In a dynasty league in a vacuum I’d be more worried about being able to breath…
11. me says:
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He only attempted a stead 55% of the time he was on base at AAA last year. I find it hard to believe that the Reds will run him THAT MUCH more than his AAA team did.
Considering his minor league numbers, 80% seems unreasonably high for attempts to steal second. Let’s look at the numbers. I’ll make some assumptions that are favorable to the 80% theory:
1. He never hit into a force out (which would drive up the number of times he’s on 1B)
2. All of his Minor League steals are attempts to steal second.
Based on those assumptions:
2013: Hamilton attempted to steal 65% of the time he was on first.
2012: 91%
2011: 70%
2010: 59%
2009: 85%
So, in five minor league seasons, he’s done it twice. I think we can throw out rookie ball, because it’s rookie ball and because he only had 180 plate appearances, so the numbers seem not sufficiently high to be meaningful. That leaves 2012, when he was running every time he was on base to try to set a record. I’m not prepared to do a full PECOTA-style analysis, but I see that as an outlier; the 59-70% of 2010, 2011, 2013 seems much more reasonable.
Even if he steals third at Ricky Henderson rates, this seems like a gross overestimation of attempted steals.
(Based on his walk drop off last year, I’d also be concerned about his OBP — one wonders if his high walk numbers at the lower levels reflect patiences against bad pitchers, which didn’t translate at AAA and won’t translate at MLB. On the other hand, I a .295 OBP seems far more feasible than an 80% steal attempt rate.)
• Tom Jacks says:
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@me: I think the difference between our views is that I’m weighing his 2012 numbers more than you are. I’ll admit that 80% is probably at or close to the best case scenario, but I see Hamilton as being the outlier player who would be able to reach that mark.
His AAA was certainly disappointing, but his very brief ML numbers were encouraging. I was especially impressed that he was able to steal so many bases in the majors when he was coming in the game as a pinch-runner, and even beating pitch outs.
12. PBMax says:
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I currently own Hamilton in an OPS Auction League for \$10 which, with a \$260 cap and all things being equal, seems a better value than Matt Moore, Boegarts, Uehara, Sano, all for \$10 as well, right?
I picked him up once he was called up and I must have seen 3/4 of his mlb steals, and it was ridiculous. So through my rose-colored glasses, I don’t see him running less than 70% of the time, and don’t see why a 285-295 OBP isnt possible.
• Tom Jacks says:
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@PBMax: I think it’s close with Boegarts. Just a tradeoff between some power and OPS compared to a lot more speed.
13. don says:
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rotochamp has him @ 0.323 obp. / steamer (these guys are genreally pessimistic) has him at .305 obp.
why are people here acting as if below .300 is the normal expectation?
• Tom Jacks says:
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@don: Sorry- I thought my reply went through a couple days ago… Thanks for bringing up those projections. I think the concern about his obp stems from him having the “all speed, no bat” label, which has led to him being perceived a little too negatively.
Comments are closed. | 2.78125 | CC-MAIN-2018-09 |
# Biology
Which of the following does not function is suspension feeding?
This makes no sense.
I am not sure what you are asking, since you did not indicate what is "following."
I searched Google under the key words "suspension feeding" to get these possible sources:
http://hypnea.botany.uwc.ac.za/marbot/kelp_beds/key_inhabitants/suspension_feeders.htm
http://www.wildsingapore.com/chekjawa/text/i021.htm
I hope this helps. If not, respond to this post with more details as to what you are seeking. Thanks for asking.
1. 👍
2. 👎
3. 👁
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# $\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} n^{1/k}$
What would you suggest here?
$$\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} n^{1/k}$$
-
looks a bit like a riemann sum for me – Dominic Michaelis Feb 18 '13 at 11:50
Seems to be really difficult, mathematica doesn't give me a result, you already know its converging ? – Dominic Michaelis Feb 18 '13 at 11:54
@DominicMichaelis: if you take $n$ finite, like $n=10^6$, W|A gives an answer that is $\approx 2$. I think the limit should be $2$. – Chris's sis Feb 18 '13 at 11:57
Is the question to prove convergence or to find the limit? – Ishan Banerjee Feb 18 '13 at 12:00
learn more approaching ways, do you mean you have one solution already? – user45099 Feb 18 '13 at 12:02
AM $\ge$ GM seems sufficient to give a simple and elementary approach.
For $k \ge 2$, we have, taking $k-2$ copies of $1$ and two copies of $\sqrt{n}$, that
$$\frac{1 + 1 + \dots + 1 + \sqrt{n} + \sqrt{n}}{k} \ge n^{1/k} \ge 1$$
i.e.
$$1 - \frac{2}{k} + \frac{2 \sqrt{n}}{k} \ge n^{1/k} \ge 1$$
Thus
$$(n-1) + 2(H_n - 1)(\sqrt{n} - 1) \ge \sum_{k=2}^{n} n^{1/k} \ge n-1$$
And so
$$2n-1 + 2(H_n - 1)(\sqrt{n} - 1) \ge \sum_{k=1}^{n} n^{1/k} \ge 2n-1$$ Where $H_n$ is the $n^{th}$ harmonic number.
Divide by $n$, and by the squeeze theorem, we have that
$$\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n} n^{1/k} = 2$$
This AM $\ge$ GM idea was also used in the answer here: Proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$ and a slight variant here: How to prove $\lim_{n \to \infty} \sqrt{n}(\sqrt[n]{n} - 1) = 0$? and
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This solution looks amazing. Such an answer I expected. Very very nice. – Chris's sis Apr 17 '13 at 8:06
Please upvote this answer. – Chris's sis Apr 17 '13 at 8:08
I love this answer so much! :-) – Chris's sis Apr 17 '13 at 8:25
It is a nice solution. I like it. – xpaul Apr 17 '13 at 12:25
@Chris'ssisterandpals/@xpaul: Glad you liked it :-) – Aryabhata Apr 18 '13 at 0:25
HINT: Looking at the sum, there are two major sources of contribution - the first few terms are large, but there are also lots of small terms on the tail that add up. So we must bound them separately, as any bounding of all terms at once will be too coarse. So separate the first $m$ terms from the rest, and estimate each part. Then look back at how to choose the $m$ so as to obtain a decent bound.
If you need more details than the hint, this page goes through the details of the method I proposed, while this takes a different approach entirely.
-
Thanks for your answer! – Chris's sis Mar 1 '13 at 14:00
The answers posted in those websites are hard to follow. Please see the following two solutions since they are much simpler. – xpaul Apr 18 '13 at 4:10
I made a mistake before. Now I want to fix it. I looked at the link posted above and the solution is very long and hard to follow. Please look at the solution below to see if it is correct. Note that $$\frac{1}{n}\sum_{k=1}^nn^{1/k}=1+\frac{1}{n}\sum_{k=2}^{n}n^{1/k}.$$ Now we will show that \begin{eqnarray} \tag{1} \lim_{n\to\infty}\frac{1}{n}\sum_{k=2}^{n}n^{1/k}=1. \end{eqnarray} Let $a_n=\sum_{k=2}^{n}n^{1/k}$ and $b_n=n$. By the Stolz-Cesaro theorem, \begin{eqnarray*} \lim_{n\to\infty}\frac{1}{n}\sum_{k=2}^{n}n^{1/k}&=&\lim_{n\to\infty}\frac{a_n}{b_n}\\ &=&\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\\ &=&\lim_{n\to\infty}(a_{n+1}-a_n)\\ &=&\lim_{n\to\infty}\sum_{k=2}^{n+1}(n+1)^{1/k}-\sum_{k=2}^{n}n^{1/k}\\ &=&\lim_{n\to\infty}\left(\sum_{k=2}^{n}\left[(n+1)^{1/k}-n^{1/k}\right]+(n+1)^{1/(n+1)}\right)\\ &=&\lim_{n\to\infty}\sum_{k=2}^{n}\left[(n+1)^{1/k}-n^{1/k}\right]+1. \end{eqnarray*} Note that by the Mean Value theorem, it is easy to check $$(n+1)^{1/k}-n^{1/k}\le \frac{1}{kn^\frac{k-1}{k}}\le\frac{1}{kn^{1/2}}$$ and hence \begin{eqnarray*} 0<\sum_{k=2}^{n}\left[(n+1)^{1/k}-n^{1/k}\right]\le\sum_{k=2}^{n}\frac{1}{kn^{1/2}}=\frac{1}{n^{1/2}}\sum_{k=2}^n\frac{1}{k}=\frac{1}{n^{1/2}}(\ln n-1+\gamma+o(1))\to 0 \end{eqnarray*} as $n\to\infty$. So (1) is true. Thus $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nn^{1/k}=2.$$ Thank you, Aryabhata, for the suggestion to simplify the solution
-
The approach looks right: you can use the Mean Value theorem to show that $$(n+1)^{1/k} - n^{1/k} \le \frac{1}{kn^{1-1/k}} \le \frac{1}{k\sqrt{n}}$$ to simplify your proof considerably. Your attempts to prove the fact I just stated are a bit hard to understand. I suggest you replace it with something simpler. +1, though. – Aryabhata Apr 17 '13 at 7:44
@xpaul Thank you. Well, if you can make things easier I have nothing against. (+1) – Chris's sis Apr 17 '13 at 8:22
@Aryabhata, I will use your suggestion to modify. Thanks. – xpaul Apr 17 '13 at 12:16
@Sasha, please look at the new proof. – xpaul Apr 19 '13 at 0:47
Using Bernoulli's Inequality, we get for $k\ge\log(n)$ $$\frac1n\ge\left(1-\frac{\log(n)}{k}\right)^k\tag{1}$$ from which it is simple to derive $$n^{1/k}-1\le\frac{\log(n)}{k-\log(n)}\tag{2}$$ Just multiplying the number of terms by the largest term, we get \begin{align} \sum_{k=2}^{\lfloor\log(n)\rfloor+1}\left(n^{1/k}-1\right) &\le(\sqrt{n}-1)\lfloor\log(n)\rfloor\\ &\le\sqrt{n}\,\log(n)\tag{3} \end{align} Using $(2)$, we get \begin{align} \sum_{k=\lfloor\log(n)\rfloor+2}^n\left(n^{1/k}-1\right) &\le\sum_{k=\lfloor\log(n)\rfloor+2}^n\frac{\log(n)}{k-\log(n)}\\ &\le\log(n)(\log(n)+\gamma)\\[8pt] &\le(\log(n)+\gamma)^2\tag{4} \end{align} Combine $(3)$ and $(4)$ to get $$\lim_{n\to\infty}\frac1n\sum_{k=2}^n\left(n^{1/k}-1\right)=0\tag{5}$$ Therefore, $$\lim_{n\to\infty}\frac1n\sum_{k=1}^nn^{1/k}=2\tag{6}$$
-
Interesting approach (+1) – Chris's sis Jul 26 '13 at 11:41 | 3.921875 | CC-MAIN-2014-41 |
# Veterinary Epidemiologic Research: GLM (part 4) – Exact and Conditional Logistic Regressions
Next topic on logistic regression: the exact and the conditional logistic regressions.
Exact logistic regression
When the dataset is very small or severely unbalanced, maximum likelihood estimates of coefficients may be biased. An alternative is to use exact logistic regression, available in R with the elrm package. Its syntax is based on an events/trials formulation. The dataset has to be collapsed into a data frame with unique combinations of predictors.
Another possibility is to use robust standard errors, and get comparable p-values to those obtained with exact logistic regression.
```### exact logistic regression
x <- xtabs(~ casecont + interaction(dneo, dclox), data = nocardia)
x
interaction(dneo, dclox)
casecont 0.0 1.0 0.1 1.1
0 20 15 9 10
1 2 44 3 5
> nocardia.coll <- data.frame(dneo = rep(1:0, 2), dclox = rep(1:0, each = 2),
+ casecont = x[1, ], ntrials = colSums(x))
nocardia.coll
dneo dclox casecont ntrials
0.0 1 1 20 22
1.0 0 1 15 59
0.1 1 0 9 12
1.1 0 0 10 15
library(elrm)
Le chargement a nécessité le package : coda
Le chargement a nécessité le package : lattice
mod5 <- elrm(formula = casecont/ntrials ~ dneo,
interest = ~dneo,
iter = 100000, dataset = nocardia.coll, burnIn = 2000)
### robust SE
library(robust)
Le chargement a nécessité le package : fit.models
Le chargement a nécessité le package : MASS
Le chargement a nécessité le package : robustbase
Le chargement a nécessité le package : rrcov
Le chargement a nécessité le package : pcaPP
Le chargement a nécessité le package : mvtnorm
Scalable Robust Estimators with High Breakdown Point (version 1.3-02)
mod6 <- glmrob(casecont ~ dcpct + dneo + dclox + dneo*dclox,
+ family = binomial, data = nocardia, method= "Mqle",
+ control= glmrobMqle.control(tcc=1.2))
> summary(mod6)
Call: glmrob(formula = casecont ~ dcpct + dneo + dclox + dneo * dclox, family = binomial, data = nocardia, method = "Mqle", control = glmrobMqle.control(tcc = 1.2))
Coefficients:
Estimate Std. Error z-value Pr(>|z|)
(Intercept) -4.440253 1.239138 -3.583 0.000339 ***
dcpct 0.025947 0.008504 3.051 0.002279 **
dneo 3.604941 1.034714 3.484 0.000494 ***
dclox 0.713411 1.193426 0.598 0.549984
dneo:dclox -2.935345 1.367212 -2.147 0.031797 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Robustness weights w.r * w.x:
89 weights are ~= 1. The remaining 19 ones are summarized as
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.1484 0.4979 0.6813 0.6558 0.8764 0.9525
Number of observations: 108
Fitted by method ‘Mqle’ (in 5 iterations)
(Dispersion parameter for binomial family taken to be 1)
No deviance values available
Algorithmic parameters:
acc tcc
0.0001 1.2000
maxit
50
test.acc
"coef"
```
Conditional logistic regression
Matched case-control studies analyzed with unconditional logistic regression model produce estimates of the odds ratios that are the square of their true value. But we can use conditional logistic regression to analyze matched case-control studies and get correct estimates. Instead of estimating a parameter for each matched set, a conditional model conditions the fixed effects out of the estimation. It can be run in R with clogit from the survival package:
```temp <- tempfile()
+ "http://ic.upei.ca/ver/sites/ic.upei.ca.ver/files/ver2_data_R.zip", temp)
essai de l'URL 'http://ic.upei.ca/ver/sites/ic.upei.ca.ver/files/ver2_data_R.zip'
Content type 'application/zip' length 1107873 bytes (1.1 Mb)
URL ouverte
==================================================
### Salmonella outbreak dataset
library(survival)
mod7 <- clogit(casecontrol ~ slt_a + strata(match_grp), data = sal_outbrk)
summary(mod7)
Call:
coxph(formula = Surv(rep(1, 112L), casecontrol) ~ slt_a + strata(match_grp),
data = sal_outbrk, method = "exact")
n= 112, number of events= 39
coef exp(coef) se(coef) z Pr(>|z|)
slt_a 1.4852 4.4159 0.5181 2.867 0.00415 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
exp(coef) exp(-coef) lower .95 upper .95
slt_a 4.416 0.2265 1.6 12.19
Rsquare= 0.085 (max possible= 0.518 )
Likelihood ratio test= 10 on 1 df, p=0.001568
Wald test = 8.22 on 1 df, p=0.004148
Score (logrank) test = 9.48 on 1 df, p=0.002075
``` | 2.671875 | CC-MAIN-2014-49 |
# Needs help in logic
Hi guys: Can any one please help me in the logic. I try first but stuck in the middle of it.
1 1 1 1 1 1
2 4 8 16 32 64
3 9 27 81 243 729
4 16 64 256 1024 4096
#include <iostream>
using namespace std;
int main()
{
int row1 = 0;
int col1 = 0;
cout << "1" << " " << "1" << " " << "1" << " " << "1" << " " << "1" << " " << "1" << endl;
for(row1 = 1; row1 <= 4; row1 ++)
{
for(col1 = 1; col1 <= 4; col1 ++)
{
cout << row1*col1;
system("pause");
return 0;
}
###### Who is Participating?
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Commented:
I'll show you how I would do row 2 and let you figure the others out.
``````const int ITERATIONS = 6;
int multiplier = 2;
int num;
num = multiplier;
for(int i = 0; i < ITERATIONS; ++i)
{
cout<<num<<'\t';
num *= multiplier;
}
``````
0
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Commented:
Then you just change multiplier for each row. Just make sure you do num = multiplier every time before you get into the loop. Note: the '\t' is the tab character to make the output look more pretty. You could also use the iomanip library and setw if you wanted to pick how wide each column should be.
Bonus: You don't need to (and shouldn't) do row 1 separately. Just set the multiplier to 1 for that row.
0
Author Commented:
Thanks. I did it try to avoid the use of tab thing and use " " but it dont work. Is there any way i can arrange them without using \t
#include <iostream>
using namespace std;
int main()
{
const int ITERATIONS = 6;
int multiplier = 1;
int num=1;
int i = 0;
int k = 0;
for(k = 0; k < 4; k++)
{
num = multiplier;
for(i = 0; i < ITERATIONS; ++i)
{
cout << num << '\t';
num *= multiplier;
}
cout << endl;
multiplier++;
}
system("pause");
return 0;
}
0
Author Commented:
Thanks.
0
Commented:
As I mentioned you could use setw (#include<iomanip>) to specify a width. You also could count the digits in the integer by finding the log base 10 and rounding down, but that's more complicated than necessary.
http://www.cplusplus.com/reference/iostream/manipulators/setw/
0
Commented:
So it would look something like this
``````#include<iostream>
#include<iomanip>
const int WIDTH = 10;
const int ITERATIONS = 6; //Note: consts usually go outside of main
int main()
{
cout << setw(WIDTH) << num;
}
``````
0
Author Commented:
ok thanks.
0
Question has a verified solution.
Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.
Have a better answer? Share it in a comment. | 2.828125 | CC-MAIN-2018-26 |
# Coin Matching
We've updated our prices on products from many online Montessori stores! If we've missed your favorite store, please let us know!
Coin Matching
Math - Money
LevelPrimary
Age3.5
PrerequisitesPicture Matching
Materials2 of each type of coin
two bowls or boxes
tray
In this activity the child matches pairs of coins. This acts as sensorial introduction to the coins themselves before any reference is made to their value. It is done the same way as in Object to Object Matching.
## Presentation
1. Take the material to a mat or table.
2. Lay out the coins from the first bowl in a column along the left edge of the mat.
3. Take a coin from the second bowl and look for its match among the first set of coins.
4. Match the other coins in the same way.
5. Continue until all the coins have been matched.
## Variations and Extensions
• Use a Mystery Bag to practice matching the coins by touch.
• Could be done with coins of other countries.
• Teach the song "Doughnut Shop." It mentions a five-cent-piece, then refers to it as a nickel later.
## Material
Dollar coins and half-dollar coins are nice to have but not necessary for this activity. Have the four most commonly used coins in the U.S. | 2.578125 | CC-MAIN-2024-30 |
## 12. Find the difference between the original number and the new number if the digits 2 and 5 of the number 42576 are interch
Question
12. Find the difference between the original
number and the new number if the digits 2
and 5 of the number 42576 are interchanged
in progress 0
1 month 2021-08-14T20:08:37+00:00 1 Answer 0 views 0
1. QUESTION
Find the difference between the original number and the new number if the digits 2 and 5 of the number 42576 are interchanged.
SOLUTION
Original Number = 52576
It is given that,
the digits 2 and 5 of the number 42576 are interchanged.
So,
The new number will be
45276.
NOW,
Difference = 45276 – 42576
= 2700
Therefore,
2700 is the difference between the original number and the new number if the digits 2 and 5 of the number 42576 are interchanged. | 4.125 | CC-MAIN-2021-39 |
PHYSICS
Mass of empty bucket of capacity 10 liter is 1 kg. find its mass when completely with a liquid of relative density .08
1. 👍 2
2. 👎 4
3. 👁 1,565
1. see other post
1. 👍 0
2. 👎 2
👨🏫
bobpursley
2. Density of bucket =Mass(1 kg) /volume(10 litres)
Rd =d of bucket /d of liquid
Mass of bucket=Mass of empty one + Mass of liquid
1. 👍 0
2. 👎 5
3. Density of bucket =Mass(1 kg) /volume(10 litres)
Rd =d of bucket /d of liquid.....a
Mass of bucket=Mass of empty one + Mass of liquid
Find Mass of liquid by using a equation
1. 👍 0
2. 👎 9
4. 12
1. 👍 0
2. 👎 8
5. Very bad explanation . I can't understand easily . Please do improvement
1. 👍 8
2. 👎 1
6. Very bad explanation . I can't understand easily . Please do improvement
😢
1. 👍 7
2. 👎 1
7. Didn’t understand anything
1. 👍 1
2. 👎 1
8. Rd =d of iquid /d of water
.8=d of liquid/1000 kg/m3
d of liquid = 800 kg/m3
density = mass/volume
mass= 800 kg/m3 * 10 litre
mass = 800* .001 m3 = 8KG
Weight of empty bucket = 1 kg
tgotal weight =8+1 = 9 kg
1. 👍 7
2. 👎 0
9. Rd =d of iquid /d of water
.8=d of liquid/1000 kg/m3
d of liquid = 800 kg/m3
density = mass/volume
mass= 800 kg/m3 * 10 litre
mass = 800* .001 m3 = 8KG
Weight of empty bucket = 1 kg
total weight =8+1 = 9 kg
1. 👍 2
2. 👎 1
10. It's not the volume it is the capacity
See the units first!
1. 👍 2
2. 👎 0
11. Rd =d of iquid /d of water
.8=d of liquid/1000 kg/m3
d of liquid = 800 kg/m3
density = mass/volume
mass= 800 kg/m3 * 10 litre
mass = 800* .001 m3 = 8KG
Weight of empty bucket = 1 kg
tgotal weight =8+1 = 9 kg
1. 👍 0
2. 👎 0
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## Alumni Liaison
Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras. | 3.8125 | CC-MAIN-2019-39 |
# Random oracle model proofs and programmability
1. Proving the security of a scheme with the random oracle model (ROM) involves two steps: first you prove that the scheme is secure in an idealized world where a random oracle exists, and then you implement this scheme in the real world by replacing the random oracle with a hash function. Why wouldn't you prove the security of the scheme with the hash function that you end up implementing the world with? Is it because no proof has been found or are there any other reasons? Could you give me some examples?
2. Could you explain to me the "programmability" feature of the random oracle model? In my book (Katz-Lindell) it says: The reduction may choose values for the output of H as it likes (as long as these values are correctly distributed, i.e. uniformly random).
3. If I understand correctly, the function $H$ that will act as a random oracle can be fixed beforehand or it can be generated "on the fly", by generating a table of inputs and outputs and when there is a new input the function generates a new output and enters it into the table. What is the theoretical difference between the two?
• To prove security (question 1) you don't have to implement. Dec 7, 2011 at 12:51
• Yes, I understand that. But let's assume that you want to prove the security of scheme A, you prove this security in the ROM. And then you implement this scheme A and instead of using the random oracle (because it doesn't exist), you implement it with using a hash function B. So why don't you prove that scheme A is secure when you use function B as opposed to proving that scheme A is secure with a random oracle?
– dira
Dec 7, 2011 at 13:18
• Right, the process you give is the process followed (typically referred to as the random oracle methodology). I just wanted to point out that both steps are not necessary to prove security. The reason you don't prove security using a hash function is that there are no practical hash functions which are provably secure. Dec 7, 2011 at 13:27
• I see what you mean about the hash functions, thank you! I now realize I worded my question wrong, I did mean to say methodology.
– dira
Dec 7, 2011 at 21:33
• Have a look at this blog series What is the Random Oracle Model and why should you care? - it explains in some more detail how a random oracle can be used in security proofs for cryptographic algorithms. (I'm thinking of adding a bit about the programmability as an answer, but it might need some time.) Dec 7, 2011 at 23:43
## 3 Answers
Your second question was about programmability. This hasn't been directly addressed yet by Thomas' answer or the comments, so I'll focus on that question only. Unfortunately I don't know of a simple primitive that is secure in the random oracle model that requires programmability, but I'll use one that is hopefully clear once I explain the background. It's called the Fiat-Shamir heuristic; it's a nice trick to make non-interactive zero knowledge proofs.
Before getting to Fiat-Shamir, consider how your favorite basic zero-knowledge proof works. Since this is Crypto SE, not CSTheory SE, hopefully you are thinking about proving knowledge of discrete logarithms and quadratic residues, not graph isomorphisms or 3-coloring graphs. ;)
[Aside: technically these are not true zero-knowledge proofs, they are honest-verifier zero-knowledge proofs (sometimes called $$\Sigma$$-protocols) but we don't care about that distinction here]
### Schnorr's proof of knowledge of a discrete logarithm
$$P$$ (for prover) comes along with two values $$g$$ and $$y$$ in some group $$\mathbb{G}_q$$ where the discrete logarithm is hard. She claims: I know the value $$x$$ such that $$y=g^x$$. As $$x$$ is the discrete logarithm of $$y$$ base $$g$$, computing $$x$$ directly is infeasible so $$V$$ (verifier) cannot initially be sure if she really knows $$x$$ or not.
The Schnorr protocol lets $$P$$ prove knowledge of $$x$$ to $$V$$ in a way that does not disclose anything about $$x$$. It goes as follows:
1. $$P$$ generates a random value $$a$$, computes $$b=g^a$$, and sends $$b$$ to $$V$$
2. $$V$$ generates a random value $$c$$ and sends $$c$$ back to $$P$$
3. $$P$$ computes $$d=a+cx$$ and sends $$d$$ to $$V$$
4. $$V$$ accepts $$\langle b,c,d\rangle$$ as proof for $$\langle g,y \rangle$$ iff $$g^d=by^c$$
### Security Analysis
We can ask ourselves, what do you we want in terms of security from such a protocol? $$V$$ is concerned that sending a bunch of numbers back and forth might not actually constitue a proof that $$P$$ knows such an $$x$$. If he can actually conclude that $$P$$ must know $$x$$ if $$P$$ can produce many accepting $$\langle b,c,d\rangle$$ transcripts, the proof is said to be sound.
$$P$$ may be concerned that $$V$$ might learn some information about $$x$$ from seeing one or more accepting transcripts. This is supposed to be a proof that leaks zero information about $$x$$ (glossing over the honest verifier technicality). If it leaks zero information, it is said to be zero-knowledge.
### Soundness (via Extraction)
To show the Schnorr protocol is sound, we are actually going to do it indirectly. We are first going to show it is something called "extractable" and then show that extractability implies soundness. I'm not going to give actual definitions or proofs, just a sketch of what is going on.
Schnorr protocols have a special soundness property (called, you guessed it, special soundness): if there are two accepting transcripts $$t_1=\langle b,c,d \rangle$$ and $$t_2=\langle b,c',d' \rangle$$ where $$t_1$$ shares the same value of $$b$$ with $$t_2$$ but $$c$$ (and thus $$d$$) are different, then it is possible to calculate the value of $$x$$: $$x=(d-d')/(c-c')$$. If $$P$$ can reliably generate accepting transcripts, then there is no reason to suppose she couldn't generate $$t_1$$. Likewise $$t_2$$. And if she can produce both, then she "knows" $$x$$ in the sense that the knowledge required to produce $$t_1$$ and $$t_2$$ is sufficient to produce $$x$$ itself.
When we eventually get to Fiat-Shamir, it will be important to have formalized this notion of "extractability" a little bit. Consider the situation where $$P$$ is a compiled binary program instead of a person. You can run $$P$$ which will perform the protocol and you can rewind $$P$$ to a previous internal state, but you can't decompile it or look at the internal state (this is called rewindable blackbox access; why these special powers are allowed in proving extractability is a topic for another time).
We say that a protocol is extractable if you can get $$x$$ from interacting with such a black box. And we say a protocol is sound if it is extractable in this way (a blackbox that you can rewind). Both of these propositions have proofs in the literature with lots of fine-print I am omitting. Note that you can prove soundness in other ways than extractability or other flavours than blackbox-rewindable extractability (extractability is sufficient but not necessary).
For Schnorr, it should be obvious how, but you do the following:
1. Let $$P$$ output $$b$$
2. Give $$P$$ a random $$c$$ as input
3. Let $$P$$ output $$d$$
4. Rewind $$P$$ to after step 1 and before step 2
5. Give $$P$$ a different random $$c'$$ as input
6. Let $$P$$ output $$d'$$
7. Compute $$x$$ from $$\langle b,c,d \rangle$$ and $$\langle b,c',d' \rangle$$
### Zero-knowledge (via Simulation)
Similarly, we can indirectly prove the protocol is zero-knowledge by showing it has a different property: simulatability. In this case, we get a compiled binary of $$V$$ and have to reliably supply it with acceptable $$b$$ and $$d$$ values for the $$c$$'s it gives us. However the protocol is for knowledge of an $$x$$ we do not actually know! If we can simulate acceptable protocol runs without knowing $$x$$, then the values in the protocol must not really be leaking any information about $$x$$. So if the protocol is simulatable in this regard, then it is zero-knowledge.
I mentioned before that Schnorr is not actually a zero-knowledge protocol. This creates some problems with simulating Schnorr transcripts that will get resolved when we use a random oracle with Fiat-Shamir. To simulate Schnorr protocols, we do the following:
1. Generate random value $$d$$
2. Guess the value of $$c$$
3. Supply $$b=g^dy^{-c }$$ as input to $$V$$
4. Let $$V$$ output $$c'$$
5. If $$c'\neq c$$ (you guessed wrong), rewind to step 2. Else continue
6. Supply $$d$$ to $$V$$ which will accept
If the values of $$c$$ are really short (say a bit), then the simulator is efficient. For longer values, you can't prove the zero-knowledgeness of Schnorr with this method. There are a handful of tricks to convert Schnorr into something that is true zero-knowledge.
### Fiat-Shamir Heuristic
Reading the above, you might do a double-take: on one hand, you can show that $$x$$ must be known if transcripts accept and on the other, you can generate transcripts that accept without $$x$$: what gives? If you look closely, you'll see that the simulated transcripts are generated out of order while the extractable ones are generated in order. In fact, by generating out of order, we cannot produce $$\langle b,c,d \rangle$$ and $$\langle b,c',d'\rangle$$ transcripts since the value of $$b$$ is no longer being chosen: it is determined by $$d$$ and $$c$$.
The idea of Fiat-Shamir is to make Schnorr (and related) protocols non-interactive. This means $$P$$ can produce all three values $$\langle b,c,d \rangle$$ instead of relying on $$V$$ to provide $$c$$. Furthermore, since we know transcripts are simulatable, $$P$$ can produce a value of $$c$$ that has to have been generated after the value $$b$$ thus ruling out any simulation. How? It is really easy actually: set $$c=\mathcal{H}(b)$$. The verifier additionally checks that $$c=\mathcal{H}(b)$$. [Aside: there is actually a neat optimization here where you don't have to send the value $$b$$ at all but leave that aside].
Finally we can introduce random oracles. It turns out that if you use regular hash functions, you can't wrestle extractability or simulation out of the protocol. We'll try but ultimately we will require a random oracle that can be programmed.
### Extraction with Fiat-Shamir heuristic
Recall that extraction relies on pairs of transcripts like $$\langle b,c,d \rangle$$ and $$\langle b,c',d' \rangle$$. With Fiat-Shamir, $$c=\mathcal{H}(b)$$ so if the values of $$b$$ between two transcripts are identical, then $$c$$ and thus $$d$$ will be as well. Therefore, we cannot get two such transcripts with a regular deterministic hash function. But if $$\mathcal{H}$$ is a programmable random oracle, we can get it to produce different values for the same input. Once again, we play the game of having rewindable blackbox access to $$P$$ but this time we also get the random oracle:
1. Let $$P$$ generate $$b$$
2. See $$P$$ query $$O$$ with $$b$$ for $$\mathcal{H}(b)$$
3. Generate random $$c$$ and program $$c=\mathcal{H}(b)$$ in $$O$$
4. Let $$O$$ answer query
5. Let $$P$$ compute $$d$$
6. Let $$P$$ output $$\langle b,c,d \rangle$$
7. Rewind to end of step 2
8. Generate random $$c'$$ and program $$c'=\mathcal{H}(b)$$ in $$O$$
9. Proceed as before and eventually let $$P$$ output $$\langle b,c',d' \rangle$$
A few notes: (1) because this is non-interactive, $$P$$ does not output $$b$$ after step 1, so we rely on the ability to see queries to the random oracle; (2) if the oracle generates answers "on the fly" (instead of entering the protocol with a codebook of all queries/responses), we don't actually have to program it with different values of $$c$$. We just rewind to before the point it is about to generate a response and let it generate a random value (which will overwhelmingly be different than in the first execution). This sheds some light on the original poster's third question.
### Simulation with Fiat-Shamir
Similarly to extraction, the use of a random oracle makes simulation a breeze. Assuming you've read this far, you can probably see how so I will just say it in a sentence: Set a random value for $$c$$, compute $$\langle b,c,d \rangle$$ by choosing $$d$$ first, and when the verifier checks with the oracle that $$c=\mathcal{H}(b)$$, program $$c$$ as the response.
• Fantastic answer. I’m not sure I understand why schnorr is not truly zkp though. Also it is not clear to me how extractability shows soundness, in practice a prover wouldn’t use the same commit b right? What if there is a way of generating accepting transcripts without being able to generate two with the same commit? Nov 19, 2019 at 16:47
• Is there any rigorous proof that plain Schnorr scheme isn't/can't be ZK, or just intuition (of course) that it might not hold?
– user16972
Mar 9, 2020 at 22:36
• @David天宇Wong did he mean that Schnorr is not true (aka general) ZK because it relies on a honest-verifier such that V will not request both d and d' for commitment b or assumption V only sends random inquiries? Afaics, FS insures the honest-verifier compliance by enabling the prover to enforce it in NIZK? Does extraction prove soundness because the protocol can employed to compute x implicating that using anything other than x would fail? In 2019 were you possibly conflating proof of soundness of the discrete logarithm problem assumption which an orthogonal concern? Oct 18, 2023 at 4:55
• If P, V don’t select uniformly at “random,” distribution leaks ZK from simulation that does. Your recent blog mentions HVZK and ‘structured.’ Is extraction insufficient if P, V collude or in replay? Oct 18, 2023 at 15:18
A random oracle is an ideal object; see this previous question for some details. What makes a random oracle convenient for proofs is the part about knowing nothing on the output for a given input if you do not try it. For instance, consider the following encryption scheme:
• $H$ is a random oracle which outputs $n$-bit values.
• The key is a $K$, a string of $k$ bits.
• A single message $m$ is encrypted by computing $c = m \oplus H(K || 1) || H(K || 2) || ...$ (you repeatedly "hash" with the random oracle the successive strings obtained by concatenating $K$ with a counter, and you concatenate the oracle outputs into a big stream which is XORed with the message to encrypt).
If $H$ is a random oracle, then it is reasonably easy to prove that the encryption is secure up to a work factor of $2^{k-1}$ invocations of $H$: for any index $i$, you can learn nothing on the bit $i$ of the generated stream ("nothing" as in "not even the slightest statistical bias") unless you do invoke $H$ on the exact input which yielded it; since there are $2^k$ possible values for $K$, the best possible attack is just trying them out (in any order), which, on average, will hit the right key after $2^{k-1}$ guesses. Each guess involve invoking $H$.
Now, with a cryptographic hash function, things are not that easy. A cryptographic hash function is defined as being resistant to preimages, second preimages, and collisions. These are much weaker properties. A function could be a good hash function and still fail to be a random oracle. This is especially true with the commonly used Merkle-Damgård function such as SHA-256 and SHA-512: these functions suffer from the so-called "length extension attack". Given $H(x)$, it is possible, under certain conditions but without knowing $x$, to compute $H(x||x')$ for some values of $x'$. A random oracle would not allow that. And this specific property totally destroys our attempts at proving the security of the encryption scheme described above. Nevertheless, the "length extension" does not seem to help in any way when you try to compute preimages or collisions. Indeed, no efficient preimage or collision attack is currently known on SHA-256 or SHA-512.
To sum up, a random oracle is an ideal object which allows for easy proofs of constructions in which they are used, proofs which rely on properties that actual hash functions do not necessarily exhibit (even if they are "secure" hash functions).
When we "implement" a random oracle, we use hash functions and other primitives in such a way to mimic the ideal properties of a random oracle. Existing hash functions are not good enough for that by themselves, as seen above. A common tool is HMAC, which uses a key and is built over an existing hash function, but invokes it twice in a specific way so as to avoid known shortcomings of concrete functions such as SHA-256. This is why, for instance, when building a cryptographically secure pseudorandom number generator as described by NIST, you may prefer the "HMAC_DRBG" construction over the faster, but "less proven" Hash_DRBG.
There is still a bit of technicality, intuition, and downright faith, in using a given construction with hash function in a protocol where a random oracle should be used. But we have nothing better: we do not know if random oracles really exist (or even secure hash functions, for that matter).
Whether a given implementation uses tables to store precomputed results has no influence whatsoever on theoretical security: a proof in the random oracle model relies on the number of times the oracle must have been invoked (on distinct inputs), but not on when the invocations took place. You can use internal tables as you wish.
There is a subtlety on the birth date of the function: if the oracle is HMAC with a key, then the oracle "exists" since the key was generated, and all oracle invocations must have used that key; on the other hand, a key-less hash function such as SHA-256 can be thought of as a kind-of random oracle which has existed since SHA-256 was first defined, more than ten years ago, and the whole World may have been busy invoking it during the last decade. So using raw SHA-256 as a random oracle (if we ignore the bit about length extensions) is equivalent to considering that the attacker could be as powerful as the whole World with a ten-years computational head start. To avoid that, it is commonplace to define protocols which use a keyed function as random oracle.
• I have one last question: All the queries to the random oracle are private, but "the reduction" can see all the queries that are made to the oracle. Does this mean that when proving with reduction, we introduce a new adversary who can see all the queries but no one else can? Doesn't this contradict the random oracle's definition if it is defined in a way that all the queries are private?
– dira
Dec 7, 2011 at 21:30
• @dira: it depends on the context. The proof ends up saying: "attacker's advantage is no more than $X$ when up to $q$ queries to the oracle are allowed". If the oracle is public (it is a hash function), then $q$ may be quite high; limit is on the computational power (e.g. $q = 2^{128}$). When the oracle is private, each query is part of an active attack, so it makes sense to disallow $q$ higher than some sensible value. Dec 7, 2011 at 21:56
• I think I put the question in a wrong way. I was reading the random oracle model chapter in the Katz-Lindell book and it says that in the random oracle, all queries that any of the parties make to the oracle are public, and then it writes that: "The reduction sees all the queries that A makes to the random oracle. This does not contradict the fact that queries are private. While that is true in the formal model itself, here we are using A as a subroutine within a reduction." I just don't quite understand how it is different and how this is "allowed".
– dira
Dec 7, 2011 at 22:12
The question of programmability of ROM, and why it's a feature which leads to "errors" when applying ROM-proved constructions with real hash-functions, is considered in details in https://link.springer.com/chapter/10.1007/3-540-45708-9_8
Answers to your questions:
1. Yes, it's often impossible to find a proof based on properties of the hash-functions, like collision-resistance preimage-resistance. Example is a proof for Schnorr signature.
2. Programmability of random oracle is often an essential property for constructing of proofs. It means, that responses of the oracle are not fixed beforehand (as it is for a real hash-function, where all hash-values are fixed), but we can choose them adaptively during an experiment considered in the proof. E.g., a proof for Schnorr signature shows that if there's an attacker A for Schnorr signature, then we can create attacker B which will use A as a black-box subroutine and will be able to break discrete logarithm problem. To run A, B should emulate its environment: respond to random oracle queries, respond to signing oracle queries. But B don't know secret key (which is a discrete log and B actually wants to find it!), so how it could respond to A on signing oracle requests? This is where random oracle programmability comes to the scene. B will assign a specific value as a random oracle response to specific request, so that B will able to create a signature (from some point of view, "forge" a signature by manipulating random oracle responses). But in a way that A can't recognize this hacking of B (can't distinguish that value from random one).
3. In reality, hash function values are known beforehand (once hash function is declared/chosen). Random oracle values are allowed to be chosen on the fly, which essentially helps in constructing of proofs. | 2.640625 | CC-MAIN-2024-22 |
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### A cylindrical beaker has a height of 16 cm and a diameter at its bottom of 10 cm. What is its volume? Give your answer to 2 decimal places.
As the beaker is a cylinder, we know the formula of its volume is (pi x (radius)2) x height.
Since the radius = 1/2 diameter and the diameter = 10 cm, the radius = 5 cm.
We know the height is 16 cm.
So the formula for the volume becomes:
(pi x (5)2) x 16 = 1256.63706144 cm3
If we round this to 2 d.p. it becomes:
1256.64 cm3 to 2 d.p.
11 months ago
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# w4-1 - Problem 5.23 Solution Known quantities Circuit shown...
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Unformatted text preview: Problem 5.23 Solution: Known quantities: Circuit shown in Figure P523, V1 =12V,C = 0.5,uF,Rl = 0.68kQ,R2 =13kfl. Find: The current through the capacitor just before and just after the switch is closed. Problem 5.26 Solution: Known quantities: Circuit shown in Figure P526, V1 =12V,RS = 0.7931: 22m; = lOOmH. Find: The voltage through the inductor just before and just after the switch is changed. IWIW'I'II Wflwllflflr. illfi- l'flfllllhfllllll [HF-ll” fill [HFTJ Problem 5.52 “”9 Solution: Known quantities: As describes in Figure P552. At t : 0, the switch closes. Find: a) rib“) for :20_ Problem 5.57 Solution: Known quantities: As describes 111 Figure P557. Find: The voltage across the C1 . ...
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# Number Theory (Interesting Facts and Algorithms)
Questions based on various concepts of number theory and different types of number are quite frequently asked in programming contests. In this article, we discuss some famous facts and algorithms:
Interesting Facts :
1. All 4 digit palindromic numbers are divisible by 11.
2. If we repeat a three-digit number twice, to form a six-digit number. The result will will be divisible by 7, 11 and 13, and dividing by all three will give your original three-digit number.
3. A number of form 2N has exactly N+1 divisors. For example 4 has 3 divisors, 1, 2 and 4.
4. To calculate sum of factors of a number, we can find the number of prime factors and their exponents. Let p1, p2, … pk be prime factors of n. Let a1, a2, .. ak be highest powers of p1, p2, .. pk respectively that divide n, i.e., we can write n as n = (p1a1)*(p2a2)* … (pkak).
```Sum of divisors = (1 + p1 + p12 ... p1a1) *
(1 + p2 + p22 ... p2a2) *
.............................................
(1 + pk + pk2 ... pkak)
We can notice that individual terms of above
formula are Geometric Progressions (GP). We
can rewrite the formula as.
Sum of divisors = (p1a1+1 - 1)/(p1 -1) *
(p2a2+1 - 1)/(p2 -1) *
..................................
(pkak+1 - 1)/(pk -1)
```
5. For a product of N numbers, if we have to subtract a constant K such that the product gets its maximum value, then subtract it from a largest value such that largest value-k is greater than 0. If we have to subtract a constant K such that the product gets its minimum value, then subtract it from the smallest value where smallest value-k should be greater than 0
6. Goldbech`s conjecture: Every even integer greater than 2 can be expressed as the sum of 2 primes.
7. Perfect numbers or Amicable numbers: Perfect numbers are those numbers which are equal to the sum of their proper divisors. Example: 6 = 1 + 2 + 3
8. Lychrel numbers: Are those numbers that cannot form a palindrome when repeatedly reversed and added to itself. For example 47 is not a Lychrel Number as 47 + 74 = 121
9. Lemoine’s Conjecture : Any odd integer greater than 5 can be expressed as a sum of an odd prime (all primes other than 2 are odd) and an even semiprime. A semiprime number is a product of two prime numbers. This is called Lemoine’s conjecture.
10. Fermat’s Last Theorem : According to the theorem, no three positive integers a, b, c satisfy the equation, for any integer value of n greater than 2. For n = 1 and n = 2, the equation have infinitely many solutions.
Number Theory Algorithms
GCD and LCM
Recent Articles on GCD and LCM!
Prime Factorization and Divisors :
Recent Articles on Prime Factors!
Fibonacci Numbers:
Recent Articles on Fibonacci Numbers!
Catalan Numbers :
Recent Articles on Catalan Numbers!
Modular Arithmetic :
Euler Totient Function:
nCr Computations :
Chinese Remainder Theorem :
Factorial :
1. Factorial
2. Legendre’s formula (Given p and n, find the largest x such that p^x divides n!)
3. Sum of divisors of factorial of a number
4. Count Divisors of Factorial
5. Compute n! under modulo p
1. Recent Articles on Factorial!
Prime numbers and Primality Tests :
Recent Articles on Prime Number!
Sieve Algorithms :
Recent Articles on Sieve!
Divisibility and Large Numbers :
Recent Articles on Divisibility!
Misc :
Recent Articles on Mathematical Algorithms!
Recent Articles on Number Theory
Above facts are contributed by Yash Kodesia. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more facts about Number Theory.
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# Cross-validation on diabetes Dataset ExerciseΒΆ
A tutorial exercise which uses cross-validation with linear models.
This exercise is used in the Cross-validated estimators part of the Model selection: choosing estimators and their parameters section of the A tutorial on statistical-learning for scientific data processing.
Out:
/home/circleci/project/sklearn/linear_model/coordinate_descent.py:492: ConvergenceWarning: Objective did not converge. You might want to increase the number of iterations. Fitting data with very small alpha may cause precision problems.
ConvergenceWarning)
/home/circleci/project/sklearn/linear_model/coordinate_descent.py:492: ConvergenceWarning: Objective did not converge. You might want to increase the number of iterations. Fitting data with very small alpha may cause precision problems.
ConvergenceWarning)
Answer to the bonus question: how much can you trust the selection of alpha?
Alpha parameters maximising the generalization score on different
subsets of the data:
[fold 0] alpha: 0.05968, score: 0.54209
[fold 1] alpha: 0.04520, score: 0.15523
[fold 2] alpha: 0.07880, score: 0.45193
Answer: Not very much since we obtained different alphas for different
subsets of the data and moreover, the scores for these alphas differ
quite substantially.
from __future__ import print_function
print(__doc__)
import numpy as np
import matplotlib.pyplot as plt
from sklearn import datasets
from sklearn.linear_model import LassoCV
from sklearn.linear_model import Lasso
from sklearn.model_selection import KFold
from sklearn.model_selection import GridSearchCV
X = diabetes.data[:150]
y = diabetes.target[:150]
lasso = Lasso(random_state=0)
alphas = np.logspace(-4, -0.5, 30)
tuned_parameters = [{'alpha': alphas}]
n_folds = 5
clf = GridSearchCV(lasso, tuned_parameters, cv=n_folds, refit=False)
clf.fit(X, y)
scores = clf.cv_results_['mean_test_score']
scores_std = clf.cv_results_['std_test_score']
plt.figure().set_size_inches(8, 6)
plt.semilogx(alphas, scores)
# plot error lines showing +/- std. errors of the scores
std_error = scores_std / np.sqrt(n_folds)
plt.semilogx(alphas, scores + std_error, 'b--')
plt.semilogx(alphas, scores - std_error, 'b--')
# alpha=0.2 controls the translucency of the fill color
plt.fill_between(alphas, scores + std_error, scores - std_error, alpha=0.2)
plt.ylabel('CV score +/- std error')
plt.xlabel('alpha')
plt.axhline(np.max(scores), linestyle='--', color='.5')
plt.xlim([alphas[0], alphas[-1]])
# #############################################################################
# Bonus: how much can you trust the selection of alpha?
# To answer this question we use the LassoCV object that sets its alpha
# parameter automatically from the data by internal cross-validation (i.e. it
# performs cross-validation on the training data it receives).
# We use external cross-validation to see how much the automatically obtained
# alphas differ across different cross-validation folds.
lasso_cv = LassoCV(alphas=alphas, cv=5, random_state=0)
k_fold = KFold(3)
"how much can you trust the selection of alpha?")
print()
print("Alpha parameters maximising the generalization score on different")
print("subsets of the data:")
for k, (train, test) in enumerate(k_fold.split(X, y)):
lasso_cv.fit(X[train], y[train])
print("[fold {0}] alpha: {1:.5f}, score: {2:.5f}".
format(k, lasso_cv.alpha_, lasso_cv.score(X[test], y[test])))
print()
print("Answer: Not very much since we obtained different alphas for different")
print("subsets of the data and moreover, the scores for these alphas differ")
print("quite substantially.")
plt.show()
Total running time of the script: ( 0 minutes 0.536 seconds)
Gallery generated by Sphinx-Gallery | 3.546875 | CC-MAIN-2024-18 |
0
# All 6 faces of this solid figure are rectangles?
Updated: 4/28/2022
Wiki User
12y ago
,kmjui
Wiki User
12y ago
Earn +20 pts
Q: All 6 faces of this solid figure are rectangles?
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Related questions
cuboid
### You have 7 faces in all. Five of your faces are rectangles. What solid figure are you?
A pentagonal prism
### What has all 6 faces of solid figure are rectangles?
A rectangular prism.
### What is a solid figure in which all 6 faces are rectangles with 3 pairs of opposite faces that are parallel and congruent?
A hexahedron, or cube.
### A solid figure in which all six faces are rectangles with three pairs of opposite faces that are parallel and congruent?
a rectangular prism
### What solid figure has 6 rectangles as faces?
The solid is a cuboid, or "right rectangular prism" in which all angles are right angles. It is a box shape, or regular hexahedron.
### What figure has 6 faces with some of them rectangles?
A rectangular prism is a hexahedron with some rectangular sides. There is no requirement that all sides be rectangles. If they are, the solid is a right rectangular prism, or cuboid.
### What solid figure has 6 faces not all the same size and 12 edges?
Any solid object with six faces is a hexahedron. A regular hexahedron (with 12 edges) is a cube where all the faces are squares An irregular hexahedron (with 12 edges) is a cuboid where two faces are equal squares and four are equal rectangles.
pentagonal prism
### Is a square a solid figure with all congruent faces?
A square is not a solid figure at all. It's a flat figure, with four congruent faces and four congruent angles.
### What is a solid figure with eight vertices and all of its faces are an equal size?
A cube is a solid figure with eight vertices and all faces of equal size.
### What kind of polygon is a cube?
A cube is a solid figure called a polyhedron. A polyhedron is a solid figure with all flat faces. So a cone would be a solid figure but not a polyhedron becasue it has a curve and does not have all flat faces. | 3.3125 | CC-MAIN-2024-22 |
# 3D Wing Geometry Creation (optional)
##### Note: If you wish to skip this topic, click "complete" on the top right of the page.
This airfoil file was created to mimic the geometry used by NASA in their CFD. First, everything was scaled down according to their scaling factor and we switched the units from meters to feet for ease. We then needed to calculate the root chord which was possible due to the knowledge of the taper ratio. We also needed to change the thickness from the scaled values that were provided to match with the NASA geometry. With some additional manipulation, we get the .txt airfoil file given below.
The airfoil file contains the coordinates of the root chord according to how we calculated the NASA geometry. There are three columns, (x,y,z) with the z column being all 0. The coordinates start at (0,0,0) and trace the coordinates to the length of the airfoil with 0 thickness at the trailing edge and then these coordinates are traced back to (0,0,0) to create a closed curve, which is required by Solidworks.
1. Open a new Solidworks Part
2. Change units so that length scale is in feet
3. Navigate to the Features Tool Bar and select Curves → Curves through X Y Z Points:
4. Browse to where you have saved the airfoil .txt file (you will need to change the file type for which Solidworks is searching)
5. Click OK
6. You should see an image similar to the one below:
7. Select the Front Plane → New Sketch
8. Select Convert Entities:
9. Select the airfoil as the entity to convert
10. Create a new plane
1. Go to Reference Geometry in the Features Tool Bar → Plane
11. In the Details Panel, select the Front Plane as the First Reference and set the Distance to 1 foot. (This is because our wing-half span is 1 foot)
12. Create New Sketch
1. Select the plane you just created
13. Select Convert Entities
14. Select the airfoil to convert. See the image below for reference:
15. Scale the second airfoil: Move Entities→Scale Entities
16. Scale Factor: .562 (This is the taper ratio of the wing, defined a the ratio of the tip chord to the root chord)
1. The point to scale about will be origin (this is the origin point of the second airfoil)
17. Move Entities: (See the above image)
18. Select the second airfoil
19. Duplicate the image below in your details window:
The change in x here is from the wing geometry described on NASA's website. Since the leading edge angle of the wing is 30 degrees and we know the span to be 1 foot we can calculate the change of the tip chord x in relation to the location of the root chord.
20. Create a new sketch on the Top Plane
21. Select Line and connect the origin from the root airfoil to the origin of the tip airfoil and make sure they are coincident with the line.
22. In the same sketch a new line and connect the airfoils via the trailing edge. You will need to use the pierce condition to make sure the line is connected with the airfoils.
23. In the Features Toolbar go to Lofted Boss/Base
1. Select the two airfoil sketches as the profiles
2. Select the two lines (leading and trailing edges) as the guide curves
3. Make sure the preview matches what you are expecting
24. Click the green checkmark and you have created the wing geometry! | 2.890625 | CC-MAIN-2023-23 |
# Characteristics, what's the inverse of (x*(x+1))/2? [closed]
In my game you can spend points to upgrade characteristics. Each characteristic has a formula like:
A) out = in : for one point spent, one pont gained (you spend 1 point on Force so your force goes from 5 to 6)
B) out = last level (starting at 1) : so the first point spent earns you 1 point, the next point spent earns you an additional 2 and so on (+3,+4,+5...)
C) The inverse of B) : You need to spend 1 point to earn one, then you need to spend 2 to earn another one and so on.
I have already found the formula for calculating the actual level of B when points spent = x :
charac = (x*(x+1))/2
But I'd like to know what the "reverse" version of B) (usable for C) is, ie. if I have spent x points, how many have I earned if 1 spent gives 1, 1+2=3 gives 2, 1+2+3=6 gives 3 and so on.
I know I can just calculate the numbers but I'd like to have the formula because its neater and so that I can stick it in an excel sheet for example...
Thanks!
ps. I think I have nailed it down to something like charac = sqrt( x*m +k) but then I'm stuck doing number guessing for k and m and I feel I might be wrong anyway as I get close but never hits the spot.
-
## closed as off topic by Byte56♦, Josh Petrie♦, RicketOct 26 '12 at 16:07
Questions on Game Development Stack Exchange are expected to relate to game development within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.
Is this what you're looking for? – William Mariager Oct 26 '12 at 13:41
Off topic and probably a little too localized as well. – Byte56 Oct 26 '12 at 14:28
I believe the answer would involve a bit of Calculus, specifically something under the topic of "Series and Sequences". Reminds me vaguely of my second-year college courses, but I can't be any more specific than that for lack of remembering the details. Math.SE could be the place to ask this, or some other math type forum. Hopefully you have some keywords to go from now. – Ricket Oct 26 '12 at 16:09
Is SO getting harder rules? I'm developing a game, I have trouble with a calculation with characteristics and that's not game-developer related? Thanks for the answers though, really on the spot (Excellent link William!) and I actually was over on math.se but it seemed like a question 'not complicated enough' for them and here on GD I got 2 fine answers :-) – Valmond Oct 26 '12 at 16:23
You should probably ask it on Math.SE ...
Short version: Your formula is generally known as "sum of integers", a special case of a power sum and generally written as follows:
Calculating the inverse requires solving the quadratic equation 0.5x2 + 0.5x - charac = 0 for x.
In the basic formula for solving such (see the link for explanation), we can set
a = 0.5
b = 0.5
c = -charac
.. and calculate that the results are ...
x = (-0.5 ± √(0.52 + 4 x 0.5 x charac)) / 2 x 0.5 = -0.5 ± √(2 x charac + 0.25)
Since one of the results is always negative it can't be true, so this leaves us with the following formula, which is positive for all positive real-valued charac:
x = √(2 x charac + 0.25) - 0.5
In most cases you'll probably want to round the result down to the nearest integer.
-
you should probably use * for multiplication denotation, because your solving of the quadratic formula is a little confusing if you aren't familiar with the quadratic formula... it looks like you've got a variable in your solution (2 x charac) – derivative Oct 26 '12 at 16:04
@derivative: Variables are in italics. – Martin Sojka Oct 26 '12 at 17:57 | 3.5625 | CC-MAIN-2016-26 |
# Non Homogenous Differential Equation
• shards5
In summary, a non-homogeneous differential equation is a type of differential equation that contains terms not equal to zero. To solve it, the method of undetermined coefficients is used. The difference between a homogeneous and non-homogeneous differential equation is that the latter also includes terms not involving the dependent variable and its derivatives. A non-homogeneous differential equation can have infinitely many solutions due to the constant of integration in its general solution. Real-life applications of these equations include modeling physical phenomena in various fields such as physics, engineering, and economics.
shards5
## Homework Statement
y"' - 9y" +18y' = 30ex
y(0) = 16
y'(0) = 14
y"(0) = 11
n/a
## The Attempt at a Solution
Factor Out
r(r2 - 9r +18)
r = 0; r = 6; r =3
General Equation
y(x) = c0 + c1e3x + c2e6x
y'(x) = 3c1e3x + 6c2e6x
y"(x) = 9c1e3x + 36c2e6x
c1 = (11 - 36c2)/9
y'(x) = 14 = 3c1e3x + 6c2e6x
y'(x) = 14 = ((11 - 36c2)/9)e3x + 6c2e6x
c2 = -31/30
c1 = (11 - 36*(-31/30))/9 = 5.35555556
y(0) = 16 = c0 + c1e3x + c2e6x
16 +31/30 - 5.35555556 = c0 = 11.6777778
Solve For A in Aex
yp = Aex
y'p = Aex
y"p = Aex
y"'p = Aex
Inputting into the original equation we get.
Aex - 9Aex +18Aex = 30ex
Simplifying we get.
10Aex = 30ex
Which gives A = 3 and since 3 is a root of the original equation we add an x to differentiate between the two.
So the final equation SHOULD BE 3xex + 11.6777778 + 5.35555556e3x -31/30e6x but of course its not.
So my question is, what am I doing wrong?
Last edited:
Most of your work is fine, but I believe you have made an error in your calculations for c_0, c_1, and c_2. I used matrix methods to solve for these constants and got c_2 = 25/18. I'm fairly confident of this value, but didn't check it.
Also, in your last paragraph you say something that isn't true. You got A = 3, which means that your particular solution is y_p = 3e^x. The fact that you got a value of 3 when you solved for A is irrelevant to anything else in this problem. Your general solution will include 3e^x, not 3xe^x.
When you say matrix method do you just make a matrix like this?
1 1 1 16
0 3 6 14
0 9 36 11
And then you row reduce?
shards5 said:
When you say matrix method do you just make a matrix like this?
1 1 1 16
0 3 6 14
0 9 36 11
And then you row reduce?
Yes, and yes.
## 1. What is a non-homogeneous differential equation?
A non-homogeneous differential equation is a type of differential equation where the terms involving the dependent variable and its derivatives are not equal to zero. This means that the equation is not in its simplest form, which is known as a homogeneous differential equation.
## 2. How do you solve a non-homogeneous differential equation?
To solve a non-homogeneous differential equation, we use a method called the method of undetermined coefficients. This involves finding a particular solution to the equation by assuming a general form for the solution and then substituting it into the equation to determine the coefficients. The general solution is then found by adding this particular solution to the general solution of the corresponding homogeneous equation.
## 3. What is the difference between a homogeneous and non-homogeneous differential equation?
A homogeneous differential equation only contains terms involving the dependent variable and its derivatives, while a non-homogeneous differential equation also includes terms that do not involve the dependent variable and its derivatives. This makes the process of solving non-homogeneous differential equations more complex, as we need to find a particular solution in addition to the general solution.
## 4. Can a non-homogeneous differential equation have more than one solution?
Yes, a non-homogeneous differential equation can have infinitely many solutions. This is because the general solution of a non-homogeneous differential equation contains a constant of integration, which can take on any value. This means that there are many different solutions that satisfy the given equation.
## 5. What are some real-life applications of non-homogeneous differential equations?
Non-homogeneous differential equations are used to model various physical phenomena in fields such as physics, engineering, and economics. Some examples include modeling the growth of a population, the spread of disease, and the motion of objects under the influence of external forces. They are also used in signal processing, control theory, and other areas of mathematics.
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495 | 4.46875 | CC-MAIN-2024-30 |
## Swaping observations
Solved
Frequent Contributor
Posts: 79
# Swaping observations
Hi Experts, Is there any way to swap two or more observations within a dataset like I want to swap 15th observation with 83rd. Please help. Regards
Accepted Solutions
Solution
08-11-2016 12:22 AM
Super User
Posts: 6,785
## Re: Swaping observations
You would need to re-create the data set. And it's clumsy. This would be one way.
data want;
do until (done1);
set have (obs=14) end=done1;
output;
end;
set have (firstobs=83 obs=83);
output;
do until (done2);
set have (firstobs=16 obs=82) end=done2;
output;
end;
set have (firstobs=15 obs=15);
output;
do until (done3);
set have (firstobs=84) end=done3;
output;
end;
run;
This would be easier (less clumsy), but probably takes longer to run:
data want;
do _n_=1 to 14, 83, 16 to 82, 15, 84 to _nobs_;
set have nobs=_nobs_ point=_n_;
output;
end;
stop;
run;
The code is untested, but should be OK.
All Replies
Super Contributor
Posts: 271
## Re: Swaping observations
You may have to write a macro to swap row.
Super User
Posts: 9,599
## Re: Swaping observations
Thats not logical. Why the 15th and why the 83rd? You can do it by using the _n_ automatic variable but I would really advise against it as any change to that data (sorting, logic etc.) would mess up that change. You would be better off looking at why you select the 15th and 83rd rows and then use that selection criteria to pull out the rows you want and swap them over. Also bear in mind that there are sort procedures and such like that sort the data based on key variables, not position in the dataset. You may be thinking in Excel terms, don't, SAS datasets are logically ordered and manipulated.
Solution
08-11-2016 12:22 AM
Super User
Posts: 6,785
## Re: Swaping observations
You would need to re-create the data set. And it's clumsy. This would be one way.
data want;
do until (done1);
set have (obs=14) end=done1;
output;
end;
set have (firstobs=83 obs=83);
output;
do until (done2);
set have (firstobs=16 obs=82) end=done2;
output;
end;
set have (firstobs=15 obs=15);
output;
do until (done3);
set have (firstobs=84) end=done3;
output;
end;
run;
This would be easier (less clumsy), but probably takes longer to run:
data want;
do _n_=1 to 14, 83, 16 to 82, 15, 84 to _nobs_;
set have nobs=_nobs_ point=_n_;
output;
end;
stop;
run;
The code is untested, but should be OK.
Frequent Contributor
Posts: 79
## Re: Swaping observations
hi
could you please explain a bit how POINT options is working here??
-Rahul
Super Contributor
Posts: 271
## Re: Swaping observations
Point options make set statement to read data from observation number mentioned with point = .
Super User
Posts: 10,784
## Re: Swaping observations
```Why not make a index variable N and proc sort it ?
data have;
do i=1 to 100;
output;
end;
run;
data want;
set have;
n=_n_;
if n=15 then n=84;
else if n=84 then n=15;
run;
proc sort data=want;by n;run;
```
Super Contributor
Posts: 271
## Re: Swaping observations
@Ksharp Simple and very effective.
☑ This topic is solved. | 2.515625 | CC-MAIN-2018-30 |
# Chemical Equations Chapter 8 Outline Chemical Reactions Evidence
## Presentation on theme: "Chemical Equations Chapter 8 Outline Chemical Reactions Evidence"— Presentation transcript:
Chemical Equations Chapter 8 Outline Chemical Reactions Evidence
Balancing Types of Reactions Combination/Synthesis Combustion Decomposition Single Replacement Double Replacement
2 C8H16 (l) + 25 O2 (g) 16 CO2 (g) + 18 H2O (l) + heat
What happens to the octane molecule when it undergoes combustion? 2 C8H16 (l) + 25 O2 (g) 16 CO2 (g) + 18 H2O (l) + heat
Evidence for Chemical Reactions
A gas is produced. A precipitate is formed. A permanent color change is observed. An energy change occurs.
Chemical Equations Must be balanced to satisfy Law of conservation of mass State Designations (g) gas (l) liquid (s) solid (aq) aqueous
What information is shown in a balanced chemical equation?
Zn (s) + 2 HCl (aq) ZnCl2 (aq) H2 (g) 1 molecule 2 molecules 1 formula unit 1 atom Zn 2 atoms H 2 atoms Cl 1 mole 2 moles 65.38 g 2(36.46 g) g 2.02 g g =
Diatomic Elements
Classifying Reactions by what Atoms Do
Classifying Reactions by what Atoms Do
Combination/Synthesis A + Z AZ
Combination Reaction
Example – Combination Reactions
Solid magnesium oxide combines with carbon dioxide gas to produce solid magnesium carbonate. What is the sum of the coefficients?
Example – Combination Reactions
Lead filings are heated with phosphorus powder, P, to yield lead(IV) phosphide solid. What is the coefficient of the lead?
Example – Combination Reactions
Chromium metal is heated with nitrogen gas to give chromium(III) nitride solid. What is the coefficient of the product?
Example – Combination Reactions Combustion Reactions
Lithium metal is heated in the presence of oxygen gas. What is the formula of the product? LiO Li2O Li O2 LiO2
Example – Combination Reactions Combustion Reactions
Copper metal is heated with oxygen gas to produce copper(I) oxide. What is the sum of the coefficients of the reactants?
Example – Combination Reactions Combustion Reactions
Chlorine gas is heated with oxygen gas to produce dichlorine trioxide gas. What is the coefficient of the product?
Example – Combination Reactions Combustion Reactions
Propane gas, C3H8, is used in gas barbecues. Write the balanced combustion reaction. C3H8 (g) + O2 (g) 3 CO (g) + 4 H2 (g) B. C3H8 (g) + O2 (g) CO2 (g) + H2O (g) C. 2 C3H8 (g) + 10 O2 (g) 6 CO2 (g) + 8 H2O (g) D. C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) E. C3H8 (g) + O2 (g) no reaction
Example – Combination Reactions Combustion Reactions
The liquid ethanol, CH3CH2OH, is often used to warm desserts by flambéing. How many moles of ethanol are needed to balance the chemical reaction?
Classifying Reactions by what Atoms Do
Decomposition AZ A + Z
Decomposition
Example – Redox Reactions Decomposition Reactions
Hydrogen peroxide is used to clean wounds. It is stored in brown bottles to prevent it’s decomposition into water and oxygen gas. What is the coefficient of the hydrogen peroxide?
Example – Redox Reactions Decomposition Reactions
Silver bicarbonate decomposes by heating to silver carbonate powder, water, and carbon dioxide gas. What is the sum of the coefficients?
Example – Redox Reactions Decomposition Reactions
Silver carbonate further decomposes upon heating to silver oxide and carbon dioxide. What is the sum of the coefficients now?
Example – Redox Reactions Decomposition Reactions
Stannous chlorate decomposes upon heating to yield stannous chloride and oxygen gas. What is the balanced chemical equation? Sn(ClO3)2 (s) SnCl2 (s) + 3 O2 (g) Sn(ClO2)2 (s) SnCl2 (s) + 4 O2 (g) Sn(ClO2)2 (s) no rxn SnCl2 (s) + 3 O2 (g) Sn(ClO3)2 (s)
Copper(II) oxide reacts with ammonia (NH3) to yield copper, nitrogen gas, and water.
Write a balanced equation for this reaction.
Lead(II) nitrate reacts with potassium chromate to form lead(II) chromate (yellow ppt.) and potassium nitrate. Hydrochloric acid reacts with sodium carbonate to form carbon dioxide, sodium chloride, and water
Zinc metal reacts with hydrochloric acid to produce zinc chloride and hydrogen gas.
Potassium chlorate when heated, decomposes to form potassium chloride and oxygen gas.
Hexane(C6H14) burns in oxygen gas to form carbon dioxide and water.
Vinegar(acetic acid) reacts with baking soda (sodium bicarbonate) to produce carbon dioxide gas, sodium acetate, and water. Ammonia reacts with oxygen gas to form nitrogen monoxide and water.
Iron(III) chloride reacts with ammonium hydroxide to form iron(III) hydroxide (brown ppt.) and ammonium chloride. Barium hydroxide and ammonium chloride react to form ammonia (NH3), water, and barium chloride.
Some more examples N2 + H2 NH3 Be2C + H2O Be(OH)2 + CH4
HCl + CaCO3 CaCl2 + H2O + CO2 C2H6 + O2 CO2 + H2O
Classifying Reactions by what Atoms Do
Single Displacement A + BZ AZ + B
Single Replacement
Example – Redox Reactions Single Replacement Reactions
Activity Series of Metals Li K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb (H) Sb Cu Ag Pd Pt Hg Au Example – Redox Reactions Single Replacement Reactions Aluminum wire is placed in a cupric nitrate solution. What is the sum of the coefficients?
Example – Redox Reactions Single Replacement Reactions
Activity Series of Metals Li K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb (H) Sb Cu Ag Pd Pt Hg Au Example – Redox Reactions Single Replacement Reactions Copper wire is placed in a solution of magnesium nitrate. What is the balanced chemical equation? Cu (s) + Mg(NO3)2 (aq) Mg (s) + Cu(NO3)2 (aq) Cu (s) + Mg(NO3)2 (aq) MgCu (s) + NO2 (g) Cu (s) + Mg(NO3)2 (aq) MgCu (s) + 2 NO3 -(aq) Cu2+ (s) + Mg(NO3)2 (aq) Mg2+ (aq) + Cu(NO3)2 (aq) No reaction
Example – Redox Reactions Single Replacement Reactions
Activity Series of Metals Li K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb (H) Sb Cu Ag Pd Pt Hg Au Example – Redox Reactions Single Replacement Reactions Magnesium ribbon is placed in a hydrochloric acid solution. What is the coefficient in front of the magnesium?
Example – Redox Reactions Single Replacement Reactions
Activity Series of Metals Li K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb (H) Sb Cu Ag Pd Pt Hg Au Example – Redox Reactions Single Replacement Reactions A piece of soft, gray lithium metal is added to water. What is the coefficient of lithium?
Example – Redox Reactions Single Replacement Reactions
Activity Series of Metals Li K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb (H) Sb Cu Ag Pd Pt Hg Au Example – Redox Reactions Single Replacement Reactions Manganese chips are added to nitric acid. What gas is formed?
Cu + AgNO3
Cr + NiCl2
Cr + Zn(NO3)2
Zn + HCl
Fe + HCl
Classifying Reactions by what Atoms Do
Double displacement AX + BZ AZ + BX Precipitation (solubility rules) Gas formation (H2S, CO2, NH3, SO2) Slightly ionizable substance (weak acid/weak base) Neutralization (special type of double displacement reaction) HX (aq) +BOH (aq) BX +H2O (l)
Double Displacement
Predicting Reactions Double Displacement
Fig. 9-13, p. 256
Common Strong Acids and Bases
Common Strong Bases Lithium hydroxide LiOH (aq) Sodium hydroxide NaOH (aq) Potassium hydroxide KOH (aq) Strontium hydroxide Sr(OH)2 (aq) Barium hydroxide Ba(OH)2 (aq) Calcium hydroxide Ca(OH)2 (aq) Common Strong Acids Hydrochloric acid HCl (aq) Hydrobromic acid HBr (aq) Hydroiodic acid HI (aq) Perchloric acid HClO4 (aq) Chloric acid HClO3 (aq) Nitric acid HNO3 (aq) Sulfuric acid H2SO4 (aq)
Example – Ion Exchange Reactions
What is the precipitate when lead(II) nitrate reacts with lithium chloride?
Example – Ion Exchange Reactions
What gas is formed when sodium sulfite reacts with hydrobromic acid? H2SO3 (g) H2S (g) SO3 (g) SO2 (g)
Example – Ion Exchange Reactions
When aqueous solutions of sodium sulfite and potassium bromide are mixed the products are: NaBr and K2SO3 K2SO3 and Na2Br NaK and SO3Br No reaction
Example – Ion Exchange Reactions
Aqueous sodium acetate is mixed with hydrobromic acid. What is formed? Precipitate Gas Slightly ionizable substance No reaction
Example – Ion Exchange Reactions
When aqueous solutions of sodium iodide and nitric acid are mixed the products are: NaI and HNO3 NaNO3 and NaI NaH and INO3 No reaction
Example – Ion Exchange Reactions
When sulfuric acid and aqueous lithium hydroxide are mixed one of the products is Water Lithium hydride Sulfur trioxide Lithium sulfide No reaction
Na2S + ZnCl2 Na2S + ZnCl2 2 NaCl + ZnS(s)
Mg(NO3)2 + NaOH Mg(NO3)2 + 2 NaOH Mg(OH)2(s)+ 2 NaNO3
AgNO3 + Na2SO4 AgNO3 + Na2SO4 NR
K2CO3 + HCl K2CO HCl (H2CO3 + 2 KCl) H2O + CO2 + 2 KCl
NH4Cl + KOH NH4Cl + KOH (NH4OH + KCl) NH3 +HOH+ KCl
HNO3 + NaC2H3O2 HNO3(aq) + NaC2H3O2(aq) HC2H3O2(aq) + NaNO3(aq)
HClO4 + NaOH HClO4 + NaOH H2O(l) + NaClO4
Similar presentations | 2.65625 | CC-MAIN-2017-51 |
### Calculating points distance (application freezes)
Hello people...!
Got a rather odd question but it is really stooping my development.
I am calculating a distance between two points for two separate center points of a bounding box.
My application works fine after I put the code to calculate the distance between points of the first center.
When I add the code to my application to calculate the distance between points of the second center point my application freezes completely.
There are no compiler errors neither run time errors the application just stops (freezes) after a second or 2 and stays like this forever.....
``1234567891011121314151617181920212223242526272829303132333435363738394041424344454647`` `````` vector Rightarm; vector Leftarm; vector Left_Arm_xy; vector Right_Arm_xy; vector Left_Point_distance; vector Right_Point_distance; Left_Arm_xy.push_back(cv::Point(center.x,center.y)); Right_Arm_xy.push_back(cv::Point(center1.x,center1.y)); double Pt1; double Pt2; for(vector::iterator iter_a = Left_Arm_xy.begin()+1; iter_a != Left_Arm_xy.end(); ++iter_a) { Pt1 = pow((double) iter_a->x - (iter_a -1)->x,2); Pt2 = pow((double) iter_a->y - (iter_a -1)->y,2); double result_a; result_a = (Pt1 + Pt2); Left_Point_distance.push_back(sqrt(result_a)); } //-----------------------------------------------------------------------------> double Pt3; double Pt4; for(vector::iterator iter_b = Right_Arm_xy.begin()+1; iter_b != Right_Arm_xy.end(); iter_b) { Pt3 = pow((double) iter_b->x - (iter_b -1)->x,2); Pt4 = pow((double) iter_b->y - (iter_b -1)->y,2); double result_b; result_b = (Pt3 + Pt4); Right_Point_distance.push_back(sqrt(result_b)); }``````
Really have no idea of why this could be happening....could anyone possibly suggest an issue that exists withing the code....?
Regards
line 36 is causing your infinite loop
should be:
` for(vector<Point>::iterator iter_b = Right_Arm_xy.begin()+1; iter_b != Right_Arm_xy.end(); ++iter_b)`
goshhhhhhhhh HEHE thank you so much Not to sure how i didn't spot this but but once again THANKS. regards
Topic archived. No new replies allowed. | 2.96875 | CC-MAIN-2017-30 |
MATH STANDARD FOR CTE - Strand
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language:
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Document Sample
``` Minnesota 6-12 Academic Standards in
Mathematics
Selected for CTE
April 14, 2007 Revision
DRAFT Minnesota K-12 Academic Standards in Mathematics DRAFT
Standards and benchmarks highlighted in yellow may be
particularly applicable to CTE Courses.
Page 2 of 13 Sorted by Grade April 14, 2007
DRAFT Minnesota K-12 Academic Standards in Mathematics DRAFT
Recognize linear, Represent and solve problems in various contexts using linear
exponential and
9.2.2.1 For example: Write a function that represents the area of a rectangular
other common
garden that can be surrounded with 32 feet of fencing, and use the function
functions in real- to determine the possible dimensions of such a garden if the area must be at
world and least 50 square feet.
mathematical
situations; Represent and solve problems in various contexts using
represent these 9.2.2.2 exponential functions, such as investment growth,
functions with depreciation and population growth.
tables, verbal Sketch graphs of linear, quadratic and exponential functions,
descriptions, and translate between graphs, tables and symbolic
symbols and 9.2.2.3
representations. Know how to use graphing technology to
graphs; solve graph these functions.
problems Express the terms in a geometric sequence recursively and by
involving these giving an explicit (closed form) formula, and express the
functions, and partial sums of a geometric series recursively.
explain results in
the original For example: A closed form formula for the terms tn in the geometric
context. sequence 3, 6, 12, 24, ... is tn = 3(2)n-1, where n = 1, 2, 3, ... , and this
9.2.2.4 sequence can be expressed recursively by writing t1 = 3 and
Recognize linear, tn = 2tn-1, for n 2.
Another example: the partial sums sn of the series 3 + 6 + 12 + 24 + ... can
10, exponential and be expressed recursively by writing s1 = 3 and
Algebra other common
11 sn = 3 + 2sn-1, for n 2.
functions in real-
world and
mathematical
situations;
represent these
functions with
tables, verbal Recognize and solve problems that can be modeled using
descriptions, finite geometric sequences and series, such as home mortgage
symbols and 9.2.2.5 and other compound interest examples. Know how to use
graphs; solve spreadsheets and calculators to explore geometric sequences
problems and series in various contexts.
involving these
functions, and
explain results in
the original
context.
Sketch the graphs of common non-linear functions such as
Generate
f x x , f x x , f x 1 , f(x) = x3, and translations of
equivalent x
9.2.2.6
algebraic these functions, such as f x x 2 4 . Know how to use
expressions
involving graphing technology to graph these functions.
Page 3 of 13 Sorted by Grade April 14, 2007
DRAFT Minnesota K-12 Academic Standards in Mathematics DRAFT
polynomials and Evaluate polynomial and rational expressions and expressions
radicals; use 9.2.3.1 containing radicals and absolute values at specified points in
algebraic their domains.
properties to
evaluate
expressions. Add, subtract and multiply polynomials; divide a polynomial
9.2.3.2
by a polynomial of equal or lower degree.
Factor common monomial factors from polynomials, factor
quadratic polynomials, and factor the difference of two
9.2.3.3 squares.
For example: 9x6 – x4 = (3x3 – x2)(3x3 + x2).
Add, subtract, multiply, divide and simplify algebraic
fractions.
9.2.3.4
1 x 1 2x x 2
For example: is equivalent to .
1 x 1 x 1 x2
Check whether a given complex number is a solution of a
quadratic equation by substituting it for the variable and
evaluating the expression, using arithmetic with complex
numbers.
9.2.3.5
1 i
9, For example: The complex number is a solution of 2x2 – 2x + 1 = 0,
2
10, Algebra 2
since 2 1 i 2 1 i 1 i 1 i 1 0 .
11 2 2
Generate Apply the properties of positive and negative rational
equivalent exponents to generate equivalent algebraic expressions,
algebraic including those involving nth roots.
9.2.3.6
expressions
2 7 2 2 7 2 14 2 14 . Rules for computing
1 1 1
involving For example:
polynomials and directly with radicals may also be used: 2 x 2x .
radicals; use Justify steps in generating equivalent expressions by
algebraic identifying the properties used. Use substitution to check the
properties to equality of expressions for some particular values of the
evaluate 9.2.3.7
variables; recognize that checking with substitution does not
expressions. guarantee equality of expressions for all values of the
Represent real- variables.
Page 4 of 13 Sorted by Grade April 14, 2007
DRAFT Minnesota K-12 Academic Standards in Mathematics DRAFT
world and Represent relationships in various contexts using quadratic
mathematical equations and inequalities. Solve quadratic equations and
situations using inequalities by appropriate methods including factoring,
equations and completing the square, graphing and the quadratic formula.
inequalities Find non-real complex roots when they exist. Recognize that
involving linear, a particular solution may not be applicable in the original
quadratic, context. Know how to use calculators, graphing utilities or
9.2.4.1
exponential, and other technology to solve quadratic equations and
nth root functions. inequalities.
Solve equations
and inequalities For example: A diver jumps from a 20 meter platform with an upward
velocity of 3 meters per second. In finding the time at which the diver hits
symbolically and the surface of the water, the resulting quadratic equation has a positive and
graphically. a negative solution. The negative solution should be discarded because of
Interpret solutions the context.
in the original Represent relationships in various contexts using equations
context. involving exponential functions; solve these equations
9.2.4.2
graphically or numerically. Know how to use calculators,
graphing utilities or other technology to solve these equations.
Recognize that to solve certain equations, number systems
need to be extended from whole numbers to integers, from
integers to rational numbers, from rational numbers to real
9.2.4.3
numbers, and from real numbers to complex numbers. In
particular, non-real complex numbers are needed to solve
some quadratic equations with real coefficients.
Represent relationships in various contexts using systems of
linear inequalities; solve them graphically. Indicate which
9.2.4.4
parts of the boundary are included in and excluded from the
solution set using solid and dotted lines.
Solve linear programming problems in two variables using
9.2.4.5
graphical methods.
9,
10, Algebra Represent real-
11 world and
mathematical
situations using Represent relationships in various contexts using absolute
equations and value inequalities in two variables; solve them graphically.
inequalities 9.2.4.6
For example: If a pipe is to be cut to a length of 5 meters accurate to within
involving linear, a tenth of its diameter, the relationship between the length x of the pipe and
quadratic, its diameter y satisfies the inequality | x – 5| ≤ 0.1y.
exponential and
nth root functions.
Solve equations
Page 5 of 13 Sorted by Grade April 14, 2007
DRAFT Minnesota K-12 Academic Standards in Mathematics DRAFT
and inequalities Solve equations that contain radical expressions. Recognize
symbolically and that extraneous solutions may arise when using symbolic
graphically. methods.
Interpret solutions
in the original For example: The equation x 9 9 x may be solved by squaring both
9.2.4.7
context. sides to obtain x – 9 = 81x, which has the solution x 9 . However, this
80
Calculate is not a solution of the original equation, so it is an extraneous solution that
measurements of should be discarded. The original equation has no solution in this case.
plane and solid
geometric figures; Another example: Solve 3 x 1 5 .
know that
physical Assess the reasonableness of a solution in its given context
measurements and compare the solution to appropriate graphical or
depend on the 9.2.4.8
numerical estimates; interpret a solution in the original
choice of a unit context.
and that they are
approximations.
Determine the surface area and volume of pyramids, cones
and spheres. Use measuring devices or formulas as
9.3.1.1 appropriate.
For example: Measure the height and radius of a cone and then use a
formula to find its volume.
Compose and decompose two- and three-dimensional figures;
use decomposition to determine the perimeter, area, surface
9.3.1.2 area and volume of various figures.
For example: Find the volume of a regular hexagonal prism by
decomposing it into six equal triangular prisms.
Understand that quantities associated with physical
measurements must be assigned units; apply such units
correctly in expressions, equations and problem solutions that
9.3.1.3 involve measurements; and convert between measurement
systems.
Calculate For example: 60 miles/hour = 60 miles/hour × 5280 feet/mile ×
1 hour/3600 seconds = 88 feet/second.
measurements of
plane and solid
geometric figures; Understand and apply the fact that the effect of a scale factor
9,
know that 9.3.1.4 k on length, area and volume is to multiply each by k, k2 and
10, Algebra
physical k3, respectively.
11
measurements
depend on the
choice of a unit Make reasonable estimates and judgments about the accuracy
and that they are of values resulting from calculations involving measurements.
approximations.
For example: Suppose the sides of a rectangle are measured to the nearest
9.3.1.5 tenth of a centimeter at 2.6 cm and 9.8 cm. Because of measurement errors,
the width could be as small as 2.55 cm or as large as 2.65 cm, with similar
errors for the height. These errors affect calculations. For instance, the
actual area of the rectangle could be smaller than 25 cm2 or larger than
26 cm2, even though 2.6 × 9.8 = 25.48.
Page 6 of 13 Sorted by Grade April 14, 2007
DRAFT Minnesota K-12 Academic Standards in Mathematics DRAFT
Understand the roles of axioms, definitions, undefined terms
9.3.2.1
and theorems in logical arguments.
Accurately interpret and use words and phrases in geometric
proofs such as "if…then," "if and only if," "all," and "not."
Recognize the logical relationships between an "if…then"
Construct logical 9.3.2.2 statement and its inverse, converse and contrapositive.
arguments, based For example: The statement "If you don't do your homework, you can't go
on axioms, to the dance" is not logically equivalent to its inverse "If you do your
Geometry & definitions and homework, you can go to the dance."
Measurement theorems, to prove
theorems and Assess the validity of a logical argument and give
other results in 9.3.2.3
counterexamples to disprove a statement.
geometry.
Construct logical arguments and write proofs of theorems and
other results in geometry, including proofs by contradiction.
Express proofs in a form that clearly justifies the reasoning,
9.3.2.4 such as two-column proofs, paragraph proofs, flow charts or
illustrations.
For example: Prove that the sum of the interior angles of a pentagon is 540˚
using the fact that the sum of the interior angles of a triangle is 180˚.
Use technology tools to examine theorems, test conjectures,
perform constructions and develop mathematical reasoning
9.3.2.5 skills in multi-step problems. The tools may include compass
and straight edge, dynamic geometry software, design
software or Internet applets.
Know and apply properties of parallel and perpendicular
Know and apply
lines, including properties of angles formed by a transversal,
properties of
to solve problems and logically justify results.
geometric figures 9.3.3.1
9,
Geometry & to solve real- For example: Prove that the perpendicular bisector of a line segment is the
10,
Measurement world and set of all points equidistant from the two endpoints, and use this fact to
11 mathematical solve problems and justify other results.
problems and to Know and apply properties of angles, including
logically justify corresponding, exterior, interior, vertical, complementary and
results in supplementary angles, to solve problems and logically justify
geometry. results.
Know and apply
properties of 9.3.3.2 For example: Prove that two triangles formed by a pair of intersecting lines
and a pair of parallel lines (an "X" trapped between two parallel lines) are
geometric figures similar.
to solve real-
world and
mathematical
Page 7 of 13 Sorted by Grade April 14, 2007
DRAFT Minnesota K-12 Academic Standards in Mathematics DRAFT
problems and to Know and apply properties of equilateral, isosceles and
logically justify scalene triangles to solve problems and logically justify
results in 9.3.3.3 results.
geometry.
For example: Use the triangle inequality to prove that the perimeter of a
quadrilateral is larger than the sum of the lengths of its diagonals.
Apply the Pythagorean Theorem and its converse to solve
problems and logically justify results.
9.3.3.4
For example: When building a wooden frame that is supposed to have a
square corner, ensure that the corner is square by measuring lengths near
the corner and applying the Pythagorean Theorem.
Know and apply properties of right triangles, including
properties of 45-45-90 and 30-60-90 triangles, to solve
problems and logically justify results.
9.3.3.5
For example: Use 30-60-90 triangles to analyze geometric figures involving
equilateral triangles and hexagons.
Another example: Determine exact values of the trigonometric ratios in
these special triangles using relationships among the side lengths.
Know and apply properties of congruent and similar figures
to solve problems and logically justify results.
For example: Analyze lengths and areas in a figure formed by drawing a
line segment from one side of a triangle to a second side, parallel to the
third side.
9.3.3.6 Another example: Determine the height of a pine tree by comparing the
length of its shadow to the length of the shadow of a person of known
height.
Another example: When attempting to build two identical 4-sided frames, a
person measured the lengths of corresponding sides and found that they
matched. Can the person conclude that the shapes of the frames are
congruent?
Use properties of polygons—including quadrilaterals and
Know and apply regular polygons—to define them, classify them, solve
properties of 9.3.3.7 problems and logically justify results.
geometric figures
For example: Recognize that a rectangle is a special case of a trapezoid.
to solve real-
world and Another example: Give a concise and clear definition of a kite.
9, Know and apply properties of a circle to solve problems and
Geometry & mathematical
10, logically justify results.
Measurement problems and to
11 9.3.3.8
logically justify
For example: Show that opposite angles of a quadrilateral inscribed in a circle are
results in supplementary.
geometry.
Solve real-world Understand how the properties of similar right triangles allow
and mathematical 9.3.4.1 the trigonometric ratios to be defined, and determine the sine,
geometric cosine and tangent of an acute angle in a right triangle.
Page 8 of 13 Sorted by Grade April 14, 2007
DRAFT Minnesota K-12 Academic Standards in Mathematics DRAFT
problems using Apply the trigonometric ratios sine, cosine and tangent to
algebraic solve problems, such as determining lengths and areas in right
methods. triangles and in figures that can be decomposed into right
9.3.4.2 triangles. Know how to use calculators, tables or other
technology to evaluate trigonometric ratios.
For example: Find the area of a triangle, given the measure of one of its
acute angles and the lengths of the two sides that form that angle.
Use calculators, tables or other technologies in connection
9.3.4.3 with the trigonometric ratios to find angle measures in right
triangles in various contexts.
Use coordinate geometry to represent and analyze line
9.3.4.4 segments and polygons, including determining lengths,
midpoints and slopes of line segments.
Know the equation for the graph of a circle with radius r and
9.3.4.5 center (h,k), (x – h)2 + (y – k)2 = r2, and justify this equation
using the Pythagorean Theorem and properties of translations.
Use numeric, graphic and symbolic representations of
transformations in two dimensions, such as reflections,
translations, scale changes and rotations about the origin by
Display and
9.3.4.6 multiples of 90˚, to solve problems involving figures on a
analyze data; use coordinate grid.
various measures
associated with For example: If the point (3,-2) is rotated 90˚ counterclockwise about the
data to draw origin, it becomes the point (2,3).
9,
Geometry & conclusions, Use algebra to solve geometric problems unrelated to
10,
Measurement identify trends coordinate geometry, such as solving for an unknown length
11
and describe 9.3.4.7 in a figure involving similar triangles, or using the
relationships. Pythagorean Theorem to obtain a quadratic equation for a
Explain the uses length in a geometric figure.
of data and
statistical thinking Describe a data set using data displays, such as box-and-
to draw whisker plots; describe and compare data sets using summary
inferences, make statistics, including measures of center, location and spread.
predictions and Measures of center and location include mean, median,
9.4.1.1
justify quartile and percentile. Measures of spread include standard
conclusions. deviation, range and inter-quartile range. Know how to use
calculators, spreadsheets or other technology to display data
and calculate summary statistics.
Page 9 of 13 Sorted by Grade April 14, 2007
DRAFT Minnesota K-12 Academic Standards in Mathematics DRAFT
Analyze the effects on summary statistics of changes in data
sets.
For example: Understand how inserting or deleting a data point may affect
9.4.1.2 the mean and standard deviation.
Another example: Understand how the median and interquartile range are
affected when the entire data set is transformed by adding a constant to
each data value or multiplying each data value by a constant.
Use scatterplots to analyze patterns and describe relationships
between two variables. Using technology, determine
9.4.1.3 regression lines (line of best fit) and correlation coefficients;
use regression lines to make predictions and correlation
coefficients to assess the reliability of those predictions.
Use the mean and standard deviation of a data set to fit it to a
normal distribution (bell-shaped curve) and to estimate
population percentages. Recognize that there are data sets for
which such a procedure is not appropriate. Use calculators,
spreadsheets and tables to estimate areas under the normal
curve.
9.4.1.4
For example: After performing several measurements of some attribute of
an irregular physical object, it is appropriate to fit the data to a normal
distribution and draw conclusions about measurement error.
Another example: When data involving two very different populations is
combined, the resulting histogram may show two distinct peaks, and fitting
the data to a normal distribution is not appropriate.
Evaluate reports based on data published in the media by
identifying the source of the data, the design of the study, and
the way the data are analyzed and displayed. Show how
graphs and data can be distorted to support different points of
9.4.2.1 view. Know how to use spreadsheet tables and graphs or
graphing technology to recognize and analyze distortions in
data displays.
For example: Shifting data on the vertical axis can make relative changes
appear deceptively large.
Identify and explain misleading uses of data; recognize when
Calculate 9.4.2.2
arguments based on data confuse correlation and causation.
probabilities and
9, Data
apply probability
10, Analysis &
concepts to solve
11 Probability
real-world and
mathematical Explain the impact of sampling methods, bias and the
problems. 9.4.2.3
phrasing of questions asked during data collection.
Page 10 of 13 Sorted by Grade April 14, 2007
DRAFT Minnesota K-12 Academic Standards in Mathematics DRAFT
Select and apply counting procedures, such as the
multiplication and addition principles and tree diagrams, to
determine the size of a sample space (the number of possible
outcomes) and to calculate probabilities.
9.4.3.1
For example: If one girl and one boy are picked at random from a class
with 20 girls and 15 boys, there are 20 × 15 = 300 different possibilities, so
the probability that a particular girl is chosen together with a particular boy
1
is .
300
Calculate experimental probabilities by performing
9.4.3.2 simulations or experiments involving a probability model and
using relative frequencies of outcomes.
Understand that the Law of Large Numbers expresses a
relationship between the probabilities in a probability model
9.4.3.3
and the experimental probabilities found by performing
simulations or experiments involving the model.
Use random numbers generated by a calculator or a
spreadsheet, or taken from a table, to perform probability
simulations and to introduce fairness into decision making.
9.4.3.4
For example: If a group of students needs to fairly select one of its
members to lead a discussion, they can use a random number to determine
the selection.
Apply probability concepts such as intersections, unions and
complements of events, and conditional probability and
independence, to calculate probabilities and solve problems.
9.4.3.5
For example: The probability of tossing at least one head when flipping a
fair coin three times can be calculated by looking at the complement of this
event (flipping three tails in a row).
Describe the concepts of intersections, unions and
Calculate
9, Data complements using Venn diagrams. Understand the
probabilities and
10, Analysis & 9.4.3.6 relationships between these concepts and the words AND,
apply probability
11 Probability OR, NOT, as used in computerized searches and
concepts to solve
real-world and
Page 11 of 13 Sorted by Grade April 14, 2007
DRAFT Minnesota K-12 Academic Standards in Mathematics DRAFT
mathematical Understand and use simple probability formulas involving
problems. intersections, unions and complements of events.
For example: If the probability of an event is p, then the probability of the
9.4.3.7 complement of an event is 1 – p; the probability of the intersection of two
independent events is the product of their probabilities.
Another example: The probability of the union of two events equals the sum
of the probabilities of the two individual events minus the probability of the
intersection of the events.
Apply probability concepts to real-world situations to make
informed decisions.
For example: Explain why a hockey coach might decide near the end of the
9.4.3.8 game to pull the goalie to add another forward position player if the team is
behind.
Another example: Consider the role that probabilities play in health care
decisions, such as deciding between having eye surgery and wearing
glasses.
Use the relationship between conditional probabilities and
relative frequencies in contingency tables.
9.4.3.9 For example: A table that displays percentages relating gender (male or
female) and handedness (right-handed or left-handed) can be used to
determine the conditional probability of being left-handed, given that the
gender is male.
9, Data
10, Analysis &
11 Probability
Page 12 of 13 Sorted by Grade April 14, 2007
DRAFT Minnesota K-12 Academic Standards in Mathematics DRAFT
Page 13 of 13 Sorted by Grade April 14, 2007
```
Related docs
Other docs by niusheng11
annualreport2009 - PDF | 4.1875 | CC-MAIN-2013-20 |
What is 70% in GPA
To convert your GPA to a 4.0 scale:
B- 80-82 2.7
C+ 77-79 2.3
C 73-76 2.0
C- 70-72 1.7
What is a UK GPA of 70%
A rough conversion of British grades to GPAs looks like this: +70% = 4.0. 60-69% = 3.3-3.9. 50-59% = 2.7-3.2.
Is a 75% a 2.5 GPA
The average GPA for graduating high school students is a 3.0, which 2.5 falls below. A 2.5 means that you received a C-average for academic performance, equally out to a percentile between 73% and 76%.
What is a 75% percentage to GPA
3
75/100*4. The GPA will be 3.
What is 70 percent in 4.0 GPA
On a 4.0 scale, 70% will equal (70/100)*4= 2.8. What is my GPA on a 4.0 scale On a 4.0 GPA scale, a 4.0 is for an A or A+ followed by 3.0 which is B, 2.0=C as well as 1.0=D.
Is A GPA of 2.8 good
A 2.8 GPA is equivalent to 83% or B letter grade. The national average GPA is 3.0 which means a 2.8 is just a bit below average.
Is 3.40 A Good GPA
A 3.4 is comfortably above the national average GPA for high school graduates and is considered competitive at most colleges and universities. It also approaches the 3.5 GPA that more selective schools use as an unofficial benchmark.
Is A 3.5 A Good GPA
Usually, a GPA of 3.0 – 3.5 is considered good enough at many high schools, colleges, and universities. Top academic institutions usually require GPAs higher than 3.5.
Is a 3.0 GPA 75%
3.0 GPA equals to 83-86% or B grade.
Does Harvard accept 2.8 GPA
A 4.0 GPA (and above) means you have a stronger chance of acceptance. A GPA between 3.75-3.99 GPA means you have a decent chance of success. A GPA between 3.50-3.74 makes Harvard a “reach” school. A GPA between 2.50-3.49 results in an extremely low chance of success.
The national average for a GPA is around 3.0 and a 1.7 GPA puts you below that average. A 1.7 GPA means that you've gotten only C-s and D+s in your high school classes so far. Since this GPA is significantly below a 2.0, it will make things very difficult for you in the college application process.
Is a 3.0+ GPA good
It indicates that you've gotten grades between B+ and B- for the majority of your classes, giving you a percentile between 83% and 86%. A 3.0 GPA is considered a straight “B” average and makes you a competitive applicant for a number of colleges and universities.
Is 3.5 an OK GPA
But, a grade point average of 3.5 and above is good in college. 3.5 GPA students normally qualify for the dean's list and honors programs. The average GPA in four-year undergraduate courses in U.S. colleges is 3.15. So, we can confidently say that yes, a 3.5 GPA is good in college.
Is A 1.5 GPA good
The answer is No. The national average for a GPA is around 3.0 and a 1.5 GPA puts you below that average. A 1.5 GPA means that you've gotten only C-s and D+s in your high school classes so far. Since this GPA is significantly below a 2.0, it will make things very difficult for you in the college application process.
Is 70% 3 GPA
What GPA is 70% On a 4.0 scale, 70% will equal (70/100)*4= 2.8. What is my GPA on a 4.0 scale On a 4.0 GPA scale, a 4.0 is for an A or A+ followed by 3.0 which is B, 2.0=C as well as 1.0=D.
Does Harvard accept 5.0 GPA
While there is no GPA requirement to apply to Harvard, the average unweighted GPA of incoming students is 4.18. Do your best to achieve a high GPA to boost your chances of acceptance!
Is a 3.70 GPA good
Overall, it's an impressive GPA and a 92% percentile grade. On a 4.0 scale, a 3.7 GPA is impressive! A 3.7 GPA on an unweighted scale will place you in good standing at many colleges.
Is 2.8 A low GPA
A 2.8 GPA is equivalent to 83% or B letter grade. The national average GPA is 3.0 which means a 2.8 is just a bit below average.
The national average GPA is 3.0; a 3.4 GPA is approximately half a point higher. This is an excellent grade point average. If you can get it above 3.5, you can brag about the GPA.
What is a 3.0+ GPA
It indicates that you've gotten grades between B+ and B- for the majority of your classes, giving you a percentile between 83% and 86%. A 3.0 GPA is considered a straight “B” average and makes you a competitive applicant for a number of colleges and universities.
Is 4.37 A Good GPA
This GPA is above a 4.0, which means it's weighted (it takes into account the difficulty of your classes in conjunction with your grades). This is a very good GPA. It most likely means you're taking high level classes and are earning As and Bs. 99.55% of schools have an average GPA below a 4.3.
Is 3.75 GPA enough for Harvard
Harvard does not have a minimum GPA requirement. However, Harvard is highly competitive, so it may be challenging to get in with a 3.7 GPA. The average GPA for entering Harvard students is 3.9.
Is 3.7 an okay GPA
A 3.7 GPA is a very good GPA, especially if your school uses an unweighted scale. This means that you've been earning mostly A-s in all of your classes. If you've been taking high level classes and earning a 3.7 unweighted GPA, you're in great shape and can expect to be accepted to many selective colleges.
Is A GPA of 5.0 good
Admission to the most prestigious schools is typically limited to students who rank in the top 5% of their class and have high GPAs. Therefore, this article is about one of the highest scores, 5.0 GPA. It is an outstanding score a handful of students can achieve.
Can you get into Harvard with a 4.5 GPA
While there is no GPA requirement to apply to Harvard, the average unweighted GPA of incoming students is 4.18. | 2.765625 | CC-MAIN-2024-33 |
# CHARGE CONTROL USING FUZZY LOGIC
CHARGE CONTROL USING FUZZY LOGIC
ABSTRACT
This work modified an existing lead-acid battery charging system by developing and simulating the conventional lead-acid battery and the fuzzy logic control of the lead-acid battery charging systems. The fuzzy logic system is developed using Mamdani inference system in Matlab toolbox and Simulink, and consequently employed in controlling the conventional charging system adopted. The output voltage of the battery is controlled using matrix laboratory (MATLAB) in creating Fuzzy Inference system (FIS). FIS is created by typing “fuzzy” in MATLAB workspace and clicking “Edit” selecting “add variable”, and click input to increase the number of inputs to two. The inputs are labeled voltage error (e) and derivative of error (De/dt). The membership function of the inputs, Mandani, and output blocks and the parameters of theses blocks are filled as shown in Figures 3.7 to 3.9, respectively. Furthermore, the Mandani rules are editted as shown in Figure 3.11. The model shows that the power outputs from the solar cells increased linearly when the voltage changes from 0 to 250V and then increased exponentially from 250 to 420V. The curves showed that the power outputs from the solar cells increased linearly when the voltage changes from 0 to 250V. The increase in voltage from 300 to 350V produces a maximum output power of about 1400W at 20 oC and 1000 W/m2. For the same range of voltage at 400C and 1000 W/m2, the maximum output power is approximately 1200W. Similarly, at 200C and 500 W/m2, and 40 0C and 500 W/m2 the maximum output power is 550W and 500W, respectively. In the case of 200C and 200 W/m2 ancd 400C and 200 W/m2 the maximum output power is 240W and 235W, respectively. The collector current increased from 0 to 800 ampere with the increase in the collector-emitter voltage from 1 to 5V. The only exception is the case where emitter-voltage (Vge) is 8V.
Disclaimer: By purchasing this Research Project Material, YOU agree to use it ONLY as a GUIDE to conduct your own academic research. | 2.6875 | CC-MAIN-2023-14 |
I do not understand tonality-series
Recommended Posts
Hello if i do
It gives me the folowing result, this i understand
```(setf chords '((c4 maj) (f4 m) (f4 m) (g4 maj) (c4 maj)))
(setf rows (gen-repeat 5 (list '(q c4 cs4 d4 ds4 e4 f4 fs4 g4 gs4 a4 as4 b4))))
(setf tonality1
(tonality-series chords
:root '(c4 f4 f4 g4 c4)))
(tonality-map tonality1 rows)
=> ((q c4 c4 c4 e4 e4 e4 g4 g4 g4 g4 c5 c5)
(q f4 f4 gs4 gs4 gs4 gs4 c5 c5 c5 c5 f5 f5)
(q f4 f4 gs4 gs4 gs4 gs4 c5 c5 c5 c5 f5 f5)
(q g4 g4 b4 b4 b4 b4 d5 d5 d5 d5 g5 g5)
(q c4 c4 c4 e4 e4 e4 g4 g4 g4 g4 c5 c5))```
but if i do
```(setf tonality1
(tonality-series chords
:root '(c4 f4 f4 g4 c4)
:closest '(up)))```
it gives
```=> ((q c4 d4 d4 e4 e4 e4 g4 g4 g4 g4 c5 c5)
(q f4 f4 gs4 gs4 gs4 b4 b4 b4 eb5 eb5 eb5 f5)
(q f4 f4 gs4 gs4 gs4 b4 b4 b4 eb5 eb5 eb5 f5)
(q g4 g4 g4 c5 c5 c5 c5 e5 e5 e5 e5 g5)
(q c4 d4 d4 e4 e4 e4 g4 g4 g4 g4 c5 c5))```
what i do not understand if :add '(2) adds an interval of two semitones from the root
why do i get
`(q f4 f4 gs4 gs4 gs4 b4 b4 b4 eb5 eb5 eb5 f5)`
and not
`(q f4 f4 g4 g4 gs4 b4 b4 b4 eb5 eb5 eb5 f5)`
There probably something i miss could you explain this to me
Thanks
Patrick
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# [EM] Conditional vote wording. CD fix.
Michael Ossipoff email9648742 at gmail.com
Fri Sep 23 18:25:53 PDT 2016
Of course the conditional vote option needs for voters to indicate their
favorite. But that isn't so onerous when conditional votes bring CD
compliance (not just Weak CD).
Besides, it's more expressive. Anyone getting a favorite mark from a
majority should automatically win.
By the way, in any Bucklin version that allows equal ranking & doesn't
assume (as Bucklin B or A does) that a voter who equal top-ranks will top
rank all hir top-set & not vote for anyone else--For any such method:
If a ballot equal-ranks n candidates at a rank-position, then the
count-rule should skip n-1 rounds before giving the votes at that ballot's
next rank.
In pairwise-count or Bucklin, the favorite-mark shouldn't be counted
in the rank count, because doing so would prevent equal top-ranking.
A better statement of the definition of the conditional votes option:
Using Approval as an example:
Say the A voters give to B some combination of conditional & unconditional
...& vice-versa from the B voters to A.
Say that the fraction of A voters voting for B (called Fab)
is greater than the fraction of the B voters voting for A (called Fba).
Then disregard just enough of the A voters' conditional votes for B, so as
to make Fab = Fba.
If the number of A voters' votes for B that are conditional isn't
sufficient to achieve that, then disregard all of the A voters' conditional
...in order to achieve it as nearly as possible.
That applies straightforwardly to Bucklin.
This can be used with rankings:
When some 1st-rankings are conditional, compare 1st-rankings just as votes
are compared in Approval. ...& disregard conditional 1st-rankings
accordingly.
When some 2nd-rankings are conditional, compare the sum of 1st + 2nd
rankings (of B by A voters, & vice-versa), just as votes are compared in
Approval. ...& disregard 2nd-rankings accordingly.
With conditional votes. any method meets CD. ...in it's strong original
version, not just Weak CD.
CD fix:
Requirement:
While voting as described in the premise, the A voters should have a way of
voting by which B doesn't win.
(end of CD requirement)
...needed because complying methods need A voters to use a special option.
Michael Ossipoff
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Data Structures and Algorithms in JavaScript
# Reviewing the Min/Max Pseudocode Part 2
Check out a free preview of the full Data Structures and Algorithms in JavaScript course:
The "Reviewing the Min/Max Pseudocode Part 2" Lesson is part of the full, Data Structures and Algorithms in JavaScript course featured in this preview video. Here's what you'd learn in this lesson:
## Bianca continues her review of the min/max deletion pseudocode. She checks the code against an edge case where BST nodes only have right children.
Get Unlimited Access Now
Transcript from the "Reviewing the Min/Max Pseudocode Part 2" Lesson
[00:00:00]
>> Bianca Gandolfo: So we're gonna talk about a couple edge cases that were brought up. How to account for them and then we're gonna look at a solution. And then we're gonna see if we can optimize our solution and then we'll keep rolling. Sound good? All right, so what David was talking about before we went and did our exercise, was what if our minimum is the root?
[00:00:24] And so there's two cases where the minimum is the root and then it has a right sub-tree. And then, what about the case when the minimum is the root and also the last element in the tree? Are there any other edge cases that come to mind, that came to mind when you were doing these exercises?
[00:00:53]
>> Bianca Gandolfo: Cool. All right. So, for this case, if we were to delete 7,
>> Bianca Gandolfo: What would we have to do to our tree? So I wrote out our tree just so we'd have a reference to what it looks like. Like, what would we have to change in this nested object for it to be correct?
[00:01:23]
>> Bianca Gandolfo: So we want.
>> Bianca Gandolfo: Sorry, that's not what we want. We want two lined up like this.
>> Student1: All the pointers aren't filled out.
>> Bianca Gandolfo: It's just the right pointers are, see?
>> Student2: Tree.
>> Student1: Which one has on the right tree, not just-
>> Bianca Gandolfo: There's no left.
>> Student1: Right.
[00:01:46]
>> Bianca Gandolfo: But the right is, this is the right.
>> Student1: So then the 8 one would be pointing to 9.
>> Student2: This is the parent.
>> Student1: I get it, I gotcha, yep.
>> Bianca Gandolfo: Actually this one would be null.
>> Bianca Gandolfo: Okay, so we have this tree on the left. Here it is as an object.
[00:02:21]
>> Bianca Gandolfo: What do we want it to look like after we run a delete method on it?
>> Student2: Just the second two lines.
>> Bianca Gandolfo: Yeah, so we would essentially delete.
>> Bianca Gandolfo: This top one, right? And we want it to look something like this.
>> Bianca Gandolfo: But how do we do that in code?
[00:02:44] We can't just delete. Can we?
>> Bianca Gandolfo: Not like that at least.
>> Student2: Yeah, we couldn't really figure out a way with our implementation. Because we don't really have the concept of a root other than it's the node that you start on.
>> Bianca Gandolfo: Mm hm.
>> Student2: And we couldn't figure out a way to.
[00:03:09]
>> Bianca Gandolfo: Yeah, so the root is just where it starts, which is going to be saved in the variable. So because it starts here, because the very first value in our object, let's save into our variable name. Has the value 7 or 7 is the root.
>> Bianca Gandolfo: So we'd want it to look like this, where our very first object in our nested data structure is, the 8.
[00:03:45]
>> Bianca Gandolfo: So essentially, what I recommend is to rewrite 7, 7s value to 8. Change its, overwrite its right value.
>> Bianca Gandolfo: To 8 dot right.
>> Bianca Gandolfo: That makes sense?
>> Student2: Yes.
>> Student1: You're just overwriting the value?
>> Bianca Gandolfo: Do you wanna over write both the value?
>> Student1: I thought we couldn't do that.
[00:04:20]
>> Bianca Gandolfo: And the right.
>> Student2: We tried to do that and we thought you couldn't do that for some reason. Something about the value being, where if in cases where the value is more complex or the node held more.
>> Bianca Gandolfo: Yeah, but in this case, do you have any other thoughts about how we could do it?
[00:04:37]
>> Student2: No.
>> Bianca Gandolfo: Anyone else?
>> Bianca Gandolfo: Me either. [LAUGH] Or I guess the other way you could do it is simply just reassign.
>> Bianca Gandolfo: Starting here, right? So if you said myBst = myBst, let's see, so we have to go to dot right, then we just assign it directly.
[00:05:16] Does that make sense? Any questions?
>> Student2: Can you do that from within? We tried setting this equal to this dot, right? In it, it wouldn't run. Can you, could we do it at three? If we had just done variable name equals variable name that would work.
>> Bianca Gandolfo: So, inside the function you probably want it to be this.
[00:05:46] And only for the case when it's the node, when it's the root node.
>> Student2: Okay, yeah, that-
>> Bianca Gandolfo: Yeah, that make sense?
>> Student2: We got an error when we tried doing that, it said-
>> Bianca Gandolfo: Just make sure-
>> Student2: Invalid left-hand side in assignment.
>> Bianca Gandolfo: Yeah, are you doing it on the prototype and everything?
[00:06:05]
>> Student2: Yes.
>> Bianca Gandolfo: Okay, I don't know.
>> Student2: The SQL stuff-
>> Bianca Gandolfo: I can look at your specific error.
>> Student1: Okay.
>> Bianca Gandolfo: Yeah, you can re-assign. Okay, questions, comments? Okay. So what about this? What do we do here? So we wanna delete the minimum, but we only have one value and because of that fact, it's also the root.
[00:06:42]
>> Student2: If we can re-ascend the value, we can just say the value equals null.
>> Bianca Gandolfo: Mm hm.
>> Student2: Or if this assignment works, we could just set so equal to null.
>> Bianca Gandolfo: So if we said this equals null, then it would look like this. Bst, like that. So both work.
[00:07:07] Just depends on what you want to do next with it, right? If you just want it to disappear and you're never going to touch it again? You know, it's reasonable to say it's a null. But if you're going to continue to add children to this? Leaves, notes, then you're going to want to make sure that you preserve the left and the right.
[00:07:27] So that you can continue growing it. So kill the tree forever, maybe you wanna use it later,
>> Bianca Gandolfo: Cool?
>> Bianca Gandolfo: All right.
>> Bianca Gandolfo: Here's a quick and dirty deleteMin.
>> Bianca Gandolfo: So let's check it out.
>> Bianca Gandolfo: So, this is the deleteMin, by the way, not the deleteMax. So if there's no left and there's no right and there's a parent.
[00:08:09] We simply set the parent left to null, otherwise we delete the value, right? And this is for the case where we want to add nodes later, or whatever. It preserves the node tree structure, doesn't delete it forever, or forever, right? Sets it to null? Cool, make sense?
>> Bianca Gandolfo: Otherwise, if there isn't a left, but there is a right and there's a parent.
[00:08:46] We set the parent's left to its right. Otherwise we can reassign the value or just like we talked about a second ago we could also say this dot Ew, you're right, we can't set this to anything.
>> Student1: Yeah you can't.
>> Bianca Gandolfo: So we would be stuck doing.
>> Student1: That's what we were stuck on.
[00:09:13]
>> Bianca Gandolfo: Yeah.
>> Bianca Gandolfo: Okay. I'm happy with that solution then. So, this the only way that you can assign this is using call and apply. That's when you can just like choose what you want this to be, but you have to do it when you're invoking the function. You can't just assign this willy nilly.
[00:09:42] You can assign properties to this, you could say like, this dot xyz, you could do that, but you can't override it. Cool. Yep, otherwise, we were gonna recurse.
>> Bianca Gandolfo: Cool? Quick and dirty, straightforward, very verbose.
>> Bianca Gandolfo: So what if we want it to just change this to delete max?
[00:10:07]
>> Bianca Gandolfo: What do we have to change?
>> Student2: Invert the lefts from the rights.
>> Bianca Gandolfo: Okay. Is that it?
>> Student2: Yeah I think so.
>> Bianca Gandolfo: Uh-oh.
>> Student2: Yeah [LAUGH] I was gonna say, wow! You have really good memory.
>> Bianca Gandolfo: I'm gonna forget it all. Okay.
>> Bianca Gandolfo: Okay, now I'm remembering.
[00:10:45]
>> Bianca Gandolfo: And you're gonna see how this.left.left works. It works because it's a nested tree.
>> Bianca Gandolfo: Double-check me here, folks.
>> Student1: Also, if not this and this, there you go.
>> Bianca Gandolfo: That's all.
>> Bianca Gandolfo: Cool. Run the code in your mind. So, if it has a right
>> Bianca Gandolfo: Where's our slides?
[00:11:25]
>> Bianca Gandolfo: Let's look at our picture. So let's say we're deleting 25. So we say if there's a right, we're gonna go all the way down right, right, right. So if there's no children, there's a parent, so we're setting to null. So leaf works great. So let's say we don't have 25 and we're doing 20.
[00:11:54] Pass it in, [SOUND] all the way through. It doesn't have a right and it has a left. It does have a parent. So we say parent dot right equals this dot left. Cool. Sounds good. Sure, question?
>> Student3: What is this value is equal to now do?
>> Bianca Gandolfo: You have to say that one more time and not have your hand in front of your mouth.
[00:12:19]
>> Student3: What's not in the else this.value is null. Yeah, what does it do?
>> Bianca Gandolfo: So this is if a, if the root, if the max value is the last node. So there's no more nodes in this tree. So we're just deleting this one node and imagining there's no other nodes here.
[00:12:48] So for a tree of size one, what do you do? You just set that value to null and it's just an empty tree at that point.
>> Student3: And then you can [INAUDIBLE] statement.
>> Bianca Gandolfo: Yeah, so this is for the case that there is, so this is for, let's see.
[00:13:13] So the max val is the root with a subtree.
>> Bianca Gandolfo: So, where is our, so this is the case that we're counting for. Where it's 7, 8, 9, this is the minimum example, but it's the same idea. So if we wanted to delete 7, it's the node, so we have to do something different, because the node doesn't have a parent.
[00:13:44] And so that's why we account for it in all these craziness nested if-else statements.
>> Bianca Gandolfo: And there are shorter ways to do this. This is the most verbose step by step.
>> Bianca Gandolfo: Cool?
>> Bianca Gandolfo: Awesome.
>> Bianca Gandolfo: Moving on.
>> Student3: Delete max.
>> Bianca Gandolfo: Hm?
>> Student3: How will you correct, if you forgot delete max.
[00:14:15]
>> Bianca Gandolfo: What's that?
>> Student3: The last statement, if, yeah delete max is.
>> Bianca Gandolfo: This one?
>> Student3: The last if.
>> Student2: Line 16, the delete min.
>> Bianca Gandolfo: Hmm, good eye.
>> Bianca Gandolfo: Very cool.
>> Student1: That would make for a confusing bug.
>> Bianca Gandolfo: Yeah.
>> Bianca Gandolfo: Yeah, especially if deleteMin also existed.
>> Bianca Gandolfo: All right, any questions?
[00:14:55]
>> Student2: Yeah, so for an empty tree, we're setting this value equals null. Does that mean we'd have to check for this value equals null in all of our other methods?
>> Bianca Gandolfo: Yeah, yep.
>> Bianca Gandolfo: Yep, so it depends. If you wanna account for an NP tree in your implementation, that's how you could do it.
[00:15:20]
>> Student2: Just making [INAUDIBLE] method.
>> Bianca Gandolfo: Hmm?
>> Student2: Probably just make them use empty method.
>> Bianca Gandolfo: Yep.
>> Student2: Insert that everywhere.
>> Bianca Gandolfo: Yeah. All right. Cool. So we talked about delete min/max. This is a simplified example of just deleting a node, more generally from a tree. So when we're deleting a node from a binary search tree, there's three cases and we already talked about the first two in our min and max, right.
[00:15:53] The first case is when a node is a leaf node. What do we do?
>> Bianca Gandolfo: What do we do if it's a leaf node, meaning it has no children?
>> Student1: We remove the pointer from the parent.
>> Bianca Gandolfo: Mm hm. If when into null, parents pointer is now null, that's pretty straightforward.
[00:16:14] And what do we do when the node has one child?
>> Student1: Move the node.
>> Bianca Gandolfo: Yep, you just move the node up to the parent. And we know that it's going to meet our requirements because the entire right sub-tree is always going to be greater than the root node.
[00:16:46] Or the node in question and the entire left sub-tree is going to be less. So if we move it up and it's just one node.
>> Bianca Gandolfo: Then we're good to go. | 3.75 | CC-MAIN-2021-17 |
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# If N is the least positive integer that is a multiple of
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If N is the least positive integer that is a multiple of [#permalink]
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27 Nov 2012, 00:58
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If N is the least positive integer that is a multiple of each of the numbers 15, 18, 40, and 50, which of the following statements about N is true?
A. N <= 200
B. 200 < N <= 2,000
C. 2,000 < N <= 20,000
D. 20,000 < N <= 200,000
E. N> 200,000
Last edited by Bunuel on 27 Nov 2012, 03:08, edited 1 time in total.
Renamed the topic and edited the tags.
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Joined: 02 Sep 2009
Posts: 39673
Re: If N is the least positive integer that is a multiple of [#permalink]
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27 Nov 2012, 03:13
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Expert's post
saxenarahul021 wrote:
If N is the least positive integer that is a multiple of each of the numbers 15, 18, 40, and 50, which of the following statements about N is true?
A. N <= 200
B. 200 < N <= 2,000
C. 2,000 < N <= 20,000
D. 20,000 < N <= 200,000
E. N> 200,000
Given that N is the least common multiple of 15, 18, 40, and 50.
15 = 3*5;
18 = 2*3^2;
40 = 2^3*5;
50 = 2*5^2.
N = 2^3*3^2*5^2 = 1,800.
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Re: If N is the least positive integer that is a multiple of [#permalink]
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20 Oct 2013, 16:07
Bunuel wrote:
saxenarahul021 wrote:
If N is the least positive integer that is a multiple of each of the numbers 15, 18, 40, and 50, which of the following statements about N is true?
A. N <= 200
B. 200 < N <= 2,000
C. 2,000 < N <= 20,000
D. 20,000 < N <= 200,000
E. N> 200,000
Given that N is the least common multiple of 15, 18, 40, and 50.
15 = 3*5;
18 = 2*3^2;
40 = 2^3*5;
50 = 2*5^2.
N = 2^3*3^2*5^2 = 1,800.
Is that from some formula for LCM of a group of numbers? I'm curious as to why the 3*5 got ignored when figuring out the answer.
edit: nevermind, google is my friend.
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Re: If N is the least positive integer that is a multiple of [#permalink]
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20 Oct 2013, 20:59
Well for the sake of the forum, and myself, could you post why exactly 3&5 were ignored?
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Re: If N is the least positive integer that is a multiple of [#permalink]
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20 Oct 2013, 21:09
ricardochavez wrote:
Well for the sake of the forum, and myself, could you post why exactly 3&5 were ignored?
Posted from my mobile device
To find the lowest common factor of a group of numbers, take the prime factorization of those numbers. For each prime number listed, take the most repeated occurrence of this number in any prime factorization. Then multiply them together.
15 = 3*5;
18 = 2*3^2; 3 occurs twice in 18 (3^2), it only appears one in 15.
40 = 2^3*5; 2 occurs thrice in 40, it only appears once in 18, and once in 50
50 = 2*5^2; 5 occurs twice in 50, it only appears once in 15.
So now that we see the highest power of each prime number that appears in these factorials we multiply them all together:
N = 2^3*3^2*5^2 = 1,800.
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Re: If N is the least positive integer that is a multiple of [#permalink]
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21 Oct 2013, 13:51
ricardochavez wrote:
Well for the sake of the forum, and myself, could you post why exactly 3&5 were ignored?
Posted from my mobile device
Least Common Multiple = the smallest multiple of two or more integers.
The LCM is the lowest product for all of the shared primes.
We ignore the 3 and 5 because we are looking to find the lowest common multiple.
LCM is not to be confused with the Greatest Common Factor- which is the common factor of ALL integers. They are easily mixed up!
Now to the problem:
Step 1. List the primes for each number
15 - 5,3
18 - 3,3,2
40 - 5,2,2,2
50 - 5,5,2
Step 2. Group common multiples - those that occur the most frequent
5,5(all but 18 have at least one 5 and you must be able to solve for each number given the primes. By having two 5's, we can still solve for 50)
3,3(18 and 15 both have 3, so we need two)
2,2,2 - we need at least three 2's to account for 40. Anything else would be redundant.
Step 3:Multiply together (2^3)(5^2)(3^2) = 1800
B is the only option that fits within the perimeters.
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Re: If N is the least positive integer that is a multiple of [#permalink] 21 Oct 2013, 13:51
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Final velocity of a ball pushed by compressed airpls. help
• brunettegurl
In summary, the final velocity of a 1.72 kg ball pushed by compressed air from a 1.50 m long tube can be found by using the formula V12 = v02+2ad, where F=ma is used to calculate the acceleration. Assuming the initial velocity to be zero, the final velocity can be determined by plugging in the calculated acceleration value and the given distance of 1.50 m.
brunettegurl
final velocity of a ball pushed by compressed air..pls. help!
Homework Statement
Compressed air is used to fire a 1.72 kg ball vertically upward from a 1.50 m long
tube. The air exerts a upward force of 28.7 N on the ball as long as it is in the tube.
With what velocity does the ball leave the top of the tube?
F=ma
The Attempt at a Solution
so i used the given mass and force to get an acceleration. then using the acceleration i plugged it into the second equation assuming vinitial to be zero and my asnwer is coming out wrong can someone point me in the right direction thanks
Resultant force= Upward force-downward force
what's the downward force in this case?
wld the downward force be the friction..but there is no mention of any friction in the question
brunettegurl said:
wld the downward force be the friction..but there is no mention of any friction in the question
if you throw a ball upwards, it does not continue to go upwards, what makes it fall back to earth? (it has a mass m and is under the influence of gravity g)
ok but that tells us what happens after it is released they're asking for velocity just before it's released from the cannon/tube
brunettegurl said:
ok but that tells us what happens after it is released they're asking for velocity just before it's released from the cannon/tube
When you find the resultant force, you can get the resultant acceleration...
and in the formula $v_1^2=v_0^2+2ad$
what do you need to find to get the value for v1?
we have the distance(1.50) and we now have an acceleration but would vinitial still be equal to zero??
brunettegurl said:
we have the distance(1.50) and we now have an acceleration but would vinitial still be equal to zero??
Yes it would be safe to assume the initial velocity as zero.
thank you so much
1. What is the "final velocity" of a ball pushed by compressed air?
The final velocity of a ball pushed by compressed air refers to the speed of the ball after it has been accelerated by the force of the compressed air. This velocity is typically measured in meters per second (m/s) or kilometers per hour (km/h).
2. How is the final velocity of a ball pushed by compressed air calculated?
The final velocity of a ball pushed by compressed air can be calculated using the equation v = √(2F/m), where v is the final velocity, F is the force of the compressed air, and m is the mass of the ball. This equation is based on Newton's second law of motion, which states that the acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass.
3. What factors can affect the final velocity of a ball pushed by compressed air?
The final velocity of a ball pushed by compressed air can be affected by several factors, including the force of the compressed air, the mass and shape of the ball, the air pressure and temperature, and any external forces acting on the ball (such as friction or air resistance).
4. How can the final velocity of a ball pushed by compressed air be increased?
The final velocity of a ball pushed by compressed air can be increased by increasing the force of the compressed air, reducing the mass of the ball, or reducing any external forces acting on the ball. Additionally, ensuring a smooth and unobstructed path for the ball to travel can also help increase its final velocity.
5. How is the final velocity of a ball pushed by compressed air used in real-world applications?
The concept of the final velocity of a ball pushed by compressed air is used in various real-world applications, such as in sports (e.g. golf, tennis, and soccer), industrial processes (e.g. pneumatic tools), and transportation (e.g. air-powered vehicles). Understanding the final velocity of a ball pushed by compressed air is crucial in designing and optimizing these systems for efficiency and performance. | 3.875 | CC-MAIN-2024-10 |
## Tuesday, October 14, 2014
Uh, oh, someone's condensate drain is clogged!
It never ceases to amaze me how people are willfully ignorant about things. If you live in America, you will own, in your lifetime, maybe a half-dozen each of refrigerators, stoves, microwaves, dishwashers, and the like. You will probably have a few toilets. You may own a dozen cars or more. And you likely will own at least one air conditioner - mostly likely three or four.
Yet most folks have little or no idea how to even operate these appliances, much less how they work. And in order to do the former, you have to know something about the latter. It is like trying to drive a 5-speed without understanding what a clutch does. You'll never get it right, no matter how much you try.
We've been gone a few months and when we came back, I found the condensate drain on our A/C system clogged. A few pitchers of hot water (and some simple green) later, it is unclogged. The outside portion got buried with lawn debris, and it backed up and formed a gelatinous goo inside. It was disgusting.
What is a condensate drain and how does it work?
Well, to understand this, you need to understand something about how A/C systems work. I'm not asking you to take Thermodynamics and understand Entropy and Enthalpy. Thermo is a hard course - I took it three times. Rather, you should understand the basics of how A/C systems work.
Basically, refrigerant (what people call "freon" sometimes) is compressed by a compressor into a liquid. This liquid is hot because it has been compressed. An outside coil and fan cool the liquid, which is then passed to an expansion valve. The expansion valve works like the nozzle of a spray can, and turns the liquid refrigerant into a gas. This phase change is what causes the cooling effect (just as ice changing phase from solid to liquid is what cools beer in your cooler). The gas cools when it expands, and this cold gas is passed through another coil inside your house, and a fan blows air over it, which then cools your house. The gas then goes back to the compressor for another ride around the loop.
OK, so that's A/C for morons. Not hard stuff. So where does the condensate line come in? Well you have to understand how air works, as well as refrigerant. And few people do, based on the number of fogged-in cars I see driving around.
Air can hold water basically dissolved in it. We call this humidity. Just as you can dissolve sugar in water, you can dissolve water in air. How much water you can dissolve in air depends on its temperature and pressure. The colder the air, the less water it can hold - which is why it gets drier in the winter. When warm, moist air passes over the cold A/C coils in your air conditioner, the water falls out of solution and condenses on the coil surface. This drips down into a pan and then is drained outside (usually) by a condensate drain.
In a car, the same thing happens. Problem is, in a car, you can choose "recirculate" or "outside air". When you do the latter, you are basically trying to air condition planet earth, with predictable results.
You see, any air conditioning system is basically also a de-humidifier. A de-humidifer is just an A/C unit with both sides in the house. If you want to "dry out" a house, (or a car) run the A/C. Funny thing, but people don't get that. They think if you turn on the heat, it will dry out the house (or a car) but usually all that does is create a hothouse effect. (Turning on the heat AND the A/C, however, works best. By heating the air, more water will dissolve (evaporate) in it. By running the A/C, this water condenses out and drains out the condensate drain. Just running the heat does nothing, as the water has no place to go!).
Few people understand the science of defrosting.
Getting back to the condensate drain, this tiny pipe can easily clog over time, as mold, mildew and even bacteria colonies grow in it. Insects may clog it. Dirt may accumulate. When this happens, the drip pan in the A/C unit will overflow, usually into your house, possibly ruining your floors or just making a puddle in the basement.
Keeping this line clean and clear is a good idea. Make sure you know where it exits the house and make sure the yard man does not cover it with lawn clippings (as happened to me). Run some water through it on occasion (it should be plumbed with an access point to pour water in). Make sure the outlet of the drain pan does not become clogged.
If all this seems too hard to do (and it is for many folks) then hire an A/C company and sign a service contract. They will come by your house 2-4 times a year and clean the condensate line, check charge levels, change filters, and clean coils. It is worth the money, as it avoids problems in advance.
In some installations, a small "condensate pump" may be used to pump the condensate outside or to a drain. If this pump clogs or fails, it may overflow and leak water all over the place. Sadly, most installations with condensate pumps are places where a water leak would mean disaster. So when the pump fails (or the line clogs) you end up having to repair floors or sheetrock or whatever. If you have a condensate pump, you could check it regularly or get a service contract.
If you don't know whether you have a condensate pump or not, maybe you need to get a service contract. | 2.640625 | CC-MAIN-2017-39 |
# Talk:Block
## Proposed FAQ question and answer
#### Won't the reward be too small?
As the reward shrinks, it may become unprofitable for anyone to create a new block, but as indicated above, the difficulty of the math problem is adjusted every two weeks in order to maintain the rate at which blocks are created. Since it takes about four years to generate the 210,000 blocks and their corresponding bitcoin rewards, it is virtually impossible to ever get to 21 million. At any point in time, it will take about four years to get halfway from the number of bitcoin currently in existence to the 21 million count. It's an unreachable goal, barring a kind of catastrophically destructive cooperation between all bitcoin miners that is untenable.
Only, I don't know if my understanding is correct. Dscotese 19:04, 29 May 2012 (GMT)
## Each block contains all recent transactions, a nonce (random number), and the hash of the previous block.
The nonce is a seed that starts at zero and is incremented until the block hashes below the target. Hence, it should not be called "random". — Preceding unsigned comment added by Henrythefifth (talkcontribs) at 10:28, 14 May 2011
## What if I'm 1% towards calculating a block and...?
"There's no such thing as being 1% towards solving a block. You don't make progress towards solving it."
Is this true? I don't know if there's a nonce which will solve any given block, but for a block that is solvable, it's possible to be 1% of the way towards finding it. Supposing it takes 10 million attempts to 'solve' a block, then after 100,000 attempts you could say you were 1% towards solving it. You're certainly closer to solving it than you were before those 100,000 failed attempts, aren't you? Dooglus 05:05, 15 January 2011 (GMT)
No one can know how many tries it will take to solve the block. After the fact you might say that at some point you had finished 1% of the necessary calculation for that block, but this was not really "progress".
If you go back in time after completing a block and don't generate for one of the days that you did originally, then you could actually end up getting the block sooner, as the work required is random. theymos 11:06, 15 January 2011 (GMT)
Probabilistically speaking, one can make a "best guess" for how long it will take to solve a block. A problem with probabilistic estimation, for example, is that it may say it will take 10 hours to solve the block, with a standard deviation of 15 hours. How much more work do you have when you've worked on it for 10 hours? The answer is that you have ~10 hours more work :).
But assuming all incorrect hashes are exhausted first, the maximal number of tries needed to solve the block is ~2256. So, there exists an upper limit (albeit, it is an obscenely large one!) on the amount of work you can do. Progress is made, just very slowly. --Dlo 20:28, 11 April 2011 (GMT)
That's not the maximum. You can't predict the hash outcome, so you could try forever and still get values above the target. You'd get repeat hash values from the same input. theymos 03:47, 18 April 2011 (GMT)
I can predict the hash outcome by calculating SHA-256 with my mighty x86 processor. :) IMO "predicting" is irrelevant here. Dlo is assuming that the hash function is injective, therefore the number of incomes (giving outcome above the target) <= 2^256. But the hash function is not injective (by the cardinality argument: the set of incomes is infinite and the set of outcomes is finite). --Shrewdwatson 13:49, 23 April 2011 (GMT) | 3.265625 | CC-MAIN-2019-09 |
# Talk:Category (mathematics)
WikiProject Mathematics (Rated B-class, Top-importance)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
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Top Importance
Field: Foundations, logic, and set theory
## Page name
I moved this page from category (category theory) back to category (mathematics) as this is the most common usage. The other usage is category (topology) which is usually referred to by Baire category, first category, or second category. -- Fropuff 17:31, 2005 Jun 1 (UTC)
## Subcategories of Set?
The examples given, Rel through Uni, don't appear to be subcategories of Set since generally a set can be made into, for example, a group in many nonisomorphic ways. I think the original author meant to point out that they are concrete categories over Set. Moreover, Rel doesn't even appear to be concrete, since many relations are not functions. Is there some way of viewing these categories as subcategories of Set that I'm not seeing? If so, it should probably be noted, even if not explained in detail. SirPeebles 03:25, 26 December 2006 (UTC)
I'm writing to concur with SirPeebles; Rel is a supercategory of Set, not a subcategory. Also, the start of the section promises to describe composition in each example, but this is not done for Rel. 66.32.207.31 20:33, 12 March 2007 (UTC)
I removed the following offending sentence:
(The following are subcategories of Set, obtained by adding some type of structure onto a set, by requiring that morphisms are functions that respect this added structure, and where morphism composition is simply ordinary function composition.)
I agree they don't seem to be subcategories; perhaps the original author wanted to say "there is a forgetful functor from these Set", which is not the same thing as saying they're subcats. However, it seems premature to try to talk about forgetful functors so early in this article, and so it seems better not to say anything at all. Err, I take that back; this article is notable in failing to use the words "concrete category", and so perhaps a few sentences to that effect should be written, and the list properly classified. linas 04:16, 14 April 2007 (UTC)
Done. Cruft patrol. Some editor unthinkingly slapped Rel at the top of the list. I simply changed "subcategory" to "concrete category" and so all should be well. linas 04:49, 14 April 2007 (UTC)
## The term "hom-class" in the Definition
The Definition contains this phrase. "... denote the hom-class of all morphisms from a to b." but hom-class is not a link to a definition. What about defining hom-class in the article on Class (set theory) and linking to that? Regards, ... PeterEasthope 18:58, 19 February 2007 (UTC)
That sentence is attempting to define the phrase "hom-class" as the "class of morphisms"; that's all that it is. "Class of morphisms"is a mouthful, so its just "hom-class" for short. linas 03:56, 14 April 2007 (UTC)
## Basic Category Theory for Computer Scientists
I was reading Basic Category Theory for Computer Scientists (Pierce, 1991) today, and the intro to the first chapter seemed... to remind me of something. After doing some history searching, it seems that anon user 63.162.153.xxx wrote the Category theory article from scratch, and that text was eventually broken up to create this article. Problem is, that the definition of a category that was used is very, very close to directly lifted from Pierce's book.
I don't want to suggest that the whole article is a copyvio, but it certainly would be good to re-write the definition section in new language that doesn't duplicate this text. -Harmil 19:10, 21 February 2007 (UTC)
I take that assertion of blame back, and appologize to whoever that anon user is. The history link only shows edits back to that revision, but the article's origin is actually not available in the history. If you keep clicking on the "older version" link, you eventually get to the automated conversion. So, we may never know who put the text in there, but we can certainly re-write it. -Harmil 19:18, 21 February 2007 (UTC)
Sigh. Can you be more specific? I don't have a copy of Pierce's book. But I can pick up, for example, Rotman "An Introduction to Algebraic Topology" and on page 6, there is a definition of a (locally small) Category that is very similar to that in this article. It differs only in punctuation, and misc "filler" words that don't change the flow. I doubt Pierce copied from Rotman's older (1988) book, but from this distance, the definition of a category seems very generic. Is it really verbatim, or just logically similar? linas 03:43, 14 April 2007 (UTC)
## Composition is a binary operation?
The article says "binary operation hom(a, b) × hom(b, c) → hom(a, c) called composition of morphisms". I think that we only call binary operation to a function S × SS, not to a function S × RT. Maybe we could call it function (but hom(a, b) × hom(b, c) and hom(a, c) need not to be sets; I don't know if an association between an unique element of B with each element of A is also called an function from A to B when A and B are proper classes.) Jayme 12:33, 12 April 2007 (UTC)
The article binary operation seems to define something called an "external binary operation" that is closer to what you want. But all this seems to be quibbling anyway. I don't believe there is any grave error committed by saying "composition is a binary operation", irrespective of whether the things being composed are classes. Right? linas 03:51, 14 April 2007 (UTC)
I agree with you. About binary operations, I believe that there is a notion of "partial binary operation" that would be a function S'S × SS. Then any function A × BC would be a partial binary operation: just take S = ABC...! Jayme 16:58, 21 April 2007 (UTC)
## Ens
In MacLane's book, there is repeated mention to a category Ens which seems to me like elements of a power set along with endomorphisms of the original set as arrows among the correct elements of the power set. Does this seem right?
## why to use small categories?
Cone_of_a_functor says:
```Let J be a small category and let C^J be the category of diagrams of type J in C (this nothing more than a functor category).
```
Is there a typical thing I'd like to do with a category, but can't if it is large. Or specific. How about the above article. Can't I define the category of type J in C if J is a large category?
Thanks, JanCK 11:43, 21 October 2007 (UTC)
```the category of small categories, denoted by Cat, is the category whose objects are all small categories
The category Cat is itself a large category, and therefore not an object of itself.
```
So what I'm supposed to get is: obj(C) of a category C is a class. So the elements have to be sets? The class article reads
```class is a collection of sets (or sometimes other mathematical objects)
```
What kind of objects are these other mathematical objects? JanCK 12:04, 21 October 2007 (UTC)
## category definition of real numbers, if available
given their importance it would be great if a definition of the reals could be given in this article. Rich Peterson4.246.233.26 (talk) 10:36, 26 December 2007 (UTC)
How about: Reals = class of all complete, ordered fields with field-isomorphisms between them? Thus, Dedekind cuts, equivalence classes of Cauchy sequences, and other models of the Reals are merely the objects of the category: Pick any one of them if you want to work on the internal details; but what makes the Reals the Reals is what all those models have in common -- which exactly what the stated class and its morphisms describe. Just a suggestion. Jmacwiki (talk) 16:39, 3 December 2009 (UTC)
## Recent revisions to the definitions
Hello, I have reverted the revamp to the definition made by COGDEN. The revamp to the intro was quite nice, but I'm not sure why the definitions were revised. I found the revised wording misleading. The revision implied that a category was just a collection of objects and a collection of morphisms; this is not true. By definition, a category must also have a composition operation that is associative. I think this needs to be stressed clearly in the definition: the composition and associativity are not derived properties of categories, they are part of the data. My other concern was the notation hom(a,b), where a and b are classes of objects. I've never seen that in print before, and I don't it's a fundamental notion. Feel free to discuss or rework if you like, though. Sam Staton (talk) 14:59, 21 February 2008 (UTC)
An arrow/morphism is usually defined as being composable and associative. Most of the literature definitions I've seen make it really simple: a category is a collection of objects and arrows (arrows being composable and associative). This is also consistent with the definition found in morphism. COGDEN 07:41, 29 February 2008 (UTC)
Composition is an operation defined at the level of a category. Morphisms are composable because the belong to a category. If you read carefully the definition in morphism you see that it refers to the category the morphisms belong to. In order to avoid a circular definition, composition should really be defined here. -- Fropuff (talk) 07:56, 29 February 2008 (UTC)
Sam Staton is right (and a professional). A collection of objects and a collection of morphisms, together with the operations 'source' and 'target', is simply a directed graph. A category is, by definition, such a graph equipped with composition and identity operations satisfying associativity and unit axioms. A single graph can carry many different category structures. You have to specify which one you want. 86.156.166.193 (talk) 02:14, 8 March 2009 (UTC)
I removed the statement "The morphisms of a category are sometimes called arrows due to the influence of commutative diagrams." I don't believe there is any evidence for the "influence of commutative diagrams". When I used it in my writing (for example, Michael Barr and Charles Wells, Category Theory for Computing Science third edition. Les Publications CRM, Montreal, 1999.) my motivation was to avoid the suggestion that "map" and "morphism" give that an arrow has to be defined on elements. But others who use the word may have other motivation. Better not to say anything. Wellsoberlin (talk) 19:46, 10 April 2009 (UTC)
## Cat category?
Why is the category of categories and functors a small category? Or even locally small? What aspect of the stated definition fails to cover the case of all categories and functors? Jmacwiki (talk) 16:44, 3 December 2009 (UTC)
Cat is not small and the "categorie of all categories" (in the sense described in this article) does not exist, since the class of all classes does not exist.Stephan Spahn (talk) 12:36, 11 May 2011 (UTC)
Well, either I misread, or the article has been fixed to remove any statement that Cat is small. Either way, my confusion is reduced.
As for it not existing: The set "S" of all sets does not exist, as those words are conventionally understood. That's because we can prove that it doesn't: The generally accepted axioms of set theory (ZFAC, presumably) require that the intersection of any set [hypothesized to be S, here] with any class C [of all things having the property that they do not contain themselves as members] be a set; only it isn't, because of the contradiction. So S itself must not exist.
I do not recall that the axioms, or others that are generally accepted, impose the same requirement on classes. (Indeed, that seems to be the value of distinguishing sets from classes in ZF.) Do they? Jmacwiki (talk) 01:20, 30 May 2011 (UTC)
Well, there is the notion of Cat which is the category of all strict small categories. See Cat in nLab for details. Seems we also have an article here on WP: Category of small categories. And, by appealing to the idea of a 2-category to hold them all (the enriched category over Cat), aka higher category theory, you can build the 2-category of all categories, without contradiction. Or so they say; I don't yet understand it. linas (talk) 04:23, 21 August 2012 (UTC)
## Does hom() mean homomorphism?
In the Definition section, it's implied that hom(a,b) means "a morphism from a to b". So first of all, is this is a correct supposition?
One reason the article leaves this in doubt is because "hom" is conspicuously similar to "homomorphism", which in turn is conspicuously different from plain "morphism". So a reader (ie: me) would be resistant to jumping to conclusions about which one is actually the same as (or a superset of) the others. For example, would it be correct to infer that "homomorphism" is a synonym for "morphism"?
Anyhow, it would be very helpful for these basics to be spelled out. Gwideman (talk) 01:11, 15 December 2009 (UTC)
Hom(a, b) is the collection of morphisms, not an individual one.
Yes, it's short for homomorphism. I don't think there's a fundamental difference. It's just that in some contexts you talk about homomorphisms, while morphism is a more general term. Peter jackson (talk) 15:03, 25 May 2010 (UTC)
## Rel and allegories
The category Rel consists of all sets, with binary relations as morphisms. Abstracting from relations instead of functions yields allegories instead of categories.
This is very confusing. I think the underlying problem is that "allegories instead of categories" implies, falsely, that allegories are not categories. 68.239.116.212 (talk) 05:31, 13 January 2010 (UTC)
## "Between" two objects
Any preordered set (P, ≤) forms a small category, where the objects are the members of P, the morphisms are arrows pointing from x to y when x ≤ y. Between any two objects there can be at most one morphism.
Is "between any two objects" ambiguous? If there is one morphism from A to B and one from B to A, is the condition "between any two objects there can be at most one morphism" violated? Not according to the meaning it must have in the above, but one might interpret it otherwise. How about saying "There can be at most one morphism from any object to any other object"? 68.239.116.212 (talk) 06:05, 13 January 2010 (UTC)
## References
I find it strange that so many basic facts are attributed to Jacobson. I think it is normal to acknowledge "Categories for the Working Mathematician" as the standard category theory text. I was thinking of changing the references from Jacobson to Mac Lane. But I actually don't think that all these individual references are necessary. I don't think we need 13 separate references all to the same 4 pages of one book. It would be better to start the article with "A standard text for this subject is CWM. All the concepts in this article can be found in the first few chapters of that book or of the other textbooks at the end of this article." Any strong feelings, before I change anything? — Preceding unsigned comment added by ComputScientist (talkcontribs) 15:51, 12 March 2011 (UTC)
I do think it is necessary to have those inline references, even if they are not from Jacobson (which was just the book I had at hand at the time I've added them). That said, I'm not convinced that is was an improvement removing the references without replacing them with some other more appropriated citations. Helder 17:48, 28 March 2011 (UTC)
Hi Helder. Do you think that inline references are necessary because they are helpful to the reader? Or because it is a mechanism for ensuring the accuracy of wikipedia? If yours is the latter reason, then I don't think any one is disputing the correctness of these basic statements. As for the first reason, I don't see why it is helpful if every single sentence has an inline citation to the same chapter of the same book, even the same page. I have put a note at the top, saying that everything is in the first few chapters of the textbooks, and in fact it is all within Chapter I of CWM. If there is anything that is not as well known, an inline citation would be fine. Or perhaps you can think of a different compromise. ComputScientist (talk) 08:19, 29 March 2011 (UTC)
Hi!
Both reasons, but with varying importance depending on the reader. For experienced mathematicians, I do agree that citations for basic facts are not really necessary, because they are familiar with those ideas and won't dispute any of those statements. For these readers the most important references will be those related to deep results (e.g. references indicating a text where a proof of some theorem can be found). On the other hand, for readers who are still learning the basic facts and consults Wikipedia to get a general picture of the subject, even citations for specific examples can be very helpful to provide references for further reading, so that the student can have a better understanding of the ideas (admitting that Wikipedia articles are not supposed to present its content with as much detail as is found in a text book, like those from Wikibooks - my "home wiki" ;-) - in which the didactics is of greater importance). The references can also be used by other editors as a source of information (for further improvements in the article).
Usually, I prefer to adopt an approach which is closer to Wikipedia:You do need to cite that the sky is blue than it is to Wikipedia:You don't need to cite that the sky is blue, but if it is needed we can merge the references which points to the same page of the book (there were 3 pointing to page 11 of the book and other 6 to page 12). Helder 20:34, 29 March 2011 (UTC)
Hi. In March I put the category of sets in the second sentence of the article, because I believe it is the most important and informative example of a category. I've reinstated it there. It's true that monoids and preorders are categories, and that is often useful, but that's not why categories were invented and it's rarely the primary motivation for them. Still, I've left them in the third sentence in case Classicalecon feels strongly about it. Best, ComputScientist (talk) 09:57, 3 June 2011 (UTC)
## File:Directed.svg Nominated for Deletion
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## Element-free definition needed.
Somewhere toward the end of this article, we should provide several alternative definitions for a category, including an element-free one. viz using diagrams. In particular, it would be nice to show how a category is a 3D simplicial set with objects being 0-D, morphisms 1-D, composition 2D and associativity 3D. I think that's a slick trick worth showing. Also at least a short blurb about enriched categories and other extensions... linas (talk) 04:41, 21 August 2012 (UTC)
I have no clue why I felt I needed to write the above. Its just another random idea, a to-do list item. linas (talk) 04:46, 21 August 2012 (UTC)
Hmm, the article on enriched category has several figures that could be recycled here... linas (talk) 05:28, 21 August 2012 (UTC)
## Large and Locally Small Categories
I think that all the examples you have given of large categories are locally small.
An example of a large category that was not locally small and stated as such would help make the distinction clearer. — Preceding unsigned comment added by 86.27.207.97 (talk) 23:04, 22 February 2013 (UTC) | 2.78125 | CC-MAIN-2017-39 |
# Heat exchange: Water to water through jacketed pipe
• Skafsgaard
In summary, the conversation discusses the need to calculate the heat exchange in a double pipe heat exchanger for a wastewater-treatment facility. The goal is to cool off the wastewater and transfer the heat back to the factory. Due to the protein content in the wastewater, a plate exchanger is not suitable. The best option is to use a jacketed pipe with counterflowing hot and cool water. The speaker is seeking a better understanding of the calculations used to determine the heat exchange before ordering a heat-exchanger. They also request quick responses as it is a time-sensitive matter.
Skafsgaard
Edit: I've realized that it's not called a 'jacketed pipe', but simply a double pipe heat exchanger. Apologies, English is not my first language.
I have been asked to scour the webs for information about how to calculate the heat exchange in a jacketed stainless steel pipe, with hot water in the inner cylinder and cool, counter-flowing water in the outer cylinder.
I need the information in order to figure out which heat exchange option is better suited for a wastewater-treatment facility.
As it is, the wastewater has a temperature of 90ºC, which cannot be cleaned by the facility. Therefore it needs to be cooled off either way. But rather than simply cooling it off, I am looking to transfer the heat back to the factory so that they can reheat it from as high a temperature as possible.
The wastewater is full of protein, so a plate exchanger is not an option (as it would clog). Seemingly, the best option is to lead the hot wastewater through a jacketed pipe with counterflowing hot wastewater and cool water (wastewater is ~90ºC, cool water is ~5ºC).
Before ordering a heat-exchanger I would like to have a better understanding of the calculations used to determine the exchange of heat, and so as to know what result I can expect.
Any responses are much appreciated - especially quick ones, as this is a bit of a rushed matter.
Thanks,
Skafsgaard
Last edited:
## 1. What is heat exchange?
Heat exchange is the transfer of thermal energy between two or more objects or substances. It can occur through conduction, convection, or radiation.
## 2. What is a jacketed pipe?
A jacketed pipe is a type of pipe that has an outer layer or "jacket" surrounding it. This jacket can be filled with a heating or cooling medium, such as water or steam, to control the temperature of the substance flowing through the pipe.
## 3. How does heat exchange occur through a jacketed pipe?
In a jacketed pipe, the heating or cooling medium in the jacket comes into direct contact with the outer surface of the pipe, causing the temperature of the pipe to change. This change in temperature is then transferred to the substance flowing through the pipe, allowing for heat exchange to occur.
## 4. What are the advantages of using a jacketed pipe for heat exchange?
Using a jacketed pipe for heat exchange allows for more precise control of the temperature of the substance being transferred. It also helps to prevent heat loss or gain during the transfer process, ensuring that the substance maintains its desired temperature.
## 5. What factors affect the efficiency of heat exchange through a jacketed pipe?
The efficiency of heat exchange through a jacketed pipe can be affected by factors such as the temperature and flow rate of the heating or cooling medium, the thickness and material of the pipe and jacket, and the type and properties of the substance being transferred.
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0
# How would you rename 2 over 5 as a decimal?
Updated: 12/24/2022
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Q: How would you rename 2 over 5 as a decimal?
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### How do you rename fractions as decimals?
Usually to rename a fraction as a decimal the little fraction symbol and the decimal stand for the word "and". Or if you have a whole number in the fraction then that number would be your "and" for example 3 1/2 would be three and one half in words right... so 3.5 is your decimal I LOVE ZAYN <3
2/8 = 0.25
### How do you rename a fraction?
Put it into decimal form (this is considered less accurate).E.G. 1/2=0.5
### How would the fraction 1 over 2 look as a decimal?
1 over 2 (½) = 0.5
1 1/2
### How do you do renaming as fractions?
Usually to rename a fraction as a decimal the little fraction symbol and the decimal stand for the word "and". Or if you have a whole number in the fraction then that number would be your "and" for example 3 1/2 would be three and one half in words right... so 3.5 is your decimal I LOVE ZAYN <3
### What would 2 over 5 be as a decimal?
2 divided by 5 = 0.4 - that is the answer.
### What is the decimal approximation for 2 over 9?
That would be .2222222222 (or .2 rounded).
### What is 19 over 2 as a decimal?
19 over 2 as a decimal = 9.5
### How do you write 2 over 9 as a decimal?
2 over 9 as a decimal = 0.2222 2/9: = 2 ÷ 9 = 0.2222 in decimal
### What is the equivalent fraction to 4 over 10 as a decimal?
An equivalent fraction would be 2/5. in decimal, this is 0.4
### What is 2 over 8 as a decimal?
8 over 2 as a decimal = 4.0 | 4 | CC-MAIN-2024-22 |
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Method: Look at the number. If the last digit or the units digit is 5 or more than 5 we add 1 to the number in the tens place and make the units digit 0. If the last digit or the units digit is less than 5 then make the units digit 0 and keep the tens digit as it is. Example: Which would be the best estimate for 108 + 32 = ? Answer: 110 + 30 Directions: Answer the following questions. Also write at least ten examples of your own.
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# insert() function doesn't add nodes to the binary search tree
## insert() function doesn't add nodes to the binary search tree
Contents
Problem Description:
I’m trying to make a binary search tree. If I start with an array my function makes a binary search tree out of that array (everything fine here). like for an array like `let a = [2,4,5,3,9,7,3,8,5];` the tree looks like the picture. my problem is with the insert() function. If I start with an empty array and add a node to it, it works. However, when I add a second node, that second node won’t be added to my tree, and my tree is shown as having only 1 node in it (the root node). Here the snippet:
``````
const Node = (data, left = null, right = null) => {
return {data, left, right};
};
const Tree = array => {
const remDupsAndSort = array => {
const mergeSort = array => {
if(array.length <= 1) return array;
let leftArr = array.slice(0, array.length / 2);
let rightArr = array.slice(array.length / 2);
return merge(mergeSort(rightArr), mergeSort(leftArr))
};
const merge = (leftArr, rightArr) => {
let sorted = [];
while(leftArr.length && rightArr.length){
if(leftArr[0] < rightArr[0]){
sorted.push(leftArr.shift());
}else{
sorted.push(rightArr.shift());
}
};
return [...sorted, ...leftArr, ...rightArr]
};
return mergeSort([... new Set(array)])
};
array = remDupsAndSort(array);
const buildTree = (array, start, end) => {
if(start > end) return null;
let mid = Math.floor((start + end) / 2);
let node = Node(array[mid]);
node.left = buildTree(array, start, mid - 1);
node.right = buildTree(array, mid + 1, end);
return node;
};
const insert = value => {
if(!root) return root = Node(value);
current = root;
while(current){
if(value < current){
current = current.left
}else{
current = current.right
}
}
current = Node(value)
// if(!current){
// current = Node(value)
// // }else{
// if(value < current){
// current.left = insert(value, current.left)
// }else{
// current.right = insert(value, current.right)
// }
// }
return root
};
const prettyPrint = (node = root, prefix = '', isLeft = true) => {
if(node){
if (node.right !== null) {
prettyPrint(node.right, `\${prefix}\${isLeft ? '│ ' : ' '}`, false);
}
console.log(`\${prefix}\${isLeft ? '└── ' : '┌── '}\${node.data}`);
if (node.left !== null) {
prettyPrint(node.left, `\${prefix}\${isLeft ? ' ' : '│ '}`, true);
}
}else{
console.log(node)
}
}
let root = buildTree(array, 0, array.length - 1);
return {root, prettyPrint, insert}
};
let a = [2,4,5,3,9,7,3,8,5];
let b = [];
let c = [1,2,4,5,6,7]
let f = Tree(a)
let d = Tree(b)
let e = Tree(c)
d.insert(4)
// d.insert(8) ---> doesn't work
// d.prettyPrint()
// f.insert(1) ---> doesn't work
f.prettyPrint()
// e.prettyPrint()
// console.log(d.root)
``````
if I run `d.prettyPrint()` I’ll get `└── 4` just as expected. But if I run `d.insert(8)` after that 8 isn’t added to the tree and the code returns `└── 4` again. To make matters more confusing if I `console.log(d.root)` it returns null even though my prettyPrint function returns `└── 4` as the root.
Clearly I expect the nodes be added to the tree. On one of my attempts I tried to write the code like this:
``````const insert = (value, current = root) => {
if(!current){
current = Node(value)
}else{
if(value < current){
current.left = insert(value, current.left)
}else{
current.right = insert(value, current.right)
}
}
return current
};
``````
even though I assigned `current = root` the code returned null for `d.insert(4)`
## Solution – 1
There are these issues in your `insert` function:
• `current` is implicitly defined as a global variable — this wouldn’t parse in strict mode. It should be declared as a local variable, using `let`
• The value is compared with a node object instead of with the data of that node. So `value < current` should be changed to `value < current.data`
• The assignment of the new node object to `current` — after the loop — will not mutate the tree. It merely assigns that object to that variable. Such assignment can never change the tree, just like `current = current.right` does not mutate the tree either. You need instead to assign the new node to either the `left` or `right` property of the parent of `current`. So this assignment should happen one step earlier.
Correction:
`````` const insert = value => {
if(!root) return root = Node(value);
let current = root; // Define with `let`
while(current){
if(value < current.data){ // Compare with the data, not the node
// Mutate the parent node when inserting
if (!current.left) return current.left = Node(value);
current = current.left
}else{
if (!current.right) return current.right = Node(value);
current = current.right
}
}
};
``````
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http://www.perlmonks.org?node_id=358150
As per discussion over here.
A few notes:
• I've used closures like this before, but i don't know if this is what is meant by "memoization". As such, i'm sure i'm doing many things wrong, but i'm trying to get a feel for what was intended. Any feedback is appreciated.
• For small numbers memory space shouldn't be a problem so the duplication from having both a hash and an array isn't a big deal. For larger values i'd forego the hash and use a binsearch on the array to emulate the exists call. For larger values don't use this algorithm since its horribly inefficient.
• Graff's version later in the thread performs the caching more suitably to the problem being solved. This is meant as a more generic algorithm.
Update: Some modifications have been made. The code now only caches primes up until the square root of the candidate. It memoizes all candidates (prime or not). A purge on extraneous values occurs whenever a candidate is proposed that's lower than the current prime threshhold. A purge on extraneous values occurs directly after a fill. This means all accesses of previously checked numbers are in O(1) time.
```{
my @primes = (2, 3);
my %primes = (2 => 2, 3 => 3);
my %memo;
my \$_is_prime = sub {
my \$posit = shift;
my \$sqrtp = sqrt(\$posit);
for my \$p (@primes) {
last if \$p > \$sqrtp;
return 0 unless \$posit % \$p;
}
return 1;
};
my \$_fill = sub {
my \$bar = int sqrt(shift);
for (my \$last = \$primes[-1] + 2; \$primes[-1] < \$bar; \$last += 2) {
if (exists \$memo{\$last} and \$memo{\$last}) {
push @primes, \$primes{\$last} = delete \$memo{\$last};
} else {
push @primes, \$primes{\$last} = \$last if \$_is_prime->(\$last);
}
}
};
my \$_purge = sub {
foreach (keys %memo) {
if (\$_ < \$primes[-1]) {
delete \$memo{\$_};
}
}
};
sub is_prime {
my \$posit = shift;
if (\$primes[-1] < \$posit) {
return \$memo{\$posit} if exists \$memo{\$posit};
\$_fill->(\$posit);
\$_purge->();
return \$memo{\$posit} = \$_is_prime->(\$posit) ? \$posit : 0;
} else {
return exists \$primes{\$posit};
}
}
}
As an amusement, here's the same algorithmic setup as above (caching all numbers <= the requested one making accessing them much faster at the expense of space), only applied to Fibonacci numbers. My guess (though i could be wrong?) is that this is just as useless as the algorithm above.
```{
my @fib = (1, 1);
sub comp_fib {
my \$posit = shift;
while (@fib < \$posit) {
push @fib, \$fib[-1] + \$fib[-2];
}
return \$fib[\$posit-1];
}
} | 2.5625 | CC-MAIN-2019-51 |
Search
Is there a more elegant way to do this in D?
Apr 08
Apr 08
Jack
Apr 08
Jack
Apr 09
ddcovery
Apr 08
Meta
Apr 08
Meta
I am trying to take an array and convert it to a string. I know that Split will let me easily go the other way. I searched for the converse of Split but have not been able to locate it. I can think of two brute force methods of doing this. I found an answer to something similar in the forum and adapted it - but it is so much code for such a simple procedure:
``````import std;
void main()
{
auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
string b = to!string(a.map!(to!string)
.chunks(a.length)
.map!join);
string f = b[2..b.length-2]; //needed to strip the first two and las two characters
writeln(f);
}
``````
I want to come out of this with a string that looks like this: 1011101011110
```On 4/7/21 8:57 PM, Brad wrote:
> auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
> I want to come out of this with a string that looks like this: 1011101011110
Me, me, me, me! :)
import std;
void main()
{
auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
string s = format!"%-(%s%)"(a);
writeln(s);
}
Ali
```
On Thursday, 8 April 2021 at 03:57:23 UTC, Brad wrote:
>
I am trying to take an array and convert it to a string. I know that Split will let me easily go the other way. I searched for the converse of Split but have not been able to locate it.
You need two functions here:
1. `joiner` [1], which lazily joins the elements of a range.
2. `array` [2], which eagerly evaluates a range and returns an array of its elements.
The final code looks like this:
``````dchar[] b = a.map!(to!string).joiner.array;
``````
You may have noticed that the type of `b` here is `dchar[]`, not `string`. This is due to a feature of D's standard library known as "auto decoding" [3]. To prevent the strings you get from `to!string` from being auto-decoded into ranges of `dchar`, you can use the function `std.utf.byCodeUnit` [4], like this:
``````string b = a.map!(to!string).map!(byCodeUnit).joiner.array;
``````
On Thursday, 8 April 2021 at 03:57:23 UTC, Brad wrote:
>
I am trying to take an array and convert it to a string. I know that Split will let me easily go the other way. I searched for the converse of Split but have not been able to locate it. I can think of two brute force methods of doing this. I found an answer to something similar in the forum and adapted it - but it is so much code for such a simple procedure:
``````import std;
void main()
{
auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
string b = to!string(a.map!(to!string)
.chunks(a.length)
.map!join);
string f = b[2..b.length-2]; //needed to strip the first two and las two characters
writeln(f);
}
``````
I want to come out of this with a string that looks like this: 1011101011110
``````string to01String(int[] x) @safe
{
auto conv = x.to!(ubyte[]); // allocates new array, so later cast to string is OK
conv[] += '0'; // assume all numbers are 0-9, then this gives the correct result
return (() @trusted => cast(string)conv)();
}
``````
```On Thursday, 8 April 2021 at 04:02:26 UTC, Ali Çehreli wrote:
> On 4/7/21 8:57 PM, Brad wrote:
>
>> auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
>
>> I want to come out of this with a string that looks like this: 1011101011110
>
> Me, me, me, me! :)
>
> import std;
> void main()
> {
> auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
>
> string s = format!"%-(%s%)"(a);
> writeln(s);
> }
>
> Ali
What does %-%s% do?
```
```On Thursday, 8 April 2021 at 16:45:14 UTC, Jack wrote:
> On Thursday, 8 April 2021 at 04:02:26 UTC, Ali Çehreli wrote:
>> On 4/7/21 8:57 PM, Brad wrote:
>>
>>> auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
>>
>>> I want to come out of this with a string that looks like this: 1011101011110
>>
>> Me, me, me, me! :)
>>
>> import std;
>> void main()
>> {
>> auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
>>
>> string s = format!"%-(%s%)"(a);
>> writeln(s);
>> }
>>
>> Ali
>
> What does %-%s% do?
nevermind, someone just asked this too[1]
```
On Thursday, 8 April 2021 at 12:19:29 UTC, WebFreak001 wrote:
>
``````string to01String(int[] x) @safe
{
auto conv = x.to!(ubyte[]); // allocates new array, so later cast to string is OK
conv[] += '0'; // assume all numbers are 0-9, then this gives the correct result
return (() @trusted => cast(string)conv)();
}
``````
The @trusted lambda can also be replaced with std.exception.assumeUnique.
On Thursday, 8 April 2021 at 18:01:56 UTC, Meta wrote:
>
On Thursday, 8 April 2021 at 12:19:29 UTC, WebFreak001 wrote:
>
``````string to01String(int[] x) @safe
{
auto conv = x.to!(ubyte[]); // allocates new array, so later cast to string is OK
conv[] += '0'; // assume all numbers are 0-9, then this gives the correct result
return (() @trusted => cast(string)conv)();
}
``````
The @trusted lambda can also be replaced with std.exception.assumeUnique.
Never mind me, assumeUnique is @system (or at least it's inferred as @system), and anyway, you can't implicitly convert `immutable(ubyte)[]` to `immutable(char)[]`.
The ascii code of 0 is 48 so I think you can add everywhere 48 (but I'm not a specialist)
```On Thu, Apr 08, 2021 at 08:28:44PM +0000, Alain De Vos via Digitalmars-d-learn wrote:
> The ascii code of 0 is 48 so I think you can add everywhere 48 (but
> I'm not a specialist)
Why bother with remembering it's 48? Just add '0', like this:
int a = [1, 0, 1, 0, 1, ...];
string s = a.map!(i => cast(char)(i + '0')).array;
writeln(s);
Or better yet, if you just want to output it and don't need to store the array, just use the range directly:
int a = [1, 0, 1, 0, 1, ...];
auto r = a.map!(i => cast(char)(i + '0'));
writeln(r);
T
--
People say I'm arrogant, and I'm proud of it.
```
« First ‹ Prev
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MY AMERICAN SCIENTIST
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MACROSCOPE
# The Music of Math Games
Video games that provide good mathematics learning should look to the piano as a model
# The Symbol Barrier
Given the effort and the expense to make a math game work, is it worth the effort? From an educational perspective, you bet it is. Though the vast majority of math video games on the market essentially capitalize on just one educationally important aspect of video games—their power to fully engage players in a single activity for long periods of time—all but a tiny number of games (fewer than 10 by my count) take advantage of another educationally powerful feature of the medium: video games’ ability to overcome the symbol barrier.
Though the name is mine, the symbol barrier has been well known in math education circles for over 20 years and is recognized as the biggest obstacle to practical mastery of middle school math. To understand the symbol barrier and appreciate how pervasive it is, you have to question the role symbolic expressions play in mathematics.
By and large, the public identifies doing math with writing symbols, often obscure symbols. Why do they make that automatic identification? A large part of the explanation is that much of the time they spent in the school mathematics classroom was devoted to the development of correct symbolic manipulation skills, and symbol-filled books are the standard way to store and distribute mathematical knowledge. So we have gotten used to the fact that mathematics is presented to us by way of symbolic expressions.
But just how essential are those symbols? After all, until the invention of various kinds of recording devices, symbolic musical notation was the only way to store and distribute music, yet no one ever confuses music with a musical score.
Just as music is created and enjoyed within the mind, so too is mathematics created and carried out (and by many of us enjoyed) in the mind. At its heart, mathematics is a mental activity—a way of thinking—one that over several millennia of human history has proved to be highly beneficial to life and society.
In both music and mathematics, the symbols are merely static representations on a flat surface of dynamic mental processes. Just as the trained musician can look at a musical score and hear the music come alive in her or his head, so too the trained mathematician can look at a page of symbolic mathematics and have that mathematics come alive in the mind.
So why is it that many people believe mathematics itself is symbolic manipulation? And if the answer is that it results from our classroom experiences, why is mathematics taught that way? I can answer that second question. We teach mathematics symbolically because, for many centuries, symbolic representation has been the most effective way to record mathematics and pass on mathematical knowledge to others.
Still, given the comparison with music, can’t we somehow manage to break free of that historical legacy?
Though the advanced mathematics used by scientists and engineers is intrinsically symbolic, the kind of math important to ordinary people in their lives—which I call everyday mathematics—is not, and it can be done in your head. Roughly speaking, everyday mathematics comprises counting, arithmetic, proportional reasoning, numerical estimation, elementary geometry and trigonometry, elementary algebra, basic probability and statistics, logical thinking, algorithm use, problem formation (modeling), problem solving, and sound calculator use. (Yes, even elementary algebra belongs in that list. The symbols are not essential.)
True, people sometimes scribble symbols when they do everyday math in a real-life context. But for the most part, what they write down are the facts needed to start with, perhaps the intermediate results along the way and, if they get far enough, the final answer at the end. But the doing-math part is primarily a thinking process—something that takes place mostly in your head. Even when people are asked to “show all their work,” the collection of symbolic expressions that they write down is not necessarily the same as the process that goes on in their minds when they do math correctly. In fact, people can become highly skilled at doing mental math and yet be hopeless at its symbolic representations.
With everyday mathematics, the symbol barrier emerges. In their 1993 book Street Mathematics and School Mathematics, Terezinha Nunes, David William Carraher and Analucia Dias Schliemann describe research carried out in the street markets of Recife, Brazil, in the early 1990s. This and other studies have shown that when people are regularly faced with everyday mathematics in their daily lives, they rapidly master it to an astonishing 98 percent accuracy. Yet when faced with what are (from a mathematical perspective) the very same problems, but presented in the traditional symbols, their performance drops to a mere 35 to 40 percent accuracy.
It simply is not the case that ordinary people cannot do everyday math. Rather, they cannot do symbolic everyday math. In fact, for most people, it’s not accurate to say that the problems they are presented in paper-and-pencil format are “the same as” the ones they solve fluently in a real life setting. When you read the transcripts of the ways they solve the problems in the two settings, you realize that they are doing completely different things. Only someone who has mastery of symbolic mathematics can recognize the problems encountered in the two contexts as being “the same.”
The symbol barrier is huge and pervasive. For the entire history of organized mathematics instruction, where we had no alternative to using static, symbolic expressions on flat surfaces to store and distribute mathematical knowledge, that barrier has prevented millions of people from becoming proficient in a cognitive skill set of evident major importance in today’s world, on a par with the ability to read and write. | 2.640625 | CC-MAIN-2017-13 |
# PatchMatch
Flowers at right bottom corner are removed using PatсhMatсh
The core PatchMatch algorithm quickly finds correspondences between small square regions (or patches) of an image. The algorithm can be used in various applications such as object removal from images, reshuffling or moving contents of images, or retargeting or changing aspect ratios of images, optical flow estimation, or stereo correspondence.
## Algorithm
The goal of the algorithm is to find the patch correspondence by defining a nearest-neighbor field (NNF) as a function ${\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} ^{2}}$ of offsets, which is over all possible matches of patch (location of patch centers) in image A, for some distance function of two patches ${\displaystyle D}$. So, for a given patch coordinate ${\displaystyle a}$ in image ${\displaystyle A}$ and its corresponding nearest neighbor ${\displaystyle b}$ in image ${\displaystyle B}$, ${\displaystyle f(a)}$ is simply ${\displaystyle b-a}$. However, if we search for every point in image ${\displaystyle B}$, the work will be too hard to complete. So the following algorithm is done in a randomized approach in order to accelerate the calculation speed. The algorithm has three main components. Initially, the nearest-neighbor field is filled with either random offsets or some prior information. Next, an iterative update process is applied to the NNF, in which good patch offsets are propagated to adjacent pixels, followed by random search in the neighborhood of the best offset found so far. Independent of these three components, the algorithm also use a coarse-to-fine approach by building an image pyramid to obtain the better result.
### Initialization
When initializing with random offsets, we use independent uniform samples across the full range of image ${\displaystyle B}$. This algorithm avoids using an initial guess from the previous level of the pyramid because in this way the algorithm can avoid being trapped in local minima[clarification needed].
### Iteration
After initialization, the algorithm attempted to perform iterative process of improving the ${\displaystyle NNF}$. The iterations examine the offsets in scan order (from left to right, top to bottom), and each undergoes propagation followed by random search.
### Propagation
We attempt to improve ${\displaystyle f(x,y)}$ using the known offsets of ${\displaystyle f(x-1,y)}$ and ${\displaystyle f(x,y-1)}$, assuming that the patch offsets are likely to be the same. That is, the algorithm will take new value for ${\displaystyle f(x,y)}$ to be ${\displaystyle \arg \min \limits _{(x,y)}{D(f(x,y)),D(f(x-1,y)),D(f(x,y-1))}}$. So if ${\displaystyle f(x,y)}$ has a correct mapping and is in a coherent region ${\displaystyle R}$, then all of ${\displaystyle R}$ below and to the right of ${\displaystyle f(x,y)}$ will be filled with the correct mapping. Alternatively, on even iterations, the algorithm search for different direction, fill the new value to be ${\displaystyle \arg \min \limits _{(x,y)}\{D(f(x,y)),D(f(x+1,y)),D(f(x,y+1))\}}$.
### Random search
Let ${\displaystyle v_{0}=f(x,y)}$, we attempt to improve${\displaystyle f(x,y)}$ by testing a sequence of candidate offsets at an exponentially decreasing distance from ${\displaystyle v_{0}}$
${\displaystyle u_{i}=v_{0}+w\alpha ^{i}R_{i}}$
where ${\displaystyle R_{i}}$ is a uniform random in ${\displaystyle [-1,1]\times [-1,1]}$, ${\displaystyle w}$ is a large window search radius which will be set to maximum picture size, and ${\displaystyle \alpha }$ is a fixed ratio often assigned as 1/2. This part of the algorithm allows the ${\displaystyle f(x,y)}$ to jump out of local minimum through random process.
### Halting criterion
The often used halting criterion is set the iteration times to be about 4~5. Even with low iteration, the algorithm works well.
## Conclusion
This is an efficient algorithm since it only takes a few second on a testing computer with Intel Core i5 CPU and Photoshop CS5. | 3.328125 | CC-MAIN-2022-21 |
### Slide 1
```Writing up results from
Structural Equation Models
What to Report, What to Omit
1
Writing up results from Structural Equation Models
•
•
Reference: Hoyle and Panter chapter in
Hoyle.
Important to note that there is a wide
variety of reporting styles (no one
“standard”).
2
Writing up results from Structural Equation Models
•
A Diagram
– Construct Equation Model
– Measurement Equation model
Some simplification may be required.
Adding parameter estimates may clutter (but for
simple models helps with reporting).
Alternatives exist (present matrices).
3
Reporting Structural Equation Models
•
“Written explanation justifying each path
and each absence of a path” (Hoyle and
Panter)
(just how much journal space is available
here? )
It might make more sense to try to identify
potential controversies (with respect to
inclusion, exclusion).
4
Controversial paths?
1
1
1
1
1
1
y1
y2
y3
y4
y5
y6
1
20
LV1
LV3
e1
e2
1
LV2
1
1
1
1
LV4
1
1
1
5
What to report and what not to report…..
•
Present the details of the statistical model
– Clear indication of all free parameters
– Clear indication of all fixed parameters
It should be possible for the reader to
reproduce the model
4. Describe the data
1. Correlations and standard errors (or
covariances) for all variables ??
Round to 3-4 digits and not just 2 if you do this
6
What to report and what not to report…
4. Describing the data (continued)
– Distributions of the data
•
•
•
•
Any variable highly skewed?
Any variable only nominally continuous (i.e., 5-6
discrete values or less)?
Report Mardia’s Kurtosis coefficient (multivariate
statistic)
Dummy exogenous variables, if any
5. Estimation Method
If the estimation method is not ML, report ML
results.
7
What to report and what not to report…
6. Treatment of Missing Data
– How big is the problem?
– Treatment method used?
•
•
•
•
•
•
Pretend there are no missing data
Listwise deletion
Pairwise deletion
FIML estimation (AMOS, LISREL, MPlus, EQS)
Nearest neighbor imputation (LISREL)
EM algorithm (covariance matrix imputation ) (SAS,
LISREL/PRELIS)
8
What to report and what not to report…
7. Fit criterion
– Hoyle and Panter suggest “.90; justify if lower”.
– Choice of indices also an issue.
There appears to be “little consensus on the
best index” (H & P recommend using multiple
indices in presentations)
Standards:
Bollen’s delta 2 (IFI)
Comparative Fit Index
RMSEA
9
Fit indices
•
Older measures:
– GFI (Joreskog & Sorbom)
– Bentler’s Normed Fit index
– Model Chi-Square
10
What to report & what not to report….
8. Alternative Models used for Nested Comparisons (if appropriate)
US South
1
1
US West
1
1
U.S. Midwest
U.S. Rust Belt
11
9. Plausible explanation for correlated errors
•
[“these things were just too darned big to ignore”]
Generally assumed when working with panel model with equivalent
indicators across time:
1
1
1
1
1
1
1
1
12
What to report
10. Interpretation of regression-based model
– Present standardized and unstandardized
coefficients (usually)
– Standard errors? (* significance test indicators?)
– R-square for equations
•
•
Measurement model too?
(expect higher R-squares)
13
What to report.
•
Problems and issues
– Negative error variances or other reasons for
non-singular parameter covariance matrices
•
How dealt with? Does the final model entail any
“improper estimates”?
– Convergence difficulties, if any
•
LISREL: can look at Fml across values of given
parameter, holding other parameters constant
– Collinearity among exogenous variables
– Factorially complex items
14
What to report & what not to report….
•
General Model Limitations, Future Research
issues:
– Where the number of available indicators
compromised the model
•
•
•
2-indicator variables? (any constraints required?)
Indicators not broadly representative of the construct
being measured?
– Where the distribution of data presented
problems
•
Larger sample sizes can help
15
What to report & what not to report….
•
General Model Limitations, Future Research
issues:
– Missing data (extent of, etc.)
– Cause-effect issues, if any (what constraints
went into non-recursive model? How reasonable
are these?)
16
``` | 3 | CC-MAIN-2016-44 |
## Citrix System Inc Interview Question for Quality Assurance Engineers
• 0
Team: Xen-Client
Country: India
Interview Type: In-Person
Comment hidden because of low score. Click to expand.
2
of 2 vote
fill 5 ltr jar and pour it in 3 ltr, so 2 ltr remains in 5 ltr jar.
Empty 3 ltr jar and pour 2 ltr from 5 ltr jar into it.
Now fill 5 ltr jar again and use it to fill remaining 1 ltr in 3 ltr jar. that keeps 4 ltr in 5 ltr jar
Comment hidden because of low score. Click to expand.
0
yes, that's the only answer possible..
Comment hidden because of low score. Click to expand.
0
@abh007
nop, thats not the only answer.. even though following has more steps, it does achieve the goal..
1> fill up 3 ltr jug -> pour it into 5 ltr jug
2> fill up 3 ltr jug -> pour as much as possible in 5 ltr jug
3> empty 5 ltr jug
4> pour 1 ltr remaining liquid from 3 ltr jug into 5 ltr jug
5> fill up 3 ltr jug again and pour it into 5 ltr jug.. now 5 ltr jug has 4 ltr liquid in it
Comment hidden because of low score. Click to expand.
1
of 1 vote
1) fill the 2
2) 2 -> 5
3) repeat step 1 and 2
You will have 4 in 5. So, what is the point of this interview question?
Comment hidden because of low score. Click to expand.
0
I am sorry question is -> You have a 3 liter & 5 liter jugs and a tank of milk. So using these 2 jugs get 4 liter milk in 5 liter jug ?
Comment hidden because of low score. Click to expand.
0
of 0 vote
i like the last line of the last answer and wonder why question and its original complicated answer
Comment hidden because of low score. Click to expand.
0
1. fill just a half of each jugs
3->1.5;
5->2.5)
2. pour jug 3 into jug 5, you will have 1.5+2.5=4
Comment hidden because of low score. Click to expand.
0
there is no measure of half MR.
try another method....
Comment hidden because of low score. Click to expand.
0
of 0 vote
i hav water of 2 jugs with 21liters and 26liters jug?how to measure 3liters of water using that two jugs...
Comment hidden because of low score. Click to expand.
0
step 1 fill 26 put it into 21
step 2 empty 21 lt jug and fill it with rem ltr of 26
do this until u get 3 ltr left in 26 ltr jug........
approx in 21 times
Comment hidden because of low score. Click to expand.
0
of 0 vote
1. Fill 2 litre
2.empty the 2 litre in 5 litre
3.again fill 2 litre
4.empty into 5 litre
Representation:
5 2
0 2
2 0
2 2
4 0
With no wastage
Name:
Writing Code? Surround your code with {{{ and }}} to preserve whitespace.
### Books
is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.
### Videos
CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance. | 2.828125 | CC-MAIN-2020-40 |
🎉 Exercism Research is now launched. Help Exercism, help science and have some fun at research.exercism.io 🎉
# amscotti's solution
## to Gigasecond in the Elm Track
Published at Jun 17 2020 · 0 comments
Instructions
Test suite
Solution
Given a moment, determine the moment that would be after a gigasecond has passed.
A gigasecond is 10^9 (1,000,000,000) seconds.
## Elm Installation
Refer to the Installing Elm page for information about installing elm.
## Writing the Code
The code you have to write is located inside the `src/` directory of the exercise. Elm automatically installs packages dependencies the first time you run the tests so we can start by running the tests from the exercise directory with:
``````\$ elm-test
``````
To automatically run tests again when you save changes:
``````\$ elm-test --watch
``````
As you work your way through the tests suite in the file `tests/Tests.elm`, be sure to remove the `skip <|` calls from each test until you get them all passing!
## Source
Chapter 9 in Chris Pine's online Learn to Program tutorial. http://pine.fm/LearnToProgram/?Chapter=09
## Submitting Incomplete Solutions
It is possible to submit an incomplete solution so you can see how others have completed the exercise.
### Tests.elm
``````module Tests exposing (tests)
import Expect
import Iso8601
import Parser
import Test exposing (..)
import Time
tests : Test
tests =
describe "Gigasecond"
[ test "2011-04-25" <|
\() ->
Expect.equal "2043-01-01T01:46:40.000Z"
(gigasecond "2011-04-25")
, skip <|
test "1977-06-13" <|
\() ->
Expect.equal "2009-02-19T01:46:40.000Z"
(gigasecond "1977-06-13")
, skip <|
test "1959-07-19" <|
\() ->
Expect.equal "1991-03-27T01:46:40.000Z"
(gigasecond "1959-07-19")
, skip <|
test "full time specified" <|
\() ->
Expect.equal "2046-10-02T23:46:40.000Z"
(gigasecond "2015-01-24T22:00:00.000Z")
, skip <|
test "full time with day roll-over" <|
\() ->
Expect.equal "2046-10-03T01:46:39.000Z"
(gigasecond "2015-01-24T23:59:59.000Z")
]
]
gigasecond : String -> String
gigasecond date =
case Iso8601.toTime date of
Ok posix ->
posix
|> Iso8601.fromTime
_ ->
""``````
``````module Gigasecond exposing (add)
import Time exposing (millisToPosix, posixToMillis)
seconds + (10 ^ 12)
timestamp
|> posixToMillis
|> millisToPosix`````` | 2.875 | CC-MAIN-2020-40 |
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Bookmark it
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# Physics II Lab w answers_Part_25 - Results The calculated...
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61 Results: The calculated value of e/m =____________________C/kg. The standard value of e/m =_____________________C/kg. % Error in e/m =_____________________
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62 Experiment 10 Charging and Discharging of a Capacitor Apparatus: Ballistic galvanometer, lamp, and scale, capacitor, resistor, solar cell, light filters, stop watch, two way switches, damping key , connecting wires . Purpose of experiment: i) To study the charging and discharging of a capacitor ii) To measure the time constant characterizing charging / discharging process iii) To understand the working of a ballistic galvanometer. Basic Methodology: A capacitor is charged by a solar cell and discharged (for different times) through a resistor and ballistic galvanometer .The deflection of the galvanometer is in proportion to the amount of charge that passes through it .A lamp and scale arrangement is used to measure the deflection. I. Introduction: I.1 The basic circuit for charging and discharging a capacitor is shown in fig 1. If switch S 1 is closed keeping S
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Ask a homework question - tutors are online | 2.640625 | CC-MAIN-2018-13 |
Explore BrainMass
# Linear programming problem
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
You need to organize your home office, especially your paperwork. You have three general kinds of paperwork: tax files, bill files, investment files. You estimate that you have 10 tax files, 30 bill files, and 15 investment files. Each month you access each tax file three times; each bill file, two times; and each investment file, eight times. You have three places to put files, a file cabinet, your desk, or a storage box. The file cabinet can hold 30 files; the desk, 10 files; and the storage box, 25 files. When a file is accessed, the time it takes for a file to be accessed depends on the type of file and where you put it. You need to decide where to put your files. Your goal is to minimize the total time you spending each month accessing your files.
Retrieval Time (minutes)
(See attached)
Hint: The minimum retrieval time will be 660 minutes per month.
## SOLUTION This solution is FREE courtesy of BrainMass!
Let Xij = No of i files stored in j storage space where i=t, b, i and j=f, d, s
For example, Xtf = No of tac files stored in file cabinate.
Decision variables
Tax Files Bill Files Investment Files
File Cabinet xtf xbf xif
Desk xtd xbd xid
Storage Box xts xbs xis
Formulation:
Minimize Z= 3*(5xtf+3xtd+10xts) + 2*(3xbf +2xbd +6xbs) + 8*(4xif +1xid +8xis)
S.t. constraints
xtf +xtd +xts = 10
xbf +xbd +xbs =30
xif +xid +xis = 15
xtf + xbf +xif <= 30
xtd + xbd +xid <= 10
xts + xbs +xis<= 25
xij >=0
Solution in attached excel file.
Tax Files Bill Files Investment Files
File Cabinet 10 15 5
Desk 0 0 10
Storage Box 0 15 0
By solving the problem using Excel solver, we get following results:
Store 10 tax files in file cabinet, 15 Bill files in file cabinet, 15 Bill files in storage box, 5 Investment files in file cabinet, and 10 investment files in Desk with minimum total retrieval time of 660 minutes/month.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here! | 3.703125 | CC-MAIN-2022-05 |
Home > Textbooks > Lessons In Electric Circuits > Vol. II - AC > Reactance and Impedance - Capacitive > Capacitor Quirks
Chapter 4: REACTANCE AND IMPEDANCE - CAPACITIVE
Capacitor Quirks
As with inductors, the ideal capacitor is a purely reactive device, containing absolutely zero resistive (power dissipative) effects. In the real world, of course, nothing is so perfect. However, capacitors have the virtue of generally being purer reactive components than inductors. It is a lot easier to design and construct a capacitor with low internal series resistance than it is to do the same with an inductor. The practical result of this is that real capacitors typically have impedance phase angles more closely approaching 90o (actually, -90o) than inductors. Consequently, they will tend to dissipate less power than an equivalent inductor.
Capacitors also tend to be smaller and lighter weight than their equivalent inductor counterparts, and since their electric fields are almost totally contained between their plates (unlike inductors, whose magnetic fields naturally tend to extend beyond the dimensions of the core), they are less prone to transmitting or receiving electromagnetic "noise" to/from other components. For these reasons, circuit designers tend to favor capacitors over inductors wherever a design permits either alternative.
Capacitors with significant resistive effects are said to be lossy, in reference to their tendency to dissipate ("lose") power like a resistor. The source of capacitor loss is usually the dielectric material rather than any wire resistance, as wire length in a capacitor is very minimal.
Dielectric materials tend to react to changing electric fields by producing heat. This heating effect represents a loss in power, and is equivalent to resistance in the circuit. The effect is more pronounced at higher frequencies and in fact can be so extreme that it is sometimes exploited in manufacturing processes to heat insulating materials like plastic! The plastic object to be heated is placed between two metal plates, connected to a source of high-frequency AC voltage. Temperature is controlled by varying the voltage or frequency of the source, and the plates never have to contact the object being heated.
This effect is undesirable for capacitors where we expect the component to behave as a purely reactive circuit element. One of the ways to mitigate the effect of dielectric "loss" is to choose a dielectric material less susceptible to the effect. Not all dielectric materials are equally "lossy". A relative scale of dielectric loss from least to greatest is given in Table below.
Dielectric loss
MaterialLoss
VacuumLow
Air-
Polystyrene-
Mica-
Glass-
Low-K ceramic-
Plastic film (Mylar)-
Paper-
High-K ceramic-
Aluminum oxide-
Tantalum pentoxidehigh
Dielectric resistivity manifests itself both as a series and a parallel resistance with the pure capacitance: (Figure below)
Real capacitor has both series and parallel resistance.
Fortunately, these stray resistances are usually of modest impact (low series resistance and high parallel resistance), much less significant than the stray resistances present in an average inductor.
Electrolytic capacitors, known for their relatively high capacitance and low working voltage, are also known for their notorious lossiness, due to both the characteristics of the microscopically thin dielectric film and the electrolyte paste. Unless specially made for AC service, electrolytic capacitors should never be used with AC unless it is mixed (biased) with a constant DC voltage preventing the capacitor from ever being subjected to reverse voltage. Even then, their resistive characteristics may be too severe a shortcoming for the application anyway. | 2.9375 | CC-MAIN-2020-10 |
# 0.6 Nonparametric regression with wavelets (Page 5/5)
Page 5 / 5
${R}_{n}\left(V,p\right)=\underset{{T}_{n}}{inf}\underset{m\in V}{sup}E{∥{T}_{n},-,m∥}_{p}^{p},$
where the infimum is taken over all measurable estimators ${T}_{n}$ of $m.$ Similarly, we define the linear ${L}_{p}-$ minimax risk as
${R}_{n}^{\text{lin}}\left(V,p\right)=\underset{{T}_{n}^{\text{lin}}}{inf}\underset{m\in V}{sup}E{∥{T}_{n}^{\text{lin}},-,m∥}_{p}^{p},$
where the infimum is now taken over all linear estimators ${T}_{n}^{\text{lin}}.$ Obviously, ${R}_{n}^{\text{lin}}\left(V,p\right)\ge {R}_{n}\left(V,p\right).$ We first state some definitions.
The sequences $\left\{{a}_{n}\right\}$ and $\left\{{b}_{n}\right\}$ are said to be asymptotically equivalent and are noted ${a}_{n}\sim {b}_{n}$ if the ratio ${a}_{n}/{b}_{n}$ is bounded away from zero and $\infty$ as $n\to \infty .$
The sequence ${a}_{n}$ is called optimal rate of convergence , (or minimax rate of convergence ) on the class $V$ for the ${L}_{p}-$ risk if ${a}_{n}\sim {R}_{n}{\left(V,p\right)}^{1/p}$ . We say that an estimator ${m}_{n}$ of $m$ attains the optimal rate of convergence if $\underset{m\in V}{sup}E{∥{m}_{n},-,m∥}_{p}^{p}\sim {R}_{n}\left(V,p\right).$
In order to fix the idea, we consider only the ${L}_{2}-$ risk in the remaining part of this section, thus $p:=2$ .
In [link] , [link] , the authors found that the optimal rate of convergence attainable by an estimator when the underlying function belongs to the Sobolev class ${W}_{q}^{s}$ is ${a}_{n}={n}^{\frac{-s}{2s+1}}$ , hence ${R}_{n}\left(V,2\right)={n}^{\frac{-2s}{2s+1}}$ . We saw in "Linear smoothing with wavelets" that linear wavelet estimators attain the optimal rate for $s-$ Hölder function in case of the ${L}_{2}-$ risk (also called `IMSE'). For a Sobolev class ${W}_{q}^{s}$ , the same result holds provided that $q\ge 2$ . More precisely, we have the two following situations.
1. If $q\ge 2,$ we are in the so-called homogeneous zone. In this zone of spatial homogeneity, linear estimators can attain the optimal rate of convergence ${n}^{-s/\left(2s+1\right)}.$
2. If $q<2,$ we are in the non-homogeneous zone, where linear estimators do not attain the optimal rate of convergence. Instead, we have:
${R}_{n}^{\text{lin}}\left(V,2\right)/{R}_{n}\left(V,2\right)\to \infty ,\phantom{\rule{4.pt}{0ex}}\text{as}\phantom{\rule{4.pt}{0ex}}n\to \infty .$
The second result is due to the spatial variability of functions in Sobolev spaces with small index $q$ . Linear estimators are based on the idea of spatial homogeneity of the function and hence do perform poorly in the presence of non-homogeneous functions. In contrast, even if $q<2$ , the SureShrink estimator attains the minimax rate [link] . The same type of results holds for more general Besov spaces, see for example [link] , Chapter 10.
## Adaptivity of wavelet estimator
We just saw that a nonlinear wavelet estimator is able to estimate in an optimal way functions ofinhomogeneous regularity. However, it may not be sufficient to know that for $m$ belonging to a given space, the estimator performs well. Indeed, in general we do not know which space the function belongs to. Hence it is ofgreat interest to consider a scale of function classes and to look for an estimator that attains simultaneously the best rates of convergence across the whole scale. For example, the ${L}_{q}-$ Sobolev scale is a set of Sobolev function classes ${W}_{q}^{s}\left(C\right)$ indexed by the parameters $s$ and $C$ , see [link] for the definition of a Sobolev class. We now formalize the notion of an adaptive estimator.
Let $A$ be a given set and let $\left\{{\mathcal{F}}_{\alpha },\alpha \in A\right\}$ be the scale of functional classes ${\mathcal{F}}_{\alpha }$ indexed by $\alpha \in A.$ Denote ${R}_{n}\left(\alpha ,p\right)$ the minimax risk over ${\mathcal{F}}_{\alpha }$ for the ${L}_{p}-$ loss:
${R}_{n}\left(\alpha ,p\right)=\underset{{\stackrel{^}{m}}_{n}}{inf}\underset{m\in {\mathcal{F}}_{\alpha }}{sup}E{∥{\stackrel{^}{m}}_{n},-,m∥}_{p}^{p}.$
The estimator ${m}_{n}^{*}$ is called rate adaptive for the ${L}_{p}-$ loss and the scale of classes ${\mathcal{F}}_{\alpha },\alpha \in A$ if for any $\alpha \in A$ there exists ${c}_{\alpha }>0$ such that
$\underset{m\in {\mathcal{F}}_{\alpha }}{sup}E{∥{m}_{n}^{*},-,m∥}_{p}^{p}\le {c}_{\alpha }{R}_{n}\left(\alpha ,p\right)\phantom{\rule{0.222222em}{0ex}}\forall n\ge 1.$
The estimator ${m}_{n}^{*}$ is called adaptive up to a logarithmic factor for the ${L}_{p}-$ loss and the scale of classes ${\mathcal{F}}_{\alpha },\alpha \in A$ if for any $\alpha \in A$ there exist ${c}_{\alpha }>0$ and $\gamma ={\gamma }_{\alpha }>0$ such that
$\underset{m\in {\mathcal{F}}_{\alpha }}{sup}E{∥{m}_{n}^{*},-,m∥}_{p}^{p}\le {c}_{\alpha }{\left(logn\right)}^{\gamma }{R}_{n}\left(\alpha ,p\right)\phantom{\rule{0.222222em}{0ex}}\forall n\ge 1.$
Thus, adaptive estimators have an optimal rate of convergence and behave as if they know in advance in which class the function to be estimated lies.
The VisuShrink procedure is adaptive up to a logarithmic factor for the ${L}_{2}-$ loss over every Besov, Hölder and Sobolev class that is contained in $C\left[0,1\right]$ , see Theorem 1.2 in [link] . The SureShrink estimator does better: it is adaptive for the ${L}_{2}-$ loss, for a large scale of Besov, Hölder and Sobolev classes, see Theorem 1 in [link] .
## Conclusion
In this chapter, we saw the basic properties of standard wavelet theory and explained how these are related to the construction of wavelet regression estimators.
#### Questions & Answers
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
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Books Books 11 - 20 of 189 on Divide the first term of the dividend by the first term of the divisor, and write....
Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply the whole divisor by the first term of the quotient, and subtract the product from the dividend.
The popular educator - Page 321
by Popular educator - 1860
## An Introduction to Algebra: With Notes and Observations : Designed for the ...
John Bonnycastle - Algebra - 1818 - 260 pages
...terms of each of them so, that the higher powers of one of the letters may stand before the lower. Then divide the first term of the dividend by the first term of the divisor, and set the result in the quotient, with its proper sign, or simply by itself, if it be affirmative....
## The young man's best companion, and book of general knowledge
L. Murray - 1821
...divided by subtracting their exponents. Thus if a5 is divided by as the quotient is as — * or a3. of the dividend by the first term of the divisor ; multiply the quotient by the whole divisor, and subtract the product from the dividend. If nothing remain, the operation...
## A Course of Mathematics: For the Use of Academies, as Well as Private ...
Charles Hutton - Mathematics - 1822
...according to the powers of some one of the letters in both, the higher powers before the lower. 2. Divide the first term of the dividend by the first term of the divisor, as in the first case, and set the result in the quotient. 3. Multiply the whole divisor by the term...
## An Elementary Treatise on Algebra, Theoretical and Practical ...
James Ryan - Algebra - 1824 - 516 pages
...letter, beginning 1 ' with the highest, and place the divisor at the right hand of the dividend ; then divide the first term of the dividend by the first term of the divisor, as in Case I., and place the result under the divisor. . Multiply the wfcole divisor by this partial...
## An Introduction to Algebra: With Notes and Observations, Designed for the ...
John Bonnycastle - Algebra - 1825 - 312 pages
...terms of each of them so, that the higher power of one of the letters may stand before the lower. Then divide the first term of the dividend by the first term of the divisor, and set the result in the quotient, with its proper sign, or simply by itself, if it be affirmative....
## Elements of Geometry
Adrien Marie Legendre - 1825 - 224 pages
...writing the terms in the order of the exponents of this letter, beginning with the highest ; 2. We divide the first term of the dividend by the first term of the divisor, and write the result in the place of the quotient ; 3. We multiply the whole divisor by the term of...
## Elements of Algebra
Silvestre François Lacroix - Algebra - 1825 - 276 pages
...writing the terms in the order of the exponents of this letter, beginning with the highest ; 2. We divide the first term of the dividend by the first term of the divisor, and write the result in the place of the quotient ; 3. We multiply the whole divisor by the term of...
## An Introduction to Algebra Upon the Inductive Method of Instruction
Warren Colburn - Algebra - 1825 - 372 pages
...division of compound numbers. Arrange the dividend and divisor according to the powers of some letter. Divide the first term of the dividend by the first term of the diviwr, and write the result in the quotient. Multiply all the terms of the divisor by the term of... | 3.34375 | CC-MAIN-2019-39 |
## Fastest Minesweeper ever! What are the chances?
For the discussion of math. Duh.
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Sableagle
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### Fastest Minesweeper ever! What are the chances?
FastestMinesweeperEver.png (4.08 KiB) Viewed 4363 times
I'm a little rusty but nCr = n!/(n-r)!r! isn't it?
That's:
64!/(64-10)!10! = 1.515 * 1011 for Beginner;
256!/(256-40)!40! = 1.049 * 1047 for Intermediate;
496!/(496-99)!99! = 2.100 * 10106 for Expert
... and ridiculously huge for a custom grid that covers the whole screen.
I'm curious.
How would you calculate the odds of getting a one-click win as shown above?
Oh, Willie McBride, it was all done in vain.
Soupspoon
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### Re: Fastest Minesweeper ever! What are the chances?
I think you have to take into account that the first square you reveal is never a mine (whether or not that means it is a neighbour-square that only reveals a number or a non-neighbour from which a flood-reveal propogates). But whether that's because it shifts any first-click boom somewhere else before you get to see it or that it waits to see where you clicked before placing any mines, I don't know. You'd have to ask Monty Hall about that, perhaps..
(But it might mean that it is 63!/(63-10)!10! = ... for Beginner, etc, if the rest of that is right...)
ThemePark
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### Re: Fastest Minesweeper ever! What are the chances?
For that particular setup, there's 376 empty cells and you could click anyone of them to get that result. So for that, it's 376/499 or 75,35 %.
But it's only possible to win in one click on those setups where all empty cells are connected in one big shape. So you "only" have to concern yourself with those, and for each of them the odds would be (board size-bombs-cells with numbers on them)/board size.
The bombs don't have to be lumped into one shape though, they can be divided. But short of iterating all ways of placing the empty cells all together, I don't see how it's possible to calculate the odds for the entire game of one particular board size.
Oh, and Soupspoon, it is very possible to hit a bomb in the first click.
I have traveled from 1979 to be a member of the unofficial board Council of Elders. Phear M3
Soupspoon
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### Re: Fastest Minesweeper ever! What are the chances?
ThemePark wrote:Oh, and Soupspoon, it is very possible to hit a bomb in the first click.
Might depend on the version of (Windows) Minesweeper, but I am sure that it quite famously hasn't been possible. Even setting to Custom parameters with whatever settings gives you most mines per grid area for maximum challenge, for ultimate testing, I've never had a first-click1-BOOM! happen and, given the number of games played at this setting (and almost inevitably lost, but only from the second reveal onwards), when I was a definite Minesweepe Junkie, I think that I'm statistically confident in that assertion.
(For Advanced difficulty, my speed-run opening move tends to be two rapid and random clicks in different halves of the board. If lucky, that floods over many non-mine spaces (never yet all of them, sometimes very very few are left to discover, though, even if still ambiguous) and I can then start using logic, or guesswork for the obvious insolubly ambiguous (behind a "wall") to get them out of the way. If unlucky, the second click detonates (if unlucky and stupid, first click reveals told me it would be so, but in my rush to get the advantage I let myself fall on my sword). Never a first click failure.)
1 First revealing click, that is. Flagging/Questionmarking, revealing elsewhere (safely!) then de-marking and clicking-for-real the original tile still possibly/likely gives a boom on the originally marked tile. Unless the true-first-reveal shows that your first choice is definitely safe, naturally.
Last edited by Soupspoon on Fri Sep 16, 2016 9:35 pm UTC, edited 4 times in total.
Eebster the Great
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### Re: Fastest Minesweeper ever! What are the chances?
The position of all the mines is set before your first click. If your first click is over a mine, that mine will be moved to the top left corner instead. If there is already a mine there, it will go one square to the right, and so on.
In early releases of Minesweeper on DOS-based Windows kernels, Beginner boards were 8x8 but one-click games were not normally possible. This is because boards were not generated from the entire space of legal boards but from a couple of small cycles which only generated 48,624 distinct boards under normal conditions. However, boards could be changed using the method mentioned above. In extremely rare cases, it is possible to change a board into a one-click board on which your first click is a solution, leading to the One Click Bug. There were also other ways to force nonstandard boards by opening two games simultaneously, but that's not really relevant here.
For releases on OSes with NT-based kernels and third-party clones, a much more sophisticated PRNG is used and a far larger of number of boards is possible. It is unknown however if every legal board is possible or if all possible boards are equally probable. In particular, clicking on a mine still moves that mine to the top left corner, ultimately making boards with a mine or multiple mines at the far left of the top row more likely than other boards. This should increase the probability of one-click games somewhat, and the page linked above gives two examples of one-click Beginner (9x9) games on XP, one "natural" game in which no mine was moved to the corner, and another "bugged" game in which a mine was moved to the corner, causing the timer bug.
The exact board you showed below is only one single board, so if the program uses a properly distributed RNG, the probability of it occurring discounting symmetry or anything else is simply the reciprocal of the number of boards. In this case, it is Expert, 16x30=480 squares with 99 mines, so the number of possible boards is simply n = 480 choose 99 = 480!/(99!(480-99)!) = 560220999337421345429058985775821108059290502723897901281458809527214479570631168198385673295159633481600 ≈ 5.6 × 10104. Since this board cannot be generated by moving a mine but must arise naturally, the probability that it arises is 1/n ≈ 1.8 × 10-105.
However, actually winning this board in one click depends where you click. For example, if you click in the bottom left corner, you will simply get a 3, moving the mine to the top left, and giving you a decent shot of simply losing on your next click. Clicking on any mine or on any number will not win immediately. Thus only 360 of the 480 spaces result in a one-click win, meaning if you play a randomly generated board and click a random space, your probability of winning with this particular board is 3/(4n) ≈ 1.4 × 10-105.
A more interesting question though is how many one-click win boards there are. Discounting first-click mines, this should be computable, though I'm not exactly sure how to go about it. Counting first-click mines makes it even more complicated.
EDIT: The board you posted is 16x31, which is super weird. Expert boards are usually 16x30 with 99 mines for modern purposes, or 24x30 with 200 (or 225) mines for super-expert. Anyway, for the 16x31 board, instead of 480 choose 99 boards, there are 496 choose 99 boards, or 2.1 × 10106, as above. Therefore the probability of getting that exact board before clicking is (480C99)-1 = 4.8 × 10-107, and to win on the first click with that exact board is 376/496 = 47/62 ≈ 0.76 times as much.
Last edited by Eebster the Great on Fri Sep 16, 2016 10:22 pm UTC, edited 3 times in total.
Tub
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### Re: Fastest Minesweeper ever! What are the chances?
ThemePark wrote:Oh, and Soupspoon, it is very possible to hit a bomb in the first click.
I've never seen it happen on the old windows-based implementations, even on custom games where the field is like 99% bombs. It seemed like the program created a random game the moment you clicked the first field, and discarded any games with a bomb under your cursor. You could notice, because as you increased the bomb count, there was a progressively longer delay between the first click and the uncovering of the first tile.
But then again, there's no authoritative implementation of minesweeper, so YMMV.
496!/(496-99)!99! = 2.100 * 10106 for Expert
That math seems sound for the exact game you pictured. There are of course other 1-click-solvable games, but counting them seems difficult, so I'm not going to try.
Allow me one side note: that chance is around 1 to 2^237, which means that any random number generator with less than 237 bits of state is unlikely to ever produce a sequence resulting in your game. The common non-cryptographic RNGs of the previous millennium all had a state of only one word (at most 32 bits back then), so for any minesweeper before win2k, it's very likely that your game cannot be generated at all.
Eebster the Great
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### Re: Fastest Minesweeper ever! What are the chances?
Tub wrote:Allow me one side note: that chance is around 1 to 2^237, which means that any random number generator with less than 237 bits of state is unlikely to ever produce a sequence resulting in your game. The common non-cryptographic RNGs of the previous millennium all had a state of only one word (at most 32 bits back then), so for any minesweeper before win2k, it's very likely that your game cannot be generated at all.
Yeah old versions used a 16 bit seed and generated fewer than 65,000 boards. Newer versions probably use 32 or 64 bit seeds, but I doubt any would use a full 256 bits, and even if they did, I doubt they would map onto all the possible boards, and even if they did, I very much doubt they would do so uniformly. But since nobody seems to know what the RNG is, statistically we have to treat all possibilities as equally likely anyway.
Soupspoon
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### Re: Fastest Minesweeper ever! What are the chances?
Entropically, though, a cluster (whether rectangular in a corner, a 'full border' that you clicked within the empty centre of or a far wavier shape) is numerically more unique.. *cough* ...than all the possible slapdash 'noisy' distributions that defeat oneclickwinability by the 'surface' of neighbourcells occupying and thus covering far more gaps behind which unrevealed (first-order) solutions may yet lurk.
(From the top of my head. The worst case scenario may well be (unbeknownst to the player) all 2x2 clusters positioned with two un-mined cells-worth of columns/rows between them all; all over the shop there's two-neighbour pairs abutting 2-neighbour pairs, 1-neighbours clustered as 2x2s at the near meeting of four corners can be open up their 1-neighbour compatriots, but the 2-neighbours don't reveal which of three positions the hindmost bomb might be in, if there's a possibility that cross-gap 2s might be referencing the same 'hindbomb', at the point in the reveal where you can't work out/guess the likelihood from the undiscovered count. Between that (unlikely high regularity) pattern and the least neighbourful combination (an entire edge lined by bombs, by whatever necessary number deep; the most ideal one-click solution) there's a whole host of godawful combinations with sufficient neighbourfulness to block even 'perfect' play from a guaranteed solution, superficially identical with countless1 kin given the initial or even continuing reveal.)
But maybe that's the way to count them. Assuming bombs, B, on an XxY grid has both X and Y as factors (for simplicity at this stage, ideally) then start with four layouts (flush and level top-lining, bottom-lining, left-lining and right-lining solutions, with both short-side layouts actually featuring least neighbour-cells), then morph to more jagged-edged variations by dislocating individual mines around the various exposed sides, pairs of mines both together and independently, and forwards with greater and greater numbers departing more combinations of origin. Include (unless too dense to leave workable space in the centre) all four-edges bordering (flatly) patterns, and even tendrils and 'minor floaters' into the interior, so long as they don't cause opposing neighbour-cells to meet and isolate separate zero-neighbour washes from each other.
Once you have done that, you (should?) have enumerated all one-click solubles. From the factorially-inclined totality of combinations, you have your proportion/likelihood thus calculated.
But this 'simple' initial count is not so trivial. Just perhaps more trivial than my other ideas.
1 Practically, that is. Not mathematically.
hujackus
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### Re: Fastest Minesweeper ever! What are the chances?
I've played well over 40,000 expert games over the last 4 years with a success rate hovering between 23% and 24% (win7 version). This thread has sniped me away from what I was planning on doing today, but unfortunately my time online has run out. I'm interested in finding out more about how the RNG is used in generating a pattern of mines. I'll be back.
lorb
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### Re: Fastest Minesweeper ever! What are the chances?
hujackus wrote:40,000 expert games over the last 4 years
so about 30 games every day? for four years. how does it never get old?
Please be gracious in judging my english. (I am not a native speaker/writer.)
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hujackus
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### Re: Fastest Minesweeper ever! What are the chances?
lorb wrote:so about 30 games every day? for four years. how does it never get old?
I checked my files again, and it's closer to 48 thousand games since mid 2013. I don't even play everyday so that equates to more 50-100 games in a single day. If I have nothing to do, or if I'm stressed out, I can mindlessly play for a few hours straight. Sometimes when I set a new time record, I don't find out about it until a few games or days later. My best time is 99 seconds, but I only have a hand full of sub 110 times.
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### Re: Fastest Minesweeper ever! What are the chances?
hujackus wrote:I've played well over 40,000 expert games over the last 4 years with a success rate hovering between 23% and 24% (win7 version).
Have you tried Tatham's minesweeper? It has a checkbox for solubility, so you don't have to guess the last 1-3 mines. (unfortunately, it doesn't listen to ctrl or shift, but at least it has the uncover-if-the-right-number-of-mines-are-marked)
hujackus
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### Re: Fastest Minesweeper ever! What are the chances?
Flumble wrote:
hujackus wrote:I've played well over 40,000 expert games over the last 4 years with a success rate hovering between 23% and 24% (win7 version).
Have you tried Tatham's minesweeper? It has a checkbox for solubility, so you don't have to guess the last 1-3 mines. (unfortunately, it doesn't listen to ctrl or shift, but at least it has the uncover-if-the-right-number-of-mines-are-marked)
I just tried it right now. It looks like a pretty good javascript version of the game, but having an undo button and a solve instantly button sitting at the top takes all the joy out of winning a game. Maybe if the solve button used a computer algorithm to actually solve the board over a period of a few seconds, it might be valuable to me. I do like the 'ensure solubility' checkbox you mentioned. If I re-write Tatham's minesweeper I'd probably keep that, add statistics and remove all traces of the the undo feature.
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### Re: Fastest Minesweeper ever! What are the chances?
Eebster the Great wrote:A more interesting question though is how many one-click win boards there are. Discounting first-click mines, this should be computable, though I'm not exactly sure how to go about it.
Hmm. Let's assume a 16x30 board (480 total tiles) with 99 mines. (So, 381 non-mines.)
Here are two facts about every one-click-win board:
• The board has a contiguous block of at least 76 "zeroes"--i.e. tiles with no adjacent mines. (Because we have to reveal 381 tiles at once; the first zero clicked reveals at most 8 tiles, and subsequent zeroes activated reveal at most 5 new tiles each.)
• Every one of the 381 non-mine tile must be adjacent to a zero. From this it can be inferred that for any given non-mile tile, at least one of the following conditions is true:
1. The tiles west, northwest, and north of it are all non-mines.
2. The tiles north, northeast, and east of it are all non-mines.
3. The tiles east, southeast, and south of it are all non-mines.
4. The tiles south, southwest, and west of it are all non-mines.
I've run out of time to think about it tonight, but maybe these facts could be used to put an upper bound on the probability of a random board being winnable in one click?
EmelinaVollmering
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### Re: Fastest Minesweeper ever! What are the chances?
It all depends on the algorithm that is created by microsoft in this case. Second, you can't draw the probability, if you don't know multiple factors. In this case, you know how many mines are there and how many are left hidden. Let say, the algorithm favors all the hidden points equally.
If the distribution of mines are equal in all the regions, we will probabilities over whole probability. | 2.984375 | CC-MAIN-2018-39 |
# One’s Complement Operator in C
The one’s complement operator is a fundamental concept in low-level programming with C. It allows for bitwise negation, which can be useful for various tasks such as toggling bits, implementing custom data types, and more. In this guide, we will explore the one’s complement operator in C, its usage, applications, and practical examples.
## Introduction to the One’s Complement Operator
The one’s complement operator, denoted as ~, is a unary operator in C that performs bitwise negation on its operand. It inverts each bit, changing 0s to 1s and 1s to 0s. Understanding how to use this operator is essential for low-level programming tasks.
## Understanding Binary Representation
To grasp the one’s complement operator’s significance, it’s crucial to have a clear understanding of binary representation:
• Binary numbers consist of 0s and 1s.
• Each digit is called a “bit.”
• Binary numbers can represent integers, characters, and other data.
## Using the One’s Complement Operator (~)
### Bitwise Negation
• The one’s complement operator negates each bit in the operand.
• Example:
int a = 5; // Binary representation: 00000101
int b = ~a; // Binary representation: 11111010
## Practical Applications
### Toggling Bits
• The one’s complement operator can toggle individual bits in a binary number.
• Example:
int num = 0b00001000; // Initial value
num ^= (1 << 3); // Toggle the 3rd bit
### Implementing Custom Data Types
• Custom data types can be implemented using bitwise operations.
• Bitfields within structures allow fine-grained control over memory usage.
• Example:
struct {
unsigned int isAvailable : 1;
unsigned int value : 4;
} customData;
## Examples and Code Snippets
Explore practical examples of the one’s complement operator in C, including bit toggling and custom data type implementation.
## Common Pitfalls
• Beware of the sign bit when working with signed integers.
• Bitwise negation might not produce the desired result for floating-point numbers.
• Be cautious when using the operator on variables with unexpected values.
## Best Practices
• Comment your code to clarify the intent of bitwise operations.
• Ensure that your code is well-documented for others to understand.
• Test your bitwise operations thoroughly, especially when dealing with complex data structures.
## Conclusion
The one’s complement operator in C is a powerful tool for bitwise negation. By understanding its usage and applications, you can enhance your low-level programming skills and tackle various tasks, from bit manipulation to custom data type implementation. Mastering this operator is valuable in the world of low-level programming and system-level development. | 2.953125 | CC-MAIN-2024-38 |
Noteworthy Differences from other Languages
Noteworthy Differences from other Languages
Noteworthy differences from MATLAB
Although MATLAB users may find Julia's syntax familiar, Julia is not a MATLAB clone. There are major syntactic and functional differences. The following are some noteworthy differences that may trip up Julia users accustomed to MATLAB:
• Julia arrays are indexed with square brackets, A[i,j].
• Julia arrays are not copied when assigned to another variable. After A = B, changing elements of B will modify A as well.
• Julia values are not copied when passed to a function. If a function modifies an array, the changes will be visible in the caller.
• Julia does not automatically grow arrays in an assignment statement. Whereas in MATLAB a(4) = 3.2 can create the array a = [0 0 0 3.2] and a(5) = 7 can grow it into a = [0 0 0 3.2 7], the corresponding Julia statement a[5] = 7 throws an error if the length of a is less than 5 or if this statement is the first use of the identifier a. Julia has push! and append!, which grow Vectors much more efficiently than MATLAB's a(end+1) = val.
• The imaginary unit sqrt(-1) is represented in Julia as im, not i or j as in MATLAB.
• In Julia, literal numbers without a decimal point (such as 42) create integers instead of floating point numbers. Arbitrarily large integer literals are supported. As a result, some operations such as 2^-1 will throw a domain error as the result is not an integer (see the FAQ entry on domain errors for details).
• In Julia, multiple values are returned and assigned as tuples, e.g. (a, b) = (1, 2) or a, b = 1, 2. MATLAB's nargout, which is often used in MATLAB to do optional work based on the number of returned values, does not exist in Julia. Instead, users can use optional and keyword arguments to achieve similar capabilities.
• Julia has true one-dimensional arrays. Column vectors are of size N, not Nx1. For example, rand(N) makes a 1-dimensional array.
• In Julia, [x,y,z] will always construct a 3-element array containing x, y and z.
• To concatenate in the first ("vertical") dimension use either vcat(x,y,z) or separate with semicolons ([x; y; z]).
• To concatenate in the second ("horizontal") dimension use either hcat(x,y,z) or separate with spaces ([x y z]).
• To construct block matrices (concatenating in the first two dimensions), use either hvcat or combine spaces and semicolons ([a b; c d]).
• In Julia, a:b and a:b:c construct AbstractRange objects. To construct a full vector like in MATLAB, use collect(a:b). Generally, there is no need to call collect though. An AbstractRange object will act like a normal array in most cases but is more efficient because it lazily computes its values. This pattern of creating specialized objects instead of full arrays is used frequently, and is also seen in functions such as range, or with iterators such as enumerate, and zip. The special objects can mostly be used as if they were normal arrays.
• Functions in Julia return values from their last expression or the return keyword instead of listing the names of variables to return in the function definition (see The return Keyword for details).
• A Julia script may contain any number of functions, and all definitions will be externally visible when the file is loaded. Function definitions can be loaded from files outside the current working directory.
• In Julia, reductions such as sum, prod, and max are performed over every element of an array when called with a single argument, as in sum(A), even if A has more than one dimension.
• In Julia, parentheses must be used to call a function with zero arguments, like in rand().
• Julia discourages the use of semicolons to end statements. The results of statements are not automatically printed (except at the interactive prompt), and lines of code do not need to end with semicolons. println or @printf can be used to print specific output.
• In Julia, if A and B are arrays, logical comparison operations like A == B do not return an array of booleans. Instead, use A .== B, and similarly for the other boolean operators like <, >.
• In Julia, the operators &, |, and (xor) perform the bitwise operations equivalent to and, or, and xor respectively in MATLAB, and have precedence similar to Python's bitwise operators (unlike C). They can operate on scalars or element-wise across arrays and can be used to combine logical arrays, but note the difference in order of operations: parentheses may be required (e.g., to select elements of A equal to 1 or 2 use (A .== 1) .| (A .== 2)).
• In Julia, the elements of a collection can be passed as arguments to a function using the splat operator ..., as in xs=[1,2]; f(xs...).
• Julia's svd returns singular values as a vector instead of as a dense diagonal matrix.
• In Julia, ... is not used to continue lines of code. Instead, incomplete expressions automatically continue onto the next line.
• In both Julia and MATLAB, the variable ans is set to the value of the last expression issued in an interactive session. In Julia, unlike MATLAB, ans is not set when Julia code is run in non-interactive mode.
• Julia's structs do not support dynamically adding fields at runtime, unlike MATLAB's classes. Instead, use a Dict.
• In Julia each module has its own global scope/namespace, whereas in MATLAB there is just one global scope.
• In MATLAB, an idiomatic way to remove unwanted values is to use logical indexing, like in the expression x(x>3) or in the statement x(x>3) = [] to modify x in-place. In contrast, Julia provides the higher order functions filter and filter!, allowing users to write filter(z->z>3, x) and filter!(z->z>3, x) as alternatives to the corresponding transliterations x[x.>3] and x = x[x.>3]. Using filter! reduces the use of temporary arrays.
• The analogue of extracting (or "dereferencing") all elements of a cell array, e.g. in vertcat(A{:}) in MATLAB, is written using the splat operator in Julia, e.g. as vcat(A...).
Noteworthy differences from R
One of Julia's goals is to provide an effective language for data analysis and statistical programming. For users coming to Julia from R, these are some noteworthy differences:
• Julia's single quotes enclose characters, not strings.
• Julia can create substrings by indexing into strings. In R, strings must be converted into character vectors before creating substrings.
• In Julia, like Python but unlike R, strings can be created with triple quotes """ ... """. This syntax is convenient for constructing strings that contain line breaks.
• In Julia, varargs are specified using the splat operator ..., which always follows the name of a specific variable, unlike R, for which ... can occur in isolation.
• In Julia, modulus is mod(a, b), not a %% b. % in Julia is the remainder operator.
• In Julia, not all data structures support logical indexing. Furthermore, logical indexing in Julia is supported only with vectors of length equal to the object being indexed. For example:
• In R, c(1, 2, 3, 4)[c(TRUE, FALSE)] is equivalent to c(1, 3).
• In R, c(1, 2, 3, 4)[c(TRUE, FALSE, TRUE, FALSE)] is equivalent to c(1, 3).
• In Julia, [1, 2, 3, 4][[true, false]] throws a BoundsError.
• In Julia, [1, 2, 3, 4][[true, false, true, false]] produces [1, 3].
• Like many languages, Julia does not always allow operations on vectors of different lengths, unlike R where the vectors only need to share a common index range. For example, c(1, 2, 3, 4) + c(1, 2) is valid R but the equivalent [1, 2, 3, 4] + [1, 2] will throw an error in Julia.
• Julia allows an optional trailing comma when that comma does not change the meaning of code. This can cause confusion among R users when indexing into arrays. For example, x[1,] in R would return the first row of a matrix; in Julia, however, the comma is ignored, so x[1,] == x[1], and will return the first element. To extract a row, be sure to use :, as in x[1,:].
• Julia's map takes the function first, then its arguments, unlike lapply(<structure>, function, ...) in R. Similarly Julia's equivalent of apply(X, MARGIN, FUN, ...) in R is mapslices where the function is the first argument.
• Multivariate apply in R, e.g. mapply(choose, 11:13, 1:3), can be written as broadcast(binomial, 11:13, 1:3) in Julia. Equivalently Julia offers a shorter dot syntax for vectorizing functions binomial.(11:13, 1:3).
• Julia uses end to denote the end of conditional blocks, like if, loop blocks, like while/ for, and functions. In lieu of the one-line if ( cond ) statement, Julia allows statements of the form if cond; statement; end, cond && statement and !cond || statement. Assignment statements in the latter two syntaxes must be explicitly wrapped in parentheses, e.g. cond && (x = value).
• In Julia, <-, <<- and -> are not assignment operators.
• Julia's -> creates an anonymous function.
• Julia constructs vectors using brackets. Julia's [1, 2, 3] is the equivalent of R's c(1, 2, 3).
• Julia's * operator can perform matrix multiplication, unlike in R. If A and B are matrices, then A * B denotes a matrix multiplication in Julia, equivalent to R's A %*% B. In R, this same notation would perform an element-wise (Hadamard) product. To get the element-wise multiplication operation, you need to write A .* B in Julia.
• Julia performs matrix transposition using the transpose function and conjugated transposition using the ' operator or the adjoint function. Julia's transpose(A) is therefore equivalent to R's t(A). Additionally a non-recursive transpose in Julia is provided by the permutedims function.
• Julia does not require parentheses when writing if statements or for/while loops: use for i in [1, 2, 3] instead of for (i in c(1, 2, 3)) and if i == 1 instead of if (i == 1).
• Julia does not treat the numbers 0 and 1 as Booleans. You cannot write if (1) in Julia, because if statements accept only booleans. Instead, you can write if true, if Bool(1), or if 1==1.
• Julia does not provide nrow and ncol. Instead, use size(M, 1) for nrow(M) and size(M, 2) for ncol(M).
• Julia is careful to distinguish scalars, vectors and matrices. In R, 1 and c(1) are the same. In Julia, they cannot be used interchangeably.
• Julia's diag and diagm are not like R's.
• Julia cannot assign to the results of function calls on the left hand side of an assignment operation: you cannot write diag(M) = fill(1, n).
• Julia discourages populating the main namespace with functions. Most statistical functionality for Julia is found in packages under the JuliaStats organization. For example:
• Julia provides tuples and real hash tables, but not R-style lists. When returning multiple items, you should typically use a tuple or a named tuple: instead of list(a = 1, b = 2), use (1, 2) or (a=1, b=2).
• Julia encourages users to write their own types, which are easier to use than S3 or S4 objects in R. Julia's multiple dispatch system means that table(x::TypeA) and table(x::TypeB) act like R's table.TypeA(x) and table.TypeB(x).
• In Julia, values are not copied when assigned or passed to a function. If a function modifies an array, the changes will be visible in the caller. This is very different from R and allows new functions to operate on large data structures much more efficiently.
• In Julia, vectors and matrices are concatenated using hcat, vcat and hvcat, not c, rbind and cbind like in R.
• In Julia, a range like a:b is not shorthand for a vector like in R, but is a specialized AbstractRange object that is used for iteration without high memory overhead. To convert a range into a vector, use collect(a:b).
• Julia's max and min are the equivalent of pmax and pmin respectively in R, but both arguments need to have the same dimensions. While maximum and minimum replace max and min in R, there are important differences.
• Julia's sum, prod, maximum, and minimum are different from their counterparts in R. They all accept an optional keyword argument dims, which indicates the dimensions, over which the operation is carried out. For instance, let A = [1 2; 3 4] in Julia and B <- rbind(c(1,2),c(3,4)) be the same matrix in R. Then sum(A) gives the same result as sum(B), but sum(A, dims=1) is a row vector containing the sum over each column and sum(A, dims=2) is a column vector containing the sum over each row. This contrasts to the behavior of R, where separate colSums(B) and rowSums(B) functions provide these functionalities. If the dims keyword argument is a vector, then it specifies all the dimensions over which the sum is performed, while retaining the dimensions of the summed array, e.g. sum(A, dims=(1,2)) == hcat(10). It should be noted that there is no error checking regarding the second argument.
• Julia has several functions that can mutate their arguments. For example, it has both sort and sort!.
• In R, performance requires vectorization. In Julia, almost the opposite is true: the best performing code is often achieved by using devectorized loops.
• Julia is eagerly evaluated and does not support R-style lazy evaluation. For most users, this means that there are very few unquoted expressions or column names.
• Julia does not support the NULL type. The closest equivalent is nothing, but it behaves like a scalar value rather than like a list. Use x === nothing instead of is.null(x).
• In Julia, missing values are represented by the missing object rather than by NA. Use ismissing(x) (or ismissing.(x) for element-wise operation on vectors) instead of is.na(x). The skipmissing function is generally used instead of na.rm=TRUE (though in some particular cases functions take a skipmissing argument).
• Julia lacks the equivalent of R's assign or get.
• In Julia, return does not require parentheses.
• In R, an idiomatic way to remove unwanted values is to use logical indexing, like in the expression x[x>3] or in the statement x = x[x>3] to modify x in-place. In contrast, Julia provides the higher order functions filter and filter!, allowing users to write filter(z->z>3, x) and filter!(z->z>3, x) as alternatives to the corresponding transliterations x[x.>3] and x = x[x.>3]. Using filter! reduces the use of temporary arrays.
Noteworthy differences from Python
• Julia requires end to end a block. Unlike Python, Julia has no pass keyword.
• In Julia, indexing of arrays, strings, etc. is 1-based not 0-based.
• Julia's slice indexing includes the last element, unlike in Python. a[2:3] in Julia is a[1:3] in Python.
• Julia does not support negative indices. In particular, the last element of a list or array is indexed with end in Julia, not -1 as in Python.
• Julia's for, if, while, etc. blocks are terminated by the end keyword. Indentation level is not significant as it is in Python.
• Julia has no line continuation syntax: if, at the end of a line, the input so far is a complete expression, it is considered done; otherwise the input continues. One way to force an expression to continue is to wrap it in parentheses.
• Julia arrays are column major (Fortran ordered) whereas NumPy arrays are row major (C-ordered) by default. To get optimal performance when looping over arrays, the order of the loops should be reversed in Julia relative to NumPy (see relevant section of Performance Tips).
• Julia's updating operators (e.g. +=, -=, ...) are not in-place whereas NumPy's are. This means A = [1, 1]; B = A; B += [3, 3] doesn't change values in A, it rather rebinds the name B to the result of the right-hand side B = B + 3, which is a new array. For in-place operation, use B .+= 3 (see also dot operators), explicit loops, or InplaceOps.jl.
• Julia evaluates default values of function arguments every time the method is invoked, unlike in Python where the default values are evaluated only once when the function is defined. For example, the function f(x=rand()) = x returns a new random number every time it is invoked without argument. On the other hand, the function g(x=[1,2]) = push!(x,3) returns [1,2,3] every time it is called as g().
• In Julia % is the remainder operator, whereas in Python it is the modulus.
Noteworthy differences from C/C++
• Julia arrays are indexed with square brackets, and can have more than one dimension A[i,j]. This syntax is not just syntactic sugar for a reference to a pointer or address as in C/C++. See the Julia documentation for the syntax for array construction (it has changed between versions).
• In Julia, indexing of arrays, strings, etc. is 1-based not 0-based.
• Julia arrays are not copied when assigned to another variable. After A = B, changing elements of B will modify A as well. Updating operators like += do not operate in-place, they are equivalent to A = A + B which rebinds the left-hand side to the result of the right-hand side expression.
• Julia arrays are column major (Fortran ordered) whereas C/C++ arrays are row major ordered by default. To get optimal performance when looping over arrays, the order of the loops should be reversed in Julia relative to C/C++ (see relevant section of Performance Tips).
• Julia values are not copied when assigned or passed to a function. If a function modifies an array, the changes will be visible in the caller.
• In Julia, whitespace is significant, unlike C/C++, so care must be taken when adding/removing whitespace from a Julia program.
• In Julia, literal numbers without a decimal point (such as 42) create signed integers, of type Int, but literals too large to fit in the machine word size will automatically be promoted to a larger size type, such as Int64 (if Int is Int32), Int128, or the arbitrarily large BigInt type. There are no numeric literal suffixes, such as L, LL, U, UL, ULL to indicate unsigned and/or signed vs. unsigned. Decimal literals are always signed, and hexadecimal literals (which start with 0x like C/C++), are unsigned. Hexadecimal literals also, unlike C/C++/Java and unlike decimal literals in Julia, have a type based on the length of the literal, including leading 0s. For example, 0x0 and 0x00 have type UInt8, 0x000 and 0x0000 have type UInt16, then literals with 5 to 8 hex digits have type UInt32, 9 to 16 hex digits type UInt64 and 17 to 32 hex digits type UInt128. This needs to be taken into account when defining hexadecimal masks, for example ~0xf == 0xf0 is very different from ~0x000f == 0xfff0. 64 bit Float64 and 32 bit Float32 bit literals are expressed as 1.0 and 1.0f0 respectively. Floating point literals are rounded (and not promoted to the BigFloat type) if they can not be exactly represented. Floating point literals are closer in behavior to C/C++. Octal (prefixed with 0o) and binary (prefixed with 0b) literals are also treated as unsigned.
• String literals can be delimited with either " or """, """ delimited literals can contain " characters without quoting it like "\"". String literals can have values of other variables or expressions interpolated into them, indicated by \$variablename or \$(expression), which evaluates the variable name or the expression in the context of the function.
• // indicates a Rational number, and not a single-line comment (which is # in Julia)
• #= indicates the start of a multiline comment, and =# ends it.
• Functions in Julia return values from their last expression(s) or the return keyword. Multiple values can be returned from functions and assigned as tuples, e.g. (a, b) = myfunction() or a, b = myfunction(), instead of having to pass pointers to values as one would have to do in C/C++ (i.e. a = myfunction(&b).
• Julia does not require the use of semicolons to end statements. The results of expressions are not automatically printed (except at the interactive prompt, i.e. the REPL), and lines of code do not need to end with semicolons. println or @printf can be used to print specific output. In the REPL, ; can be used to suppress output. ; also has a different meaning within [ ], something to watch out for. ; can be used to separate expressions on a single line, but are not strictly necessary in many cases, and are more an aid to readability.
• In Julia, the operator (xor) performs the bitwise XOR operation, i.e. ^ in C/C++. Also, the bitwise operators do not have the same precedence as C/++, so parenthesis may be required.
• Julia's ^ is exponentiation (pow), not bitwise XOR as in C/C++ (use , or xor, in Julia)
• Julia has two right-shift operators, >> and >>>. >>> performs an arithmetic shift, >> always performs a logical shift, unlike C/C++, where the meaning of >> depends on the type of the value being shifted.
• Julia's -> creates an anonymous function, it does not access a member via a pointer.
• Julia does not require parentheses when writing if statements or for/while loops: use for i in [1, 2, 3] instead of for (int i=1; i <= 3; i++) and if i == 1 instead of if (i == 1).
• Julia does not treat the numbers 0 and 1 as Booleans. You cannot write if (1) in Julia, because if statements accept only booleans. Instead, you can write if true, if Bool(1), or if 1==1.
• Julia uses end to denote the end of conditional blocks, like if, loop blocks, like while/ for, and functions. In lieu of the one-line if ( cond ) statement, Julia allows statements of the form if cond; statement; end, cond && statement and !cond || statement. Assignment statements in the latter two syntaxes must be explicitly wrapped in parentheses, e.g. cond && (x = value), because of the operator precedence.
• Julia has no line continuation syntax: if, at the end of a line, the input so far is a complete expression, it is considered done; otherwise the input continues. One way to force an expression to continue is to wrap it in parentheses.
• Julia macros operate on parsed expressions, rather than the text of the program, which allows them to perform sophisticated transformations of Julia code. Macro names start with the @ character, and have both a function-like syntax, @mymacro(arg1, arg2, arg3), and a statement-like syntax, @mymacro arg1 arg2 arg3. The forms are interchangeable; the function-like form is particularly useful if the macro appears within another expression, and is often clearest. The statement-like form is often used to annotate blocks, as in the distributed for construct: @distributed for i in 1:n; #= body =#; end. Where the end of the macro construct may be unclear, use the function-like form.
• Julia has an enumeration type, expressed using the macro @enum(name, value1, value2, ...) For example: @enum(Fruit, banana=1, apple, pear)
• By convention, functions that modify their arguments have a ! at the end of the name, for example push!.
• In C++, by default, you have static dispatch, i.e. you need to annotate a function as virtual, in order to have dynamic dispatch. On the other hand, in Julia every method is "virtual" (although it's more general than that since methods are dispatched on every argument type, not only this, using the most-specific-declaration rule). | 2.8125 | CC-MAIN-2019-13 |
Chemistry mass problems
HOME > Chemistry > Chemistry mass problems
# Chemistry mass problems
[From: ] [author: ] [Date: 11-04-26] [Hit: ]
b) what mass of oxygen will be consumed?c) what mass of sulfur dioxide will be produced?And finally the last one I cant get, What mass of potassium chlorate would be required to produce 45 g of oxygen?Thanks a lot for any help, out of like 11 questions,......
Your Open QuestionShow me another »
Chemistry problems for mass relationships?
Hello, I have a lot of questions that need to be done for homework, but so far, there are a couple I can't do, and I need help please.
How many grams of chlorine could be obtained by heating 70 g of manganese dioxide with an excess of hydrochloric acid?
MnO2+4HCl --> MnCl2 + 2H20 + Cl2
The action of dilute nitric acid on copper is represented by the equation: 3Cu +8HNO3 --> 3Cu(NO3)2 + 4H2O + 2NO
What mass of copper would be required to react completely with 250 g of nitric acid? (for this part of the question I got it right) The mass turned out to 94.6 gCu.
But I can't get the second part, What mass of cupric nitrate would be formed?
For this question I don't know how to do etheir: The burning of iron pyrites (FeS2) in air may be expressed by the following equation:
4FeS2 +11O2 --> 2Fe2O3 + 8SO2
a) what mass of iron pyrites must be burned in air to produce 400 g of ferric oxide?
b) what mass of oxygen will be consumed?
c) what mass of sulfur dioxide will be produced?
And finally the last one I can't get, What mass of potassium chlorate would be required to produce 45 g of oxygen?
Thanks a lot for any help, out of like 11 questions, I can't get these 4. I have a test the day after tommorow and need to understand how to do them. THANKS!
-
How many grams of chlorine could be obtained by heating 70 g of manganese dioxide with an excess of hydrochloric acid?
MnO2+4HCl --> MnCl2 + 2H20 + Cl2
1. Since HCl is the excess reagent (ER), MNO2 is the limiting reagent (LR). You will need the limiting reagent as a basis for the next step.
2. Get Molar Mass of MnO2
2. Convert 70 g of MnO2 into mol.
3. Use stoichiometric ratio to get the the amount of Cl2 in mol.
4. Get Molar Mass of Cl2
5. Convert the amount of Cl2 from mol to g.
Molar Mass of MnO2 = 54.94 + (2 x 16) = 86.94 g of MnO2
70 g of MnO2 x (1 mol of MnO2 / Molar Mass of MnO2) = 0.8 mol of MnO2
0.8 mol of MnO2 x (1 mol of Cl2 / 1 mol of MnO2) = 0.8 mol of Cl2
Molar Mass of Cl2 = (2 x 35.45) = 70.9 g of Cl2
0.8 mol of Cl2 x (Molar Mass of Cl2 / 1 mol of Cl2) = 56.72 g of Cl2
or you can do the shortcut altogether:
70g MnO2 x [1 mol MnO2 / 54.94 + (2x16)](1 mol Cl2 / 1 mol MnO2)[(35.45x2) g Cl2 / 1 mol Cl2]
= 57.08 g of Cl2
1
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Frequent Visitor
## Why does this measure work?
This is more of a "Why does this work" question rather than a "how do I" question.
Here is some context of the scenario using dummy data:
I have a fact table that looks like this:
Each column ID points to its respective dimension table
Within those dimension tables, the corresponding ID determines a value. For example the Risk Score Value column in the Risk Score dimension (Relationship = Risk Score_ID -> AnalysisCodeValueID)
I have created two measures:
1. Risk Score SUM = CALCULATE(SUM('Risk Score'[Risk Score]))
2. Total Risk Score = CALCULATE(SUMX('Analysis Codes',RELATED('Risk Score'[Risk Score])))
When I bring these elements into a table visual why does the Measure: "Risk Score SUMXRelated" return the correct sum but the Risk Score SUM measure does not?
Any help would be appreciated
1 ACCEPTED SOLUTION
Solution Specialist
Hi,
I see the issue here, in AnalysisCodes Table - you have two rows but only one row in Risk Score for each for RiskScoreID. When you calculate the Risk Score Measure in Risk Score, it's taking consideration of only one row. However, you are already getting correct data from
Total Risk Score as the relationship between the two are met. Hence, the measure you are creating for the Risk Score Measure based on the Risk Score table will never match with the other measure if you are not including relationship between the AnalysisCodes table.
6 REPLIES 6
Solution Specialist
Would you able to share the pbix file with dummy data?
Frequent Visitor
Solution Specialist
Hi,
The Risk Score SUM measure you’ve created uses the SUM function within a CALCULATE statement. This measure sums up the ‘Risk Score’[Risk Score] column, but it doesn’t iterate over each row of your ‘Analysis Codes’ table. As a result, it might not consider the row context properly when calculating totals, especially if there are filters or slicers affecting the visual.
On the other hand, the Total Risk Score measure uses SUMX with a RELATED function, which iterates over each row in the ‘Analysis Codes’ table and sums the related ‘Risk Score’[Risk Score] for each row. This row-by-row operation ensures that the context is considered for each individual calculation, leading to a correct total.
To fix the issue with the Risk Score SUM measure, you might need to adjust it to ensure it respects the row context. One way to do this is by using SUMX over a virtual table that includes all the necessary rows and relationships. Here’s an example of how you might modify your measure:
Risk Score SUM Corrected =
SUMX(
RELATEDTABLE('Risk Score'),
'Risk Score'[Risk Score]
)
Let me know the result.
Frequent Visitor
Hi Kaviraj,
Thanks for your response, unfortunately the right number still wasn't shown when the new dax was used:
I think I see the problem:
I have replicated the slicers in the data view. A RiskScore_ID of 17 = 5 and a RiskScore_ID of 317 = 15.
If I have two rows of each then the sum should total 40. However, it's summing the distinct values instead. (One row of 5 and one row of 15)
Solution Specialist
Hi,
I see the issue here, in AnalysisCodes Table - you have two rows but only one row in Risk Score for each for RiskScoreID. When you calculate the Risk Score Measure in Risk Score, it's taking consideration of only one row. However, you are already getting correct data from
Total Risk Score as the relationship between the two are met. Hence, the measure you are creating for the Risk Score Measure based on the Risk Score table will never match with the other measure if you are not including relationship between the AnalysisCodes table.
Frequent Visitor
Ah I see, that makes sense. Thanks a lot
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# How to take the average of a list in Python
Educative Answers Team
The goal here is to find the average/mean of a list of numbers. The average is the sum of all the numbers in a list divided by its length.
## Algorithms
Let’s have a look at a few of the algorithms used to compute the average for a list of numbers.
### 1. Using sum() and len() functions
An average can be computed using the sum() and len() functions on the list. sum() will return the sum of all the values in the list, which can be divided by the number of elements returned by the len() function. Take a look at the code below:
def Average(l):
avg = sum(l) / len(l)
return avg
my_list = [2,4,6,8,10]
average = Average(my_list)
print("Average of my_list is", average)
### 2. Using the mean() function
The mean() function in the python statistics library can be used to directly compute the average of a list. Take a look at the code below:
from statistics import mean
def Average(l):
avg = mean(l)
return avg
my_list = [2,4,6,8,10]
average = Average(my_list)
print "Average of my_list is", average
The statistics library needs to be installed in order to use the mean() function.
### 3. Using reduce() and lambda functions
The reduce method can be used to loop through the list and the sum can be computed in the lambda function. The number of elements can be obtained using len().
from functools import reduce
def Average(l):
avg = reduce(lambda x, y: x + y, l) / len(l)
return avg
my_list = [2,4,6,8,10]
average = Average(my_list)
print "Average of my_list is", average
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# Oil Deposits
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17683 Accepted Submission(s): 10172
Problem Description
The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.
Input
The input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*’, representing the absence of oil, or `@’, representing an oil pocket.
Output
For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.
Sample Input
1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5
****@
*@@*@
*@**@
@@@*@
@@**@
0 0
Sample Output
0
1
2
2
Source
## 作者： CrazyKK
ex-ACMer@hust，stackoverflow-engineer@sensetime | 2.546875 | CC-MAIN-2018-51 |
## closed form expression for Rényi entropy for multivariate Gaussian distributions
Is there any closed form expression for Rényi entropy of a set variables with multivariate Gaussian distribution?
-
Yes. First, do a change of variable in the integral to convert it to the Renyi entropy of a set of uncorrelated Gaussians with standard deviation $1$. The integral now splits into a product of 1-dimensional integrals, where each one is the Renyi entropy of a 1-dimensional Gaussian.
If $X$ is $\mathcal{N}(\mu, K)$ random vector then $K$ can be written as $K=U\Lambda U^T$ where $\Lambda$ is a diagonal matrix with the eigen values of $K$ and $U$ is orthonormal. If we define $Y=U^T(X-\mu)$, then $Y$ is a Gaussian vector of independent random variables with mean $0$ and covariance matrix $\Lambda$. Do you mean to say that Renyi entropy of $X$ is equal to The Renyi entropy of $Y$? – Ashok Sep 29 2011 at 5:12 Yes. Just use the definition of Renyi entropy and do the change of variables. – Deane Yang Sep 29 2011 at 14:33 | 2.640625 | CC-MAIN-2013-20 |
# Top 10 Java Programs for Freshers
August 20, 2014
Welcome to GUVI Blogs. This is Prashanth, a Java and Selenium Expert. Here are my list of 10 Java coding interview questions and answers, which is good to prepare before appearing on any Java interviews. As I said Java coding questions are mostly based on programming, logical analysis and problem solving skill , so better to get it right in first place. Any way you may be able to solve and find answers of these Java coding questions by yourself, but if you stuck do a google, and you can get many alternative ways to solve these problem. Some times knowing more than one way to solve any programming question or coding problem in Java also helps to impress interviewer. This list mainly contains basic programs asked on Interviews.
Verify a number is Even/Odd
Output
Swapping Numbers without using 3rd variable
Output
Factorial of a number
Output
How to get the prime numbers between a given range.
Output
Check a number is prime or not. Note- A number is prime if it is not divisible by any other number except itself.
Output
Check if a number is Armstrong or not. Note- A number is armstrong if the sum of the cubes of digit of number is equal to the number. ex- 407 = 4*4*4 + 0*0*0 + 7*7*7
Output
Floyd Triangle Note- Floyd Triangle is like 1 2 3 4 5 6 7 8 9 10 ———— Code-
Output
Palindrome of String or reverse a String.
Output
Binary Search
Bubble Sort
## 10 thoughts on “Top 10 Java Programs for Freshers”
1. simhachalam
import java.lang.*;
import java.util.*;
class Palindrome
{
public static void main(String args[])
{
Scanner s1=new Scanner(System.in);
System.out.println(“enter rows”);
int n=s1.nextInt();
int c=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<=i;j++)
{
System.out.print(++c);
}
System.out.println();
}
}
} | 2.640625 | CC-MAIN-2018-51 |
## Savings Bonds 101: How Savings Bonds Work
### Introduction
Savings bonds have long been a popular investment option for those seeking a safe and reliable way to grow their wealth.
Pexels
According to the U.S. Department of the Treasury, as of December 2021, there were over \$200 billion in savings bonds outstanding. But for many, the question remains: how do savings bonds work? In this article, we’ll delve into the intricacies of savings bonds, providing a clear and concise understanding of this investment vehicle.
### Understanding Savings Bonds
At their core, savings bonds are debt securities issued by the U.S. government to finance its borrowing needs. When you purchase a savings bond, you are essentially lending money to the government, which promises to pay you back with interest over a specified period. There are two primary types of savings bonds: Series EE and Series I.
Series EE Savings Bonds
Series EE savings bonds are the more traditional of the two. They are purchased at half their face value, meaning a \$100 bond costs \$50. These bonds earn a fixed rate of interest, which is determined at the time of purchase and remains the same throughout the bond’s life. As of January 2022, the fixed rate for Series EE bonds was 0.10% for bonds issued between November 2021 and April 2022.
For those wondering “how do savings bonds work for dummies,” it’s crucial to understand that Series EE bonds have a maturity period of 30 years. However, if redeemed within the first five years, they can be redeemed after one year, with a penalty of three months’ interest. If held until maturity, Series EE bonds are guaranteed to double in value, providing a secure return for patient investors.
To illustrate the potential of Series EE bonds, consider a hypothetical scenario. Imagine a grandparent, Robert, who wants to invest in his granddaughter’s future education. He decides to purchase a \$1,000 Series EE bond for \$500 when his granddaughter, Emily, is born. Assuming a fixed interest rate of 0.10%, after 30 years, the bond will be worth at least \$1,000. While this may not cover the entire cost of Emily’s education, it provides a solid foundation for her future.
Chinese philosopher Confucius, who lived from 551 to 479 BC, once said, “The man who moves a mountain begins by carrying away small stones.” This wisdom can be applied to investing in Series EE bonds. One can gradually build a substantial portfolio by consistently investing small amounts over time. Regular investments, even in modest amounts, can grow significantly over the long term, thanks to the power of compound interest. This approach to investing in Series EE bonds exemplifies the value of patience, perseverance, and a long-term perspective.
Series I Savings Bonds
Series I savings bonds, introduced in 1998, offer a unique blend of fixed and variable interest rates. The fixed rate remains constant throughout the life of the bond, while the variable rate changes twice a year based on inflation. This combination helps protect investors from the erosive effects of inflation. As of January 2022, the composite rate for Series I bonds was 7.12%, with a fixed rate of 0.00% and an inflation rate of 7.12%.
For those wanting a “For Dummies" guide to how savings bonds work, it’s essential to understand that Series I bonds are designed to provide a return that keeps pace with inflation. The variable rate is based on the Consumer Price Index for Urban Consumers (CPI-U), which measures the average change in prices paid by urban consumers for goods and services. When the CPI-U increases, the variable rate on Series I bonds also rises, ensuring that the purchasing power of the investor’s money remains stable over time.
To illustrate the potential benefits of Series I bonds, consider a hypothetical scenario. Imagine an investor, Sarah, who purchased a \$10,000 Series I bond in January 2010 when the composite rate was 3.36%. By January 2022, Sarah’s bond would have grown to approximately \$14,887, reflecting a compound annual growth rate of 3.36% over the 12-year period. This growth would have helped Sarah’s investment keep pace with inflation, preserving her purchasing power.
Legendary trader Jesse Livermore, who made and lost millions during the early 20th century, once quipped, “The market is never wrong, but opinions often are.” This sentiment underscores the importance of understanding market dynamics and not getting swayed by popular opinion. With their inflation-adjusted returns, Series I bonds offer a way to navigate market uncertainties. By investing in these bonds, individuals can protect their wealth from the erosive effects of inflation, regardless of market fluctuations or popular investment trends.
### Purchasing and Redeeming Savings Bonds
Savings bonds can be purchased directly from the U.S. Treasury through their website, TreasuryDirect.gov. They are also available through many banks and financial institutions. Bonds can be held in electronic or paper form, although paper bonds are no longer being issued as of January 2012.
When it comes to redeeming savings bonds, there are a few key points to keep in mind. Series EE and I bonds must be held for at least one year before they can be redeemed. If redeemed within the first five years, there is a penalty of three months’ interest. Bonds earn interest for up to 30 years, after which they stop earning interest and should be redeemed.
Philosopher Immanuel Kant, who lived from 1724 to 1804, famously said, “Science is organized knowledge. Wisdom is organized life.” In the context of savings bonds, this reminds us that while understanding the technical aspects of these investments is important, it’s equally crucial to consider how they fit into our overall financial plan and life goals.
### Market Psychology and Savings Bonds
Understanding market psychology can be valuable when considering savings bonds. The bandwagon effect, where people tend to follow the crowd, can lead to irrational exuberance or pessimism in the markets. However, with their fixed returns and government backing, savings bonds offer a way to avoid getting caught up in market hype.
Legendary investor Warren Buffett, known for his contrarian approach, has long advocated for a long-term, value-oriented investing strategy. In his 2013 letter to Berkshire Hathaway shareholders, Buffett wrote, “The stock market is a device for transferring money from the impatient to the patient.” This wisdom can be applied to savings bonds, which reward patience and a long-term perspective.
Technical analysis, which involves studying past market data to identify trends and make predictions, can also be useful when considering savings bonds. By analyzing historical interest rates and inflation data, investors can decide when to purchase bonds and how long to hold them.
Real-World Examples: To illustrate the potential benefits of savings bonds, consider the following real-world example. Imagine a couple, John and Sarah, who decided to invest \$5,000 in Series EE savings bonds for their newborn daughter’s college education. Assuming a fixed interest rate of 0.10%, after 18 years, their initial investment would have grown to approximately \$5,090. While this may not seem like a substantial return, it’s important to remember that savings bonds offer a safe and reliable way to save for the future.
Another example involves a retiree, Michael, who allocated a portion of his portfolio to Series I savings bonds to help protect against inflation. With a composite rate of 7.12%, Michael’s investment would keep pace with rising prices, ensuring that his purchasing power remains stable.
### Conclusion
Savings bonds may not be the most exciting investment option, but they can be an excellent choice for those seeking a safe and reliable way to grow their wealth. By understanding how savings bonds work, investors can make informed decisions about when and how to incorporate them into their financial plans.
As the ancient Greek philosopher Heraclitus once said, “The only constant in life is change.” In a world of economic uncertainty and market volatility, savings bonds offer a beacon of stability. Investors can confidently navigate the ever-changing financial landscape by taking a long-term perspective and focusing on the fundamentals.
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Math Calculators, Lessons and Formulas
It is time to solve your math problem
You are here:
# Calculators :: Other :: "Work" Problems Calculator
This calculators will solve three types of 'work' word problems. Also, it will provide a detailed explanation.
## Two Workers Problem
Complete the sentence by entering a positive numbers (integers, fractions or decimals) into text boxes. Important: The form will NOT let you enter wrong characters ( - , y, x, p, ...)
First worker can do the job in and the second worker takes . How long would it take the two workers together to finish the job?
## "How many workers" problem
Complete the sentence by entering a positive numbers (integers, fractions or decimals) into text boxes. Important: The form will NOT let you enter wrong characters ( - , y, x, p, ...)
If men can do a job in days, how many would it take to do the job in days?
## The pipe problem.
Complete the sentence by entering a positive numbers (integers, fractions or decimals) into text boxes. Note: Leave third text box empty, if outlet pipe doesn't exist. Important: The form will NOT let you enter wrong characters ( - , y, x, p, ...)
A swimming pool has 2 inlet pipes. One fills the pool in hours, the other in hours. The outlet pipe empties the pool in hours. If the pool is initially empty and all three pipes are open, how many hours will it take to fill up the pool?
Quick Calculator Search | 3.1875 | CC-MAIN-2017-39 |
## ››Convert newton/square metre to centipascal
newton/square meter centipascal
How many newton/square meter in 1 centipascal? The answer is 0.01.
We assume you are converting between newton/square metre and centipascal.
You can view more details on each measurement unit:
newton/square meter or centipascal
The SI derived unit for pressure is the pascal.
1 pascal is equal to 1 newton/square meter, or 100 centipascal.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between newtons/square meter and centipascals.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of newton/square meter to centipascal
1 newton/square meter to centipascal = 100 centipascal
2 newton/square meter to centipascal = 200 centipascal
3 newton/square meter to centipascal = 300 centipascal
4 newton/square meter to centipascal = 400 centipascal
5 newton/square meter to centipascal = 500 centipascal
6 newton/square meter to centipascal = 600 centipascal
7 newton/square meter to centipascal = 700 centipascal
8 newton/square meter to centipascal = 800 centipascal
9 newton/square meter to centipascal = 900 centipascal
10 newton/square meter to centipascal = 1000 centipascal
## ››Want other units?
You can do the reverse unit conversion from centipascal to newton/square meter, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Centipascal
The SI prefix "centi" represents a factor of 10-2, or in exponential notation, 1E-2.
So 1 centipascal = 10-2 pascals.
The definition of a pascal is as follows:
The pascal (symbol Pa) is the SI unit of pressure.It is equivalent to one newton per square metre. The unit is named after Blaise Pascal, the eminent French mathematician, physicist and philosopher.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 3.40625 | CC-MAIN-2022-21 |
JackJobs - 3 years ago 345
C Question
# quic sort char array in c
I copy a int array quick sort algorithmn from http://www.comp.dit.ie/rlawlor/Alg_DS/sorting/quickSort.c
Then change it to char array quick sort algorithmn.
Here is the code
``````extern void quickSortChar(char a[], int left, int right)
{
int j;
if (left < right)
{
j = partition(a, left, right);
quickSort(a, left, j - 1);
quickSort(a, j + 1, right);
}
}
static int partition(char a[], int left, int right)
{
int pivot, i, j, t;
pivot = a[left];
i = left; j = right + 1;
while (1)
{
do i++; while (a[i] <= pivot && i <= right);
do j--; while (a[j] > pivot);
if (i >= j) break;
t = a[i]; a[i] = a[j]; a[j] = t;
}
t = a[left]; a[left] = a[j]; a[j] = t;
return j;
}
void main()
{
char arr[] = "bacgd";
puts(arr);
quickSortChar(arr, 0, strlen(arr) - 1);
puts(arr);
system("pause");
}
``````
I expecte the output to be: abcdg, but output is: abcgd
You're calling `quickSort()` recursively (wrong function name) inside `quickSortChar()`.
``````\$ ./program | 2.546875 | CC-MAIN-2021-21 |
# 5dimes NCAAF Odds Tuesday March 13, 2018
Published on March 13, 2018 Updated on March 13th, 2018
5Dimes Football College Odds Updated 3/13/2018 6:19 PM
teamspreadmoneylinetotal
away / homepointsoddsedgeoddsedgepointsoddsedge
Rotation Number - Rotation Numbers are a shorthand used by sportsbooks to refer to a side or team in a game. They are standardized throughout North American sportsbooks and casinos. Also, the rotation number allows each book to list the games in similar order and format. It keeps all of the games that are posted each day and throughout the week organized. That makes it easy for the bettor and the book, particularly in noisy casinos, to efficiently place a bet.
Spread/Puckline/Runline - Spreads are the number of points/goals/runs to add or subtract from the final score to determine if the bet is won. Spreads help give the bettor a different look at the odds and encourage betting on longshot teams or lower the chalk on big favorites. here are videos/articles explaining Spreads, Runline and Puckline.
Implied Probability - Implied Probability is best explained as odds of any type converted to percent likelihood. It can give the bettor a general idea the chance of the outcome occurring. There is an important caveat: Implied Probability always contains the sportsbooks edge, sometimes called their profit. Implied Probabilities for a single match always add up to more than 100%. The extra percent is the edge, as explained below. Here is a good article on Implied Probability and a useful Implied Probability Calculator.
Sportsbook Edge - Edge represents the percent advantage that the sportsbook has over the bettor. A lower Sportsbook Edge is always better for gamblers. The best way to think about it, is that if you flip coins but the person you are playing against has a 2.5% advantage you are going to lose to them over time. If that percentage is 4.75% you are going to lose much sooner.
Totals or Over/Under - Totals betting is not specific to a particular team. The sportsbook sets the total number of points it expects to be scored in any given match and bettors choose to bet that the score is going to be over or under this mark. The most common over/under market is total combined points for both teams, but sometimes can be a total points scored for a single team or side. Here is a video/article that explains Over Under or Totals Betting | 3.203125 | CC-MAIN-2019-18 |
# Math
posted by .
I cannot figure this out I keep getting the order pair(1.24,-3.44), but it does not work when I check. Can you help me?
5x+5y=-11
7x-3y=19
The other oneI needed to solve is this Question: 4y+3y<-35
I got -5>-35...is that right?
• Math -
the first one does check.
7y<-35
y<-5
• Mathquick??? -
Are you sure that the sigh does not change In the second problem? I thought that when you divide by neg then the signs switch?
My instructor told me that the first one was wrong, so I kept going over it and over it, and she insists that it is wrong, unless she wants me to use fractions, but an answer in a decimal form is still considered an integer right?
• Math -
I agree with bobpursley, your first equation solution is correct. It should not matter if you answer as a decimal or fraction of the decimal is exact, which it is.
Yes, you do reverse the inequality sign if you multiply or divide by a negative, BUT...
you did not divide by a negative, you divided both sides by +7, so it stays the same.
## Similar Questions
1. ### Algebra
- ok... we keep getting this one wrong: Solve the following equation for x. Write your answer as a fraction in simplest form. -6(x-5)=-2x-8-6(3x-5) What is it you don't understand about this?
(x)/(x-2) - (x+1)/(x) = (8)/(x^2-2x) I keep getting x = 6 can someone show me how you guys are getting x = -10 The lowest commond denominator is x(x-2) x/x *(x)/(x-2) - (x+1)/(x) *(x-2)/(x-2)=(8)/x(x-2) multiplying both sides by the …
4. ### college math
solve: (3/x)/1=(7/x-20) i keep getting different answers for this question. the last answer i got was 20 (which would make it not a solution) but im not sure if that is right
5. ### Chemistry
One other question that I cannot figure out what I'm doing wrong is this.. If .002 L of .650 M NaOH are added to 1.0 L of 0.9 M CaCl2 what is the Q value?
6. ### algebra
well i'm stumped. i can't figure out how to solve this. solve for (x,y): 3/x + 2/y = 0 5/x - 1/y = 13 i can't graph it, and i can't use a calculator. i keep getting stuck on the second equation b/c i get 5y -13xy -x=0 and since 13 …
7. ### MATH PLZ HELP
well i'm stumped. i can't figure out how to solve this. solve for (x,y): 3/x + 2/y = 0 5/x - 1/y = 13 i can't graph it, and i can't use a calculator. i keep getting stuck on the second equation b/c i get 5y -13xy -x=0 and since 13 …
8. ### algebra -
#21 -can you help me solve this equation: 3x^2 (2x - 2) i keep getting: 6x ^2 -6 #22 can you check this one: (2x ^2) (6x^2) i got 12x #23 can you check this one: (4m^2 + m -2) + (-m^2 - 3m) i got: -4m ^2 + 4m-2 #25 can you check this …
9. ### Math
How do I work this out? Corporate bonds issued by Johnson Corporation currently yield 11%. Municipal bonds of equal risk currently yield 6.5%. At what tax rate would an investor be indifferent between these two bonds?
10. ### math
The center of a circle is located at (–5, 2) and a point on the circle is located at (5, -22). Which other points are also on the circle?
More Similar Questions | 3.921875 | CC-MAIN-2018-05 |
NumWords.com
# How to write One thousand seven hundred ninety-seven in numbers in English?
We can write One thousand seven hundred ninety-seven equal to 1797 in numbers in English
< One thousand seven hundred ninety-six :||: One thousand seven hundred ninety-eight >
Three thousand five hundred ninety-four = 3594 = 1797 × 2
Five thousand three hundred ninety-one = 5391 = 1797 × 3
Seven thousand one hundred eighty-eight = 7188 = 1797 × 4
Eight thousand nine hundred eighty-five = 8985 = 1797 × 5
Ten thousand seven hundred eighty-two = 10782 = 1797 × 6
Twelve thousand five hundred seventy-nine = 12579 = 1797 × 7
Fourteen thousand three hundred seventy-six = 14376 = 1797 × 8
Sixteen thousand one hundred seventy-three = 16173 = 1797 × 9
Seventeen thousand nine hundred seventy = 17970 = 1797 × 10
Nineteen thousand seven hundred sixty-seven = 19767 = 1797 × 11
Twenty-one thousand five hundred sixty-four = 21564 = 1797 × 12
Twenty-three thousand three hundred sixty-one = 23361 = 1797 × 13
Twenty-five thousand one hundred fifty-eight = 25158 = 1797 × 14
Twenty-six thousand nine hundred fifty-five = 26955 = 1797 × 15
Twenty-eight thousand seven hundred fifty-two = 28752 = 1797 × 16
Thirty thousand five hundred forty-nine = 30549 = 1797 × 17
Thirty-two thousand three hundred forty-six = 32346 = 1797 × 18
Thirty-four thousand one hundred forty-three = 34143 = 1797 × 19
Thirty-five thousand nine hundred forty = 35940 = 1797 × 20
Sitemap | 2.921875 | CC-MAIN-2019-26 |
# Simple Roulette Betting Method
#### Ping
##### Dormant account
Got this from another forum.... what do you think about this?
Pick any two Dozens and place one unit on all the numbers of a same color (12 numbers). Now, place a 12-unit outside bet on the opposite color.
Ex: with a \$1 unit, cover all red numbers within the first two Dozens (\$12 total). Now place a \$12 bet on Black.
If a black number shows up, you will get your bets back.
If a red number from your two Dozens shows up, you make \$12 in profit.
If a red number from the other Dozen shows up, you lose all your bets.
Without changing anything to the position of your bets, keep placing the same bets over and over again. Now see how often you will win and how often you will lose.
You may want to cover the zero(s) with one or half a unit.
Ping
#### chalupa
##### Experienced Member
It sounds like a bad idea to me.
Keeping in mind that Roulette is a high casino edge game, why would you want to play any strategy where half the time you only break even on your bet -- and the most you can win is half your bet?
If you're betting big for small wins, the casino is getting maximum advantage of their 2.7% edge with little risk on their part. It is going to be difficult for you to put together a string of significant wins that hurts them.
Essentially what the strategy is doing is covering 12 red numbers and 18 black numbers with slightly varying amounts. That's a whopping 30 out of 37 numbers. Or 31 out of 37 if you stick something on the 0 too.
Take it to a slightly further extreme and you'll see the fallacy. Cover 36 numbers with \$1 each. With each spin you either break even or lose \$36. I'll cover that action for you.
#### razormedia
##### Dormant account
tried it and ended up down by a few k.
My favourite tactic is to simply choose one number (23) and surround it with chips like the attached.
#### Simmo!
##### Paleo Meister (means really, really old)
razormedia said:
tried it and ended up down by a few k.
My favourite tactic is to simply choose one number (23) and surround it with chips like the attached.
Spooky - 23 is my lucky number too!!
#### nafanny29
##### Dormant account
Simmo! said:
Spooky - 23 is my lucky number too!!
and mine!!!! my daughters birthday!!!!!
#### chalupa
##### Experienced Member
Ok, as long as we're all confessing to playing this horrible (edge-wise) game...
Here's my strategy: I decide how what payoff I want (2:1 at minimum for me to play this game) and determine how many numbers I have to cover to get that.
Then I put an equal number of chips on the 0 and several numbers on either side of it, as they appear on the wheel.
That makes watching the ball spin around much more interesting... the bright green 0 is easy to spot, and adds suspense as the ball starts to slow, slow... drop... and... BINGO! (Well, not usually the last part, hehe).
I usually like about 9 numbers, which is almost a 25% chance of winning.
Cover numbers starting with zero and continue as long as you want, in this order (see attached picture and it will make sense):
0 26 32 3 15 35 19 12 4 28 21 7 2 29 25 18 17 ...
Try it, you'll like it! Then go play a better game.
Last edited:
#### chalupa
##### Experienced Member
Here's what a 9-number Green Zone* bet looks like. \$9 pays \$36.
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# From those who have taken the gmat...?
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
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### Hide Tags
Director
Joined: 01 May 2007
Posts: 793
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From those who have taken the gmat...? [#permalink]
### Show Tags
15 Apr 2008, 16:09
1
KUDOS
When you take the test...Do you remember the problems looking similar to those in the OG?
Manager
Joined: 25 Jan 2008
Posts: 166
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Schools: Wharton, Chicago
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Re: From those who have taken the gmat...? [#permalink]
### Show Tags
15 Apr 2008, 16:40
Similar, yes. Although, I think you're going to get unique Q looking questions that you haven't seen the format of, but that's kind of the nature of Q.
Senior Manager
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Re: From those who have taken the gmat...? [#permalink]
### Show Tags
15 Apr 2008, 16:46
1
KUDOS
More similar to GMATPrep than to OG. OG has easier questions on average.
One of my Quant questions resembled a crossward puzzle. I'd never seen such format before.
_________________
Chinese Democracy is misunderstood...at your nearest BestBuy.
Best AWA guide here: http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html
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Re: From those who have taken the gmat...? [#permalink]
### Show Tags
15 Apr 2008, 17:13
It really depends on your level - the actual thing adapts to your level - so if you get the initial questions correct then you'll get the tougher questions (like the ones on the last 2 pages of OG) ...
Senior Manager
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Location: Newport, RI
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Re: From those who have taken the gmat...? [#permalink]
### Show Tags
16 Apr 2008, 04:15
1
KUDOS
I think the difficulty of problems in the OG are in the 500-700 range with less in the 650-700 range. So if you are looking for 700+, then it will be different but not substantially. If you get the framework down from the OG then the raised diffculty shouldn't throw you on RC and CR...SC is another beast though. I found the SC on the real one more idioms than actual grammar.
Manager
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Re: From those who have taken the gmat...? [#permalink]
### Show Tags
16 Apr 2008, 06:10
In recently studying the OG and GMATPrep, I have noticed that the OG questions seem a bit easier and not of the level of GMATPrep questions...
The OG Math guide questions are even easier, IMO.
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Re: From those who have taken the gmat...? [#permalink]
### Show Tags
16 Apr 2008, 20:54
1
KUDOS
jimmyjamesdonkey wrote:
When you take the test...Do you remember the problems looking similar to those in the OG?
ohh, good topic. I am so curious, too.. Someone said GMAT tests become harder and harder. This really scares me.
It seems that there are more questions on SC than on RC, I guess.
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Re: From those who have taken the gmat...? [#permalink]
### Show Tags
17 Apr 2008, 05:50
Actually, I was a little thrown b/c there were some questions that were different (maybe experimental?) that tripped up my timing. The first quant question took me about 4 minutes b/c though it was easy, it was a type that I had not seen before.
Senior Manager
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Re: From those who have taken the gmat...? [#permalink]
### Show Tags
17 Apr 2008, 06:06
The test definately is geared to determine your ability so it will get harder but don't just think because you have easy questions that you are doing poorly. On the quant, I remember getting absolutely blasted in the middle of my test (first 10 were easy, next 25 brutal) and my last 10 or so were super easy. I ended up with a 48 but thought I totally bombed it.
As for the Verbal breakdown, the books I've read said be prepared for an even breakdown of RC, SC and CR. You will typically see 4 RC (three short, one long passage of 12-14 questions) so if you are watching time remember how many RC you have left so you don't panic at the end.
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Re: From those who have taken the gmat...? [#permalink]
### Show Tags
17 Apr 2008, 06:53
1
KUDOS
yea that happened to me, I originally thought 3 RC passages so when I saw the 4th one I was a little shocked, started to panic and heart started racing because it killed my timing.
Yea my last ten were super easy too on quant, was behind on time the whole test and then breezed through them, ended with a 46 with 3 minutes to spare. Going for round 2 on Saturday. I will be more comfortable and not as shocked if I see crazy questions.
Senior Manager
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Re: From those who have taken the gmat...? [#permalink]
### Show Tags
18 Apr 2008, 04:35
good luck terp!
Re: From those who have taken the gmat...? [#permalink] 18 Apr 2008, 04:35
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# From those who have taken the gmat...?
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2.765625 | CC-MAIN-2017-22 |
# Effective exchange rate
Effective Exchange Rate is an index that describes the relative strength of a currency relative to a basket of other currencies.[citation needed] Suppose a country has $N$ trading partners and denote $Trade_i$ and $E_i$as the trade and exchange rate with country $i$ respectively. Then the effective exchange rate is calculated as:
$E_{effective} = E_1 \frac{Trade_1}{Trade} + ... + E_N \frac{Trade_N}{Trade}$
Although typically that basket is trade-weighted, the trade-weighted effective exchange rate index is not the only way to derive a meaningful effective exchange rate index. Ho(2012) proposed a new approach to compiling effective exchange rate indices. It defines the effective exchange rate as a ratio if the "normalized Exchange Value of Currency i against the US dollar" to the normalized exchange value of the "benchmark currency basket" against the US dollar. The US dollar is here used as numeraire for convenience, and since it cancels out, in principle any other currency can be used instead without affecting the results. The benchmark currency basket is a GDP-weighted basket of the major fully convertible currencies of the world.
Bilateral exchange rate involves a currency pair, while an effective exchange rate is a weighted average of a basket of foreign currencies, and it can be viewed as an overall measure of the country's external competitiveness. A nominal effective exchange rate (NEER) is weighted with the inverse of the asymptotic trade weights. A real effective exchange rate (REER) adjusts NEER by appropriate foreign price level and deflates by the home country price level. There are four aspects for alternative measures of REER which are (a) Using end-of-period or period averages of the nominal exchange rate. (b) Choosing price indexes. (c) In obtaining the real effective exchange rates, deciding upon the number of trading partners in calculating the weights. (d) Deciding upon the formula to use in aggregation. Considering all these aspects together led to the calculation of a great number of alterative series. [1] | 2.9375 | CC-MAIN-2014-52 |
# Your question: How do you describe a congruence transformation?
Contents
## How do you describe a transformation in geometry?
A translation moves a shape up, down or from side to side but it does not change its appearance in any other way. A transformation is a way of changing the size or position of a shape. … Every point in the shape is translated the same distance in the same direction.
## Which transformation makes congruent?
Rotations, reflections, and translations are isometric. That means that these transformations do not change the size of the figure. If the size and shape of the figure is not changed, then the figures are congruent.
## What describes a transformation?
A transformation changes the size, shape, or position of a figure and creates a new figure. A geometry transformation is either rigid or non-rigid; another word for a rigid transformation is “isometry”. An isometry, such as a rotation, translation, or reflection, does not change the size or shape of the figure.
## What is an example of a congruent shape?
Congruent shapes can be said as identical shapes in terms of sides and angles. Two bricks and two playing dice are always congruent to each other.
## How do you identify transformations?
The function translation / transformation rules:
1. f (x) + b shifts the function b units upward.
2. f (x) – b shifts the function b units downward.
3. f (x + b) shifts the function b units to the left.
4. f (x – b) shifts the function b units to the right.
5. –f (x) reflects the function in the x-axis (that is, upside-down).
THIS IS IMPORTANT: Which is more efficient full wave rectifier or bridge rectifier?
## What do you know about transformations?
A transformation is a process that manipulates a polygon or other two-dimensional object on a plane or coordinate system. Mathematical transformations describe how two-dimensional figures move around a plane or coordinate system. A preimage or inverse image is the two-dimensional shape before any transformation.
## What are the different types of transformations in geometry?
The four main types of transformations are translations, reflections, rotations, and scaling.
• Translations. A translation moves every point by a fixed distance in the same direction. …
• Reflections. …
• Rotations. …
• Scaling. …
• Vertical Translations. …
• Horizontal Translations. …
• Reflections. …
• Learning Objectives.
## How do you identify congruent figures?
Two polygons are congruent if they are the same size and shape – that is, if their corresponding angles and sides are equal. Move your mouse cursor over the parts of each figure on the left to see the corresponding parts of the congruent figure on the right.
## What do congruent figures look like?
If two figures are congruent, then they’re exactly the same shape, and they’re exactly the same size. They may appear different because one is shifted or rotated a certain way, but they’re still the same shape, and all the sides of one are the same length as the corresponding sides of the other. | 4.25 | CC-MAIN-2022-33 |
### Other Sites
Front Page Titles (by Subject) THE MOTION OF PROJECTILES - Dialogues Concerning Two New Sciences
#### Also in the Library:
Subject Area: Science
## THE MOTION OF PROJECTILES - Galileo Galilei, Dialogues Concerning Two New Sciences [1638]
##### Edition used:
Dialogues Concerning Two New Sciences by Galileo Galilei. Translated from the Italian and Latin into English by Henry Crew and Alfonso de Salvio. With an Introduction by Antonio Favaro (New York: Macmillan, 1914).
Liberty Fund, Inc. is a private, educational foundation established to encourage the study of the ideal of a society of free and responsible individuals.
### THE MOTION OF PROJECTILES
In the preceding pages we have discussed the properties of uniform motion and of motion naturally accelerated along planes of all inclinations. I now propose to set forth those properties which belong to a body whose motion is compounded of two other motions, namely, one uniform and one naturally accelerated; these properties, well worth knowing, I propose to demonstrate in a rigid manner. This is the kind of motion seen in a moving projectile; its origin I conceive to be as follows:
Imagine any particle projected along a horizontal plane without friction; then we know, from what has been more fully explained in the preceding pages, that this particle will move along this same plane with a motion which is uniform and perpetual, provided the plane has no limits. But if the plane is limited and elevated, then the moving particle, which we imagine to be a heavy one, will on passing over the edge of the plane acquire, in addition to its previous uniform and perpetual motion, a downward propensity due to its own weight; so that the resulting motion which I call projection [projectio], is compounded of one which is uniform and horizontal and of another which is vertical and naturally accelerated. We now proceed to demonstrate some of its properties, the first of which is as follows:
### Theorem I, Proposition I[269]
A projectile which is carried by a uniform horizontal motion compounded with a naturally accelerated vertical motion describes a path which is a semi-parabola.
#### Sagr.
Here, Salviati, it will be necessary to stop a little while for my sake and, I believe, also for the benefit of Simplicio; for it so happens that I have not gone very far in my study of Apollonius and am merely aware of the fact that he treats of the parabola and other conic sections, without an understanding of which I hardly think one will be able to follow the proof of other propositions depending upon them. Since even in this first beautiful theorem the author finds it necessary to prove that the path of a projectile is a parabola, and since, as I imagine, we shall have to deal with only this kind of curves, it will be absolutely necessary to have a thorough acquaintance, if not with all the properties which Apollonius has demonstrated for these figures, at least with those which are needed for the present treatment.
#### Salv.
You are quite too modest, pretending ignorance of facts which not long ago you acknowledged as well known—I mean at the time when we were discussing the strength of materials and needed to use a certain theorem of Apollonius which gave you no trouble.
#### Sagr.
I may have chanced to know it or may possibly have assumed it, so long as needed, for that discussion; but now when we have to follow all these demonstrations about such curves we ought not, as they say, to swallow it whole, and thus waste time and energy.
#### Simp.
Now even though Sagredo is, as I believe, well equipped for all his needs, I do not understand even the elementary terms; for although our philosophers have treated the motion of projectiles, I do not recall their having described the path of a projectile except to state in a general way that it is always a curved line, unless the projection be vertically upwards. But[270] if the little Euclid which I have learned since our previous discussion does not enable me to understand the demonstrations which are to follow, then I shall be obliged to accept the theorems on faith without fully comprehending them.
#### Salv.
On the contrary, I desire that you should understand them from the Author himself, who, when he allowed me to see this work of his, was good enough to prove for me two of the principal properties of the parabola because I did not happen to have at hand the books of Apollonius. These properties, which are the only ones we shall need in the present discussion, he proved in such a way that no prerequisite knowledge was required. These theorems are, indeed, given by Apollonius, but after many preceding ones, to follow which would take a long while. I wish to shorten our task by deriving the first property purely and simply from the mode of generation of the parabola and proving the second immediately from the first.
Beginning now with the first, imagine a right cone, erected upon the circular base ibkc with apex at l. The section of this cone made by a plane drawn parallel to the side lk is the curve which is called a parabola. The base of this parabola bc cuts at right angles the diameter ik of the circle ibkc, and the axis ad is parallel to the side lk; now having taken any point f in the curve bfa draw the straight line fe parallel to bd; then, I say, the square of bd is to the square of fe in the same ratio as the axis ad is to the portion ae. Through the point e pass a plane parallel to the circle ibkc, producing in the cone a circular section whose diameter is the line geh. Since bd is at right angles to ik in the circle ibk, the square of bd is equal to the rectangle formed by id and dk; so also in the upper circle which passes through the points gfh the square of fe is equal to the rectangle formed by ge and eh; hence the square of bd is to the square of fe as the rectangle id.dk is to the rectangle ge.eh. And since the line ed is parallel to hk, the line eh, being parallel to dk, is equal to it; therefore the rectangle id.dk is to the rectangle ge.eh as id is to[271] ge, that is, as da is to ae; whence also the rectangle id.dk is to the rectangle ge.eh, that is, the square of bd is to the square of fe, as the axis da is to the portion ae.
q. e. d.
Fig. 106
The other proposition necessary for this discussion we demonstrate as follows. Let us draw a parabola whose axis ca is prolonged upwards to a point d; from any point b draw the line bc parallel to the base of the parabola; if now the point d is chosen so that da = ca, then, I say, the straight line drawn through the points b and d will be tangent to the parabola at b. For imagine, if possible, that this line cuts the parabola above or that its prolongation cuts it below, and through any point g in it draw the straight line fge. And since the square of fe is greater than the square of ge, the square of fe will bear a greater ratio to the square of bc than the square of ge to that of bc; and since, by the preceding proposition, the square of fe is to that of bc as the line ea is to ca, it follows that the line ea will bear to the line ca a greater ratio than the square of ge to that of bc, or, than the square of ed to that of cd (the sides of the triangles deg and dcb being proportional). But the line ea is to ca, or da, in the same ratio as four times the rectangle ea.ad is to four times the square of ad, or, what is the same, the square of cd, since this is four times the square of ad; hence four times the rectangle ea.ad bears to the square of cd a greater ratio than the square of ed to the square of cd; but that would make four times the rectangle ea.ad greater than the square of ed; which is false, the fact being just the opposite, because the two portions ea and ad of the line ed are not equal. Therefore the line db touches the parabola without cutting it.
q. e. d.
Fig. 107
#### Simp.
Your demonstration proceeds too rapidly and, it seems to me, you keep on assuming that all of Euclid’s theorems are[272] as familiar and available to me as his first axioms, which is far from true. And now this fact which you spring upon us, that four times the rectangle ea.ad is less than the square of de because the two portions ea and ad of the line de are not equal brings me little composure of mind, but rather leaves me in suspense.
#### Salv.
Indeed, all real mathematicians assume on the part of the reader perfect familiarity with at least the elements of Euclid; and here it is necessary in your case only to recall a proposition of the Second Book in which he proves that when a line is cut into equal and also into two unequal parts, the rectangle formed on the unequal parts is less than that formed on the equal (i. e., less than the square on half the line), by an amount which is the square of the difference between the equal and unequal segments. From this it is clear that the square of the whole line which is equal to four times the square of the half is greater than four times the rectangle of the unequal parts. In order to understand the following portions of this treatise it will be necessary to keep in mind the two elemental theorems from conic sections which we have just demonstrated; and these two theorems are indeed the only ones which the Author uses. We can now resume the text and see how he demonstrates his first proposition in which he shows that a body falling with a motion compounded of a uniform horizontal and a naturally accelerated [naturale descendente] one describes a semi-parabola.
Let us imagine an elevated horizontal line or plane ab along which a body moves with uniform speed from a to b. Suppose this plane to end abruptly at b; then at this point the body will, on account of its weight, acquire also a natural motion downwards along the perpendicular bn. Draw the line be along the plane ba to represent the flow, or measure, of time; divide this line into a number of segments, bc, cd, de, representing equal intervals of time; from the points b, c, d, e, let fall lines which are parallel to the perpendicular bn. On the first of these lay off any distance ci, on the second a distance four times as long, df; on[273] the third, one nine times as long, eh; and so on, in proportion to the squares of cb, db, eb, or, we may say, in the squared ratio of these same lines. Accordingly we see that while the body moves from b to c with uniform speed, it also falls perpendicularly through the distance ci, and at the end of the time-interval bc finds itself at the point i. In like manner at the end of the time-interval bd, which is the double of bc, the vertical fall will be four times the first distance ci; for it has been shown in a previous discussion that the distance traversed by a freely falling body varies as the square of the time; in like manner the space eh traversed during the time be will be nine times ci; thus it is evident that the distances eh, df, ci will be to one another as the squares of the lines be, bd, bc. Now from the points i, f, h draw the straight lines io, fg, hl parallel to be; these lines hl, fg, io are equal to eb, db and cb, respectively; so also are the lines bo, bg, bl respectively equal to ci, df, and eh. The square of hl is to that of fg as the line lb is to bg; and the square of fg is to that of io as gb is to bo; therefore the points i, f, h, lie on one and the same parabola. In like manner it may be shown that, if we take equal time-intervals of any size whatever, and if we imagine the particle to be carried by a similar compound motion, the positions of this particle, at the ends of these time-intervals, will lie on one and the same parabola.
q. e. d.
Fig. 108
#### Salv.
This conclusion follows from the converse of the first of the two propositions given above. For, having drawn a parabola through the points b and h, any other two points, f and i, not falling on the parabola must lie either within or without; consequently the line fg is either longer or shorter than the line which terminates on the parabola. Therefore the square of hl will not bear to the square of fg the same ratio as the line lb to bg, but a greater or smaller; the fact is, however, that the square of hl does bear this same ratio to the square of fg. Hence the point f does lie on the parabola, and so do all the others.
#### Sagr.
One cannot deny that the argument is new, subtle and conclusive, resting as it does upon this hypothesis, namely, that the horizontal motion remains uniform, that the vertical motion continues to be accelerated downwards in proportion to the square of the time, and that such motions and velocities as these combine without altering, disturbing, or hindering each other,* so that as the motion proceeds the path of the projectile does not change into a different curve: but this, in my opinion,[274] is impossible. For the axis of the parabola along which we imagine the natural motion of a falling body to take place stands perpendicular to a horizontal surface and ends at the center of the earth; and since the parabola deviates more and more from its axis no projectile can ever reach the center of the earth or, if it does, as seems necessary, then the path of the projectile must transform itself into some other curve very different from the parabola.
#### Simp.
To these difficulties, I may add others. One of these is that we suppose the horizontal plane, which slopes neither up nor down, to be represented by a straight line as if each point on this line were equally distant from the center, which is not the case; for as one starts from the middle [of the line] and goes toward either end, he departs farther and farther from the center [of the earth] and is therefore constantly going uphill. Whence it follows that the motion cannot remain uniform through any distance whatever, but must continually diminish. Besides, I do not see how it is possible to avoid the resistance of the medium which must destroy the uniformity of the horizontal motion and change the law of acceleration of falling bodies. These various difficulties render it highly improbable that a result derived from such unreliable hypotheses should hold true in practice.
#### Salv.
All these difficulties and objections which you urge are so well founded that it is impossible to remove them; and, as for me, I am ready to admit them all, which indeed I think our Author would also do. I grant that these conclusions proved in the abstract will be different when applied in the concrete and will be fallacious to this extent, that neither will the horizontal motion be uniform nor the natural acceleration be in the ratio assumed, nor the path of the projectile a parabola, etc. But, on the other hand, I ask you not to begrudge our Author that which other eminent men have assumed even if not strictly true. The authority of Archimedes alone will satisfy everybody. In his Mechanics and in his first quadrature of the parabola he takes for granted that the beam of a balance or steelyard is a straight line, every point of which is equidistant from the common center of all heavy bodies, and that the cords by which heavy bodies are suspended are parallel to each other.
Some consider this assumption permissible because, in practice, our instruments and the distances involved are so small in comparison with the enormous distance from the center of the earth that we may consider a minute of arc on a great circle as a straight line, and may regard the perpendiculars let fall from its two extremities as parallel. For if in actual practice one had to[275] consider such small quantities, it would be necessary first of all to criticise the architects who presume, by use of a plumbline, to erect high towers with parallel sides. I may add that, in all their discussions, Archimedes and the others considered themselves as located at an infinite distance from the center of the earth, in which case their assumptions were not false, and therefore their conclusions were absolutely correct. When we wish to apply our proven conclusions to distances which, though finite, are very large, it is necessary for us to infer, on the basis of demonstrated truth, what correction is to be made for the fact that our distance from the center of the earth is not really infinite, but merely very great in comparison with the small dimensions of our apparatus. The largest of these will be the range of our projectiles—and even here we need consider only the artillery—which, however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth; and since these paths terminate upon the surface of the earth only very slight changes can take place in their parabolic figure which, it is conceded, would be greatly altered if they terminated at the center of the earth.
As to the perturbation arising from the resistance of the medium this is more considerable and does not, on account of its manifold forms, submit to fixed laws and exact description. Thus if we consider only the resistance which the air offers to the motions studied by us, we shall see that it disturbs them all and disturbs them in an infinite variety of ways corresponding to the infinite variety in the form, weight, and velocity of the projectiles. For as to velocity, the greater this is, the greater will be the resistance offered by the air; a resistance which will be greater as the moving bodies become less dense [men gravi]. So that although the falling body ought to be displaced [andare accelerandosi] in proportion to the square of the duration of its motion, yet no matter how heavy the body, if it falls from a very considerable height, the resistance of the air will be such as to prevent any increase in speed and will render the motion[276] uniform; and in proportion as the moving body is less dense [men grave] this uniformity will be so much the more quickly attained and after a shorter fall. Even horizontal motion which, if no impediment were offered, would be uniform and constant is altered by the resistance of the air and finally ceases; and here again the less dense [più leggiero] the body the quicker the process. Of these properties [accidenti] of weight, of velocity, and also of form [figura], infinite in number, it is not possible to give any exact description; hence, in order to handle this matter in a scientific way, it is necessary to cut loose from these difficulties; and having discovered and demonstrated the theorems, in the case of no resistance, to use them and apply them with such limitations as experience will teach. And the advantage of this method will not be small; for the material and shape of the projectile may be chosen, as dense and round as possible, so that it will encounter the least resistance in the medium. Nor will the spaces and velocities in general be so great but that we shall be easily able to correct them with precision.
In the case of those projectiles which we use, made of dense [grave] material and round in shape, or of lighter material and cylindrical in shape, such as arrows, thrown from a sling or crossbow, the deviation from an exact parabolic path is quite insensible. Indeed, if you will allow me a little greater liberty, I can show you, by two experiments, that the dimensions of our apparatus are so small that these external and incidental resistances, among which that of the medium is the most considerable, are scarcely observable.
I now proceed to the consideration of motions through the air, since it is with these that we are now especially concerned; the resistance of the air exhibits itself in two ways: first by offering greater impedance to less dense than to very dense bodies, and secondly by offering greater resistance to a body in rapid motion than to the same body in slow motion.
Regarding the first of these, consider the case of two balls having the same dimensions, but one weighing ten or twelve times as much as the other; one, say, of lead, the other of oak, both allowed to fall from an elevation of 150 or 200 cubits.
Experiment shows that they will reach the earth with slight difference in speed, showing us that in both cases the retardation caused by the air is small; for if both balls start at the same moment and at the same elevation, and if the leaden one be slightly retarded and the wooden one greatly retarded, then the former ought to reach the earth a considerable distance in advance of the latter, since it is ten times as heavy. But this[277] does not happen; indeed, the gain in distance of one over the other does not amount to the hundredth part of the entire fall. And in the case of a ball of stone weighing only a third or half as much as one of lead, the difference in their times of reaching the earth will be scarcely noticeable. Now since the speed [impeto] acquired by a leaden ball in falling from a height of 200 cubits is so great that if the motion remained uniform the ball would, in an interval of time equal to that of the fall, traverse 400 cubits, and since this speed is so considerable in comparison with those which, by use of bows or other machines except fire arms, we are able to give to our projectiles, it follows that we may, without sensible error, regard as absolutely true those propositions which we are about to prove without considering the resistance of the medium.
Passing now to the second case, where we have to show that the resistance of the air for a rapidly moving body is not very much greater than for one moving slowly, ample proof is given by the following experiment. Attach to two threads of equal length—say four or five yards—two equal leaden balls and suspend them from the ceiling; now pull them aside from the perpendicular, the one through 80 or more degrees, the other through not more than four or five degrees; so that, when set free, the one falls, passes through the perpendicular, and describes large but slowly decreasing arcs of 160, 150, 140 degrees, etc.; the other swinging through small and also slowly diminishing arcs of 10, 8, 6, degrees, etc.
In the first place it must be remarked that one pendulum passes through its arcs of 180°, 160°, etc., in the same time that the other swings through its 10°, 8°, etc., from which it follows that the speed of the first ball is 16 and 18 times greater than that of the second. Accordingly, if the air offers more resistance to the high speed than to the low, the frequency of vibration in the large arcs of 180° or 160°, etc., ought to be less than in the small arcs of 10°, 8°, 4°, etc., and even less than in arcs of 2°, or 1°; but this prediction is not verified by experiment; because if two persons start to count the vibrations, the one the large, the other the small, they will discover that after counting tens and even hundreds they will not differ by a single vibration, not even by a fraction of one.
#### Sagr.
On the contrary, since we cannot deny that the air hinders both of these motions, both becoming slower and finally vanishing, we have to admit that the retardation occurs in the same proportion in each case. But how? How, indeed, could the resistance offered to the one body be greater than that offered to the other except by the impartation of more momentum and speed [impeto e velocità] to the fast body than to the slow? And if this is so the speed with which a body moves is at once the cause and measure [cagione e misura] of the resistance which it meets. Therefore, all motions, fast or slow, are hindered and diminished in the same proportion; a result, it seems to me, of no small importance.
#### Salv.
We are able, therefore, in this second case to say that the errors, neglecting those which are accidental, in the results which we are about to demonstrate are small in the case of our machines where the velocities employed are mostly very great and the distances negligible in comparison with the semidiameter of the earth or one of its great circles.
#### Simp.
I would like to hear your reason for putting the projectiles of fire arms, i. e., those using powder, in a different class from the projectiles employed in bows, slings, and crossbows, on the ground of their not being equally subject to change and resistance from the air.
#### Salv.
I am led to this view by the excessive and, so to speak, supernatural violence with which such projectiles are launched; for, indeed, it appears to me that without exaggeration one might say that the speed of a ball fired either from a musket or from a piece of ordnance is supernatural. For if such a ball be allowed to fall from some great elevation its speed will, owing to the resistance of the air, not go on increasing indefinitely; that which happens to bodies of small density in falling through short distances—I mean the reduction of their motion to uniformity—will also happen to a ball of iron or lead after it has fallen a few thousand cubits; this terminal or final speed [terminata velocità] is the maximum which such a heavy body can naturally acquire[279] in falling through the air. This speed I estimate to be much smaller than that impressed upon the ball by the burning powder.
An appropriate experiment will serve to demonstrate this fact. From a height of one hundred or more cubits fire a gun [archibuso] loaded with a lead bullet, vertically downwards upon a stone pavement; with the same gun shoot against a similar stone from a distance of one or two cubits, and observe which of the two balls is the more flattened. Now if the ball which has come from the greater elevation is found to be the less flattened of the two, this will show that the air has hindered and diminished the speed initially imparted to the bullet by the powder, and that the air will not permit a bullet to acquire so great a speed, no matter from what height it falls; for if the speed impressed upon the ball by the fire does not exceed that acquired by it in falling freely [naturalmente] then its downward blow ought to be greater rather than less.
This experiment I have not performed, but I am of the opinion that a musket-ball or cannon-shot, falling from a height as great as you please, will not deliver so strong a blow as it would if fired into a wall only a few cubits distant, i. e., at such a short range that the splitting or rending of the air will not be sufficient to rob the shot of that excess of supernatural violence given it by the powder.
The enormous momentum [impeto] of these violent shots may cause some deformation of the trajectory, making the beginning of the parabola flatter and less curved than the end; but, so far as our Author is concerned, this is a matter of small consequence in practical operations, the main one of which is the preparation of a table of ranges for shots of high elevation, giving the distance attained by the ball as a function of the angle of elevation; and since shots of this kind are fired from mortars [mortari] using small charges and imparting no supernatural momentum [impeto sopranaturale] they follow their prescribed paths very exactly.
But now let us proceed with the discussion in which the Author invites us to the study and investigation of the motion of a body [impeto del mobile] when that motion is compounded of two others; and first the case in which the two are uniform, the one horizontal, the other vertical.
### Theorem II, Proposition II[280]
When the motion of a body is the resultant of two uniform motions, one horizontal, the other perpendicular, the square of the resultant momentum is equal to the sum of the squares of the two component momenta.*
Let us imagine any body urged by two uniform motions and let ab represent the vertical displacement, while bc represents the displacement which, in the same interval of time, takes place in a horizontal direction. If then the distances ab and bc are traversed, during the same time-interval, with uniform motions the corresponding momenta will be to each other as the distances ab and bc are to each other; but the body which is urged by these two motions describes the diagonal ac; its momentum is proportional to ac. Also the square of ac is equal to the sum of the squares of ab and bc. Hence the square of the resultant momentum is equal to the sum of the squares of the two momenta ab and bc.
q. e. d.
Fig. 109
#### Simp.
At this point there is just one slight difficulty which needs to be cleared up; for it seems to me that the conclusion just reached contradicts a previous proposition* in which it is claimed that the speed [impeto] of a body coming from a to b is equal to that in coming from a to c; while now you conclude that the speed [impeto] at c is greater than that at b.
#### Salv.
Both propositions, Simplicio, are true, yet there is a great difference between them. Here we are speaking of a body urged by a single motion which is the resultant of two uniform motions, while there we were speaking of two bodies each urged with naturally accelerated motions, one along the vertical ab the other along the inclined plane ac. Besides the time-intervals were there not supposed to be equal, that along the incline ac being greater than that along the vertical ab; but the motions of which we now speak, those along ab, bc, ac, are uniform and simultaneous.
#### Simp.
Pardon me; I am satisfied; pray go on.
Our Author next undertakes to explain what happens when a body is urged by a motion compounded of one which is horizontal and uniform and of another which is vertical but naturally accelerated; from these two components results the path of a projectile, which is a parabola. The problem is to determine the speed [impeto] of the projectile at each point. With this purpose in view our Author sets forth as follows the manner, or rather the method, of measuring such speed [impeto] along the path which is taken by a heavy body starting from rest and falling with a naturally accelerated motion.
### Theorem III, Proposition III
Let the motion take place along the line ab, starting from rest at a, and in this line choose any point c. Let ac represent the time, or the measure of the time, required for the body to fall through the space ac; let ac also represent the velocity [impetus seu momentum] at c acquired by a fall through the distance ac. In the line ab select any other point b. The problem now is to determine the velocity at b acquired by a body in falling through the distance ab and to express this in terms of the velocity at c, the measure of which is the length ac. Take as a mean proportional between ac and ab. We shall prove that the velocity at b is to that at c as the length as is to the length ac. Draw the horizontal line cd, having twice the length of ac, and be, having twice the length of ba. It then follows, from the preceding theorems, that a body falling through the distance ac, and turned so as to move along the horizontal cd with a uniform speed equal to that acquired on reaching c[282] will traverse the distance cd in the same interval of time as that required to fall with accelerated motion from a to c. Likewise be will be traversed in the same time as ba. But the time of descent through ab is as; hence the horizontal distance be is also traversed in the time as. Take a point l such that the time as is to the time ac as be is to bl; since the motion along be is uniform, the distance bl, if traversed with the speed [momentum celeritatis] acquired at b, will occupy the time ac; but in this same time-interval, ac, the distance cd is traversed with the speed acquired in c. Now two speeds are to each other as the distances traversed in equal intervals of time. Hence the speed at c is to the speed at b as cd is to bl. But since dc is to be as their halves, namely, as ca is to ba, and since be is to bl as ba is to sa; it follows that dc is to bl as ca is to sa. In other words, the speed at c is to that at b as ca is to sa, that is, as the time of fall through ab.
Fig. 110
The method of measuring the speed of a body along the direction of its fall is thus clear; the speed is assumed to increase directly as the time.
But before we proceed further, since this discussion is to deal with the motion compounded of a uniform horizontal one and one accelerated vertically downwards—the path of a projectile, namely, a parabola—it is necessary that we define some common standard by which we may estimate the velocity, or momentum [velocitatem, impetum seu momentum] of both motions; and since from the innumerable uniform velocities one only, and that not selected at random, is to be compounded with a velocity acquired by naturally accelerated motion, I can think of no simpler way of selecting and measuring this than to assume another of the same kind.* For the sake of clearness, draw the vertical line ac to meet the horizontal line bc. Ac is the height and bc the amplitude of the semi-parabola ab, which is the resultant of the two motions, one that of a body falling[283] from rest at a, through the distance ac, with naturally accelerated motion, the other a uniform motion along the horizontal ad. The speed acquired at c by a fall through the distance ac is determined by the height ac; for the speed of a body falling from the same elevation is always one and the same; but along the horizontal one may give a body an infinite number of uniform speeds. However, in order that I may select one out of this multitude and separate it from the rest in a perfectly definite manner, I will extend the height ca upwards to e just as far as is necessary and will call this distance ae the “sublimity.” Imagine a body to fall from rest at e; it is clear that we may make its terminal speed at a the same as that with which the same body travels along the horizontal line ad; this speed will be such that, in the time of descent along ea, it will describe a horizontal distance twice the length of ea. This preliminary remark seems necessary.
Fig. 111
The reader is reminded that above I have called the horizontal line cb the “amplitude” of the semi-parabola ab; the axis ac of this parabola, I have called its “altitude”; but the line ea the fall along which determines the horizontal speed I have called the “sublimity.” These matters having been explained, I proceed with the demonstration.
#### Sagr.
Allow me, please, to interrupt in order that I may point out the beautiful agreement between this thought of the Author and the views of Plato concerning the origin of the various uniform speeds with which the heavenly bodies revolve. The latter chanced upon the idea that a body could not pass from rest to any given speed and maintain it uniformly except by passing through all the degrees of speed intermediate between the given speed and rest. Plato thought that God, after having created the heavenly bodies, assigned them the proper and uniform speeds with which they were forever to revolve; and that He made them start from rest and move over definite distances under a natural and rectilinear acceleration such as governs the motion of terrestrial bodies. He added that once these bodies had gained their proper and permanent speed, their rectilinear motion was converted into a circular one, the only[284] motion capable of maintaining uniformity, a motion in which the body revolves without either receding from or approaching its desired goal. This conception is truly worthy of Plato; and it is to be all the more highly prized since its underlying principles remained hidden until discovered by our Author who removed from them the mask and poetical dress and set forth the idea in correct historical perspective. In view of the fact that astronomical science furnishes us such complete information concerning the size of the planetary orbits, the distances of these bodies from their centers of revolution, and their velocities, I cannot help thinking that our Author (to whom this idea of Plato was not unknown) had some curiosity to discover whether or not a definite “sublimity” might be assigned to each planet, such that, if it were to start from rest at this particular height and to fall with naturally accelerated motion along a straight line, and were later to change the speed thus acquired into uniform motion, the size of its orbit and its period of revolution would be those actually observed.
#### Salv.
I think I remember his having told me that he once made the computation and found a satisfactory correspondence with observation. But he did not wish to speak of it, lest in view of the odium which his many new discoveries had already brought upon him, this might be adding fuel to the fire. But if any one desires such information he can obtain it for himself from the theory set forth in the present treatment.
We now proceed with the matter in hand, which is to prove:
### Problem I, Proposition IV
To determine the momentum of a projectile at each particular point in its given parabolic path.
Let bec be the semi-parabola whose amplitude is cd and whose height is db, which latter extended upwards cuts the tangent of the parabola ca in a. Through the vertex draw the horizontal line bi parallel to cd. Now if the amplitude cd is equal to the entire height da, then bi will be equal to ba and also to bd; and if we take ab as the measure of the time required for fall through the distance ab and also of the momentum acquired at b in consequence of its fall from rest at a, then if. we turn into a horizontal direction the momentum acquired by fall through ab [impetum ab] the space traversed in the same interval of time will be represented by dc which is twice bi. But a body which falls from rest at b along the line bd will during the same time-interval fall through the height of the parabola[285] bd. Hence a body falling from rest at a, turned into a horizontal direction with the speed ab will traverse a space equal to dc. Now if one superposes upon this motion a fall along bd, traversing the height bd while the parabola bc is described, then the momentum of the body at the terminal point c is the resultant of a uniform horizontal momentum, whose value is represented by ab, and of another momentum acquired by fall from b to the terminal point d or c; these two momenta are equal. If, therefore, we take ab to be the measure of one of these momenta, say, the uniform horizontal one, then bi, which is equal to bd, will represent the momentum acquired at d or c; and ia will represent the resultant of these two momenta, that is, the total momentum with which the projectile, travelling along the parabola, strikes at c.
With this in mind let us take any point on the parabola, say e, and determine the momentum with which the projectile passes that point. Draw the horizontal ef and take bg a mean proportional between bd and bf. Now since ab, or bd, is assumed to be the measure of the time and of the momentum [momentum velocitatis] acquired by falling from rest at b through the distance bd, it follows that bg will measure the time and also the momentum [impetus] acquired at f by fall from b. If therefore we lay off bo, equal to bg, the diagonal line joining a and o will represent the momentum at the point e; because the length ab has been assumed to represent the momentum at b which, after diversion into a horizontal direction, remains constant; and because bo measures the momentum at f or e, acquired by fall, from rest at b, through the height bf. But the square of ao equals the sum of the squares of ab and bo. Hence the theorem sought.
Fig. 112
#### Sagr.
The manner in which you compound these different momenta to obtain their resultant strikes me as so novel that my mind is left in no small confusion. I do not refer to the composition of two uniform motions, even when unequal, and when one takes place along a horizontal, the other along a vertical direction; because in this case I am thoroughly convinced that the resultant is a motion whose square is equal to the sum of the squares of the two components. The confusion arises when one undertakes to compound a uniform horizontal motion with a vertical one which is naturally accelerated. I trust, therefore, we may pursue this discussion more at length.
And I need this even more than you since I am not yet as clear in my mind as I ought to be concerning those fundamental propositions upon which the others rest. Even in the case of the two uniform motions, one horizontal, the other perpendicular, I wish to understand better the manner in which you obtain the resultant from the components. Now, Salviati, you understand what we need and what we desire.
#### Salv.
Your request is altogether reasonable and I will see whether my long consideration of these matters will enable me to make them clear to you. But you must excuse me if in the explanation I repeat many things already said by the Author.
Concerning motions and their velocities or momenta [movimenti e lor velocità o impeti] whether uniform or naturally accelerated, one cannot speak definitely until he has established a measure for such velocities and also for time. As for time we have the already widely adopted hours, first minutes and second minutes. So for velocities, just as for intervals of time, there is need of a common standard which shall be understood and accepted by everyone, and which shall be the same for all. As has already been stated, the Author considers the velocity of a freely falling body adapted to this purpose, since this velocity increases according to the same law in all parts of the world; thus for instance the speed acquired by a leaden ball of a pound weight starting from rest and falling vertically through the height of, say, a spear’s length is the same in all places; it is therefore excellently adapted for representing the momentum [impeto] acquired in the case of natural fall.
It still remains for us to discover a method of measuring momentum in the case of uniform motion in such a way that all who discuss the subject will form the same conception of its size and velocity [grandezza e velocità]. This will prevent one person from imagining it larger, another smaller, than it really is; so that in the composition of a given uniform motion with one which is accelerated different men may not obtain different values for the resultant. In order to determine and represent such a momentum and particular speed [impeto e velocità particolare] our Author has found no better method than to use the momentum acquired by a body in naturally accelerated motion.[287] The speed of a body which has in this manner acquired any momentum whatever will, when converted into uniform motion, retain precisely such a speed as, during a time-interval equal to that of the fall, will carry the body through a distance equal to twice that of the fall. But since this matter is one which is fundamental in our discussion it is well that we make it perfectly clear by means of some particular example.
Let us consider the speed and momentum acquired by a body falling through the height, say, of a spear [picca] as a standard which we may use in the measurement of other speeds and momenta as occasion demands; assume for instance that the time of such a fall is four seconds [minuti secondi d’ora]; now in order to measure the speed acquired from a fall through any other height, whether greater or less, one must not conclude that these speeds bear to one another the same ratio as the heights of fall; for instance, it is not true that a fall through four times a given height confers a speed four times as great as that acquired by descent through the given height; because the speed of a naturally accelerated motion does not vary in proportion to the time. As has been shown above, the ratio of the spaces is equal to the square of the ratio of the times.
If, then, as is often done for the sake of brevity, we take the same limited straight line as the measure of the speed, and of the time, and also of the space traversed during that time, it follows that the duration of fall and the speed acquired by the same body in passing over any other distance, is not represented by this second distance, but by a mean proportional between the two distances. This I can better illustrate by an example. In the vertical line ac, lay off the portion ab to represent the distance traversed by a body falling freely with accelerated motion: the time of fall may be represented by any limited straight line, but for the sake of brevity, we shall represent it by the same length ab; this length may also be employed as a measure of the momentum and speed acquired during the motion; in short, let ab be a measure of the various physical quantities which enter this discussion.
Fig. 113
Having agreed arbitrarily upon ab as a measure of these three different quantities, namely, space, time, and momentum, our next task is to find the time required for fall through a[288] given vertical distance ac, also the momentum acquired at the terminal point c, both of which are to be expressed in terms of the time and momentum represented by ab. These two required quantities are obtained by laying off ad, a mean proportional between ab and ac; in other words, the time of fall from a to c is represented by ad on the same scale on which we agreed that the time of fall from a to b should be represented by ab. In like manner we may say that the momentum [impeto o grado di velocità] acquired at c is related to that acquired at b, in the same manner that the line ad is related to ab, since the velocity varies directly as the time, a conclusion, which although employed as a postulate in Proposition III, is here amplified by the Author.
This point being clear and well-established we pass to the consideration of the momentum [impeto] in the case of two compound motions, one of which is compounded of a uniform horizontal and a uniform vertical motion, while the other is compounded of a uniform horizontal and a naturally accelerated vertical motion. If both components are uniform, and one at right angles to the other, we have already seen that the square of the resultant is obtained by adding the squares of the components [p. 257] as will be clear from the following illustration.
Let us imagine a body to move along the vertical ab with a uniform momentum [impeto] of 3, and on reaching b to move toward c with a momentum [velocità ed impeto] of 4, so that during the same time-interval it will traverse 3 cubits along the vertical and 4 along the horizontal. But a particle which moves with the resultant velocity [velocità] will, in the same time, traverse the diagonal ac, whose length is not 7 cubits—the sum of ab (3) and bc (4)—but 5, which is in potenza equal to the sum of 3 and 4, that is, the squares of 3 and 4 when added make 25, which is the square of ac, and is equal to the sum of the squares of ab and bc. Hence ac is represented by the side—or we may say the root—of a square whose area is 25, namely 5.
Fig. 114
As a fixed and certain rule for obtaining the momentum which[289] results from two uniform momenta, one vertical, the other horizontal, we have therefore the following: take the square of each, add these together, and extract the square root of the sum, which will be the momentum resulting from the two. Thus, in the above example, the body which in virtue of its vertical motion would strike the horizontal plane with a momentum [forza] of 3, would owing to its horizontal motion alone strike at c with a momentum of 4; but if the body strikes with a momentum which is the resultant of these two, its blow will be that of a body moving with a momentum [velocità e forza] of 5; and such a blow will be the same at all points of the diagonal ac, since its components are always the same and never increase or diminish.
Let us now pass to the consideration of a uniform horizontal motion compounded with the vertical motion of a freely falling body starting from rest. It is at once clear that the diagonal which represents the motion compounded of these two is not a straight line, but, as has been demonstrated, a semi-parabola, in which the momentum [impeto] is always increasing because the speed [velocità] of the vertical component is always increasing. Wherefore, to determine the momentum [impeto] at any given point in the parabolic diagonal, it is necessary first to fix upon the uniform horizontal momentum [impeto] and then, treating the body as one falling freely, to find the vertical momentum at the given point; this latter can be determined only by taking into account the duration of fall, a consideration which does not enter into the composition of two uniform motions where the velocities and momenta are always the same; but here where one of the component motions has an initial value of zero and increases its speed [velocità] in direct proportion to the time, it follows that the time must determine the speed [velocità] at the assigned point. It only remains to obtain the momentum resulting from these two components (as in the case of uniform motions) by placing the square of the resultant equal to the sum of the squares of the two components. But here again it is better to illustrate by means of an example.
On the vertical ac lay off any portion ab which we shall employ as a measure of the space traversed by a body falling freely along the perpendicular, likewise as a measure of the time and also of the speed [grado di velocità] or, we may say, of the momenta [impeti]. It is at once clear that if the momentum of a[290] body at b, after having fallen from rest at a, be diverted along the horizontal direction bd, with uniform motion, its speed will be such that, during the time-interval ab, it will traverse a distance which is represented by the line bd and which is twice as great as ab. Now choose a point c, such that bc shall be equal to ab, and through c draw the line ce equal and parallel to bd; through the points b and e draw the parabola bei. And since, during the time-interval ab, the horizontal distance bd or ce, double the length ab, is traversed with the momentum ab, and since during an equal time-interval the vertical distance bc is traversed, the body acquiring at c a momentum represented by the same horizontal, bd, it follows that during the time ab the body will pass from b to e along the parabola be, and will reach e with a momentum compounded of two momenta each equal to ab. And since one of these is horizontal and the other vertical, the square of the resultant momentum is equal to the sum of the squares of these two components, i. e., equal to twice either one of them.
Fig. 115
Therefore, if we lay off the distance bf, equal to ba, and draw the diagonal af, it follows that the momentum [impeto e percossa] at e will exceed that of a body at b after having fallen from a, or what is the same thing, will exceed the horizontal momentum [percossa dell’impeto] along bd, in the ratio of af to ab.
Suppose now we choose for the height of fall a distance bo which is not equal to but greater than ab, and suppose that bg represents a mean proportional between ba and bo; then, still retaining ba as a measure of the distance fallen through, from rest at a, to b, also as a measure of the time and of the momentum which the falling body acquires at b, it follows that bg will be the measure of the time and also of the momentum which the body acquires in falling from b to o. Likewise just as the momentum ab during the time ab carried the body a distance along the horizontal equal to twice ab, so now, during the time-interval bg, the body will be carried in a horizontal direction through a distance which is greater in the ratio of bg to ba. Lay off lb equal to bg and draw the diagonal al, from which we have a quantity compounded of two velocities [impeti] one horizontal, the other vertical; these determine the parabola. The horizontal and uniform velocity is that acquired at b in falling from a; the other is that acquired at o, or, we may say, at i, by a body falling through the distance bo, during a time measured by the line bg,[291] which line bg also represents the momentum of the body. And in like manner we may, by taking a mean proportional between the two heights, determine the momentum [impeto] at the extreme end of the parabola where the height is less than the sublimity ab; this mean proportional is to be drawn along the horizontal in place of bf, and also another diagonal in place of af, which diagonal will represent the momentum at the extreme end of the parabola.
To what has hitherto been said concerning the momenta, blows or shocks of projectiles, we must add another very important consideration; to determine the force and energy of the shock [forza ed energia della percossa] it is not sufficient to consider only the speed of the projectiles, but we must also take into account the nature and condition of the target which, in no small degree, determines the efficiency of the blow. First of all it is well known that the target suffers violence from the speed [velocità] of the projectile in proportion as it partly or entirely stops the motion; because if the blow falls upon an object which yields to the impulse [velocità del percuziente] without resistance such a blow will be of no effect; likewise when one attacks his enemy with a spear and overtakes him at an instant when he is fleeing with equal speed there will be no blow but merely a harmless touch. But if the shock falls upon an object which yields only in part then the blow will not have its full effect, but the damage will be in proportion to the excess of the speed of the projectile over that of the receding body; thus, for example, if the shot reaches the target with a speed of 10 while the latter recedes with a speed of 4, the momentum and shock [impeto e percossa] will be represented by 6. Finally the blow will be a maximum, in so far as the projectile is concerned, when the target does not recede at all but if possible completely resists and stops the motion of the projectile. I have said in so far as the projectile is concerned because if the target should approach the projectile the shock of collision [colpo e l’incontro] would be greater in proportion as the sum of the two speeds is greater than that of the projectile alone.
Moreover it is to be observed that the amount of yielding in the target depends not only upon the quality of the material, as regards hardness, whether it be of iron, lead, wool, etc., but[292] also upon its position. If the position is such that the shot strikes it at right angles, the momentum imparted by the blow [impeto del colpo] will be a maximum; but if the motion be oblique, that is to say slanting, the blow will be weaker; and more and more so in proportion to the obliquity; for, no matter how hard the material of the target thus situated, the entire momentum [impeto e moto] of the shot will not be spent and stopped; the projectile will slide by and will, to some extent, continue its motion along the surface of the opposing body.
All that has been said above concerning the amount of momentum in the projectile at the extremity of the parabola must be understood to refer to a blow received on a line at right angles to this parabola or along the tangent to the parabola at the given point; for, even though the motion has two components, one horizontal, the other vertical, neither will the momentum along the horizontal nor that upon a plane perpendicular to the horizontal be a maximum, since each of these will be received obliquely.
#### Sagr.
Your having mentioned these blows and shocks recalls to my mind a problem, or rather a question, in mechanics of which no author has given a solution or said anything which diminishes my astonishment or even partly relieves my mind.
My difficulty and surprise consist in not being able to see whence and upon what principle is derived the energy and immense force [energia e forza immensa] which makes its appearance in a blow; for instance we see the simple blow of a hammer, weighing not more than 8 or 10 lbs., overcoming resistances which, without a blow, would not yield to the weight of a body producing impetus by pressure alone, even though that body weighed many hundreds of pounds. I would like to discover a method of measuring the force [forza] of such a percussion. I can hardly think it infinite, but incline rather to the view that it has its limit and can be counterbalanced and measured by other forces, such as weights, or by levers or screws or other mechanical instruments which are used to multiply forces in a manner which I satisfactorily understand.
#### Salv.
You are not alone in your surprise at this effect or in obscurity as to the cause of this remarkable property. I studied this matter myself for a while in vain; but my confusion merely increased until finally meeting our Academician I received from[293] him great consolation. First he told me that he also had for a long time been groping in the dark; but later he said that, after having spent some thousands of hours in speculating and contemplating thereon, he had arrived at some notions which are far removed from our earlier ideas and which are remarkable for their novelty. And since now I know that you would gladly hear what these novel ideas are I shall not wait for you to ask but promise that, as soon as our discussion of projectiles is completed, I will explain all these fantasies, or if you please, vagaries, as far as I can recall them from the words of our Academician. In the meantime we proceed with the propositions of the author.
### Proposition V, Problem
Having given a parabola, find the point, in its axis extended upwards, from which a particle must fall in order to describe this same parabola.
Fig. 116
### COROLLARY
Hence it follows that half the base, or amplitude, of the semi-parabola (which is one-quarter of the entire amplitude) is a mean proportional between its altitude and the sublimity from which a falling body will describe this same parabola.
### Proposition VI, Problem
Given the sublimity and the altitude of a parabola, to find its amplitude.
Let the line ac, in which lie the given altitude cb and sublimity ab, be perpendicular to the horizontal line cd. The problem is to find the amplitude, along the horizontal cd, of the semi-parabola which is described with the sublimity ba and altitude bc. Lay off cd equal to twice the mean proportional between cb and ba. Then cd will be the amplitude sought, as is evident from the preceding proposition.
Fig. 117
### Theoremposition. Pro VII
If projectiles describe semi-parabolas of the same amplitude, the momentum required to describe that one whose amplitude is double its altitude is less than that required for any other.
Let gm be a mean proportional between ab and gl; then gm will [Prop. IV] represent the time and momentum acquired at g by a fall from l; for ab has been assumed as a measure of both time and momentum. Again let gn be a mean proportional between bc and cg; it will then represent the time and momentum which the body acquires at c in falling from g. If now we join m and n, this line mn will represent the momentum at d of the projectile traversing the parabola dg; which momentum is, I say, greater than that of the projectile travelling along the parabola bd whose measure was given by ae. For since gn has been taken as a mean proportional between bc and gc; and since bc is equal to be and also to kg (each of them being the half of dc) it follows that cg:gn=gn:gk, and as cg or (hg) is to gk so is to to : but by construction hg:gk=gk:gl. Hence :=gk:gl. But gk:gl=:, since gm is a mean proportional between kg and gl. Therefore the three squares ng, kg, mg form a continued proportion, :=: And the sum of the two extremes which is equal to the square of mn is greater than twice the square of gk; but the square of ae is double the square of gk. Hence the square of mn is greater than the square of ae and the length mn is greater than the length ae.
q. e. d.
### COROLLARY[296]
The force of rigid demonstrations such as occur only in mathematics fills me with wonder and delight. From accounts given by gunners, I was already aware of the fact that in the use of cannon and mortars, the maximum range, that is the one in which the shot goes farthest, is obtained when the elevation is 45° or, as they say, at the sixth point of the quadrant; but to understand why this happens far outweighs the mere information obtained by the testimony of others or even by repeated experiment.
#### Salv.
What you say is very true. The knowledge of a single fact acquired through a discovery of its causes prepares the mind to understand and ascertain other facts without need of recourse to experiment, precisely as in the present case, where by argumentation alone the Author proves with certainty that the maximum range occurs when the elevation is 45°. He thus demonstrates what has perhaps never been observed in experience, namely, that of other shots those which exceed or fall short of 45° by equal amounts have equal ranges; so that if the balls have been fired one at an elevation of 7 points, the other at 5, they will strike the level at the same distance: the same is true if the shots are fired at 8 and at 4 points, at 9 and at 3, etc. Now let us hear the demonstration of this.
### Theorem. Proposition VIII[297]
The amplitudes of two parabolas described by projectiles fired with the same speed, but at angles of elevation which exceed and fall short of 45° by equal amounts, are equal to each other.
In the triangle mcb let the horizontal side bc and the vertical cm, which form a right angle at c, be equal to each other; then the angle mbc will be a semi-right angle; let the line cm be prolonged to d, such a point that the two angles at b, namely mbe and mbd, one above and the other below the diagonal mb, shall be equal. It is now to be proved that in the case of two parabolas described by two projectiles fired from b with the same speed, one at the angle of ebc, the other at the angle of dbc, their amplitudes will be equal. Now since the external angle bmc is equal to the sum of the internal angles mdb and dbm we may also equate to them the angle mbc; but if we replace the angle dbm by mbe, then this same angle mbc is equal to the two mbe and bdc: and if we subtract from each side of this equation the angle mbe, we have the remainder bdc equal to the remainder ebc. Hence the two triangles dcb and bce are similar. Bisect the straight lines dc and ec in the points h and f: and draw the lines hi and fg parallel to the horizontal cb, and choose l such that dh:hi=ih:hl. Then the triangle ihl will be similar to ihd, and also to the triangle egf; and since ih and gf are equal, each being half of bc, it follows that hl is equal to fe and also to fc; and if we add to each of these the common part fh, it will be seen that ch is equal to fl.
Fig. 119
Let us now imagine a parabola described through the points h and b whose altitude is hc and sublimity hl. Its amplitude will be cb which is double the length hi since hi is a mean proportional between dh (or ch) and hl. The line db is tangent to the parabola at b, since ch is equal to hd. If again we imagine a parabola described through the points f and b, with a sublimity fl and altitude fc, of which the mean proportional is fg, or one-half of cb, then, as before, will cb be the amplitude and the line eb a tangent at b; for ef and fc are equal.
The amplitudes of two parabolas are equal when their altitudes and sublimities are inversely proportional.
### Lemma for the following Proposition
If a straight line be cut at any point whatever and mean proportionals between this line and each of its parts be taken, the sum of the squares of these mean proportionals is equal to the square of the entire line.
Let the line ab be cut at c. Then I say that the square of the mean proportional between ab and ac plus the square of the mean proportional between ab and cb is equal to the square of the whole line ab. This is evident as soon as we describe a semicircle upon the entire line ab, erect a perpendicular cd at c, and draw da and db. For da is a mean proportional between ab and ac while[299] db is a mean proportional between ab and bc: and since the angle adb, inscribed in a semicircle, is a right angle the sum of the squares of the lines da and db is equal to the square of the entire line ab. Hence follows the proposition.
Fig. 121
### Theorem. Proposition X
The momentum [impetus seu momentum] acquired by a particle at the terminal point of any semi-parabola is equal to that which it would acquire in falling through a vertical distance equal to the sum of the sublimity and the altitude of the semi-parabola.*
And since dc has been cut at some point a and since cf and ce are mean proportionals between the whole of cd and its parts, da and ac, it follows, from the preceding lemma, that the sum of the squares of these mean proportionals is equal to the square of the whole: but the square of ef is also equal to the sum of these same squares; whence it follows that the line ef is equal to dc.
Accordingly the momentum acquired at c by a particle in falling from d is the same as that acquired at b by a particle traversing the parabola ab.
q. e. d.
### COROLLARY
Hence it follows that, in the case of all parabolas where the sum of the sublimity and altitude is a constant, the momentum at the terminal point is a constant.
### Problem. Proposition XI
Given the amplitude and the speed [impetus] at the terminal point of a semi-parabola, to find its altitude.
Let the given speed be represented by the vertical line ab, and the amplitude by the horizontal line bc; it is required to find the sublimity of the semi-parabola whose terminal speed is ab and amplitude bc. From what precedes [Cor. Prop. V] it is clear that half the amplitude bc is a mean proportional between[300] the altitude and sublimity of the parabola of which the terminal speed is equal, in accordance with the preceding proposition, to the speed acquired by a body in falling from rest at a through the distance ab. Therefore the line ba must be cut at a point such that the rectangle formed by its two parts will be equal to the square of half bc, namely bd. Necessarily, therefore, bd must not exceed the half of ba; for of all the rectangles formed by parts of a straight line the one of greatest area is obtained when the line is divided into two equal parts. Let e be the middle point of the line ab; and now if bd be equal to be the problem is solved; for be will be the altitude and ea the sublimity of the parabola. (Incidentally we may observe a consequence already demonstrated, namely: of all parabolas described with any given terminal speed that for which the elevation is 45° will have the maximum amplitude.)
Fig. 123
But suppose that bd is less than half of ba which is to be divided in such a way that the rectangle upon its parts may be equal to the square of bd. Upon ea as diameter describe a semicircle efa, in which draw the chord af, equal to bd: join fe and lay off the distance eg equal to fe. Then the rectangle bg.ga plus the square of eg will be equal to the square of ea, and hence also to the sum of the squares of af and fe. If now we subtract the equal squares of fe and ge there remains the rectangle bg.ga equal to the square of af, that is, of bd, a line which is a mean proportional between bg and ga; from which it is evident that the semi-parabola whose amplitude is bc and whose terminal speed [impetus] is represented by ba has an altitude bg and a sublimity ga.
If however we lay off bi equal to ga, then bi will be the altitude of the semi-parabola ic, and ia will be its sublimity. From the preceding demonstration we are able to solve the following problem.
### Problem. Proposition XII
To compute and tabulate the amplitudes of all semi-parabolas which are described by projectiles fired with the same initial speed [impetus].
From the foregoing it follows that, whenever the sum of the altitude and sublimity is a constant vertical height for any set of parabolas, these parabolas are described by projectiles having the same initial speed; all vertical heights thus[301] obtained are therefore included between two parallel horizontal lines. Let cb represent a horizontal line and ab a vertical line of equal length; draw the diagonal ac; the angle acb will be one of 45°; let d be the middle point of the vertical line ab. Then the semi-parabola dc is the one which is determined by the sublimity ad and the altitude db, while its terminal speed at c is that which would be acquired at b by a particle falling from rest at a. If now ag be drawn parallel to bc, the sum of the altitude and sublimity for any other semi-parabola having the same terminal speed will, in the manner explained, be equal to the distance between the parallel lines ag and bc. Moreover, since it has already been shown that the amplitudes of two semi-parabolas are the same when their angles of elevation differ from 45° by like amounts, it follows that the same computation which is employed for the larger elevation will serve also for the smaller. Let us also assume 10000 as the greatest amplitude for a parabola whose angle of elevation is 45°; this then will be the length of the line ba and the amplitude of the semi-parabola bc. This number, 10000, is selected because in these calculations we employ a table of tangents in which this is the value of the tangent of 45°. And now, coming down to business, draw the straight line ce making an acute angle ecb greater than acb: the problem now is to draw the semi-parabola to which the line ec is a tangent and for which the sum of the sublimity and the altitude is the distance ba. Take the length of the tangent*be from the table of tangents, using the angle bce as an argument: let f be the middle point of be; next find a third proportional to bf and bi (the half of bc), which is of necessity greater than fa. Call this fo. We have now discovered that, for the parabola inscribed[302] in the triangle ecb having the tangent ce and the amplitude cb, the altitude is bf and the sublimity fo. But the total length of bo exceeds the distance between the parallels ag and cb, while our problem was to keep it equal to this distance: for both the parabola sought and the parabola dc are described by projectiles fired from c with the same speed. Now since an infinite number of greater and smaller parabolas, similar to each other, may be described within the angle bce we must find another parabola which like cd has for the sum of its altitude and sublimity the height ba, equal to bc.
Fig. 124
Therefore lay off cr so that, ob:ba=bc:cr; then cr will be the amplitude of a semi-parabola for which bce is the angle of elevation and for which the sum of the altitude and sublimity is the distance between the parallels ga and cb, as desired. The process is therefore as follows: One draws the tangent of the given angle bce; takes half of this tangent, and adds to it the quantity, fo, which is a third proportional to the half of this tangent and the half of bc; the desired amplitude cr is then found from the following proportion ob:ba=bc:cr. For example let the angle ecb be one of 50°; its tangent is 11918, half of which, namely bf, is 5959; half of bc is 5000; the third proportional of these halves is 4195, which added to bf gives the value 10154 for bo. Further, as ob is to ab, that is, as 10154 is to 10000, so is bc, or 10000 (each being the tangent of 45°) to cr, which is the amplitude sought and which has the value 9848, the maximum amplitude being bc, or 10000. The amplitudes of the entire parabolas are double these, namely, 19696 and 20000. This is also the amplitude of a parabola whose angle of elevation is 40°, since it deviates by an equal amount from one of 45°.
In order to thoroughly understand this demonstration I need to be shown how the third proportional of bf and bi is, as the Author indicates, necessarily greater than fa.
#### Salv.
This result can, I think, be obtained as follows. The square of the mean proportional between two lines is equal to the rectangle formed by these two lines. Therefore the square of bi (or of bd which is equal to bi) must be equal to the rectangle formed by fb and the desired third proportional. This third proportional is necessarily greater than fa because the rectangle formed by bf and fa is less than the square of bd by an amount equal to the square of df, as shown in Euclid, II. 1. Besides it is to be observed that the point f, which is the middle point of the tangent eb, falls in general above a and only once at a; in which cases it is self-evident that the third proportional to the half of the tangent and to the sublimity bi lies wholly above a. But the Author has taken a case where it is not evident that the third proportional is always greater than fa, so that when laid off above the point f it extends beyond the parallel ag.
Now let us proceed. It will be worth while, by the use of this table, to compute another giving the altitudes of these semi-parabolas described by projectiles having the same initial speed. The construction is as follows:
70° 6428 20° 26° 1922 71° 8940 71 6157 19 27 2061 72 9045 72 5878 18 28 2204 73 9144 73 5592 17 29 2351 74 9240 74 5300 16 30 2499 75 9330 75 5000 15 31 2653 76 9415 76 4694 14 32 2810 77 9493 77 4383 13 33 2967 78 9567 78 4067 12 34 3128 79 9636 79 3746 11 35 3289 80 9698 80 3420 10 36 3456 81 9755 81 3090 9 37 3621 82 9806 82 2756 8 38 3793 83 9851 83 2419 7 39 3962 84 9890 84 2079 6 40 4132 85 9924 85 1736 5 41 4302 86 9951 86 1391 4 42 4477 87 9972 87 1044 3 43 4654 88 9987 88 698 2 44 4827 89 9998 89 349 1 45 5000 90 10000
### Problem. Proposition XIII[305]
From the amplitudes of semi-parabolas given in the preceding table to find the altitudes of each of the parabolas described with the same initial speed.
It will be worth while to add a third table giving the altitudes and sublimities for parabolas in which the amplitude is a constant.
#### Sagr.
I shall be very glad to see this; for from it I shall learn the difference of speed and force [degl’ impeti e delle forze] required to fire projectiles over the same range with what we call mortar shots. This difference will, I believe, vary greatly with the elevation so that if, for example, one wished to employ an elevation of 3° or 4°, or 87° or 88° and yet give the ball the same range which it had with an elevation of 45° (where we have shown the initial speed to be a minimum) the excess of force required will, I think, be very great.
#### Salv.
You are quite right, sir; and you will find that in order to perform this operation completely, at all angles of elevation, you will have to make great strides toward an infinite speed. We pass now to the consideration of the table.
39° 4049 6174 84° 47572 525 40 4196 5959 85 57150 437 41 4346 5752 86 71503 349 42 4502 5553 87 95405 262 43 4662 5362 88 143181 174 44 4828 5177 89 286499 87 45 5000 5000 90 infinita
### Proposition XIV[308]
To find for each degree of elevation the altitudes and sublimities of parabolas of constant amplitude.
Here we see, first of all, how very true is the statement made above, that, for different angles of elevation, the greater the deviation from the mean, whether above or below, the greater the initial speed [impeto e violenza] required to carry the projectile over the same range. For since the speed is the resultant of two motions, namely, one horizontal and uniform, the other vertical and naturally accelerated; and since the sum of the altitude and sublimity represents this speed, it is seen from the preceding table that this sum is a minimum for an elevation of 45° where the altitude and sublimity are equal, namely, each 5000; and their sum 10000. But if we choose a greater elevation, say 50°, we shall find the altitude 5959, and the sublimity 4196, giving a sum of 10155; in like manner we shall find that this is precisely the value of the speed at 40° elevation, both angles deviating equally from the mean.
Secondly it is to be noted that, while equal speeds are required for each of two elevations that are equidistant from the mean, there is this curious alternation, namely, that the altitude and sublimity at the greater elevation correspond inversely to the sublimity and altitude at the lower elevation. Thus in the[309] preceding example an elevation of 50° gives an altitude of 5959 and a sublimity of 4196; while an elevation of 40° corresponds to an altitude of 4196 and a sublimity of 5959. And this holds true in general; but it is to be remembered that, in order to escape tedious calculations, no account has been taken of fractions which are of little moment in comparison with such large numbers.
#### Sagr.
I note also in regard to the two components of the initial speed [impeto] that the higher the shot the less is the horizontal and the greater the vertical component; on the other hand, at lower elevations where the shot reaches only a small height the horizontal component of the initial speed must be great. In the case of a projectile fired at an elevation of 90°, I quite understand that all the force [forza] in the world would not be sufficient to make it deviate a single finger’s breadth from the perpendicular and that it would necessarily fall back into its initial position; but in the case of zero elevation, when the shot is fired horizontally, I am not so certain that some force, less than infinite, would not carry the projectile some distance; thus not even a cannon can fire a shot in a perfectly horizontal direction, or as we say, point blank, that is, with no elevation at all. Here I admit there is some room for doubt. The fact I do not deny outright, because of another phenomenon apparently no less remarkable, but yet one for which I have conclusive evidence. This phenomenon is the impossibility of stretching a rope in such a way that it shall be at once straight and parallel to the horizon; the fact is that the cord always sags and bends and that no force is sufficient to stretch it perfectly straight.
#### Salv.
In this case of the rope then, Sagredo, you cease to wonder at the phenomenon because you have its demonstration; but if we consider it with more care we may possibly discover some correspondence between the case of the gun and that of the string. The curvature of the path of the shot fired horizontally appears to result from two forces, one (that of the weapon) drives it horizontally and the other (its own weight) draws it vertically downward. So in stretching the rope you have the force which pulls it horizontally and its own weight which acts downwards. The circumstances in these two cases are, therefore, very similar. If then you attribute to the weight of the rope a power and[310] energy [possanza ed energia] sufficient to oppose and overcome any stretching force, no matter how great, why deny this power to the bullet?
Besides I must tell you something which will both surprise and please you, namely, that a cord stretched more or less tightly assumes a curve which closely approximates the parabola. This similarity is clearly seen if you draw a parabolic curve on a vertical plane and then invert it so that the apex will lie at the bottom and the base remain horizontal; for, on hanging a chain below the base, one end attached to each extremity of the base, you will observe that, on slackening the chain more or less, it bends and fits itself to the parabola; and the coincidence is more exact in proportion as the parabola is drawn with less curvature or, so to speak, more stretched; so that using parabolas described with elevations less than 45° the chain fits its parabola almost perfectly.
#### Sagr.
Then with a fine chain one would be able to quickly draw many parabolic lines upon a plane surface.
#### Salv.
Certainly and with no small advantage as I shall show you later.
#### Simp.
But before going further, I am anxious to be convinced at least of that proposition of which you say that there is a rigid demonstration; I refer to the statement that it is impossible by any force whatever to stretch a cord so that it will lie perfectly straight and horizontal.
#### Sagr.
I will see if I can recall the demonstration; but in order to understand it, Simplicio, it will be necessary for you to take for granted concerning machines what is evident not alone from experiment but also from theoretical considerations, namely, that the velocity of a moving body [velocità del movente], even when its force [forza] is small, can overcome a very great resistance exerted by a slowly moving body, whenever the velocity of the moving body bears to that of the resisting body a greater ratio than the resistance [resistenza] of the resisting body to the force [forza] of the moving body.
#### Simp.
This I know very well for it has been demonstrated by Aristotle in his Questions in Mechanics; it is also clearly seen in the lever and the steelyard where a counterpoise weighing not more than 4 pounds will lift a weight of 400 provided that the distance of the counterpoise from the axis about which the steelyard rotates be more than one hundred times as great as the distance between this axis and the point of support for[311] the large weight. This is true because the counterpoise in its descent traverses a space more than one hundred times as great as that moved over by the large weight in the same time; in other words the small counterpoise moves with a velocity which is more than one hundred times as great as that of the large weight.
#### Sagr.
You are quite right; you do not hesitate to admit that however small the force [forza] of the moving body it will overcome any resistance, however great, provided it gains more in velocity than it loses in force and weight [vigore e gravità]. Now let us return to the case of the cord. In the accompanying figure ab represents a line passing through two fixed points a and b; at the extremities of this line hang, as you see, two large weights c and d, which stretch it with great force and keep it truly straight, seeing that it is merely a line without weight. Now I wish to remark that if from the middle point of this line, which we may call e, you suspend any small weight, say h, the line ab will yield toward the point f and on account of its elongation will compel the two heavy weights c and d to rise. This I shall demonstrate as follows: with the points a and b as centers describe the two quadrants, eig and elm; now since the two semidiameters ai and bl are equal to ae and eb, the remainders fi and fl are the excesses of the lines af and fb over ae and eb; they therefore determine the rise of the weights c and d, assuming of course that the weight h has taken the position f. But the weight h[312] will take the position f, whenever the line ef which represents the descent of h bears to the line fi—that is, to the rise of the weights c and d—a ratio which is greater than the ratio of the weight of the two large bodies to that of the body h. Even when the weights of c and d are very great and that of h very small this will happen; for the excess of the weights c and d over the weight of h can never be so great but that the excess of the tangent ef over the segment fi may be proportionally greater. This may be proved as follows: Draw a circle of diameter gai; draw the line bo such that the ratio of its length to another length c, c>d, is the same as the ratio of the weights c and d to the weight h. Since c>d, the ratio of bo to d is greater than that of bo to c. Take be a third proportional to ob and d; prolong the diameter gi to a point f such that gi:if=oe:eb; and from the point f draw the tangent fn; then since we already have oe:eb=gi:if, we shall obtain, by compounding ratios, ob:eb=gf:if. But d is a mean proportional between ob and be; while nf is a mean proportional between gf and fi. Hence nf bears to fi the same ratio as that of cb to d, which is greater than that of the weights c and d to the weight h. Since then the descent, or velocity, of the weight h bears to the rise, or velocity, of the weights c and d a greater ratio than the weight of the bodies c and d bears to the weight of h, it is clear that the weight h will descend and the line ab will cease to be straight and horizontal.
Fig. 126
And now this which happens in the case of a weightless cord ab when any small weight h is attached at the point e, happens also when the cord is made of ponderable matter but without any attached weight; because in this case the material of which the cord is composed functions as a suspended weight.
#### Simp.
I am fully satisfied. So now Salviati can explain, as he promised, the advantage of such a chain and, afterwards, present the speculations of our Academician on the subject of impulsive forces [forza della percossa].
#### Salv.
Let the preceding discussions suffice for to-day; the hour is already late and the time remaining will not permit us to clear up the subjects proposed; we may therefore postpone our meeting until another and more opportune occasion.
#### Sagr.
I concur in your opinion, because after various conversations with intimate friends of our Academician I have concluded that this question of impulsive forces is very obscure, and I think that, up to the present, none of those who have treated[313] this subject have been able to clear up its dark corners which lie almost beyond the reach of human imagination; among the various views which I have heard expressed one, strangely fantastic, remains in my memory, namely, that impulsive forces are indeterminate, if not infinite. Let us, therefore, await the convenience of Salviati. Meanwhile tell me what is this which follows the discussion of projectiles.
#### Salv.
These are some theorems pertaining to the centers of gravity of solids, discovered by our Academician in his youth, and undertaken by him because he considered the treatment of Federigo Comandino to be somewhat incomplete. The propositions which you have before you would, he thought, meet the deficiencies of Comandino’s book. The investigation was undertaken at the instance of the Illustrious Marquis Guid’ Ubaldo Dal Monte, a very distinguished mathematician of his day, as is evidenced by his various publications. To this gentleman our Academician gave a copy of this work, hoping to extend the investigation to other solids not treated by Comandino. But a little later there chanced to fall into his hands the book of the great geometrician, Luca Valerio, where he found the subject treated so completely that he left off his own investigations, although the methods which he employed were quite different from those of Valerio.
#### Sagr.
Please be good enough to leave this volume with me until our next meeting so that I may be able to read and study these propositions in the order in which they are written.
#### Salv.
It is a pleasure to comply with your request and I only hope that the propositions will be of deep interest to you.
end of fourth day.
[* ]A very near approach to Newton’s Second Law of Motion. [Trans.]
[* ]In the original this theorem reads as follows:
Si aliquod mobile duplici motu æquabili moveatur, nempe orizontali et perpendiculari, impetus seu momentum lationis ex utroque motu compositæ erit potentia æqualis ambobus momentis priorum motuum.
For the justification of this translation of the word “potentia” and of the use of the adjective “resultant” see p. 266 below. [Trans.]
[* ]See p. 169 above. [Trans.]
[* ]Galileo here proposes to employ as a standard of velocity the terminal speed of a body falling freely from a given height. [Trans.]
[* ]In modern mechanics this well-known theorem assumes the following form: The speed of a projectile at any point is that produced by a fall from the directrix. [Trans.]
[* ]The reader will observe that the word “tangent” is here used in a sense somewhat different from that of the preceding sentence. The “tangent ec” is a line which touches the parabola at c; but the “tangent eb” is the side of the right-angled triangle which lies opposite the angle ecb, a line whose length is proportional to the numerical value of the tangent of this angle. [Trans.]
[]This fact is demonstrated in the third paragraph below. [Trans.] | 3.078125 | CC-MAIN-2014-10 |
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InterferenceExamples
# InterferenceExamples - Examples 1.A frequency generator in...
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Examples 1 .A frequency generator in DL creates a sinusoidal signal: Signal t ( )= A sin 2 ! t 0.00025 s ( ) + ! 2 " # \$ \$ % . This generator is connected to two speakers. However, speaker 2 has its wires switched, with respect to the correctly wired speaker 1. Assume that the sound waves generated by these speakers are 1D. (a) What are the harmonic sound wave functions for the sounds produced by speaker 1 and for speaker 2? (b) At time t = 0, what are the total phases ! 1 and ! 2 , and what is the difference in total phase ΔΦ , at the location of the microphone? Is there constructive or destructive interference? Use a phase chart to organize your answer. Solution Since these are sound waves, the wavelength l of the sound waves created by speakers 1 and 2 is: ! = v wave f = v wave " = 340 m s # \$ % ( 0.00025 s ( ) = 0.085 m . The harmonic wave functions for the sounds coming from speakers 1 and 2 are then given by: ! P 1 x , t ( )= A sin 2 " t 0.00025 s ( ) 2 " x 0.085 m ( ) + " 2 # \$ % % ( ( ; ! P 2 x , t ( ) = A sin 2 " t 0.00025 s ( ) 2 " x 0.085 m
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# Almost 50% SC questions use <ing>, non-ing forms in
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Almost 50% SC questions use , non-ing forms in [#permalink]
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15 Sep 2007, 00:46
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Almost 50% SC questions use <ing>, non-ing forms in the various options.
Lets get the clarification about the usage of various <ing> forms.
Start the discussions with the following example:
"Consultants engage in preliminary discussions with clients, complete an initial analysis of ......."
So where do you feel "engage" like above has to be used with <ing> form, and where we need <ing> forms.
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16 Sep 2007, 09:19
Some verbs can take either the gerund or the infinitive and the sentence still means the same thing.
but other times it does,
* With the verb remember - "I remembered to do my homework". or "I remembered doing my homework."
In the first sentence (I remembered to do my homework), the person speaking remembered they had some homework first and then carried out the action and did it. In the second sentence (I remembered doing my homework.), the person speaking carried out the action (their homework) first and then remembered doing it.
It looks like it will not be easy to spot whether the meaning of the sentence will change.
If anyone has more on this it would be appreciated.
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18 Sep 2007, 04:57
how commas....vlad cud u elaborate pls
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19 Sep 2007, 06:27
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An example of -ing participle
762. The growth of the railroads led to the abolition of local times, which was determined by when the sun reached the observer’s meridian and differing from city to city, and to the establishment of regional times.
(A) which was determined by when the sun reached the observer’s meridian and differing
(B) which was determined by when the sun reached the observer’s meridian and which differed
(C) which were determined by when the sun reached the observer’s meridian and differing
(D) determined by when the sun reached the observer’s meridian and differed
(E) determined by when the sun reached the observer’s meridian and differing
IMO E
"Determined" is participle whereas "differed" is past form of verb. So, "differing" , which is a participle, is appropriate.
Here, "the abolition of local times" if followed by which/ that would modify the noun before the "of" preposition structure (of + preposition object) therefore GMAT would like us to use a modifying phrase. Therefore, we let the A, B and C go.
Now the tough part. To understand which option is correct, we must first try to construct how the clauses would have modified "local time"
"local times, which were determined by when the sun reached the observer’s meridian and which differed"
When we reduce an adjective clause to a modifying phrase the rule is:
1. If there is a "be" verb after the "pronoun" (which/ that), then we simply remove the pronoun and the "be" verb.
2. If "be" verb is absent, then we remove the "pronoun" (which/ that) and change the verb to verb-ing (participle)
Using the rules the sentence becomes:
"local times, determined by when the sun reached the observer’s meridian and differing"
and this is the choice E.
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19 Sep 2007, 07:12
msrinath wrote:
An example of -ing participle
762. The growth of the railroads led to the abolition of local times, which was determined by when the sun reached the observer’s meridian and differing from city to city, and to the establishment of regional times.
(A) which was determined by when the sun reached the observer’s meridian and differing
(B) which was determined by when the sun reached the observer’s meridian and which differed
(C) which were determined by when the sun reached the observer’s meridian and differing
(D) determined by when the sun reached the observer’s meridian and differed
(E) determined by when the sun reached the observer’s meridian and differing
IMO E
"Determined" is participle whereas "differed" is past form of verb. So, "differing" , which is a participle, is appropriate.
Here, "the abolition of local times" if followed by which/ that would modify the noun before the "of" preposition structure (of + preposition object) therefore GMAT would like us to use a modifying phrase. Therefore, we let the A, B and C go.
Now the tough part. To understand which option is correct, we must first try to construct how the clauses would have modified "local time"
"local times, which were determined by when the sun reached the observer’s meridian and which differed"
When we reduce an adjective clause to a modifying phrase the rule is:
1. If there is a "be" verb after the "pronoun" (which/ that), then we simply remove the pronoun and the "be" verb.
2. If "be" verb is absent, then we remove the "pronoun" (which/ that) and change the verb to verb-ing (participle)
Using the rules the sentence becomes:
"local times, determined by when the sun reached the observer’s meridian and differing"
and this is the choice E.
thank you for explaining these rules.
However, could you please clarify what it means when you say:
"When we reduce an adjective clause to a modifying phrase"
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19 Sep 2007, 07:39
DTM wrote:
msrinath wrote:
An example of -ing participle
762. The growth of the railroads led to the abolition of local times, which was determined by when the sun reached the observer’s meridian and differing from city to city, and to the establishment of regional times.
(A) which was determined by when the sun reached the observer’s meridian and differing
(B) which was determined by when the sun reached the observer’s meridian and which differed
(C) which were determined by when the sun reached the observer’s meridian and differing
(D) determined by when the sun reached the observer’s meridian and differed
(E) determined by when the sun reached the observer’s meridian and differing
IMO E
"Determined" is participle whereas "differed" is past form of verb. So, "differing" , which is a participle, is appropriate.
Here, "the abolition of local times" if followed by which/ that would modify the noun before the "of" preposition structure (of + preposition object) therefore GMAT would like us to use a modifying phrase. Therefore, we let the A, B and C go.
Now the tough part. To understand which option is correct, we must first try to construct how the clauses would have modified "local time"
"local times, which were determined by when the sun reached the observer’s meridian and which differed"
When we reduce an adjective clause to a modifying phrase the rule is:
1. If there is a "be" verb after the "pronoun" (which/ that), then we simply remove the pronoun and the "be" verb.
2. If "be" verb is absent, then we remove the "pronoun" (which/ that) and change the verb to verb-ing (participle)
Using the rules the sentence becomes:
"local times, determined by when the sun reached the observer’s meridian and differing"
and this is the choice E.
thank you for explaining these rules.
However, could you please clarify what it means when you say:
"When we reduce an adjective clause to a modifying phrase"
Hi msrinath,
Pls try to provide small examples supporting ur explainaiton of the rules. It become difficult to understand,without small examples.
Thanks for such efforts and nice explainations.
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19 Sep 2007, 18:05
msrinath wrote:
An example of -ing participle
When we reduce an adjective clause to a modifying phrase the rule is:
1. If there is a "be" verb after the "pronoun" (which/ that), then we simply remove the pronoun and the "be" verb.
2. If "be" verb is absent, then we remove the "pronoun" (which/ that) and change the verb to verb-ing (participle)
Can you elaborate this? Maybe using some examples.
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19 Sep 2007, 20:37
bkk145 wrote:
msrinath wrote:
An example of -ing participle
When we reduce an adjective clause to a modifying phrase the rule is:
1. If there is a "be" verb after the "pronoun" (which/ that), then we simply remove the pronoun and the "be" verb.
2. If "be" verb is absent, then we remove the "pronoun" (which/ that) and change the verb to verb-ing (participle)
Can you elaborate this? Maybe using some examples.
As I understand....
For example:
the cat, which was chased by your dog and escaped by jumping on a high wall, is mine (sorry for my poor example )
1.since after "which" is "was chased" , the modification is to eliminate "which was"
2. Since "escaped" is without a be-verb, the modification is to transform "escaped" into "escaping"
==> The overall modification is:
the cat, chased by your dog and escaping by jumping on a high wall, is mine
Hope you all catch it
BTW, thank to the author who posted this interesting grammatical point
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19 Sep 2007, 20:43
Hi, msrinath, could you please divulge the source of this grammatical point to us all? Thank you
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20 Sep 2007, 08:40
Thanks for your wonderful explanation ...Msrinath ...buty could you please let us know ...from where did you learn this rule ..I mean the source ...
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20 Sep 2007, 11:57
Check this interesting observation by vlad...
http://www.gmatclub.com/forum/p372412#372412
lets combine these 2 posts together for discussions.
Will surely get 2-3 SC questions based on this rule.
Cheers!!
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Re: SC Rule: use of <-ing> forms [#permalink]
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20 Sep 2007, 12:16
younggun044 wrote:
:idea: Almost 50% SC questions use <ing>, non-ing forms in the various options.
Lets get the clarification about the usage of various <ing> forms.
Start the discussions with the following example:
"Consultants engage in preliminary discussions with clients, complete an initial analysis of ......."
So where do you feel "engage" like above has to be used with <ing> form, and where we need <ing> forms.
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20 Sep 2007, 22:52
One notable point
If a participial phrase comes at the end of a sentence, a comma usually precedes the phrase if it modifies an earlier word in the sentence but not if the phrase directly follows the word it modifies.
The local residents often saw Ken wandering through the streets.
(The phrase modifies Ken, not residents.)
Tom nervously watched the woman, alarmed by her silence.
(The phrase modifies Tom, not woman.)
Younggun, you can just type "participial phrase" in google to get a bunch of useful elaborations
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20 Sep 2007, 23:34
According to this URL http://owl.english.purdue.edu/handouts/ ... rbals.html
If a participial phrase comes at the end of a sentence, a comma usually precedes the phrase if it modifies an earlier word in the sentence but not if the phrase directly follows the word it modifies.
The local residents often saw Ken wandering through the streets.
(The phrase modifies Ken, not residents.)
Tom nervously watched the woman, alarmed by her silence.
(The phrase modifies Tom, not woman.)
Then, back to the example provided by msrinath:
"The growth of the railroads led to the abolition of local times, determined by when the sun reached the observer’s meridian and differing"
If we follow the rule provided by the authoritative page, "determined...." will be understood to refer to "the growth of..."
Apart from the meaning and logic of the original sentence, is there a more universal rule governing adjective phrase?
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21 Sep 2007, 14:16
laxieqv wrote:
According to this URL http://owl.english.purdue.edu/handouts/ ... rbals.html
If a participial phrase comes at the end of a sentence, a comma usually precedes the phrase if it modifies an earlier word in the sentence but not if the phrase directly follows the word it modifies.
The local residents often saw Ken wandering through the streets.
(The phrase modifies Ken, not residents.)
Tom nervously watched the woman, alarmed by her silence.
(The phrase modifies Tom, not woman.)
Then, back to the example provided by msrinath:
"The growth of the railroads led to the abolition of local times, determined by when the sun reached the observer’s meridian and differing"
If we follow the rule provided by the authoritative page, "determined...." will be understood to refer to "the growth of..."
Apart from the meaning and logic of the original sentence, is there a more universal rule governing adjective phrase?
I guess its not contradiction, since determined by does not come at the end of the sentence.
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21 Sep 2007, 18:46
oops, my mistake
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24 Sep 2010, 09:54
tone of kudos msrinath!!!
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Re: SC Rule: use of <-ing> forms [#permalink]
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24 Sep 2010, 17:34
(D) determined by when the sun reached the observer’s meridian and differed
(E) determined by when the sun reached the observer’s meridian and differing
I would pick "differing" over "differed" because the "differing" phrase provides more information on the time at which the sun reaches an observer's median, whereas "differed" makes the phrase parallel to "the sun.." which is inappropriate.
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Re: SC Rule: use of <-ing> forms [#permalink]
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19 Dec 2010, 07:07
In E, could determined and differing refer to "The growth" instead of "local times"?
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Re: SC Rule: use of <-ing> forms [#permalink]
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19 Dec 2010, 10:35
1
KUDOS
To come to the original question where one should use a word like ‘engage’ and where one should use the ‘+ ing' form’:
1. engage: a verb when we say the consultants ‘engage’. – A description of an action that normally occurs
2. Engaging: A present progressive when we say consultants ‘are engaging’ – a description of action that is happening at the time of writing,
3. Consultants, engaging in preliminary …. A present participle used as post –subject modifier.
4. Engaging in preliminary discussions with clients, consultants; a present participle used a pre- subject modifier.
5. In addition ‘engaging’ can also be used as gerund in certain cases although ‘engagement ‘will be largely preferred in such cases. : Engaging adolescent children is a tough task. Engaging describes a noun
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# Raising a number to the second power?
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2011-09-07 22:33:38
Raising a number to the second power means you are squaring the number, multiplying it by itself. 2 squared= 2*2=4
3 squared=3*3=9
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## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
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it is called squared | 3.484375 | CC-MAIN-2023-23 |
Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > csbexg Structured version Visualization version GIF version
Theorem csbexg 5179
Description: The existence of proper substitution into a class. (Contributed by NM, 10-Nov-2005.) (Revised by NM, 17-Aug-2018.)
Assertion
Ref Expression
csbexg (∀𝑥 𝐵𝑊𝐴 / 𝑥𝐵 ∈ V)
Proof of Theorem csbexg
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 df-csb 3829 . . 3 𝐴 / 𝑥𝐵 = {𝑦[𝐴 / 𝑥]𝑦𝐵}
2 abid2 2932 . . . . . . . 8 {𝑦𝑦𝐵} = 𝐵
3 elex 3459 . . . . . . . 8 (𝐵𝑊𝐵 ∈ V)
42, 3eqeltrid 2894 . . . . . . 7 (𝐵𝑊 → {𝑦𝑦𝐵} ∈ V)
54alimi 1813 . . . . . 6 (∀𝑥 𝐵𝑊 → ∀𝑥{𝑦𝑦𝐵} ∈ V)
6 spsbc 3733 . . . . . 6 (𝐴 ∈ V → (∀𝑥{𝑦𝑦𝐵} ∈ V → [𝐴 / 𝑥]{𝑦𝑦𝐵} ∈ V))
75, 6syl5 34 . . . . 5 (𝐴 ∈ V → (∀𝑥 𝐵𝑊[𝐴 / 𝑥]{𝑦𝑦𝐵} ∈ V))
8 nfcv 2955 . . . . . 6 𝑥V
98sbcabel 3807 . . . . 5 (𝐴 ∈ V → ([𝐴 / 𝑥]{𝑦𝑦𝐵} ∈ V ↔ {𝑦[𝐴 / 𝑥]𝑦𝐵} ∈ V))
107, 9sylibd 242 . . . 4 (𝐴 ∈ V → (∀𝑥 𝐵𝑊 → {𝑦[𝐴 / 𝑥]𝑦𝐵} ∈ V))
1110imp 410 . . 3 ((𝐴 ∈ V ∧ ∀𝑥 𝐵𝑊) → {𝑦[𝐴 / 𝑥]𝑦𝐵} ∈ V)
121, 11eqeltrid 2894 . 2 ((𝐴 ∈ V ∧ ∀𝑥 𝐵𝑊) → 𝐴 / 𝑥𝐵 ∈ V)
13 csbprc 4313 . . . 4 𝐴 ∈ V → 𝐴 / 𝑥𝐵 = ∅)
14 0ex 5176 . . . 4 ∅ ∈ V
1513, 14eqeltrdi 2898 . . 3 𝐴 ∈ V → 𝐴 / 𝑥𝐵 ∈ V)
1615adantr 484 . 2 ((¬ 𝐴 ∈ V ∧ ∀𝑥 𝐵𝑊) → 𝐴 / 𝑥𝐵 ∈ V)
1712, 16pm2.61ian 811 1 (∀𝑥 𝐵𝑊𝐴 / 𝑥𝐵 ∈ V)
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ∧ wa 399 ∀wal 1536 ∈ wcel 2111 {cab 2776 Vcvv 3441 [wsbc 3720 ⦋csb 3828 ∅c0 4243 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2113 ax-9 2121 ax-10 2142 ax-11 2158 ax-12 2175 ax-ext 2770 ax-nul 5175 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-tru 1541 df-fal 1551 df-ex 1782 df-nf 1786 df-sb 2070 df-clab 2777 df-cleq 2791 df-clel 2870 df-nfc 2938 df-v 3443 df-sbc 3721 df-csb 3829 df-dif 3884 df-nul 4244 This theorem is referenced by: csbex 5180 abfmpeld 30431
Copyright terms: Public domain W3C validator | 3.359375 | CC-MAIN-2024-33 |
# R/rstar.R In generalCorr: Generalized Correlations, Causal Paths and Portfolio Selection
#### Documented in rstar
#' Function to compute generalized correlation coefficients r*(x,y).
#'
#' Uses Vinod (2015) definition of generalized (asymmetric) correlation
#' coefficients. It requires kernel regression of x on y obtained by using the np' package.
#' It also reports usual Pearson correlation coefficient r and p-value for testing
#' the null hypothesis that (population r)=0.
#' @param x {Vector of data on the dependent variable}
#' @param y {Vector of data on the regressor}
#' @importFrom stats cor.test
#' @return Four objects created by this function are:
#' \item{corxy}{r*x|y or regressing x on y}
#' \item{coryx}{r*y|x or regressing y on x}
#' \item{pearson.r}{Pearson's product moment correlation coefficient}
#' \item{pv}{The p-value for testing the Pearson r}
#' @note This function needs the kern function which in turn needs the np package.
#' @author Prof. H. D. Vinod, Economics Dept., Fordham University, NY
#' @references Vinod, H. D. Generalized Correlation and Kernel Causality with
#' Applications in Development Economics' in Communications in
#' Statistics -Simulation and Computation, 2015,
#' \url{https://doi.org/gffn86}
#'
#' @references Vinod, H. D. 'Matrix Algebra Topics in Statistics and Economics
#' Using R', Chapter 4 in Handbook of Statistics: Computational Statistics
#' with R, Vol.32, co-editors: M. B. Rao and C.R. Rao. New York:
#' North Holland, Elsevier Science Publishers, 2014, pp. 143-176.
#' @concept asymmetric p-values
#' @examples
#' x=sample(1:30);y=sample(1:30); rstar(x,y)
#'
#' @export
rstar <- function(x, y) {
c1 = cor.test(x, y)
sig = sign(c1$estimate) pv = c1$p.value
pearson.r = c1$estimate mod.1 = kern(dep.y = x, reg.x = y) mod.2 = kern(dep.y = y, reg.x = x) corxy = sqrt(mod.1$R2) * sig
coryx = sqrt(mod.2\$R2) * sig
list(corxy = corxy, coryx = coryx, pearson.r = pearson.r, pv = pv)
}
## Try the generalCorr package in your browser
Any scripts or data that you put into this service are public.
generalCorr documentation built on Jan. 4, 2022, 1:08 a.m. | 2.59375 | CC-MAIN-2022-21 |
# 4 Simple Steps to Calculating Your Social Security Benefit: Educator Edition
If you have a pension from work where you did not pay Social Security taxes, but qualified for SS benefits from other work, your SS benefits formula is NOT THE SAME AS IT IS FOR EVERYONE ELSE! Here’s how it is different…
Transcript Follows:
Hey everyone this is Devin Carroll with Social Security Intelligence.com. If you’ve qualified for a Social Security benefit from prior work, and also qualified for a pension from work where you did not pay Social Security tax, you need to understand the differences in how your Social Security benefit is going to be calculated.
There’s only four simple steps to calculating this. And it’s the same steps used for the individual who does not have this non-covered pension. So if you worked in accounting or as a nurse the same formula is used with one slight difference in it.
So I want to go through these four steps and make it make sense for you so you can do the calculation for yourself. Let’s dive in.
So the first step is that your earnings are all going to be adjusted upwards for inflation.
The second step is the calculation of your average index monthly earnings. And this is meant to give them the average amount that you earned on a monthly basis after it was adjusted for inflation.
Once they have that amount they will run that through the formula to calculate your primary insurance amount which is also your full retirement age benefit amount.
And the last step there are going to adjust it out for your filing age. So let’s dive into each of these steps and show you how you can calculate your own benefit.
Step one we’re going to adjust your earnings. So let’s take an individual who went to work in 1976 and began earning fifteen thousand dollars per year. The earnings went up every year thereafter until 1981 at which point he became a teacher in the great state of Texas. Now this could be one of the other 15 states where teachers don’t pay into Social Security.
But you’ll notice that all the way up until 2005 under the actual earnings there are zeroes. Now this doesn’t mean that he wasn’t earning anything. It just means that when the Social Security Administration looks at your historical earnings if you were in a job where you did not pay Social Security tax, they see a zero there. And then this individual goes back at age 55 into accounting or he makes \$36000 a year. And you can see that there were some annual increases thereafter. And at 66 he’s finished. He’s retired and now it’s time to file for Social Security. Now when the Social Security Administration looks at these historical earnings the actual earnings aren’t that important to them. What’s much more important are the inflation adjusted earnings. They maintain a table on their website that shows how to adjust these earnings.
We’ve taken the actual earnings here and adjusted those. So the fifteen thousand dollars in earnings in 1976 would be adjusted up to sixty six thousand one hundred eighty seven dollars. That’s meant to reflect the cost of living adjustments that have happened over the years to make sure that your Social Security benefit is going to be enough to keep up with that amount. If they used your actual earnings instead your benefit amount would actually be much lower when they finish this calculation.
And you can go on through here and see I’ve actually added the adjusted earnings in here from the Social Security tables. And those are the numbers that the Social Security ministration is going to use when they’re moving to step two which is to calculate the averaged index monthly earnings.
They’ll also refer to that by its acronym name, AIME. So they’ll take all of the adjusted earnings and they will sum up the highest thirty five years. So what happens if you don’t have 35 years? Well they’ll still use 35 years in their calculation. But if you have 20 years of earnings they’re simply going to use 15 years of zeroes. And so in this case if you add up the 35 highest years you come up to an amount of \$874,698 and then they divide that by 420 which is simply the number of months in 35 years because remember they are wanting to get to the average indexed monthly earnings and that gives them the amount of \$2,082. That is your AIME or your averaged index monthly earnings. Once they have this amount they can move to step three which is to calculate your primary insurance amount.
Now this is where the formula gets slightly different for individuals who have a non-covered pension and are subject to the windfall elimination provision and those who do not have a non-covered pension and are not subject to the WEP. So I will cover this formula first without considering the windfall elimination provision and then I’ll show you the slight change in the formula that they make that adds in the penalty amount for that windfall elimination provision. So they take your average index monthly earnings and they start to run through this grid. For the first \$896 of average indexed monthly earnings it’s applied to your full retirement age benefit at 90 percent. So in this case it would be. Eight hundred and six dollars and the remainder would be applied to your primary insurance amount at thirty to it now if you had a very high aim it would be applied at 15 percent at amounts beyond \$5,300 and\$40.
But in this case it did not exceed that fifty three hundred ninety nine dollars. And so it’s only into those first two bands and then you simply add up that last column that says PIA. And that gives you your full retirement age benefit. And in this case it would be \$1,185. Now here’s where it gets different. If you have a pension from work where you did not pay Social Security tax and this is the only difference between the calculations instead of 90 percent in that first band it’s 40 percent. And so in this case it would reduce down the first amount from \$806 down to \$358. And instead of an \$1,185 primary insurance amount that would be reduced down to \$737.
Now those amounts and the bend points change every year so from that zero to eight hundred ninety six next year it will likely be from zero to \$905-\$910 or somewhere in that range. And all of the other points will change with that. So that’s the way the reduction works. And then once that penalty amount is applied with that lower been point it’s time then and only then to reduce or increase your benefit because of filing age. If you’re born between 1943 and 1954 your full retirement age benefit is 66 and every year thereafter it increases by two months all the way up until 1960 at which point it is set at 67. That could certainly change in the future. But right now that’s unknown.
So using the full retirement age amount of 66 years of age that is when you can get 100 percent of your pay. So that \$737 if that was your benefit amount you would be able to get 100 percent of that. However if you filed early you would get less and you can see how it goes all the way back to age 62. You would only get 75 percent of that amount. However if you delayed and you waited all the way up until age 70 you would get a 132% of that amount. And though charts like these are fairly common where we express the increases or decreases on an annual basis they actually occur on a monthly basis. And here you can see that the 36 period prior to your full retirement age your benefit decreases by point .555% on a monthly basis and then anything beyond that it decreases by .417% and at your full retirement age and beyond it increases by .667% on a monthly basis.
So that’s it. That’s how you calculate your Social Security benefit if you have a pension from work or you did not pay Social Security tax.
Hope you’ve enjoyed this video. Check out my other videos on my YouTube channel. And also don’t forget to go to my Web site. Social security intelligence.com.
Thanks for watching! | 2.59375 | CC-MAIN-2019-30 |
Continue to Site
am my steps about my question solution about bjt correct?
circuit975
New Member
Hello there
I am faced with such a question
1) Calculate and determine the value of the resistor RE that will ensure that VC=6 V.
2) Redraw the transistor model of the equivalent circuit for small amplitude AC signals. Determine the voltage gain Av of the circuit.
I solved the question as follows ;
VB = [24K / 24K+82K] x 16V = 3,62 Volt
RB = 24Kx82K / 24k+82k = 18,56K
6 = 16 - ICx5,6k
Ic = 1,78mA
IC = ßxIb = 150xIb
Ib = 11,86 microamper
IE = 11,86x10^-6 x 151 = 1,79mA
re = 26mV / 1,79mA = 14,5
VB - VBE - IBxRB - IExRE = 0
3,62 - 0,7 - [ (11,86 x 10^6) x 18,56K ] - 1,79mA x RE = 0
RE = 402K
Av = -ß x Rc / ßre + (ß+1)RE
Av = -150 x 5600 / 2,175 + 0 (The reason why it is 0 is because the Re resistor and capacitor are in parallel, I know that it is an ineffective element in AC analysis.)
This is my solution, are my solution steps correct? or if there is a point where I made a mistake when converting units, can you warn me, I need to correct my shortcomings and learn the right one. thanks a lot in advance
RE will be less than 5.6 kOhms, as the voltage is less than the collector resistance and the current is slightly more, so you've got something wrong.
As a first estimate, ignore the base current. The emitter voltage will be 3.62 - 0.7 = 2.92 V. The emitter current will be 1.78 mA, the same as the collector current, so the emitter resistor needs to be 2.92 / .00178 = 1.6 kOhm
If you want to be more accurate, you can add in the base current to the emitter current, making it 1.792 mA. Also the base current will reduce the base voltage by 18560 * 0.00001186 = 0.22 V, so the emitter voltage becomes 2.70 V and the emitter resistor is 1.5 kOhm
However, the base-emitter voltage of 0.7 V will depend on temperature, so adjustments of 0.2 V or so probably don't matter on a real circuit.
RE will be less than 5.6 kOhms, as the voltage is less than the collector resistance and the current is slightly more, so you've got something wrong.
As a first estimate, ignore the base current. The emitter voltage will be 3.62 - 0.7 = 2.92 V. The emitter current will be 1.78 mA, the same as the collector current, so the emitter resistor needs to be 2.92 / .00178 = 1.6 kOhm
If you want to be more accurate, you can add in the base current to the emitter current, making it 1.792 mA. Also the base current will reduce the base voltage by 18560 * 0.00001186 = 0.22 V, so the emitter voltage becomes 2.70 V and the emitter resistor is 1.5 kOhm
However, the base-emitter voltage of 0.7 V will depend on temperature, so adjustments of 0.2 V or so probably don't matter on a real circuit.
2,92 / 0,00178 = 1,6k
How right is it to do it directly like this?
Because according to this equation; VB - VBE - IBxRB - IExRE = 0
It is as if we made a transaction without subtracting IBxRB. isn't this wrong?
2,92 / 0,00178 = 1,6k
How right is it to do it directly like this?
Because according to this equation; VB - VBE - IBxRB - IExRE = 0
It is as if we made a transaction without subtracting IBxRB. isn't this wrong?
The first estimate is done in a more simple way by ignoring the base current, so it's somewhat less accurate, but it is more simple and it gets an answer that is close to what will happen in reality. It is always good to have a simple estimate, just as a reality check. It was clear to me that 402 kOhm for RE was very wrong before I did any calculations. The more complications you add, the more possible places to make a huge error. If you have a simple and reliable estimate, it is a good basis and you would expect your more complicated and more accurate estimate to be similar.
In you equation
VB - VBE - IBxRB - IExRE = 0
VB is the base voltage before the base current is taken into account.
VBE is the base-emitter voltage drop
IB is the base current
RB is the effective base resistance
IE is the emitter current
RE is the emitter resistance
Ignoring IB x RB is an inaccuracy of the model, but it's only about 0.22 V in this case so it's the smallest item in the list and only changed RE from 1.6 k to 1.5 k, so it's not a big deal. As I mentioned the model of VBE = 0.7 V is not all that accurate anyhow.
You can have a more accurate representation of the situation, but whatever you do, it will only be a representation as there will always be aspects of the transistor that are not simulated correctly. You have a gain of 150 in the model. In practice that will change with current, temperature and between different examples of the transistor.
Also by estimating the base current as being zero for the simple model, that is modelling the transistor having a much larger (infinite) gain. If there is a large change in conditions, between normal gain and infinite gain, the circuit is probably badly designed.
The first estimate is done in a more simple way by ignoring the base current, so it's somewhat less accurate, but it is more simple and it gets an answer that is close to what will happen in reality. It is always good to have a simple estimate, just as a reality check. It was clear to me that 402 kOhm for RE was very wrong before I did any calculations. The more complications you add, the more possible places to make a huge error. If you have a simple and reliable estimate, it is a good basis and you would expect your more complicated and more accurate estimate to be similar.
In you equation
VB - VBE - IBxRB - IExRE = 0
VB is the base voltage before the base current is taken into account.
VBE is the base-emitter voltage drop
IB is the base current
RB is the effective base resistance
IE is the emitter current
RE is the emitter resistance
Ignoring IB x RB is an inaccuracy of the model, but it's only about 0.22 V in this case so it's the smallest item in the list and only changed RE from 1.6 k to 1.5 k, so it's not a big deal. As I mentioned the model of VBE = 0.7 V is not all that accurate anyhow.
You can have a more accurate representation of the situation, but whatever you do, it will only be a representation as there will always be aspects of the transistor that are not simulated correctly. You have a gain of 150 in the model. In practice that will change with current, temperature and between different examples of the transistor.
Also by estimating the base current as being zero for the simple model, that is modelling the transistor having a much larger (infinite) gain. If there is a large change in conditions, between normal gain and infinite gain, the circuit is probably badly designed.
Yes, RE should not be that high, now I understand exactly what you are saying, thank you.
So, is there a mistake here? Is there a wrong value or unit conversion or something about RE being shorted in the AC state of the circuit or apart from all this is there anything I need to know about the points to be considered when calculating Av or are all my steps here correct? ;
note: RE is short-circuited in AC analysis because it is connected in parallel with the capacitor, hence (ß+1)RE = 0
-ß x Rc / ßre + (ß+1)RE = -150 x 5600 / 2,175 + 0 = -386,206
The equation that you have doesn't work for very low values of RE, so it won't work on AC where you have put in a value of zero for the impedance of the capacitor.
The circuit is stable and the gain is set by the resistor values. That is preferable because the transistor gain varies so much. There is effectively feedback, because if the transistor conducts more, the emitter voltage rises, which reduces the base current.
If the emitter voltage, or the AC component of the voltage, is zero, you need a more complicated model. When the input goes up, there will be a large increase in base current, and it is difficult to predict that. The model of having a fixed base-emitter voltage of 0.7 V is not sophisticated enough to predict the behaviour for low emitter impedance.
The equation that you have doesn't work for very low values of RE, so it won't work on AC where you have put in a value of zero for the impedance of the capacitor.
The circuit is stable and the gain is set by the resistor values. That is preferable because the transistor gain varies so much. There is effectively feedback, because if the transistor conducts more, the emitter voltage rises, which reduces the base current.
If the emitter voltage, or the AC component of the voltage, is zero, you need a more complicated model. When the input goes up, there will be a large increase in base current, and it is difficult to predict that. The model of having a fixed base-emitter voltage of 0.7 V is not sophisticated enough to predict the behaviour for low emitter impedance.
Then the Av value (voltage gain) that I found is -386 and my steps and result when calculating the value are correct?
Then the Av value (voltage gain) that I found is -386 and my steps and result when calculating the value are correct?
It's possible that the voltage gain is as high as that, but the input impedance would be very low, the source impedance would have to be considered.
you said ...
VB - VBE - IBxRB - IExRE = 0
then ... RE = 402K which is the wrong result
Re= (Vb-Vbe)/Ie - Rb/(ß+1). . . . .(1) ok
for level 1 approximations
Hint #1
Since ß >>1 has a high tolerance of 50% ~ 300% replace (ß+1) with ß and let Ie=Ic = 10V/5k6.
#2 if you expect Ic ~ 1mA then use Vbe = 0.6V
Solve Re in (1) becomes (Vb-0.6)*5k6/10V. So you expect Re to be much smaller than Rc.
#3 For reality check on Zc=0, compute 1/(2pifC)= Zc(f), for Ce, let Zc (f) = re , you would need > 1 mF (huge) at 100 Hz. and a 50-ohm source is not low enough for Av > 300.
#4 Do not expect Av > 100 for a linear large signal unless negative feedback is used then Av is reduced. H biased designs are primitive and very nonlinear - due to the square law of Vbe vs Ic with a bypassed Re. Gain of 50 is more reasonable per stage with Rc/Re ratios
#5 When estimating THD harmonic distortion on a sine wave, use AC coupling and compare the peak difference to Vpp or (Vp+ - Vp-)/Vpp *100%. Another method is just to scope a triangle wave, which is more visual.
Next design improvement shorts out Re and bypass cap, reduces Vcc to 12V, uses DC bias to Vcc/2 on collector or 6V/10k = 600 uAdc. Negative feedback (NFB) ratio of 50% of Rcb/Rsb (series Rb) will be your approximate Av.
web SIM on Falstad
Gain and Distortion are tradeoffs. Here back to H bias but raised input impedance by moving Re to series R and thus reducing Vbe variations but this is with constant hFE = 150 which is unrealistic.
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Hello there
I am faced with such a question
1) Calculate and determine the value of the resistor RE that will ensure that VC=6 V.
2) Redraw the transistor model of the equivalent circuit for small amplitude AC signals. Determine the voltage gain Av of the circuit.
View attachment 146355
I solved the question as follows ;
VB = [24K / 24K+82K] x 16V = 3,62 Volt
RB = 24Kx82K / 24k+82k = 18,56K
6 = 16 - ICx5,6k
Ic = 1,78mA
IC = ßxIb = 150xIb
Ib = 11,86 microamper
IE = 11,86x10^-6 x 151 = 1,79mA
re = 26mV / 1,79mA = 14,5
VB - VBE - IBxRB - IExRE = 0
3,62 - 0,7 - [ (11,86 x 10^6) x 18,56K ] - 1,79mA x RE = 0
RE = 402K
Av = -ß x Rc / ßre + (ß+1)RE
Av = -150 x 5600 / 2,175 + 0 (The reason why it is 0 is because the Re resistor and capacitor are in parallel, I know that it is an ineffective element in AC analysis.)
This is my solution, are my solution steps correct? or if there is a point where I made a mistake when converting units, can you warn me, I need to correct my shortcomings and learn the right one. thanks a lot in advance
Hi,
I get around 1500 for RE.
That's considering Vbe=0.65v which is typical, and Beta can spread a bit starting with 100, maybe 50 to 200.
For increasing Vbe the ideal value for RE would go down to as low as 1400.
The base current comes out somewhere around 18ua with Beta=100.
This is after doing a full network analysis which includes Vbe and Beta.
Usually these questions include more information like the values to use for Vbe and Beta.
Note: a more exact value for R4 is 1464 Ohms.
Also, sensitivity due to any internal rb is minimal.
Solution comes from Nodal Analysis.
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# Math Geek Mom: Can’t We Find a Better Way to Pay for College?
Reality TV is not the answer.
## By
May 16, 2013
The field of Economics includes several strong assumptions that are called upon occasionally when modeling economic activity. The most severe of these is the assumption of “perfect competition,” an assumption that claims that there is free entry and exit in markets, allowing firms to enter a potentially lucrative area and therefore provide competition to those already acting in that realm, potentially driving down prices and driving up the cost of inputs. The proliferation of “Silly Bands” competitors a few years ago is somewhat of a good example of such entry, but in general, such pure competition does not exist in the real world. Another assumption that is often invoked in creating economic models is that of “perfect capital markets,” the assumption that money will always be available to those who want to borrow to fund a plan that appears will be a successful investment opportunity. I thought of this later assumption a few weeks ago when I heard of a new television program that claims to be offering scholarships to young children based on how correctly their parents predict how they will act under some very contrived situations. It is called “Bet on your Baby.”
When my grieving parents found themselves all but housebound as my mother’s cancer came out of remission last fall following the death of my sister, they found themselves exploring the multitude of channels on the cable TV in their home. As they surfed the stations, they stumbled upon a show called “Bet on your Baby.” In this show, a couple with a young child is separated, with one member going backstage with the child to try to get the toddler to perform some very contrived activity, such as making a soccer goal or dancing using particular moves. Meanwhile, the other parent is on stage and makes a wager as to how successful the first parent will be in getting the child to perform as expected. Correct bets yield a range of payoffs, all called “scholarships”, and may lead to an opportunity to earn even more money, this time significant money that may actually pay for a large portion of a year of college today. At my parents’ request, I watched the show, and I must admit that I thought the kids on it were cute, but in the end, I was appalled to think that this is what we have been reduced to in trying to find a way to pay for college.
I recalled the assumption of “perfect capital markets”, an assumption that would claim that anyone wanting to make an investment in their future by going to college should be able to find someone willing to lend them the money to make that investment, as, upon graduating, they would easily be able to pay back the loan (assuming they find a job, that is.) Although I momentarily thought of selling stock in myself to fund college when I was a teenager, we all know that the real world does not work that way, and as a concession to the real world, alternative ways to pay for college must be found. Apparently, bringing one’s child onto a game show has become the newest option available to parents these days. When I described the show to a fellow mother from our neighborhood, she was also appalled.
And so, I am left with the question of how we can find less disturbing ways for parents to find the money to pay for college. Does anyone have any ideas that don’t involve a toddler mimicking dance moves on national TV?
--- I want to congratulate my students from Ursuline College who will graduate tonight. May your future hold wonderful opportunities! I look forward to seeing you walk across the stage in a few hours. ---
+ - | 2.78125 | CC-MAIN-2018-26 |
# Lid-driven cavity problem
(Difference between revisions)
Revision as of 00:49, 15 September 2005 (view source)Michail (Talk | contribs) (→References)← Older edit Latest revision as of 12:44, 9 April 2013 (view source)m (17 intermediate revisions not shown) Line 4: Line 4: [[Image:ldc_geom.png]] [[Image:ldc_geom.png]] - Since this case has been solved many times, there is a great deal of data to compare with. A good set of data for comparison is the data of Ghia, Ghia, and Shin (1982), since it includes tabular results for a number of Reynolds numbers. Other researchers have solved the problem under laminar and turbulent conditions, and aspect ratios other than one have also been used. This problem is also sometimes referred to as the shear-driven cavity problem, though the shear-driven cavity is technically a different problem. + Similar simulations have also been done at various aspect ratios, and it can also be done with the lid replaced with a moving fluid. This problem is a somewhat different situation, and is usually referred to as the shear-driven cavity. You may see the two names (lid-driven and shear-driven) used interchangeably in spite of the fact that they are distinct (and different) problems. - == Sample Results == + This problem has been solved as both a laminar flow and a turbulent flow, and many different numerical techniques have been used to compute these solutions. Since this case has been solved many times, there is a great deal of data to compare with. A good set of data for comparison is the data of [[#References|Ghia, Ghia, and Shin (1982)]], since it includes tabular results for various of Reynolds numbers. These simulation results are obtained using a non-primitive variable approach. + + This problem is a nice one for testing for several reasons. First, as mentioned above, there is a great deal of literature to compare with. Second, the (laminar) solution is steady. Third, the boundary conditions are simple and compatible with most numerical approaches. Note that this is not necessarily the case for finite element methods, in which difficulties may arise at the corner intersections of the moving wall and the stationary wall. + + == Sample results == Shown below are a few results from a simulation of the lid-driven cavity (for a Reynolds number of one hundred) using the commercial Fluent code. Shown below are a few results from a simulation of the lid-driven cavity (for a Reynolds number of one hundred) using the commercial Fluent code. - [[Image:ldc_fluentu.png]] + [[Image:v&tc_liddriven_fluentu.png]] - The 32x32 grid used here is somewhat coarse, as can be seen in the upper left and right corners of the contour plot. Increasing the grid resolution would improve the look of the contour plot somewhat, but since the boundary conditions are not smooth at these corners we will probably always have some issues there. In spite of this issues, we do a reasonable job matching the Ghia et al (1983) results, as shown below. + The 32x32 uniform grid used here is somewhat coarse, as can be seen in the upper left and right corners of the contour plot. Increasing the grid resolution would improve the look of the contour plot somewhat, but since the boundary conditions are not smooth at these corners we will probably always have some issues there. In spite of these issues, we do a reasonable job matching the [[#References|Ghia et al (1982)]] results, as shown below. - [[Image:ldc_ggsu.png]] + [[Image:v&tc_liddriven_ggsu.png]] Here we have plotted u/U on the vertical line x/L = 0.5. We have reasonably good agreement with the previous results. Here we have plotted u/U on the vertical line x/L = 0.5. We have reasonably good agreement with the previous results. == References == == References == + {{reference-paper|author=Ghia, Ghia, and Shin|year=1982|title=High-Re solutions for incompressible flow using the Navier-Stokes equations and a multigrid method|rest=Journal of Computational Physics, Vol. 48, pp. 387-411}} - * '''Ghia, Ghia, and Shin''', ''"High-Re solutions for incompressible flow using the Navier-Stokes equations and a multigrid method"'', Journal of Computational Physics, Vol. 48, pp. 387-411, 1982. + == Additional references == + {{reference-paper|author=Bozeman and Dalton|year=1973|title=Numerical Study of viscous flow in a cavity|rest=Journal of Computational Physics, Vol. 12, pp. 348-363}} - ''Need more references that involve this problem'' + {{reference-paper|author=Burgaff|year=1966|title=Analytical and Numerical studies of the structures of steady separated flows|rest=JFM, Vol. 24, pp. 113-151}} - Please look this link. - Michail + {{reference-paper|author=Erturk, Corke, and Gokcol|year=2005|title=Numerical Solutions of 2-D Steady Incompressible Driven Cavity Flow at High Reynolds Numbers|rest=International Journal for Numerical Methods in Fluids, Vol. 48, pp. 747-774}} + + ''Need more references that involve this problem'' - http://www.gyte.edu.tr/enerji/ercanerturk/drivencavity/cavityflow.htm + == Additional resources == + [http://www.cavityflow.com http://www.cavityflow.com]
## Introduction
The lid-driven cavity problem has long been used a test or validation case for new codes or new solution methods. The problem geometry is simple and two-dimensional, and the boundary conditions are also simple. The standard case is fluid contained in a square domain with Dirichlet boundary conditions on all sides, with three stationary sides and one moving side (with velocity tangent to the side).
Similar simulations have also been done at various aspect ratios, and it can also be done with the lid replaced with a moving fluid. This problem is a somewhat different situation, and is usually referred to as the shear-driven cavity. You may see the two names (lid-driven and shear-driven) used interchangeably in spite of the fact that they are distinct (and different) problems.
This problem has been solved as both a laminar flow and a turbulent flow, and many different numerical techniques have been used to compute these solutions. Since this case has been solved many times, there is a great deal of data to compare with. A good set of data for comparison is the data of Ghia, Ghia, and Shin (1982), since it includes tabular results for various of Reynolds numbers. These simulation results are obtained using a non-primitive variable approach.
This problem is a nice one for testing for several reasons. First, as mentioned above, there is a great deal of literature to compare with. Second, the (laminar) solution is steady. Third, the boundary conditions are simple and compatible with most numerical approaches. Note that this is not necessarily the case for finite element methods, in which difficulties may arise at the corner intersections of the moving wall and the stationary wall.
## Sample results
Shown below are a few results from a simulation of the lid-driven cavity (for a Reynolds number of one hundred) using the commercial Fluent code.
The 32x32 uniform grid used here is somewhat coarse, as can be seen in the upper left and right corners of the contour plot. Increasing the grid resolution would improve the look of the contour plot somewhat, but since the boundary conditions are not smooth at these corners we will probably always have some issues there. In spite of these issues, we do a reasonable job matching the Ghia et al (1982) results, as shown below.
Here we have plotted u/U on the vertical line x/L = 0.5. We have reasonably good agreement with the previous results.
## References
Ghia, Ghia, and Shin (1982), "High-Re solutions for incompressible flow using the Navier-Stokes equations and a multigrid method", Journal of Computational Physics, Vol. 48, pp. 387-411.
Bozeman and Dalton (1973), "Numerical Study of viscous flow in a cavity", Journal of Computational Physics, Vol. 12, pp. 348-363.
Burgaff (1966), "Analytical and Numerical studies of the structures of steady separated flows", JFM, Vol. 24, pp. 113-151.
Erturk, Corke, and Gokcol (2005), "Numerical Solutions of 2-D Steady Incompressible Driven Cavity Flow at High Reynolds Numbers", International Journal for Numerical Methods in Fluids, Vol. 48, pp. 747-774.
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imaginary number
(redirected from Imaginary numbers)
Also found in: Dictionary, Thesaurus.
imaginary number:
see numbernumber,
entity describing the magnitude or position of a mathematical object or extensions of these concepts. The Natural Numbers
Cardinal numbers describe the size of a collection of objects; two such collections have the same (cardinal) number of objects if their
.
Imaginary Number
a number of the form x + iy, where i = x and y are real numbers and y ≠ 0; that is, a complex number that is not real. Imaginary numbers of the form iy are called pure imaginary; sometimes only the latter are referred to as imaginary numbers. The term “imaginary number” appeared after such numbers had already entered general use although their real meaning had not been ascertained.
imaginary number
[ə′maj·ə‚ner·ē ′nəm·bər]
(mathematics)
A complex number of the form a + bi, with b not equal to zero, where a and b are real numbers, and i = √(-1); some mathematicians require also that a = 0. Also known as imaginary quantity.
References in periodicals archive ?
Imaginary time is a mathematical construct, like imaginary numbers, and does not have a physical realisation.
Beginning with a discussion of cultural impacts on mathematics, he presents esoteric but plausible interpretations of imaginary numbers and the quantum wavefunction.
Much of the book is a journey through non-consensus reality (NCR)--the world of imaginary numbers, complex wavefunctions, virtual particles, and individual subjective experiences such as dreams, feelings, emotions, telepathy, fleeting thoughts, fantasies, sudden intuitions, altered states, and near-death experiences.
The imaginary numbers are a wonderful flight of God's spirit.
Pappas delivers a new way to enjoy and learn some sublimely abstract notions, such as tessellations, fractals, googols, imaginary numbers, and much more.
Figure 1 shows how imaginary numbers are are plotted in the complex plane.
Learning about fractal music is a rich context for learning fractals, mapping, iterative equations, self-similarity, the Mandelbrot and Julia sets, and aspects of number theory such as: real and imaginary numbers, prime numbers and bases other than 10.
An econometrist at the Massachusetts Institute of Technology, he cites papers proposing that prices can be negative or can incorporate so-called imaginary numbers, based on the square root of negative one.
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# Question: How Are Pressure And Volume Related?
The pressure of a gas is inversely proportional to its volume when temperature is constant.
The product of pressure and volume is constant when temperature is constant.
This relationship is known as Boyle’s law or Mariotte’s law .
A constant temperature process is said to be isothermal ..
## What are 3 ways to increase the pressure of a gas?
Three Ways to Increase the Pressure of a GasIncrease the amount of gas. This is represented by the “n” in the equation. … Increase the temperature of the gas. This is represented by “T” in the equation. … Decrease the volume of the gas. This is the “V” in the equation.
We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then P and T are directly proportional (again, when volume and moles of gas are held constant); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor.
At constant temperature and pressure the volume of a gas is directly proportional to the number of moles of gas. At constant temperature and volume the pressure of a gas is directly proportional to the number of moles of gas.
## Do temperature and volume have a direct relationship?
If two letters are touching each other, they have a direct relationship. That is, pressure and temperature have a direct relationship, and volume and temperature have a direct relationship. That means if one of them goes up, the other will go up, assuming the third variable is held constant.
## Are pressure and volume directly proportional?
The relationship between pressure and volume is inversely proportional. … It is summarized in the statement now known as Boyle’s law: The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured.
## Does volume decrease as pressure increases?
For a fixed mass of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional. Or Boyle’s law is a gas law, stating that the pressure and volume of a gas have an inverse relationship. If volume increases, then pressure decreases and vice versa, when the temperature is held constant.
the amount of gas, volume and temperature are all factors that affect gas pressure. If the temperature is constant, as the pressure of a gas increases, the volume decreases. … As the temperature of an enclosed gas increases, the pressure increases, if the volume is constant.
## What are the 6 gas laws?
Gas Laws: Boyle’s Law, Charle’s Law, Gay-Lussac’s Law, Avogadro’s Law.
## Does increasing temperature increase volume?
The volume of the gas increases as the temperature increases. As temperature increases, the molecules of the gas have more kinetic energy. They strike the surface of the container with more force. If the container can expand, then the volume increases until the pressure returns to its original value.
## What is the R in PV NRT?
In the equation PV=nRT, the term “R” stands for the universal gas constant. The universal gas constant is a constant of proportionality that relates the energy of a sample of gas to the temperature and molarity of the gas.
## Which three factors can affect gas pressure?
Temperature, pressure, volume and the amount of a gas influence its pressure.
## Why do temperature and volume have a direct relationship?
Gay Lussac’s Law – states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. If you heat a gas you give the molecules more energy so they move faster. This means more impacts on the walls of the container and an increase in the pressure.
## What is the T in PV NRT?
In SI units, p is measured in pascals, V is measured in cubic metres, n is measured in moles, and T in kelvins (the Kelvin scale is a shifted Celsius scale, where 0.00 K = −273.15 °C, the lowest possible temperature).
## What is the pressure law?
Gay-Lussac’s law, Amontons’ law or the pressure law was found by Joseph Louis Gay-Lussac in 1808. It states that, for a given mass and constant volume of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.
## How does volume change with pressure?
When the volume decreases, the pressure increases. This shows that the pressure of a gas is inversely proportional to its volume. … So increasing pressure from pressure 1 to pressure 2 means that volume 1 will change to volume 2, providing the temperature remains constant.
## What are the standard conditions of temperature and pressure?
Since 1982, STP is defined as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).
## What happens to the pressure when the volume is cut in half?
Boyle’s Law simply describes the relationship between the pressure and volume of an enclosed gas when Temperature remains constant. … Tripling the pressure will reduce its volume to 1/3, and so on. Alternatively, if you double the volume available to an enclosed gas, pressure is halved. | 4.125 | CC-MAIN-2021-04 |
# variance
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Related to additive genetic variance: Quantitative genetics, narrow sense heritability
## Variance
A measure of dispersion of a set of data points around their mean value. The mathematical expectation of the average squared deviations from the mean. The square root of the variance is the standard deviation.
## Variance
1. In accounting, the difference between the estimated and actual cost of a project or other operation.
2. In risk, the average deviation of a set of data points from their mean.
## variance
A statistical measure of the variability of measured datum from the average value of the set of data. A high variance, indicating relatively great variability, also indicates that the average is of minimal use in projecting future values for the data. Standard deviation is the square root of variance. Financial analysts use both statistical measures to weigh investment risk. Compare covariance. See also risk.
## variance
1. the difference between budgeted and actual results (see BUDGETING), or between STANDARD COSTS/revenues and actual costs/revenues. Variances can be:
1. adverse or negative when actual revenues fall short of budget or standard, or when actual costs exceed budget or standard;
2. favourable or positive when actual revenues exceed budget or standard, or when actual costs are less than budget or standard.
3. a measure of variation within a group of numerical observations, specifically the average of the squared deviations of the observations from the group AVERAGE.
See STANDARD DEVIATION.
## variance
Permission to use a property in a manner that does not meet current zoning requirements.In order to gain a variance,the property owner usually has to show a hardship on the property—not on the owner—if the requested use is not allowed. It is considered a hardship if the property will otherwise remain vacant or if a structurally sound improvement must be demolished to allow some other use. Buyers with a signed purchase contract can usually petition for a variance; this is commonly one of the steps in a due diligence plan that must be completed in a satisfactory manner before the buyer will purchase property.
References in periodicals archive ?
Analyzing Table 2 we can observe that the estimatives for additive genetic variances obtained on cycle 1 for PH and EH traits were of 162.5 and 85.05 [(cm [plant.sup.-1]).sup.2], respectively, and on cycle 2 were of 91.20 and 50.52 [(cm [plant.sup.-1]).sup.2], being considered high when compared to values obtained by Silva et al.
Maternal additive genetic variances ([Mathematical Expression Omitted]) are two to three times as large as direct additive genetic variances ([Mathematical Expression Omitted]).
The efficiency measures used are described below: (i) the molecular heritability was given by [h.sup.2] = [[sigma].sup.2.sub.a]/[[sigma].sup.2.sub.y], where [[sigma].sup.2.sub.y] is the phenotypic variance and is the additive genetic variance estimated by REML in G-BLUP and estimated through [mathematical expression not reproducible].
Epistasis and the effect of founder events on the additive genetic variance. Evolution 42:441-454.
The absence of environmental effects on the additive genetic variance for morphological traits may be relevant to comparisons of heritabilities in field and laboratory environments.
In other words, additive genetic variance rates in females are increasingly greater than those in males as from the second biometry, providing a ratio between increasing variances.
Traits with moderate to high additive genetic variance and heritability were ALP, BET, and hop storability.
Epistasis and the increase in additive genetic variance: implication for phase 1 of Wright's shifting-balance process.
Heritability trends for daily milk, fat, and proteins yields were similar to trends for additive genetic variance but were less extreme at the start and end of the curve because of higher permanent environmental variances.
Furthermore, high heritability combined with high expected genetic advance proved the involvement of additive genetic variance, therefore simple selection may be effective for improvement of these traits.
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# Area of a general n-sided polygon
• I_am_learning
In summary, there are various methods for finding the area of irregular polygons, including using Heron's formula for (n-2) triangles and using matrices. However, if the given measurements do not result in (n-2) triangles, it can be difficult to calculate the area. Land surveyors are professionals who are responsible for determining the boundaries and area of a plot of land, and their formula is often the best approach. It is also important to determine the proper orientation of the vertex coordinates, typically counterclockwise, in order to accurately calculate the area.
I_am_learning
Finding the area of an irregular polygon with n side is quite easy when we are given the length of all of the n sides and the length of (n-3) specific diagonals. This way, we get (n-2) triangles whose areas can be calculated using Heron's formula and then added up.
But what if the length of the (n-3) diagonals provided doesn't make (n-2) triangles, such as this case:
The polygon is still fully determined by the given measurements, but calculating the area is difficult.
Is there some sort of generic formula for such cases? Like maybe using matrices. :D
I thought of making a mobile application to help real-estates peoples calculate the area of lands, and came-up with this question.
Thank you.
#### Attachments
• easy.png
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There are several different methods of finding areas of general polygons:
http://en.wikipedia.org/wiki/Polygon
When you say 'real estate peoples', are you referring to land surveyors or someone else?
SteamKing said:
There are several different methods of finding areas of general polygons:
http://en.wikipedia.org/wiki/Polygon
When you say 'real estate peoples', are you referring to land surveyors or someone else?
There is no formula for finding area when n sides and n-3 diagonals are known.
The surveyors formula seems to be the best way to go. So from the given information, I should somehow find the coordinates of all the vertices and also sort them counter clockwise or clockwise.
By real state peoples I just meant anyone who is involved in buying or selling of lands.
Well, in the US, the land surveyor is the professional who confirms and measures the boundaries of a particular plot of land. There is usually a legal description of the land produced for a deed of title to the land, which would contain the area enclosed by these boundaries.
The proper orientation for your vertex coordinates is counterclockwise to calculate the positive area of the figure. If you use a clockwise orientation, the result will be a negative area. This comes in handy if you want to evaluate the area of complex, non-convex polygons, say a polygon with a hole in it.
I would like to first commend you on your idea for a mobile application to help real estate professionals calculate the area of lands. It is important to have accurate measurements in this field, and your idea could definitely be helpful.
In regards to your question about a generic formula for calculating the area of a general n-sided polygon, unfortunately, there is no one formula that can be applied in all cases. The formula you mentioned, using Heron's formula and dividing the polygon into (n-2) triangles, is a commonly used method for finding the area of irregular polygons. However, as you mentioned, it may not always work if the given diagonals do not form (n-2) triangles.
In such cases, one approach could be to use trigonometry and break the polygon into smaller, simpler shapes such as triangles, rectangles, and parallelograms, and then calculate their individual areas and add them up. Another approach could involve using coordinate geometry and dividing the polygon into smaller polygons with known area formulas, and then adding them up.
Alternatively, as you suggested, using matrices could also be a possible approach. This would involve representing the polygon as a matrix and using mathematical operations to calculate its area. However, this may be a more complex and time-consuming method compared to some of the other techniques mentioned above.
In conclusion, there is no one generic formula for calculating the area of a general n-sided polygon. It depends on the given measurements and the shape of the polygon. I would suggest exploring different methods and techniques, and using the one that is most suitable for the given polygon. Best of luck with your mobile application!
## 1. What is the formula for finding the area of a general n-sided polygon?
The formula for finding the area of a general n-sided polygon is A = (1/2) * n * s * h, where n is the number of sides, s is the length of a side, and h is the height of the polygon.
## 2. How do you calculate the height of a general n-sided polygon?
To calculate the height of a general n-sided polygon, you can use the formula h = (s/2) * cot(180/n), where s is the length of a side and n is the number of sides. Alternatively, you can use trigonometry to find the height by dividing the polygon into triangles and using the Pythagorean theorem.
## 3. Can the area of a general n-sided polygon be negative?
No, the area of a polygon cannot be negative. It represents the amount of space enclosed by the polygon and therefore must be a positive value.
## 4. How is the area of a general n-sided polygon related to its perimeter?
The area of a general n-sided polygon is not directly related to its perimeter. However, as the number of sides increases, the perimeter also increases, and therefore the area also increases.
## 5. Can the area of a general n-sided polygon be calculated without knowing the length of a side?
Yes, the area of a general n-sided polygon can be calculated without knowing the length of a side by using the formula A = (1/2) * n * a * h, where a is the apothem (the distance from the center of the polygon to the midpoint of a side) and h is the height. Alternatively, the area can be calculated by dividing the polygon into triangles and using the formula A = (1/2) * base * height for each triangle.
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1K | 3.96875 | CC-MAIN-2024-22 |
Case 8) P, Q, R, S, T, U, V and W are sitting around a circular table, facing the centre. P sits third to the right of W and third to the left of Q. S sits second to the right of T. V sits second to the left of R. T is not the neighbour of Q while U is neither a neighbour of T nor of W. Which of the following pairs has only one person sitting between them, if the counting is done in the clockwise direction?
Case 8)
P, Q, R, S, T, U, V and W are sitting around a circular table, facing the centre. P sits third to the right of W and third to the left of Q. S sits second to the right of T. V sits second to the left of R. T is not the neighbour of Q while U is neither a neighbour of T nor of W.
Which of the following pairs has only one person sitting between them, if the counting is done in the clockwise direction?
1)T ,V
2)V, Q
3)W, P
4)R, P
5)None of these
• : 284
• : 16
R, P | 2.625 | CC-MAIN-2021-39 |
user2833876 - 1 year ago 147
C++ Question
# Deallocate dynamic memory
I have written some code like this:
`````` double *psi1 = multiplyscalar(
multiplyscalar(submatrix( multiplyscalar(I3, ttt2, 3, 3), multiplyscalar(multiply(teta0, teta0t, 3, 1, 1, 3), pow(*ttt, -1.5), 3, 3),3,3), 2, 3, 3)
,( (*multiply(A3t, mu, 1, m, m, 1))-( *multiply(multiply(multiply(transpose(mu, m, 1), G1, 1, m, m, m),transpose(G, m, m), 1, m, m, m), mu, 1,m,m,1) ) )
,3
,3
);
``````
Where the functions called are similar to this:
``````static double* multiply(double A[], double B[], int ra, int ca, int rb, int cb){
if(ca != rb){
hal.console->printf("\n multiplication error\n");
return nullptr;
}
double* C = new double[ra*cb];
double sum = 0;
for(int i = 0; i < ra; ++i){
for(int j=0; j<cb; j++){
for(int x=0; x<ca; x++){
sum += A[(i*ca)+x]*B[(x*cb)+j];
}
C[(i*cb)+j] = sum;
sum = 0;
}
}
return C;}
``````
So I have memory leak because those pointers that are returned are never "erased".
Is there a way to
`delete`
them afterward?
I know doing every operation separately would bypass the issue, but I was searching for a more elegant solution?
Return a proper container, e.g. some std::vector (or your `class` using it). Read also about the rule of five.
Maybe define a `template` function (with the template arguments being the sizes, perhaps using `std::array`).
Consider also smart pointers (like `std::unique_ptr`), avoiding explicit `new` and `delete`. | 2.71875 | CC-MAIN-2019-18 |
Tagged Questions
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Definition of dimension
Let us consider Euclidean space $\mathbb{R}^n$. We say it is $n$-dimensional because each vector in it is an $n$-tuple $(x_1,...,x_n)$. However, it is possible to represent this exact same space using ...
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I'm considering vector spaces over $\mathbb{R}$ or $\mathbb{C}$ : $(E_n)_n$ is a sequence of vector space (all included in $\mathbb{R}^p$ for example, with $p$ fixed). What meaning(s) do we usually ...
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On the canonical isomorphism between $V$ and $V^{**}$
I am trying to understand more about the Bidualspace (or double dual space). The whole idea is that $V$ and $V^{**}$ are canonically isomorphic to one another, which means that they are isomorphic ...
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About definition of “direct sum of $p$-vector subspaces”
In the books 1 and 2, in Somme directe d'une famille de sous-espaces vectoriels, I am reading the following: 1) let $E,F$ two vector subspaces of $V$, $E+F$ is direct sum, $E+F \doteq E\oplus F$, if ...
48 views
Equivalent definitions of isometry
Consider a map $T:\mathbb{R}^2\to\mathbb{R}^2$ such that $\lVert T(x)\rVert=\lVert x\rVert$. Is this equivalent to stating that $\langle x, y\rangle=\langle T(x), T(y)\rangle$ for all ...
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Subspace with different vector space operations
Let $A,B$ be vector spaces such that $A\subseteq B$. Is it true that $A$ is a subspace of $B$? I claim that the answer is no, because it is possible that $A$ and $B$ might be equipped with different ...
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Has anyone succeeded in formalizing the notion of a complete vector space? (Not using topological ideas).
In order theory, we have the concept of a lattice, which is defined as consisting of an underlying set $L$ together with two binary operations $\wedge$ and $\vee$. Now when $L$ is finite, the concept ...
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Quotient spaces in linear algebra
There's a statement in some notes I'm reading that goes like this: "...$V/U$ is a 'simplified version' of $V$ where the elements of $U$ are ignored" ($V$ and $U$ are vector spaces). I'm still ...
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Similar linear operators and change of coordinates
Let $S, T$ be operators in $\mathcal{L}(V)$, the space of all linear maps from $V$ to itself. In my lecture notes, I have the definition of similar: "We say that operators $S,T \in \mathcal{L}(V)$ ...
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What does it mean when a set is said to be a “finitely generated vector space”?
I've somehow managed to go through 3 years of my Maths degree without truly understanding what the term "finitely generated vector space" means. Generated by what? Generating what? And for what? I ...
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Why are vector spaces sometimes called linear spaces?
I have never come across the term 'linear space' as a synonym for 'vector space' and it seems from the book I am using (Linear Algebra by Kostrikin and Manin) that the term linear space is more ...
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Suppose I said “$X$ spans $W$”…
So I've seen two definitions of this: Let $V$ be a vector space with subspace $W$. We say that $X \subseteq V$ spans $W$ if and only if (Definition 1): Every $\vec{w} \in W$ can be written as ...
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I have seen the statement "Every finite dimensional vector space has a basis." (Here on page 5) I'm confused about what this tells me. It seems to tell me nothing: by definition, the dimension of a ...
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Why is there implied an equality between vectors and $n$-tuples?
Are they considered equal in some sense? For instance, I always write "...for vector $\mathbf{x} \in {\mathbb{R}}^n$ we have ...". I have a small problem with this (not a big one). The problem comes ...
27 views
Definition clarifications on functions of vectors and their derivatives
Definition clarifications would be appreciated: How do I interpret the following ? For $f: R^n\to R^m$, $Df(\vec\xi)(\vec{x})$ in differentiation of a vector function. I know it as a function that ...
i have question from vector mathematics,i know that if there is given two vector, for instance $a=\{a_1,a_2,a_3\}$,$b={b_1,b_2,b_3}$; then so called exterior product is determined as \$a\wedge ... | 3.296875 | CC-MAIN-2014-35 |
Search a number
7701 = 317151
BaseRepresentation
bin1111000010101
3101120020
41320111
5221301
655353
731311
oct17025
911506
107701
115871
124559
133675
142b41
152436
hex1e15
7701 has 8 divisors (see below), whose sum is σ = 10944. Its totient is φ = 4800.
The previous prime is 7699. The next prime is 7703. The reversal of 7701 is 1077.
Added to its reverse (1077) it gives a triangular number (8778 = T132).
It is an interprime number because it is at equal distance from previous prime (7699) and next prime (7703).
It is a sphenic number, since it is the product of 3 distinct primes.
It is not a de Polignac number, because 7701 - 21 = 7699 is a prime.
7701 is a lucky number.
It is a plaindrome in base 12.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (7703) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 25 + ... + 126.
It is an arithmetic number, because the mean of its divisors is an integer number (1368).
27701 is an apocalyptic number.
7701 is the 51-st octagonal number.
7701 is the 56-th centered pentagonal number.
It is an amenable number.
7701 is a deficient number, since it is larger than the sum of its proper divisors (3243).
7701 is a wasteful number, since it uses less digits than its factorization.
7701 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 171.
The product of its (nonzero) digits is 49, while the sum is 15.
The square root of 7701 is about 87.7553417178. The cubic root of 7701 is about 19.7476630253.
Adding to 7701 its product of nonzero digits (49), we get a triangular number (7750 = T124).
Adding to 7701 its reverse (1077), we get a palindrome (8778).
The spelling of 7701 in words is "seven thousand, seven hundred one", and thus it is an iban number.
Divisors: 1 3 17 51 151 453 2567 7701 | 3.625 | CC-MAIN-2023-14 |
# Tagged Questions
Use this tag for questions about differential and integral calculus with more than one independent variable. Some related tags are (differential-geometry), (real-analysis), and (differential-equations).
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### Reference for the fact that a smooth function analytic on every line is itself analytic
Let $f \in \mathcal C^\infty(\mathbb R^p)$ ($p \geq 2$) be a smooth function such that the functions $g_d(t) := f(td)$ are all analytic for all $t \in \mathbb R$ and all $d \in \mathbb R^p.$ (i.e. $f$ ...
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### partial derivative of a facet normal wrt to one of its vertex
I am struggling to understand the derivation of an equation in a paper (A Bayesian Method for Probable Surface Reconstruction and Decimation, specifically Eqn. 16). Basically they define three ...
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### On the form of injective function $f:\mathbb R^+ \to \mathbb R^+$ such that $\|u\|:=f^{-1}\Big(\sum_{i=1}^nf(|u_i|)\Big)$ is a norm
Let $f:\mathbb R^+ \to \mathbb R^+$ be injective such that on $\mathbb R^n , n>1)$ , $\|u\|:=f^{-1}\Big(\sum_{i=1}^nf(|u_i|)\Big)$ , where $u=(u_1,u_,\ldots,u_n)$ , defines a norm ; then is it true ...
61 views
### Manifold, exist smooth nonnegative function with regular value at 0?
If $X$ is any manifold with boundary, then does there exist a smooth nonnegative function $f$ on $X$, with a regular value at $0$, such that $\partial X = f^{-1}(0)$?
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### Second Variation of Area Functional
This is a follow up question from this one. I have proved that given a parametrized surface ${\bf x}$, the mean curvature is zero if and only if it is a critical point of the area functional. Then ...
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I found an exercise in a book, where one was supposed to transform the differential equation $$y\frac{\partial f}{\partial x}-x\frac{\partial f}{\partial y}=xyf(x,y)$$ by using the substitutions $x^2+... 0answers 85 views ### Conditions for Taylor formula I know that, if$F:X\to Y$, where$X,Y$are Banach spaces, is a map whose$n$-th Fréchet derivative$x\mapsto F^{(n)}(x)$is continuous as a function of$x$in a neighbourhood of$x_0\in X$, then the ... 0answers 78 views ### How do I do this change of variables? Use a change of variables to evaluate: $$\iiint\limits_{D}xy\,\mathrm{d}V$$$D$is bounded by the planes$y-x=0$,$y-x = 2$,$z-y = 0$,$z-y = 1$,$z=0$,$z=3$. I set $$u = y-x$$ $$v = z-y$$ $$w = ... 0answers 188 views ### Differential forms and determinants 2-forms are defined as du^{j} \wedge du^{k}(v,w) = v^{j}w^{k}-v^{k}w^{j} = \begin{vmatrix} du^{j}(v) & du^{j}(w) \\ du^{k}(v) & du^{k}(w) \end{vmatrix} But what if I have two concret 1-... 0answers 143 views ### Why is a_{n}(x,y)=a_{n}(y)? This particular question is connected (with a slight variation in the definition of g) to an earlier question. The link is here. The specifics are: Given that u(x,y) is the solution of a PDE (x ... 0answers 147 views ### Showing some complicated integral expression is bounded In my research, I come across some expression I need to bound. I wish to show that the following integral is bounded in t and x, i.e. the following supremum is finite:$$\sup_{t,x\in \mathbb{R}}\,... 0answers 121 views ### Question on using Leibniz formula to derive thin-film equation from Navier-Stokes I am trying to work through the derivation in this paper by Petr Vita, which derives a thin-film simplification of the Navier-Stokes equation, similar to the Reynolds or Lubrication Equation, but ... 0answers 75 views ### Inhomogeneous Wave Equation in 3 dimensions From section 7.5 in this source, I see that, for$\vec{x} \in \mathbb{R}^3$, if $$\frac{\partial^2 u\left(\vec{x},t\right)}{\partial t^2} - \nabla^2u\left(\vec{x},t\right) =f\left(\vec{x},t\... 0answers 227 views ### How to prove following integral equality? Let's have the equality$$ \int \limits_{-\infty}^{\infty} \left[ [\nabla_{\mathbf r'} \times \mathbf A (\mathbf r' )] \times \frac{\mathbf r' - \mathbf r}{|\mathbf r' - \mathbf r |^{3}}\right]d^{3}\... 0answers 160 views ### the spectrum and determinant of the Laplacian on$S^3$I came across the following statement in a paper: On$S^3$, the eigenvalues of the vector Laplacian on divergenceless vector fields is$(\ell + 1)^2$with degeneracy$2\ell(\ell+2)$with$\ell \...
One of the major difficulties of student in advanced calculus (including myself when student) is to obtain the extremes of repeated integrals to calculate the volume integral in $R^n$ i.e. transform ... | 3.328125 | CC-MAIN-2016-30 |
# [S] Box's M calculation problem (long!)
Marc Feldesman ([email protected])
Mon, 08 Jun 1998 09:39:01 -0700
As you know, I have been working on a function to calculate Box's M and
test its significance against both the chisquare and F distributions. This
is the first SPlus function I've written and there are, no doubt, many
inefficiencies in it. I've run up against a peculiar problem. I've tested
the function with published data from Cooley and Lohnes(1971) and Morrison
(1990), both of whom give test data and "correct" answers for the
algorithm. My function, these authors, SPSS 8.0, NCSS97, and SAS 6.12 all
give identical answers, leading me to conclude that I have distilled the
essence of the algorithm correctly.
Here is my problem. Abridged and annotated output from my function applied
to some of my own test data is below:
\$"Group Sizes":
1 2 3
75 66 59 <===== IDENTICAL in SAS, SPSS, NCSS
\$"Num Cases":
[1] 200 <====== IDENTICAL in SAS, SPSS, NCSS
\$"Num Groups":
[1] 3 <======= IDENTICAL in SAS, SPSS, NCSS
\$"Box's M":
[1] 22.7283 <===== ???? answer from SPSS and NCSS 25.681 (not reported
in SAS)
\$"Chi Square Probability":
[1] 0.3372944 <===== ???? answer from SAS 0.2043 & NCSS, not reported in
SPSS
\$"Log Determinants":
[1] 27.67216 27.59350 27.35155 <======IDENTICAL to SPSS, SAS, NCSS
\$"Pooled Within Groups Determinant":
[1] 27.66719 <======== IDENTICAL to SPSS, SAS, NCSS
\$"Chi Square Statistic":
[1] 22.05943 <====== ?????? 24.924802 in SAS and NCSS, not reported in SPSS
\$DF1:
[1] 20 <=======IDENTICAL in SAS, SPSS, NCSS
\$DF2:
[1] 129299.2 <=======IDENTICAL in SAS, SPSS, and NCSS
\$"F Statistic":
[1] 1.102796 <=====????? 1.246 in SPSS and NCSS, not reported in SAS
\$"F Probability":
[1] 0.3375029 <=====????? 0.204 in SPSS and NCSS, not reported in SAS
I've manually calculated the M statistic using the determinantal
information provided by all of the programs. My hand calculations give me
the same answers as I report above (for the SPlus function), suggesting
that SAS, SPSS, and NCSS are all wrong, something I find impossible to
believe.
The wrong answers are all interrelated. The figure that is incorrect is
the Box's M value itself. All other incorrect values depend on it.
For those unfamiliar with Box's M, it is calculated as:
Let ni=n-1 for the i th population
S = pooled within groups covariance matrix
Si = covariance matrix for ith group
M = (Sum(ni) * ln|S|) - Sum(ni * ln|Si|)
So, for the example above
M = (197 * 27.66719) - [(74*27.67216) + (65*27.59350) + (58*27.35155)]
M = 22.7292
Sorry to be such a nuisance, but I'm completely stumped here.
Dr. Marc R. Feldesman
Professor and Chairman
Anthropology Department
Portland State University
P.O. Box 751
Portland, Oregon 97207
email: [email protected]
phone: 503-725-3081
fax: 503-725-3905
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ROOT Reference Guide
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rf312_multirangefit.C
Go to the documentation of this file.
1/// \file
2/// \ingroup tutorial_roofit
3/// \notebook -nodraw
4/// Multidimensional models: performing fits in multiple (disjoint) ranges in one or more dimensions
5///
6/// \macro_code
7/// \macro_output
8///
9/// \date July 2008
10/// \author Wouter Verkerke
11
12#include "RooRealVar.h"
13#include "RooDataSet.h"
14#include "RooGaussian.h"
15#include "RooProdPdf.h"
17#include "RooPolynomial.h"
18#include "TCanvas.h"
19#include "TAxis.h"
20#include "RooPlot.h"
21#include "RooFitResult.h"
22using namespace RooFit;
23
25{
26
27 // C r e a t e 2 D p d f a n d d a t a
28 // -------------------------------------------
29
30 // Define observables x,y
31 RooRealVar x("x", "x", -10, 10);
32 RooRealVar y("y", "y", -10, 10);
33
34 // Construct the signal pdf gauss(x)*gauss(y)
35 RooRealVar mx("mx", "mx", 1, -10, 10);
36 RooRealVar my("my", "my", 1, -10, 10);
37
38 RooGaussian gx("gx", "gx", x, mx, 1.0);
39 RooGaussian gy("gy", "gy", y, my, 1.0);
40
41 RooProdPdf sig("sig", "sig", gx, gy);
42
43 // Construct the background pdf (flat in x,y)
44 RooPolynomial px("px", "px", x);
45 RooPolynomial py("py", "py", y);
46 RooProdPdf bkg("bkg", "bkg", px, py);
47
48 // Construct the composite model sig+bkg
49 RooRealVar f("f", "f", 0., 1.);
50 RooAddPdf model("model", "model", RooArgList(sig, bkg), f);
51
52 // Sample 10000 events in (x,y) from the model
53 std::unique_ptr<RooDataSet> modelData{model.generate({x, y}, 10000)};
54
55 // D e f i n e s i g n a l a n d s i d e b a n d r e g i o n s
56 // -------------------------------------------------------------------
57
58 // Construct the SideBand1,SideBand2,Signal regions
59 //
60 // |
61 // +-------------+-----------+
62 // | | |
63 // | Side | Sig |
64 // | Band1 | nal |
65 // | | |
66 // --+-------------+-----------+--
67 // | |
68 // | Side |
69 // | Band2 |
70 // | |
71 // +-------------+-----------+
72 // |
73
74 x.setRange("SB1", -10, +10);
75 y.setRange("SB1", -10, 0);
76
77 x.setRange("SB2", -10, 0);
78 y.setRange("SB2", 0, +10);
79
80 x.setRange("SIG", 0, +10);
81 y.setRange("SIG", 0, +10);
82
83 x.setRange("FULL", -10, +10);
84 y.setRange("FULL", -10, +10);
85
86 // P e r f o r m f i t s i n i n d i v i d u a l s i d e b a n d r e g i o n s
87 // -------------------------------------------------------------------------------------
88
89 // Perform fit in SideBand1 region (RooAddPdf coefficients will be interpreted in full range)
90 std::unique_ptr<RooFitResult> r_sb1{model.fitTo(*modelData, Range("SB1"), Save(), PrintLevel(-1))};
91
92 // Perform fit in SideBand2 region (RooAddPdf coefficients will be interpreted in full range)
93 std::unique_ptr<RooFitResult> r_sb2{model.fitTo(*modelData, Range("SB2"), Save(), PrintLevel(-1))};
94
95 // P e r f o r m f i t s i n j o i n t s i d e b a n d r e g i o n s
96 // -----------------------------------------------------------------------------
97
98 // Now perform fit to joint 'L-shaped' sideband region 'SB1|SB2'
99 // (RooAddPdf coefficients will be interpreted in full range)
100 std::unique_ptr<RooFitResult> r_sb12{model.fitTo(*modelData, Range("SB1,SB2"), Save(), PrintLevel(-1))};
101
102 // Print results for comparison
103 r_sb1->Print();
104 r_sb2->Print();
105 r_sb12->Print();
106}
#define f(i)
Definition RSha256.hxx:104
Option_t Option_t TPoint TPoint const char GetTextMagnitude GetFillStyle GetLineColor GetLineWidth GetMarkerStyle GetTextAlign GetTextColor GetTextSize void char Point_t Rectangle_t WindowAttributes_t Float_t Float_t Float_t Int_t Int_t UInt_t UInt_t Rectangle_t Int_t Int_t Window_t TString Int_t GCValues_t GetPrimarySelectionOwner GetDisplay GetScreen GetColormap GetNativeEvent const char const char dpyName wid window const char font_name cursor keysym reg const char only_if_exist regb h Point_t winding char text const char depth char const char mx
Efficient implementation of a sum of PDFs of the form.
RooArgList is a container object that can hold multiple RooAbsArg objects.
Definition RooArgList.h:22
Plain Gaussian p.d.f.
Definition RooGaussian.h:24
RooPolynomial implements a polynomial p.d.f of the form.
Efficient implementation of a product of PDFs of the form.
Definition RooProdPdf.h:33
Variable that can be changed from the outside.
Definition RooRealVar.h:37
RooCmdArg Save(bool flag=true)
RooCmdArg PrintLevel(Int_t code)
Double_t y[n]
Definition legend1.C:17
Double_t x[n]
Definition legend1.C:17
The namespace RooFit contains mostly switches that change the behaviour of functions of PDFs (or othe...
Definition JSONIO.h:26
Ta Range(0, 0, 1, 1) | 2.6875 | CC-MAIN-2024-26 |
# Quickly Estimate the True Cost of Ongoing Spending
by on Oct 21, 2013 · 7 comments
Today I want to share a rule of thumb to help you quickly estimate the long-term impact of your ongoing spending decisions.
If you’re having trouble keeping your spending in check, this tip will hopefully motivate you to plug some of the holes in your budget. And it may help keep the rest of you out of trouble.
In short, to estimate the true cost of an ongoing indulgence, simply multiply the monthly expense by 200. Or, for the math challenged, you can simply double the amount in question and tack on a couple of zeroes.
Seems too easy, right? Well, it’s not. 200x the monthly cost of your habit is (roughly) the future value of that monthly expenditure ten years down the road. I’ll explain more below. But first, some examples.
## The high cost of habits
Consider your morning coffee. Yes, the latte factor. Again. It may be trite, but it’s useful for illustrating the point.
Let’s say that you spend \$3 on a high-end coffee on the way to work each day. With 20 work days in a typical month (and taking weekends off), you’re looking at \$60 per month. Not so bad, right?
Well… After multiplying by 200, you’ll discover that you’re foregoing \$12k in wealth over the next decade in order to get your caffeine fix.
Irony alert: I’m sitting in a local coffee shop drinking a latte as I write this. This is an occasional indulgence that’s well worth it in my book. But it’s nowhere near a daily occurrence. I usually brew my own with an Aeropress at home.
What about cable TV? That \$100/month package will set you back a whopping \$20k over the next ten years. That’s not quite the deal you thought you were getting when you signed the contract, is it?
This isn’t to say that you shouldn’t enjoy things in life. As noted above, I enjoy an occasional latte, and we do have satellite TV service. But these have been conscious decisions as opposed to mindless indulgences.
The point here is to encourage you to stop and think — ever so briefly — about the impact of your spending decisions before you reach for your wallet.
## Where does 200x come from?
To arrive at the 200x value, I simply computed what \$X/month over the next ten years would be worth if you had invested the money instead of spending it.
Investment returns were pegged at 7%/year, which is somewhat below the historical average for a standard 60/40 portfolio. Seems like a reasonable assumption.
Of course, thanks to our friend inflation, a dollar ten years from now will be worth less than a dollar today. Thus, I also accounted for inflation.
To do this, I pegged inflation at 3%/year (near the historical average) and did two things. First, I ratcheted up the spending 3% per year to account for the increased cost of your habit over time. Second, I docked the investment return by 3%/year to arrive at a real return of 4%/year.
And at the end of all this, I learned that a dollar in monthly spending has a ten year value of roughly \$200. So please, the next time you’re thinking about taking on a new expense, think about the long-term effect…
Would you rather incrementally ratchet up your standard of living or ratchet up your savings and have thousands (and thousands!) of extra dollars down the road?
1 Little House October 21, 2013 at 9:46 am
This is why I’ve changed some of my spending habits over the past year; I now make my coffee at home, we’ve ditched cable, and found a cell phone plan that’s almost \$100 less per month than our old one. I’d rather have that money for paying off debt then investing!
2 Michael October 21, 2013 at 11:30 am
Your future self will thank you for making these decisions. 🙂
3 Tie the Money Knot October 22, 2013 at 12:49 am
I like it! I think like this as well, at least some of the time anyway. When put into perspective like this, it can give a person reason to pause and think before making such purchases. Small amounts today can add up to a lot of money down the line.
4 Edward Antrobus October 22, 2013 at 8:27 am
On the flip side, I’m on track to have a fully funded retirement when I am ready to retire. Why forgo a pleasure now to have more money than I need in the future?
5 Michael October 22, 2013 at 9:35 am
Edward: I agree completely. As stated above, the point here is just to get you to think about the impact of your spending decisions.
In other words, this is more about combatting mindless spending than it is about enforcing arbitrary austerity.
6 Money Beagle October 22, 2013 at 10:29 am
It’s funny because I am a long-term thinker when it comes to money, and my wife often gets crazy when I offer the opinion that ‘no, we can’t afford it’ on various things that I know would only have short-term impact and that I feel we can do without. However, the payoff comes when we are looking at a bigger goal and find that we can easily afford it, pay cash, and not have to worry about it.
7 thepotatohead October 29, 2013 at 10:10 pm
I’ve been trying to cut down on my recurring bills as much as possible for this reason. I came to the realization that If I can get rid of some of the pointless bills each month that in the next 10,15,20 yrs I’d have a boatload more money. Easy concept but definitely takes some self motivation and willpower to implement.
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Giant Green Giraffe
A giant, goofy, Green giraffe, Had a ten-foot neck, Don't laugh!!! Liked to eat, chocolate chips, Melt them on his purple lips... Once he saw a chocolate pie, Floating high up, In the sky... 20-feet above the ground, Floating as it made no sound!!!! Truly, really wanna' chew, Chocolate pie, With chocolate goo!!! I can't reach, No can do, Need that pie, Thought it through... Took all day, Thought some more, It was solved, Brain was sore! 10-foot neck, Stretched 10 feet!!!!!!!!!! This was not an easy feat! 10 + 10, Gave him 20, Hurt his neck, Wasn't funny... But that pie, Oh-so-sweet, Chocolate pie, He got to eat...
10 + 10 = 20
Math Challenges If you have 2 dimes, how many cents do you have? If a giraffe's neck is 20 feet long, but it shrinks 5 feet, how long will the giraffe's neck be? If the pie in the poem had been floating 25-feet in the air, how far would the giraffe have had to stretch his 10-foot neck?
Watch Mr. R.'s Double Numbers Music Video Watch Mr. R.'s Music Video of This Poem
Listen to the song version of this poem green giraffe!
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