<p>The answers to <a href="https://astronomy.stackexchange.com/questions/9/could-someone-explain-ra-dec-in-simple-terms">this question</a> are very clear and much better than I could have written.</p>

<p>I'm not sure what you mean by <code>"the stellar time at Greenwich at 0h UTC is 22h20min"</code>, but I'm assuming that you mean that that is the amount of time since the culmination of Aries over the Prime Meridian (Greenwich). That being the case, the Right Ascension (RA) of your body can be easily calculated. All we need to do is figure out what RA is passing your meridian at the time in question. This is probably a good time to point out that the longitude of Mt Teide is 16° 30'W not E.</p>

<p>We already know that RA 22H 20m is passing over Greenwich at the time of interest (or, at least that is what I am assuming as your explanation is not at all clear.), as we are to the West of Greenwich, then we know that any RA on our meridian is earlier and we can calculate how much earlier simply by dividing our longitude by 15 (as the earth spins at a nice steady 15°/hour):-</p>

<pre><code>diff in RA = Long/15 
           = 16° 30'/15 
           = 1H 06m
</code></pre>

<p>Then:-</p>

<pre><code>RA = RA at Greenwich - diff in RA 
   = 22H 20m - 1H 06m 
   = 21H 14m
</code></pre>

<p>Now for Declination:-</p>

<p>The first stage is to take a piece of ruled or graph paper and near the middle mark a short, horizontal line. This is your horizon, and is marked 'H'. Now draw upwards 9 lines and mark the top ,'Z' and downwards 9 lines and mark the bottom 'N'. You should end up with something that looks like this:-</p>

<p><img src="https://i.sstatic.net/DtOhK.png" alt="enter image description here"></p>

<p>Which is, basically a picture of your meridian as you sit and gaze south at it. 'Z' is your zenith, the point above you in the sky and 'N' is the nadir, it's opposite out on the celestial sphere through the earth beneath your feet.</p>

<p>Now we need to add a little more detail. Imagine that the celestial sphere is actually a glass sphere surrounding the earth with the stars and stuff stuck to it that we can project anything we want on to (we'll ignore porn and coke adverts for now), where would the celestial equator show and where would it go on our picture of the meridian? Well, it would be a distance above the horizon equal to our latitude (lets round it off to 30°), so we now have:-</p>

<p><img src="https://i.sstatic.net/6A532.png" alt="enter image description here"></p>

<p>Now we put in the body we are looking at. We know it's elevation, just over 43°, we can just call it x as you didn't specify a body. We can also label some of the distances we know, to end up with:-</p>

<p><img src="https://i.sstatic.net/TMkp7.png" alt="enter image description here"></p>

<p>We already knew that Lat was H -> Q, that's how we place Q, we know that elevation, or altitude is between the horizon and the body and that declination is measured from the celestial equator. We want to find dec and the diagram shows us that:-</p>

<pre><code>dec = Alt - Lat 
    = 43° 40' - 28° 18' 
    = 15° 18'N
</code></pre>

<p>As 'x' is to the north of 'Q', then Declination is north.</p>

<p>So now we know that we are looking at a body with RA 21H 14m and Dec N 15° 18'.</p>

<p>This is much more complex to explain here than it is to do, or to teach in person. With a bit of practice you'll be doing it in your head. I hope this has helped a bit.</p>

<p>I should point out that I have ignored observation errors, such as height of eye, which may be significant from the place you specify and parallax but I wanted to keep it as simple as possible.</p>

CC BY-SA 3.0
2013-12-04T21:33:02.170Z
Your answer is very diligently written, but it only answers half of the question.
Alexey Bobrick
CC BY-SA 3.0
2013-12-04T22:41:50.920Z
Yes, I give a reason for that in the second paragraph.
vascowhite
CC BY-SA 3.0
2013-12-04T23:40:29.180Z
Yes, absolutely, it is fair and I have read it. However, the other half of the question still needs to be answered by someone.
Alexey Bobrick
CC BY-SA 3.0
2013-12-05T12:57:33.533Z
Oh no! Sorry for misunderstanding. I put a downvote on your post, and then I wrote why. I think thew answer is good, but another answer is needed to complement yours.
Alexey Bobrick
CC BY-SA 3.0
2013-12-05T14:20:40.710Z
Oh, I see. Sorry. Obviously your votes are yours to dish out as you see fit, but I feel it is a little unfair to down vote for something that is clearly explained in the answer. Will you also down vote the person who provides the other half of the solution?
vascowhite
CC BY-SA 3.0
2013-12-05T18:55:30.327Z
actually, yes, I would. My current impression of SE system is that users tend to read one or two first replies only, on average upvoting them somewhat, and hence making the next users think that the other posts are bad or wrong. In the case of your post, I wouldn't like it to end up as the only answer to the question, which is upvoted high enough and hence considered sufficient. This is my thinking. However, if you complete your answer, I will of course turn the donwvote to upvote.
Alexey Bobrick
CC BY-SA 3.0
2013-12-05T19:05:06.313Z
One way of making sure it is not the only answer is to offer your own :)
vascowhite
CC BY-SA 3.0
2013-12-05T19:08:41.200Z
true, but I don't find the question interesting enough :)
Alexey Bobrick
CC BY-SA 3.0
2013-12-07T10:38:00.200Z
@AlexeyBobrick I've had a stab at explaining the RA calculation too. I'd be interested to know if it is clear/correct.
vascowhite
CC BY-SA 3.0
2013-12-07T14:44:06.857Z
I can surely take a look at it if you like.
Alexey Bobrick
CC BY-SA 3.0
2013-12-04T15:27:02.477Z
2013-12-07T10:35:40.550Z
2017-04-13T12:59:06.910Z
vascowhite
<p>I am confused about this problem:</p>

<p>If I see an object from Mount Teide (longitude is 16"30'E and latitude is 28"18'N) that passes the meridian (azimuth=0) at 5h (am) UTC, and I also know that the elevation of the star is 43"40', the stellar time at Greenwich at 0h UTC is 22h20min.</p>

<p>How can I calculate the right ascension and declination?</p>

CC BY-SA 3.0
2013-11-27T17:32:46.693Z
Have you read [this question and its answers](http://astronomy.stackexchange.com/questions/9/could-someone-explain-ra-dec-in-simple-terms) yet? If so, is there something still unclear to you which you could state more explicitly?
called2voyage
CC BY-SA 3.0
2013-11-27T18:10:13.977Z
When you say elevation, do you mean altitude?
astromax
CC BY-SA 3.0
2013-11-27T20:27:24.453Z
@astromax I think that the elevation is synonymous with altitude.
Jackson Hart
CC BY-SA 3.0
2013-11-27T20:28:19.037Z
@called2voyage I am very confused with the basic equations in which to use to solve this problem. I do not have a background in this field. I guess the task just seems daunting right off the bat, but I understand that it is just a regular coordinate transformation
Jackson Hart
CC BY-SA 3.0
2013-11-30T01:50:11.870Z
Also, are you using arcminutes/arcseconds to describe latitude and longitude?
astromax
CC BY-SA 3.0
2013-11-30T09:43:58.357Z
@astromax I am not really sure. The latitude and longitude are expressed as I expressed them above. I am not sure how else they might be expressed. Do you have any ideas on this problem?
Jackson Hart
CC BY-SA 3.0
2013-12-05T01:50:40.487Z
I could possibly help, but first, a few things have to be properly defined. Where did you get this problem from? Textbook, class, somewhere else? I'm trying to understand the context so I can try to piece together some things I'm still confused about: namely your use of the word "elevation" and the single and double quotes to describe this coordinate.
astromax
<fundamental-astronomy><ascension><declination>
Calculation of right ascension and declination
Jackson Hart
# Calculation of right ascension and declination

(Asked by: username_0 on 2013-11-27)

I am confused about this problem:

If I see an object from Mount Teide (longitude is 16"30'E and latitude is 28"18'N) that passes the meridian (azimuth=0) at 5h (am) UTC, and I also know that the elevation of the star is 43"40', the stellar time at Greenwich at 0h UTC is 22h20min.

How can I calculate the right ascension and declination?

**Tags:** fundamental-astronomy, ascension, declination
> **Comments:**
> **username_1**: Have you read [this question and its answers](http://astronomy.stackexchange.com/questions/9/could-someone-explain-ra-dec-in-simple-terms) yet? If so, is there something still unclear to you which you could state more explicitly?  
> **username_2**: When you say elevation, do you mean altitude?  
> **username_0**: @astromax I think that the elevation is synonymous with altitude.  
> **username_0**: @called2voyage I am very confused with the basic equations in which to use to solve this problem. I do not have a background in this field. I guess the task just seems daunting right off the bat, but I understand that it is just a regular coordinate transformation  
> **username_2**: Also, are you using arcminutes/arcseconds to describe latitude and longitude?  
> **username_0**: @astromax I am not really sure. The latitude and longitude are expressed as I expressed them above. I am not sure how else they might be expressed. Do you have any ideas on this problem?  
> **username_2**: I could possibly help, but first, a few things have to be properly defined. Where did you get this problem from? Textbook, class, somewhere else? I'm trying to understand the context so I can try to piece together some things I'm still confused about: namely your use of the word "elevation" and the single and double quotes to describe this coordinate.  

---

## 1 answers

### Answer by username_3 on 2013-12-04. Score: 5 | Accepted answer

The answers to this question are very clear and much better than I could have written.
I'm not sure what you mean by "the stellar time at Greenwich at 0h UTC is 22h20min", but I'm assuming that you mean that that is the amount of time since the culmination of Aries over the Prime Meridian (Greenwich). That being the case, the Right Ascension (RA) of your body can be easily calculated. All we need to do is figure out what RA is passing your meridian at the time in question. This is probably a good time to point out that the longitude of Mt Teide is 16° 30'W not E.
We already know that RA 22H 20m is passing over Greenwich at the time of interest (or, at least that is what I am assuming as your explanation is not at all clear.), as we are to the West of Greenwich, then we know that any RA on our meridian is earlier and we can calculate how much earlier simply by dividing our longitude by 15 (as the earth spins at a nice steady 15°/hour):-
diff in RA = Long/15 
           = 16° 30'/15 
           = 1H 06m

Then:-
RA = RA at Greenwich - diff in RA 
   = 22H 20m - 1H 06m 
   = 21H 14m

Now for Declination:-
The first stage is to take a piece of ruled or graph paper and near the middle mark a short, horizontal line. This is your horizon, and is marked 'H'. Now draw upwards 9 lines and mark the top ,'Z' and downwards 9 lines and mark the bottom 'N'. You should end up with something that looks like this:-

Which is, basically a picture of your meridian as you sit and gaze south at it. 'Z' is your zenith, the point above you in the sky and 'N' is the nadir, it's opposite out on the celestial sphere through the earth beneath your feet.
Now we need to add a little more detail. Imagine that the celestial sphere is actually a glass sphere surrounding the earth with the stars and stuff stuck to it that we can project anything we want on to (we'll ignore porn and coke adverts for now), where would the celestial equator show and where would it go on our picture of the meridian? Well, it would be a distance above the horizon equal to our latitude (lets round it off to 30°), so we now have:-

Now we put in the body we are looking at. We know it's elevation, just over 43°, we can just call it x as you didn't specify a body. We can also label some of the distances we know, to end up with:-

We already knew that Lat was H -> Q, that's how we place Q, we know that elevation, or altitude is between the horizon and the body and that declination is measured from the celestial equator. We want to find dec and the diagram shows us that:-
dec = Alt - Lat 
    = 43° 40' - 28° 18' 
    = 15° 18'N

As 'x' is to the north of 'Q', then Declination is north.
So now we know that we are looking at a body with RA 21H 14m and Dec N 15° 18'.
This is much more complex to explain here than it is to do, or to teach in person. With a bit of practice you'll be doing it in your head. I hope this has helped a bit.
I should point out that I have ignored observation errors, such as height of eye, which may be significant from the place you specify and parallax but I wanted to keep it as simple as possible.

> **Comments:**
> **username_4**: Your answer is very diligently written, but it only answers half of the question.  
> **username_3**: Yes, I give a reason for that in the second paragraph.  
> **username_4**: Yes, absolutely, it is fair and I have read it. However, the other half of the question still needs to be answered by someone.  
> **username_4**: Oh no! Sorry for misunderstanding. I put a downvote on your post, and then I wrote why. I think thew answer is good, but another answer is needed to complement yours.  
> **username_3**: Oh, I see. Sorry. Obviously your votes are yours to dish out as you see fit, but I feel it is a little unfair to down vote for something that is clearly explained in the answer. Will you also down vote the person who provides the other half of the solution?  
> **username_4**: actually, yes, I would. My current impression of SE system is that users tend to read one or two first replies only, on average upvoting them somewhat, and hence making the next users think that the other posts are bad or wrong. In the case of your post, I wouldn't like it to end up as the only answer to the question, which is upvoted high enough and hence considered sufficient. This is my thinking. However, if you complete your answer, I will of course turn the donwvote to upvote.  
> **username_3**: One way of making sure it is not the only answer is to offer your own :)  
> **username_4**: true, but I don't find the question interesting enough :)  
> **username_3**: @AlexeyBobrick I've had a stab at explaining the RA calculation too. I'd be interested to know if it is clear/correct.  
> **username_4**: I can surely take a look at it if you like.  

---

