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Air Force Stress Manual:
Column Analysis
Simple Columns
Column Data
Straight Line Constants
Crippling of Columns
Complex Columns
Air Force Stress Manual
Beam Analysis
Frames and Rings
Lug Analysis
Shaft Analysis
Bearing Stresses
Related References
Properties of Metals
Fastener Design Manual (NASA)
Preloaded Joint Methodology (NASA)
Column Buckling Calculator
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Relevant Textbooks
Analysis of Simple Columns
This page provides the sections on the analysis of simple columns from the "Stress Analysis Manual," Air Force Flight Dynamics Laboratory, October 1986.
Other related chapters from the Air Force "Stress Analysis Manual" can be seen to the right.
Constants for Straight Line Equation
Analysis of Complex Columns
2.1 Introduction to Column Analysis
The stresses that a structural element can sustain in compression are functions of several parameters. These parameters are:
the length of the element along its loading axis,
the moment of inertia of the element normal to its loading axis,
the cross-sectional variation of the element with length,
the eccentricity of the applied load,
the continuity of the integral parts of the element,
the cross-sectional characteristics of the element,
the homogeneity of the element material,
the straightness of the element, and
the end fixity of the element.
The effects of these parameters can be categorized by first establishing certain necessary assumptions. For the following analysis, it is assumed that the material is homogeneous and isotropic. It is further assumed that the element is initially straight and, if it is composed of several attached parts, that the parts act as integral components of the total structural configuration.
The remainder of the previously mentioned parameters dictate more general classifications of compression elements. If a compression element is of uniform cross section and satisfies the previously mentioned assumptions, it is referred to as a simple column and is treated in the first part of this chapter. On the other hand, compression members having variable cross-sectional properties are called complex columns and are covered in the latter part of this chapter. Stepped and latticed columns are included in the treatment of complex columns.
The possible basic types of failure defined for columns are primary and secondary failure. Primary failure occurs when a column fails as a whole and may be defined by the fact that cross sections of the element retain their original shape although they may be translated and/or rotated with respect to their original position. If cross sections are translated but not rotated, the primary failure is of the bending type. Failures for which cross sections of a column are either rotated or rotated and translated are treated in the section on torsional instability.
If a column experiences a failure due to lateral bending at a stress level below the elastic limit of the material, it is defined to be a long column while failures at a maximum stress greater than the elastic limit are characteristic of short columns.
Secondary failures occur when buckling or crippling occur in sections of a column before it is loaded enough to produce a primary failure.
A column failure of a selected element is influenced by the eccentricity of the applied load and by the end fixity of the element. Both of these dictate boundary conditions that modify the solutions to the differential equations governing column response.
In general, a column must be designed to prevent both the bending and torsional types of primary failure as well as crippling. Crippling is likely to occur in columns having thin portions in their cross sections. The torsional type of primary failure is likely to occur at a lower load than the bending type in columns having cross sections of relatively low torsional stiffness. Closed sections have enough torsional stiffness to insure that any primary failure will be of the bending type so they must only be designed against this and crippling.
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buckling analysis of long and intermediate-length columns loaded in compression
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2.2 Nomenclature for Column Analysis
A = area
a = linear dimension as indicated in diagrams
b = linear dimension as indicated in diagrams
C = coefficient of constraint = (L/L' )2
CBT = torsion - bending constant
C = distance from neutral axis to the concave side of loaded column
D = diameter
E = modulus of elasticity
Er = reduced modulus of elasticity
Es = secant modulus of elasticity
Et = tangent modulus of elasticity
e = eccentricity of loading
e = strain
ec/ρ2 = eccentric ratio
Facol = working concentrically loaded column stress
Fb = working bending stress
Fc = allowable compressive stress
Fcb = working compressive stress in bending
Fcc = allowable crippling stress
Fco = empirical constant in Johnson parabolic equation (column yield stress)
Fcol = maximum fiber stress for primary failure of a column
Fcb = proportional limit in compression
Fcy = compressive yield stress
f = calculated stress
fc = calculated compressive stress
f1 = stress at which the secant modulus of elasticity is equal to 0.7
G = modulus of elasticity in shear
I = moment of inertia
Ip = polar moment of inertia
J = torsion constant
K = spring constant
K = empirical constant
L' = effective length = L/√C
L/ρ = slenderness ratio
L'/ρ = effective slenderness raho
(L'/ρ)cr = critical effective slenderness ratio
M = empirical constant in straight line column equation
N = empirical constant in straight line column equation
n = empirical constant in Ramberg-Osgood equation
P = axial load
Pa = allowable load
Pcc = crippling load
Pcr = critical load
Pe = Euler critical load
T = torque
t = thickness
x, y, z = rectangular coordinates
ϵ = Ee/f1 where e is strain
μ = Poisson's ratio
μ = torsional spring constant
ρ = radius of gyration = \( \sqrt{I/A} \)
Σ = summation
σ = fc/f1
θ = angular deflection
φv; = angular deflection
2.3 Simple Columns
A simple column acts as a single unit and has a uniform cross section along its length. Such columns are treated in the following material.
2.3.1 Primary Failure of Simple Columns
A simple column has a primary failure when its cross sections are translated and/or rotated while retaining their original shape, that is, when the column fails as a whole without local instability. If the column cross sections are translated but not rotated as shown to the left of Figure 2-1, it is said to fail by bending while pure rotation or a combination of rotation and translation are characteristic of torsional failures.
2.3.1.1 Column Data Applicable to Both Long and Short Columns
A stable section (not subject to crippling) testing for various lengths will generate data of the form shown in Figure 2-2. The stress Fcol is the stress at failure, and L'/ρ is the ratio of the effective column length to the radius of gyration of the section. This L'/ρ ratio is called the effective slenderness ratio of the column.
From the figure, it is apparent that the Euler column curve is quite accurate beyond a critical L'/ρ which defines the separation between long and short columns. A great amount of test data, collected for particular materials, is available and eliminates the need to determine whether a long or a short column curve is applicable. A summary of column allowable curves that are applicable to both long and short columns is outlined and presented on the following pages. These curves are based upon the tangent modulus equation which is discussed in Section 2.3.1.11.1. The column allowables are based on minimum guaranteed properties, Basis A, or probability properties, Basis B, if the latter are available. The pertinent basis is indicated in the figures.
The figures for the column data can be found on this page.
2.3.1.2 Sample Problem - Column Data Applicable to Both Long and Short Columns
Given: The 0.6 in. square concentrically loaded column shown in Figure 2-35.
Find: The critical load, Pcr, by using column curves applicable to both long and short columns.
Solution: Since the column is pinned at both ends; L' = L = 6 in. For a square, I = b4/12 and A = b2. Thus,
$$ \rho = \sqrt{ ~I \over A } = \sqrt{ b^2 \over 12 } = \sqrt{ (0.6)^2 \over 12 } = 0.173 ~\text{in} $$ $$ { L' \over \rho } = { 6 ~\text{in} \over 0.173 ~\text{in} } = 34.6 $$
From Figure 2-3, curve 2, find Fcol = 56,200 psi. Thus,
$$ P_{cr} = F_{col} A = 56,200 (0.6)^2 = 20,200 ~\text{lb} $$
2.3.1.3 Bending Failure of Concentrically Loaded Long Columns
In the process of describing column behavior in this chapter, the simplest cases are covered first and then various complications are covered. Historically, the first type of column to be successfully studied was the long concentrically loaded one for which Euler developed an equation giving the buckling load in terms of column parameters. This is also the simplest case.
The Euler formula, which is perhaps the most familiar of all column formulas, is derived with the assumptions that loads are applied concentrically and that stress is proportional to strain. Thus, it is valid for concentrically loaded columns that have stable (not subject to crippling) cross sections and fail at a maximum stress less than the proportional limit, that is, concentrically loaded long columns. The form of the Euler formula is
$$ { P_{cr} \over A } = { C \pi^2 E \over (L / \rho)^2 } $$ or
$$ { P_{cr} \over A } = { \pi^2 E \over (L' / \rho)^2 } $$
Here, P/A is the ratio of the axially applied load at failure to the cross-sectional area of the column. E is the modulus of elasticity of the column and L/ρ is the ratio of the column length to the least radius of gyration of its section. The radius of gyration is defined to be equal to \(\sqrt{I/A}\), where I is the moment of inertia of the section. The constant, C, which is called the coefficient of constraint, is dependent upon end restraints and is discussed in Section 2.3.1.4. In the second form of Equation (2-1), L' is an effective length which takes into account end restraint conditions.
2.3.1.4 Coefficient of Constraint for End Loaded Columns
In the discussion of columns, the coefficient of constraint C often occurs. As was mentioned before, this coefficient depends upon the manner in which the ends of a column are restrained, which, in turn, determines the boundary conditions that must be satisfied by the equations describing the column.
Sometimes, the use of a coefficient is avoided by using an effective length L' instead of the actual length L in formulas derived for a column with both ends pinned of length L. The term L' is then the distance between points of inflection of the loaded column curve. For example, the effective length of a column of length L that is rigidly supported at both ends is L/2 as can be seen in Figure 2-36.
The relationship between L' and L, and C is
$$ \left({ L \over L' }\right)^2 = C $$
By assuming various combinations of idealized end restraints, a table of theoretical end constraint factors may be constructed. This is shown in Table 2-1.
Table 2-1 Coefficients of Constraint for Idealized End Conditions
Type of Fixity
C 0.25 1.00 2.05 4.00
The end restraints of an actual column are never exactly equivalent to pinned or fixed ends, but lie somewhere between the two extremes. This discrepancy is due to the fact that a pinned joint is never entirely frictionless and a member to which a column is fixed is never perfectly rigid.
Cases for which one or both of the ends of a column are fixed to nonrigid members may be treated by considering these ends to be restrained by a torsional spring of spring constant μ. The constant μ is defined to be dT/dθ where T is a torque applied to the support at the point where the column is attached and θ is the angle of twist at this point. Given this definition, μ may be calculated by applying formulas from strength of materials to the member to which the column is attached before attachment. For example, consider the column shown in Figure 2-37. The torsional spring constant for the end is found by considering the beam supporting the column and calculating dT/dθ where T and θ are as shown in the middle diagram. dT/dθ is a constant for small deflections so T/θ may be found from beam formulas. The column is then redrawn with the attached beam replaced by an equivalent torsional spring.
Equations have been developed that give the end fixity coefficient, C, as a function of μ, L, E, and I for columns equally elastically restrained at both ends and for those pinned at one end. These equations are given and plotted in Figure 2-38. From these plots, it can be seen that C approaches the value corresponding to a fixed end rather than an elastically restrained end as μ increases. Likewise, C approaches the value corresponding to a pinned end as the spring constant approaches zero.
At times, a column may be supported along its length as well as at the ends. For example, the column shown in Figure 2-39 is supported at its midsection by another member.
In order to treat such a case, namely that illustrated in Figure 2-39, we may consider the member supporting the column at its midsection to be a spring of spring constant K lbs/in which may be found by applying strength of materials to the support. Charts that give the coefficients of restraint of a pinned column with a single lateral support along its midsection or two lateral supports symmetrically placed on its midsection are available. Figure 2-40 shows a simply supported column with one lateral restraint and gives the coefficient of constant of such a column as a function of column parameters. Figure 2-41 shows a simply supported column with two symmetrically placed lateral restraints and gives the coefficient of constraint as a function of column parameters.
The previous discussion of coefficients of constraint was limited to relatively simple cases. At times, however, a column may be attached to members for which an equivalent torsional spring constant is not easily found or it may be attached in more complicated ways than those previously discussed. In such cases, certain general rules may be applied.
In normal practice, the coefficient of constraint is less than two and the effect of end fixity is smaller for short columns than for long ones.
The ends of a compression member in a welded truss of steel tubes, like those that are often used in aircraft structures, cannot rotate without bending all of the other members at the end joints. Such a truss is shown in Figure 2-42. It is difficult to obtain the true end fixity of a compressive member in such a truss since the member may buckle either horizontally or vertically and iv restrained by the torsional and bending rigidity of many other members. It is usually conservative to assume C = 2.0 for all members. A smaller coefficient of constraint might be used for a heavy compressive member restrained by comparatively light members. Likewise, a larger coefficient of constraint may hold for a light compressive member restrained by heavier members. A coefficient of constraint of one should be used if all of the members at a joint are in compression. Steel tube engine mounts are usually designed with the conservative assumption of a coefficient of constraint of unity.
Stringers which act as compression members in semimonocoque wing or fuselage structure, such as that shown in Figure 2-43, are usually supported by comparatively flexible ribs or bulkheads. The ribs or bulkheads are usually free to twist as shown so that their restraining effect may be neglected and the value of C is taken to be unity, where L is the length between bulkheads. If the bulkheads are rigid enough to provide restraint and clips are provided to attach the stringers to the bulkheads, a value of 1.5 is sometimes used for C.
2.3.1.5 Distributed Axial Loads
Columns subjected to distributed axial loads may be treated by formulas developed for end leaded columns if a coefficient of constraint is used that takes into account both the load condition and end fixities. Figure 2-44 shows columns under a uniformly distributed axial load of P/L lbs/in with various end restraints and their corresponding coefficients of constraint. The values P, L, and C are used in the formulas for end loaded columns.
2.3.1.6 Sample Problem - Concentrically Loaded Long Columns in Bending
Given: The concentrically loaded rectangular bar shown in Figure 2-45 is fixed at one end and attached to a round bar at the other end. Both bars are made of steel for which E = 30×106 psi and G = 11.5×106 psi.
Find: Pcr
Solution: From elementary strength of materials, θ = TL/IpG for a torsion bar. In effect, there are two 7.25-in -long bars attached to the end of the column so that the equivalent torsional spring constant is
$$ \mu = 2 { T \over \theta } = { 2 J G } \over L $$
Substituting the appropriate values into the expression for μ gives
$$ \mu = { (2) { \pi (1)^4 \over 32 } (11.5 \times 10^6) \over 7.25 } = 3.12 \times 10^5 ~\text{in-lb/rad} $$
The column may now be redrawn as Figure 2-46.
$$ { \mu L \over E I } = { (3.12 \times 10^5) (40) \over (30 \times 10^6) \left[{ 2 (0.5)^3 \over 12 }\right] } = 20 $$
From Figure 2-38, we find that C = 1.88. Solving Equation (2-2) for L' gives
$$ L' = { L \over \sqrt{C} } = { 40 \over \sqrt{1.88} } = 29.2 ~\text{in} $$
The radius of gyration of the bar is
$$ \sqrt{I/A} = \sqrt{ {2(0.5)^2 \over 12} / 2(0.5) } = 0.144 ~\text{in} $$
The slenderness ratio
$$ { L' \over \rho } = { 30.5 \over 0.144 } = 212 $$
From Section 2.3.1.11.7, it is found that a steel column for which L/ρ is greater than 120 is a long column. Thus, the Euler formula, Equation (2-1), becomes
$$ P_{cr} = { A \pi^2 E \over (L' / \rho)^2 } $$
Substituting the values for the given column into this equation gives
$$ P_{cr} = { (0.5 \times 2) \pi^2 (30 \times 10^6) \over 150^2 } = 13,100 ~\text{lb} $$
In general, a column such as this must also be checked for buckling in the plane through the bar and the column. In this case, however, the column will not fail by this mode.
2.3.1.7 Bending Failure of Eccentrically Loaded Long Columns
A theoretically correct formula that holds for eccentrically loaded columns is the secant formula:
$$ { P \over A } = { F_{col} \over 1 + {ec \over \rho^2} \sec{ \left[ {L' \over 2 \rho} \sqrt{P \over AE} \right] } } $$
In this formula, P, A, ρ, L, C, and E are defined as before, and e is the eccentricity of the load. The distance between the central axis and the concave side of the loaded column is designated as c. In the case of a long eccentrically loaded column, Fcol may be taken to be the value of Pcr/A found from the Euler formula in Section 2.3.1.3. If a factor of safety, FS, is applied, the corresponding formula for allowable load, Pa, becomes
$$ { (FS) P_a \over A } = { F_{col} \over 1 + {ec \over \rho^2} \sec{ \left[ {L' \over 2 \rho} \sqrt{ (FS) P_a \over AE} \right] } } $$
The secant formula may also be used for short columns by finding Fcol differently as will be shown in the material on short columns.
All physical columns have some accidental initial curvature due to imperfections and some eccentricity of loading. In these cases, an equivalent eccentricity may be used to approximate the effects of the imperfections. Data may also be found for the equivalent eccentric ratio which is the ec/ρ2 term in the secant formula. Values for these may be found in Section 2.3.1.8.
The secant formula applies when the eccentricity is in the plane of the bending.
Unfortunately, the secant formula is difficult to solve and must be solved by either trial and error or charts.
2.3.1.8 Equivalent Eccentricity for Imperfect Columns
As was mentioned previously, no column is perfectly straight and concentrically loaded. In order to allow for these initial imperfections in a column whose loads are concentrically applied, an equivalent eccentricity of loading may be assumed. The column may then be treated by methods used for an eccentrically loaded column.
Opinion is divided as to whether the equivalent eccentricity assumed for imperfections is dependent upon or independent of the length of a column. If we assume that it is independent of column length, an equivalent eccentric ratio ec/ρ2 ranging from 0.1 to 0.25 may be used. The mean value of such an eccentric ratio is approximately equal to 0.2. We may also assume that the equivalent eccentricity, e, is proportional to the effective length of the column. If this procedure is used, e may be taken to be equal to KL where K is a constant. Values of K ranging from 0.001 to 0.0025 may be used with the latter yielding conservative results.
2.3.1.9 Sample Problem – Long Eccentrically Loaded Columns and Equivalent Eccentricity
Given: The round column shown in Figure 2-47 with nominally concentric loading.
Find: The column diameter, D, for a factor of safety of 1.5, considering the initial imperfections of the column.
Solution: For a round section, I = πD4/64 and A = πD2/4. Thus, \(\rho = \sqrt{I/A} = D/4\). Since the column is pin ended, L = L' = 50 in. Try D = 1.215.
$$ { L' \over \rho } = { 50 \over 1.215/4 } = 164.5 $$
According to Section 2.3.1.11.7, steel columns for which L'/ρ is greater than 120 may be treated as long columns. Thus Fcol in the secant formula may be found by the Euler formula:
$$ F_{col} = { \pi^2 E \over (L'/\rho)^2 } = { \pi^2 (30 \times 10^6) \over (164.5)^2 } = 10,950 ~\text{psi} $$
The secant formula, Equation (2-4), may now be written as
$$ \begin{eqnarray} { (FS) P_a \over A } &=& { F_{col} \over 1 + {ec \over \rho^2} \sec{ \left[ {L' \over 2 \rho} \sqrt{ (FS) P_a \over AE} \right] } } \nonumber \\ &=& { 10,590 \over 1 + {ec \over \rho^2} \sec{ \left[ {L' \over 2 \rho} \sqrt{ (FS) P_a \over AE} \right] } } \end{eqnarray} $$
According to Section 2.3.1.8, equivalent eccentric ratio due to initial imperfections is between 0.1 and 0.25. To be on the conservative side, use an equivalent eccentric ratio of 0.25. Substituting this and column parameters into the secant formula, the expression below is obtained.
$$ { (1.5)(5000) \over { \pi (1.215)^2 \over 4 } } = { 10,950 \over 1 + 0.25 \sec{ \left[ {50 \over { 2 (1.215) \over 4 } } \sqrt{ (1.5)(5000) \over { \pi (1.215)^2 \over 4 } (30 \times 10^6) } \right] } } $$
or 6450 = 6450
Thus, the original guess of D = 1.215 was correct. If this were not true, different values of D would have to be chosen until one was found that would make both sides of the secant formula equal.
2.3.1.10 Bending Failure of Short Columns
In the previous discussion of long columns, it was assumed that the column material was in the elastic range at the time of buckling. This assumption, however, is not true for columns having an effective slenderness ratio of less than a certain critical value for a given material. This value of the critical effective slenderness ratio, L'/ρ, is discussed in Section 2.3.1.11.7. Since the Euler formula no longer applies for short columns, one of the formulas used to fit short column data must be used to treat them.
2.3.1.11 Bending Failure of Concentrically Loaded Short Columns
Several formulas are available to treat short columns. These have no theoretical justification as did the Euler formula, but fit column data to a degree of accuracy depending upon the material and column parameters. The equations most commonly used for short columns are the tangent modulus, Johnson Parabolic, and straight-line equations.
2.3.1.11.1 Tangent Modulus Equation
If the slenderness ratio of a column is low enough that some of its fibers are no longer in the elastic range at the time of failure, the Euler formula no longer holds. However, this case may be treated by defining a tangent modulus of elasticity, Et to be the slope of the stress-strain curve at a given point. This tangent modulus of elasticity may then be substituted for the modulus of elasticity in the Euler equation to obtain the tangent modulus equation (modified Euler equation).
$$ { P \over A } = { \pi^2 E_r \over (L / \rho)^2 } $$
Since Et is equal to E in the elastic range, the tangent modulus equation reduces to the Euler equation for long columns and is thus valid for both long and short columns.
The value of stress at which Et is found is the maximum stress in the column. In the case of a concentrically loaded column, this is equal to P/A. Since Et is a function of loading, the tangent modulus equation must be solved by trial and error if Et is found from stress-strain diagrams or tables.
The tangent modulus equation has been solved for a number of different materials and these solutions are shown in Section 2.3.1.1.
The main disadvantage of the previously described procedure for solving the tangent modulus equation is the trial and error method required. This disadvantage may be eliminated if an equation for the stress-strain curve is available. Such an equation is the Ramberg-Osgood equation,
$$ \varepsilon = \sigma + {3 \over 7} ~\sigma^n $$
Here, ϵ = Ee/f1 and σ = fc /f1 where e is the strain, fc is the compressive stress, and f1 is the stress at which the slope of a line from the origin to a point on the stress-strain curve is 0.7. E and n are constants determined experimentally for a given material. The Ramberg-Osgood equation may be used to obtain the following expression for the tangent modulus of elasticity.
$$ E_t = { E \over 1 + 3/7 ~n \sigma^{n-1} } $$
This expression may in turn be substituted into the tangent modulus equation with fc equal to Fcol to obtain
$$ F_{col} = { \pi^2 E \over (L' / \rho)^2 } \left( { 1 \over 1 + 3/7 ~n \sigma^{n-1} } \right) $$
In the case of a column under concentric loading, Fcol is equal to P/A. A nondimensional plot of this equation is shown in Figure 2-48. Values of n, f1 and E that are needed for the Ramberg-Osgood equation are shown in Table 2-2.
Table 2-2: Properties of Various Materials for Ramberg-Osgood Equation
E, ksi
f1, ksi
24 S-T Sheet 10,700 10 41
24 S-T Extrusion 10,700 10 37
Clad 2024-T3 Longitudinal 10,000 10 38.1
Normalized 29,000 20 75
Ftu = 100,000 29,000 25 80
Ftu = 125,000 29,000 35 100
Titanium 6Al-4V Bar Stock,
Longitudinal Ftu = 145 ksi
at room temp, 1/2 hr
exposure to temperature
at room temperature 17,500 10 164
at 500°F 16,000 17 108
at 700°F 15,000 10 93.4
at 900°F 13,800 9 85.7
One great advantage of using the Ramberg-Osgood relation is that the necessary constants may be obtained for new materials without extensive testing.
In general, the tangent modulus theory will yield conservative results. However, this theory yields values for the critical load that are too high for very short columns for which Et may be less than 0.2 at failure.
2.3.1.11.2 Sample Problem - Use of Tangent Modulus Equation for Concentrically Loaded Short Columns
Given: The 1-in. square concentrically loaded column shown in Figure 2-49.
Find: The maximum value of P by using the tangent modulus equation, with Et obtained from the Ramberg-Osgood relation.
Solution: Ultimately, Equation (2-7) must be solved. This equation is shown graphically in Figure 2-48. From Table 2-2, find E = 10.7 x 106 psi, n = 10, and f1 = 37,000 psi for the given material. Consulting Section 2.3.1.4, find that C = 2.05 for the given end constraints. Solving Equation (2-2), find that L' = L/√C. In this case,
$$ L' = {15 \over \sqrt{2.05}} = 10.55 ~\text{in} $$
For a square, I = b4/12 and A = b2.
$$ \rho = \sqrt{I \over A} = \sqrt{b^2 \over 12} $$
In this case,
$$ \rho = \sqrt{1 \over 12} = 0.289 $$
$$ { L' / \rho \over \pi \sqrt{E / f_1} } = { 10.55/0.289 \over \pi \sqrt{ 10.7 \times 10^6 / 3.7 \times 10^4 } } $$
From Figure 2-48,
$$ { F_{col} \over f_1 } = 0.88 $$
$$ F_{col} = {P_{cr} \over A} = 0.88 ~f_1 $$ $$ P_{cr} = 0.88 ~A ~f_1 = 0.88 (1) (37,000) $$ $$ P_{cr} = 32,600 ~\text{lb} $$
2.3.1.11.3 Reduced Modulus Equation
An equation that is theoretically more correct than the tangent modulus equation is the reduced modulus equation. In this case, a reduced modulus of elasticity, Er, is used to replace E in the Euler equation. This reduced modulus is a value between E and Et, one suggested value being
$$ E_r = { 4 ~E ~E_t \over \left( \sqrt{E} + \sqrt{E_t} \right)^2 } $$
It can be seen from this equation that Er approaches E as Et approaches E so that the reduced modulus equation reduces to the Euler equation for long columns as does the tangent modulus equation.
The reduced modulus equation is accurate for specimens in which extreme care in manufacturing and testing is used but yields high values of critical load for other columns. The more conservative tangent modulus equation is preferred to the reduced modulus equation for this reason as well as for its greater simplicity.
2.3.1.11.4 Johnson-Euler Equation
For many materials a parabola may be used to fit column test data in the short column range. The equation of such a parabola may be written as
$$ F_{col} = F_{co} - K \left({ L' \over \rho }\right)^2 $$
where Fco and K are constants chosen to fit the parabola to test data for a particular material. The Johnson-Euler column curve consists of the Euler curve for high L'/ρ ratios combined with a parabola that is tangent to this curve and covers short column ranges. If Equation (2-9) is adjusted so that it is tangent to the Euler curve, we obtain the Johnson equation
$$ F_{col} = F_{co} \left[ 1 - { F_{co} \left({L' \over \rho}\right)^2 \over 4 \pi^2 E } \right] $$
Here, Fcol is the maximum column stress which is given as P/A for concentrically loaded long columns and by more complicated formulas for eccentrically loaded columns. The single experimentally determined constant, Fco, is called the column yield stress but has little physical significance for columns with stable cross sections since very short columns for which L'/ρ is less than approximately 12 fail by block compression. In cases where local crippling and primary bending failure interact, Fco may be taken to be equal to the section crippling allowable. This case will be discussed more extensively in Section 2.3.2. A typical Johnson-Euler curve is shown in Figure 2-50. The coordinates that are marked show the point of tangency between the Johnson parabola and the Euler curve. Thus, the critical effective slenderness ratio separating long columns from short columns is \( \sqrt{2} ~\pi \sqrt{E/F_{co}} \) for the Johnson-Euler curve.
The main advantages of the Johnson-Euler curve are its ability to fit data when there is an interaction between crippling and primary bending failure and simplicity of computation. For columns having stable cross sections, one of the other short column curves is normally perferred.
2.3.1.11.5 Straight Line Equation
Short column curves for most aluminum alloys and several other materials are best represented by straight lines. A typical straight line is shown in Figure 2-51.
As can be seen from Figure 2-51, the straight line is not necessarily tangent to the Euler curve as is the case for the Johnson parabola but this is often the case. The straight line shown may be given by the equation
$$ F_{col} = N - M \left({ L' \over \rho }\right) $$
where N and M are constants chosen to best fit column data for a given material. As in the case of the Johnson parabola, this straight line does not always hold for very short columns that fail by block compression so it is cut off at Fcy as shown in the figure.
The critical effective slenderness ratio may be found by equating Fcol as found by the Euler formula to that value as found by Equation (2-11). This procedure involves solving a cubic equation and will not be presented here since values of the critical slenderness ratio are tabulated in Table 2-3.
Values of the constants N, M and the corresponding critical slenderness ratio (L'/ρ)cr are available for a large number of aluminum alloys and manufacturing processes. These values are shown in Table 2-3.
2.3.1.11.6 Sample Problem - Use of Straight Line Equation for Concentrically Loaded Short Columns
Given: The concentrically loaded rectangular bar shown in Figure 2-52.
Find: The critical load, Pcr.
Solution: Since the column is made of aluminum alloy, a straight-line equation should be accurate if the column is short. From Section 2.3.1.4, C = 4 for columns having both ends fixed. Since C = (L/L')2, L' = L/2. The radius of gyration,
$$ \rho = \sqrt{ I \over A } = \sqrt{ { bh^3 \over 12 } / bh } $$
$$ \rho = \sqrt{ h^2 \over 12 } = \sqrt{ (0.5)^2 \over 12 } = 0.144 $$ $$ { L' \over \rho } = { 6 \over 0.144 } = 41.6 $$
From Table 2-3, we find that (L'/ρ)cr = 128 for this material. Since L'/ρ is less than this critical value, a straight line equation may be applied. Substituting the values of N and M, given for 6061-T4 extruded bars in Table 2-3, into Equation (2-11), we obtain
$$ F_{col} = \left[ 15.7 - 0.074 \left({ L' \over \rho }\right) \right] ~\text{ksi} $$
$$ { P_{cr} \over A } = F_{col} = \left[ 15.7 - 0.074 (41.6) \right] ~\text{ksi} $$
$$ \begin{eqnarray} P_{cr} &=& 12.55 ~A = 12.55(0.5) = 6.27 ~\text{kip} \nonumber \\ P_{cr} &=& 6,270 ~\text{lb} \end{eqnarray} $$
2.3.1.11.7 Critical Effective Slenderness Ratio
So far, long columns and concentrically loaded short columns have been discussed. It has been mentioned that a long column is one for which the Euler equation holds in the concentrically loaded case. This equation holds for columns having an effective slenderness ratio greater than a value called the critical effective slenderness ratio of the column. In some cases, this critical effective slenderness ratio need not be determined. For example, the tangent modulus equation, reduced modulus equation, or plotted data may be applied to either long or short columns. In other cases, this ratio must be known in order to decide which of two equations is to be applied to a column. A critical effective slenderness ratio must be found if the Johnson-Euler or a straight line equation is to be used.
The critical effective slenderness ratio is, in general, a function of the column material. If the Johnson-Euler formula is to be used, we obtain
$$ \left({ L' \over \rho }\right)_{cr} = \pi \sqrt{ 2 E \over F_{co} } $$
as our critical effective slenderness ratio. If a straight line formula is to be used, the critical slenderness ratio is the slenderness ratio for which the Euler curve and the straight line either intersect or are tangent. A general formula is not given here since the critical effective slenderness ratio for each straight-line equation whose parameters are given in Table 2-3 is also given there.
In general, a steel column having a critical effective slenderness ratio of greater than 120 or an aluminum column having one of greater than 220 may be immediately treated as a long column. Columns having lower effective slenderness ratios must be checked by Equation (2-12) or Table 2-3 or treated by a method applicable to both long and short columns. If Equation (2-12) is used, Fco may be assumed to be approximately equal to Fcy for a rough estimate of L'/ρ.
2.3.1.11.8 Bending Failure of Eccentrically Loaded Short Columns
The column formulas in the previous section do not apply if a column is eccentrically loaded or if initial imperfections are large enough to have the effect of an appreciable initial eccentricity. In such cases, either another formula must be used or adjustments must be made to existing equations.
The secant formula that was given in Section 2.3.1.7 may also be applied to short columns if its parameters are chosen correctly. This equation is given again below for convenient reference.
Fcol is the maximum fiber stress at failure as before. Different references suggest various ways of choosing Fcol. Reasonable results may be obtained for very short columns if Fcol is assumed to be equal to the compressive yield point for steel or the compressive yield stress for light alloys. In the case of intermediate length columns, Fcol should be taken to be the stress obtained from one of the formulas for a concentrically loaded short column. If the formula used for this purpose is the tangent modulus equation, E should be replaced by Et in the secant equation. The secant equation must either be solved by trial and error or through use of a chart such as that shown in Figure 2-53.
Unfortunately, the secant formula is inconvenient for computation. A simpler approach may be used if the column deflection is small compared to the eccentricity of loading. This assumption is true for very short columns and to a lesser extent for intermediate length columns.
Using the previously mentioned assumption, the basic design equation for an eccentrically loaded short column becomes
$$ { F_{col} \over FS } = { P \over A } + { P_{ec} \over I } $$
Here Fcol is the maximum column stress as computed from one of the equations for a short concentrically loaded column.
A more refined design equation based on the assumption of a small column deflection relative to the eccentricity of loading is
$$ 1 = { P \over A ~F_{acol} } + { P_{ec} \over I ~F_b } $$
where Facol is the working concentrically loaded column stress and Fb is the working compressive stress in bending.
In conclusion, the secant formula is theoretically more correct and yields better results for eccentrically loaded short columns. However, due to the difficulty of applying this formula, Equation (2-14) or (2-15) may at times be applied, especially for shorter columns.
2.3.1.11.9 Sample Problem - Eccentrically Loaded Short Column in Bending
Given: The eccentrically loaded round column shown in Figure 2-54.
Find: The allowable load, Pa, if a factor of safety of 1.3 is used. Use (a) the secant modulus equation, and (b) Equation (2-14).
Solution: (a) The secant modulus equation may be written as
Since the column is made of aluminum alloy, the straight-line column Equation (2-11) may be used to find Fcol if it is short. From Table 2-3, find that B, D, and (L'/ρ)cr for this material are 15,700 lb, 15 lb, and 128 respectively. Thus,
$$ F_{col} = 15,700 - 15 \left({ L' \over \rho }\right) $$
Since the column is pin ended, L' = L = 12 in. For a circular section, ρ = D/4 = 0.25. Inserting these values into the above equation gives Fcol = 14,980 psi. Since L'/ρ is equal to 48 and (L'/ρ)cr is equal to 128, the assumption of a short column is valid. All of the parameters for the secant formula are now known except Pa. Inserting these values in the secant modulus equation gives
$$ { 1.3 ~P_a \over {\pi(1)^2 \over 4} } = { 14,980 \over 1 + {0.1(0.5) \over (0.25)^2} \sec{ \left[ {12 \over 2 (0.25)} \sqrt{ 1.3 ~P_a \over {\pi (1)^2 \over 4} (10 \times 10^6)} \right] } } $$
Simplifying this gives
$$ P_a = { 9,090 \over 1 + 0.8 \sec{[ 0.00975 \sqrt{P_a} ]} } $$
Try Pa = 5,000 lb. Substituting this value, we find that it solves this equation. Thus, Pa = 5,000. If our original guess was incorrect, other values of Pa would have to be tried until a value was found that solved the equation.
(b) Inserting the equation for Fcol from part (a) into Equation (2-14) gives
$$ { 15,700 - 15 (L'/\rho) \over FS } = {P \over A} + {P_{ec} \over I} $$
$$ { 15,700 - 15 (12/0.25) \over 1.3 } = {P \over \pi/4} + {P (0.1)(0.5) \over \pi/64} $$
Solving this for P gives Pcr = 6,540 lb. Notice that although this procedure eliminates the trial and error methods used with the secant equation, it yields less conservative results.
2.3.1.12 Torsional Failure of Simple Columns
The previous sections discussed the failure of long and short columns by bending. It was assumed throughout this treatment that the sections of the column are translated but not rotated as it fails. However, primary failure may occur at loads lower than those predicted in the section on bending failure if the sections should rotate as well as translate as shown in Figure 2-55.
In general, columns whose cross sections have a low torsional rigidity, GJ, must be examined for torsional instability. Columns having closed sections or solid ones, such as round or square sections, have such a high torsional rigidity that there is little possibility of them failing by torsional instability. However, torsional instability must be considered in the case of columns composed of thin sections. Failure by twisting is unlikely for flanged columns whose cross sections are symmetrical about a point such as I, H, and Z sections. Twisting-type failures are most apt to occur in the case of torsionally weak sections that are unsymmetrical or have only one axis symmetry such as angles, tees, and thin-walled channels.
The basic equation for a column that fails by a combination of bending and twisting gives the critical stress as
$$ F_{col} = { GJ \over I_p } + { C_{BT} ~\pi^2 E \over I_p (L')^2 } $$
In this case, CBT is a sectional property defined below and the parameters are defined as usual.
The torsion bending constant is dependent upon the axis of rotation and defined as
$$ C_{BT} = \int_A w^2 dA - {1 \over A} \left[ \int_A w da \right]^2 $$
$$ w = \int_o^u r_t du $$
The parameters used above are shown on an arbitrary cross section in Figure 2-56. Values of the torsional bending constant are given in graphical form for various cross sections in Figures 2-57 through 2-60. The torsion constant, J, may be obtained in Figures 2-61 for bulb angles or from the following equation for formed sections:
$$ J = st^3 / 3 $$
Here, s is the developed length of the median line.
If a column is not attached to a sheet, the twisting failure is by rotation of sections about shear center of cross section. In this case, Ip and CBT are taken about this shear center. If the column is attached to a sheet, sections may be assumed to rotate about a point in the plane of the sheet. This procedure gives rough results but unfortunately littLe specific information is available on this subject. In general, columns are stronger when they are used as stiffeners than when they stand free. However, a column having an unsymmetrical cross section may be weaker when it is attached to a sheet.
2.3.1.13 Sample Problem- Torsional Failure of Simple Columns
Given: A column with cross section and material properties as shown in Figure 2-62.
Find: Fcol for failure by a combination of bending and twisting.
Solution: Since t is small, the shear center may be assumed to be at the corner of the angle.
The polar moment of inertia about the shear center is given by
$$ I_p = I_x + I_y $$
$$ I_p = \left\{ \left[{ 1.5(0.1)^3 \over 12 }\right] + \left[ { (0.1)(3)^3 \over 12 } + (0.1)(3)(1.5)^2 \right] \right\} = 0.9 $$
From Equation (2-19)
$$ J = { st^3 \over 3 } = { 4.5 (0.1)^3 \over 3 } = 1.5 \times 10^{-3} $$
$$ C_{bt} = { t^3 (h^3 + b^3) \over 36 } = { (0.1)^3 (27 + 3.48) \over 36 } = 8.46 \times 10^{-4} $$
Inserting these values into Equation (2-16) gives
$$ F_{col} = { (3.85 \times 10^6)(1.5 \times 10^{-3}) \over 0.9 } + { (8.46 \times 10^{-4}) \pi^2 (10 \times 10^6) \over (0.9)(10^4) } $$
Thus, Fcol = 6,419 psi.
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Tear gas. 10 reasons for Bastian Schweinsteiger's greatness
April 24, 2021 by raj
German midfielder Bastian Schweinsteiger retires. What is the phenomenon of the footballer who is considered the most iconic player in Germany in recent years?
The world champion playing for the American MLS Chicago Fire, on Twitter thanked the teams for which he played, as well as his family: "Dear fans, it's time to announce that at the end of this season I will end my active career as an athlete. Thanks to Bayern Munich, Manchester United, Chicago Fire and the German national team, as well as my wife Ana Ivanovic and other family members for their support. "
1. Shvaini – Football God
Schweinsteiger's unchanging nickname in the football world is Fussballgott, literally meaning Football God. For 15 years in the main team of "Bavaria" the second name of the German has taken root so much that even after moving to "Manchester United" he heard the cherished words from the English fans, although he did not appear on the field so often. Bayern fans had reasons to call the midfielder that way. In his debut for the Munich club in 2002, Basti entered the field against Stuttgart, replacing current Bayern coach Niko Novac. Since then, he has played 342 matches for Munich in the Bundesliga, scoring 45 goals and 70 assists.
2. Schweinsteiger has great friends
What other environment can a great player have? The best friend of a football player from early childhood was and remains the alpine skier Felix Neureuther. It was he who was invited by the German TV channel as a commentator on the last match of Schweinsteiger. Racer Nico Rosberg, basketball player Dirk Nowitzki and actor Til Schweiger are also sent to the collection of Basti's friends. And no, this is not fan fiction with the participation of all eminent Germans, but an ordinary feast in Bastian's house. Do not forget that Angela Merkel and Frank-Walter Steinmeier are considered fans of the Bavarian.
3. Bastian's wife – Ana Ivanovich
Almost all players marry girls whose main profession is the spouse of a football player. Schweinsteiger married Serbian tennis star Ana Ivanovic in 2016. The couple celebrated their wedding in Venice, invited a bunch of eminent guests, and then moved away from sports. Now they are much more common in a coffee shop in downtown Chicago, but not on the field or court. According to media reports, Ana Ivanovich became the reason for the unexpected separation of Schweini and the German model Sarah Brandner. The German clearly has excellent taste.
4. He loves other sports too.
If he had not become a symbol of German football, then with great probability he would have become famous thanks to another sport. Bastian's father owned a sporting goods store and was a big cross-country skiing fan. In early childhood, the future Bayern legend did not even think about football, but conquered the snow-capped mountains. Another German passion is basketball. He can be constantly seen at games both in Germany and in the USA. Sometimes it even seems that basketball has long attracted Schweinsteiger much more than football.
"I'm not the kind of coach who will fill in about philosophy." 10 facts about coach Steven Gerrard
5. Basti has a cupboard with titles
In the last match for Bayern, all the cups won by the German team with Schweinsteiger were brought from the club's museum to the Allianz Arena. Champions League Cup, eight silver plates, seven German Cups, two German Super Cups and one UEFA Super Cup. This does not include the coveted 2014 World Cup and the FA Cup. Bastian never boasted of his awards, although he won more of them than the average football club.
6. Schweinsteiger – a symbol of stability
For the past four years, Basti hasn't played as often as in his early youth, but one thing about him has always remained the same. When he is on the field, the team has character. This applied to both Bayern and the national team. When Schweini appeared half-lame in a match against even the strongest opponent, his partners got a second wind and fought to the last. He is the heart of Bayern Munich and the German national team, without which both the one and the other frankly began to fever due to the lack of an obvious leader on the field, able to fight back in any game.
7. Shweini is a good and interesting person
How many cases were there when an excellent player turned out to be rude and stupid outside the field, could not connect a couple of words, and there was no question of good deeds. Interviews with Bastian are valued by German journalists worth their weight in gold – this interlocutor can fascinatingly tell stories from his life and at the same time be thoughtful enough. Schweini, after joining Manchester United, sent autographed T-shirts to fans who bought a uniform with the player's old number. In addition, the most revealing case is an encouraging letter to Judge Babak Rafati, who was trying to commit suicide.
8. Bastian is always "in style"
Classical Germans are not distinguished by a special sense of taste: what are the "beautiful" images of Thomas Müller. Schweinsteiger is a completely different story. The German has never been famous for his outstanding appearance, but his style of dress and gentlemanly habits are something that cannot be taken away from him. Every now and then, photos of Schweini strolling around Chicago in a classic suit with Ana Ivanovich on the arm are popping up on the Internet. Definitely working with Hugo Boss under the contract with the national team went to Bastian's advantage.
10 quotes from Aurelio De Laurentiis – President of Napoli
9. The word of the family for Schweini is law
The Schweinsteiger family has never been behind the dark curtain. It's very difficult to do when your older brother also plays for Bayern and is like you like two peas. Tobias Schweinsteiger – Bastian's older brother – has always had influence over the youngest member of the family, but at the same time protected him from emerging critics. In 2015, it was rumored that Tobias was the initiator of his brother's move to Manchester. Later, Basti himself confirmed that Toby is a fan of Manchester United. After signing the contract, they both went to test the lawn at Old Trafford and take a couple of photos in front of the red stands. The big brother's dream has come true.
10. Schweinsteiger – the face of the Bundesliga
Even those who hate Bayern will admit that Schweini is the player worthy of representing the German championship in the eyes of the world. "I gave myself 100%, my heart and soul belong to Bavaria, – Basti is not even dissembling. Cheerful, moderately rude, ready to fight to the very end, getting along well with both veterans and young people – it's all about Schweini and the Bundesliga. Schweinsteiger ceased to be just a player, but became a household name for all German football and made all non-football representatives of Germany pay attention to the sport with the ball.

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Did Kirk's captaincy of the Enterprise have a canon starting date in TOS? Among the 79 episodes of the original series, is there ever mention of the specific date on which James T. Kirk was given command of the Enterprise?
I'm quite sure a date has been retcon'd through later works, novels, etc., but I am wondering if such date was ever mentioned, either on-screen in one or more of the original series episodes, or in background materials authored at the time (e.g. original series writer's "bible").
Motivating my question is the situation with the currently-running Star Trek Discovery, stated to be set ten years prior to the events of TOS (obviously excluding the original pilot, the events of which Spock declares to be 13 years prior). Piecing this timeline together, the Talos IV incident (The Cage, The Menagerie flashbacks) occurred about 3 years before the Discovery stories pick up.

 In the Discovery season finale, Discovery comes to the aid of Enterprise, still under the command of Pike, but for how much longer?

Depending on how earlier canon has established things, we might reasonably expect to encounter Kirk, either as a junior officer on the way up, or even assuming command of Enterprise as her captain during the Discovery show run... or not?

A: 2264, according to the official Star Trek web site and the Star Trek Encyclopedia
According to James T. Kirk's file on StarTrek.com (the official site of the franchise), he was promoted to captain and took command of the Enterprise in 2264. His first five-year mission ended in 2269. (Source)
The above is the final word on this, but it is worth noting that this is corroborated by the Star Trek Encyclopedia, compiled by Michael and Denise Okuda, long-time Star Trek production designers (especially during the TNG era):

*

*Google Books preview of ST Encyclopedia (linked to relevant section)

As Discovery is set in 2256, it would be another 8 years before Kirk takes command of the Enterprise.

A: Pike will have command for, based on assumptions regarding when Kirk took over, another few years. Kirk was born in 2233, so he'd only be 25 or so at this point in "Discovery". Although never stated explicitly, he's assumed to have assumed command of the Enterprise in his early 30s, so sometime after 2263.

A: Short Answer, no definite statements.  Long answer, a few possible hints or clues
In "Shore Leave", sometime after stardate 3025.8, Kirk meets Finnegan from when Kirk was at the Academy:

FINNEGAN: I never answer questions from plebes, Jimmy boy.
KIRK: I'm not a plebe. This is today, fifteen years later. What are you doing here?

So apparently Kirk was a plebe, in his first year at the Academy, about 15.0 to 16.0 years before "Shore Leave" and should have graduated and been commissioned an ensign about 12.0 to 13.0 years before "Shore Leave".
In "A private Little War", around stardate 4211.4, Kirk says he made a planetary survey as a lieutenant thirteen years earlier.
In "Obsession", stardate 3619.2, it is said that Kirk was a lieutenant eleven years earlier.
Presumably it took Kirk a number of years to rise from lieutenant to commander of the starship Enterprise.
Thus there is an earliest possible date indication relative to the second season.
There are also indications of the latest possible date Kirk could have become commander of the Enterprise.
TOS episodes may have happened in:
1) production order.
2) stardate order.
3) broadcast order.
Or possibly in some other order or no order at all.
The first episode to mention the five year mission in the introduction was:
1) in production order "The Corbomite Maneuver" beginning on stardate 1512.2 and in 2266 according to the official but not necessarily correct chronology.
2) in stardate order "Mudd's Women" beginning on stardate 1329.8 and in 2266 according to the official but not necessarily correct chronology.
3) in broadcast order "The Man Trap" beginning on stardate 1513.1 and in 2266 according to the official but not necessarily correct chronology.
The first episode where Kirk was in command of the Enterprise was the second pilot "Where No Man Has Gone Before" - which does not have an introduction mentioning the five year mission - beginning on stardate 1312.4 and in 2265 according to the official but not necessarily correct chronology.
In "Where No Man Has Gone Before" Mitchell makes a tombstone for Kirk inscribed
"James R. KIrk  C 1277.1 to 1313.7".  It has been speculated that 1277.1 was the stardate of Kirk's birth, thus making about one year per stardate unit in the stardate system used at the time. So Kirk would have commanded the Enterprise for thousands of years during TOS if that stardate system continued for the rest of the series.
It has also been speculated that 1277.1 was the stardate when Kirk took command of the Enterprise probably a relatively short time before "Where No Man Has Gone Before".
Someone who accepts the theory that 1277.1 was the stardate when Kirk took command of the Enterprise and also has a theory about the relationship between Earth time and stardates can calculate when they think that Kirk took command relative to other episodes.
If the TOS episodes happen in stardate order and the "five year mission" lasted between 5.0 and 6.0 years from "Mudd's Women" stardate 1329.8 to stardate 5943.9, the second stardate in "All Our Yesterdays", then there would be 769.016 to 922.82 stardates per year and 2.105 to 2.526 stardates per day.
If the TOS episodes happen in stardate order and the "five year mission" lasted between 4.5 and 5.5 years from "Mudd's Women" stardate 1329.8 to stardate 5943.9, the second stardate in "All Our Yesterdays", then there would be 838.927 to 1025.355 stardates per year and 2.296 to 2.807 stardates per day.
Using such stardate ranges it would be easy to calculate how long stardate 1277.1 was before any particular episode with a stardate.
What if someone considers TAS to be part of the five year mission? Then known stardates of the five year mission would range from 1254.4 (before 1277.1) in "The Magicks of Megas-Tu" to 7403.6 in 'Bem", 6149.2 stardates.  In 4.5 to 6.0 years there would be 1024.86 to 1366.488 stardates per year and 2.805 to 3.741 stardates per day.
Of course if the five year mission began before the first episode's stardate and and ended after the last episode's stardate there might be many more stardates per year and per day.
In "Menagerie part 1", stardate 3012.4 and 3012.6, and "Menagerie Part 2", stardate 3013.1 and 3013.2 Spock says several times that the events on Talos IV were thirteen years earlier - and thus between 13.0 and 14.0 years.  Spock also says that he served with Captain Pike for:

SPOCK: Eleven years, four months, five days.

Right after Kirk says:

KIRK: I took over the Enterprise from him. Spock served with him for several years.

If "Eleven years, four months, five days" equal approximately 11.3415 Earth years, and if "thirteen years" equal about 13.0 to 14.0 Earth years, then Kirk must have commanded the Enterprise for at least 1.6585 to 2.6585 years before "Menagerie".  If Spock began serving with Pike during the mission to Rigel VIII and Talos IV.  If Spock began serving with Pike before the events of "The Cage" then kirk would have commanded the Enterprise for a longer period before "Menagerie".
In "Who Mourns for Adonais?" stardate 3468.1,  Kirk asks Chekov's age:

KIRK: How old are you?
CHEKOV: Twenty two, sir.

The Making of Star Trek  Stephen E. Whitfield, 1968, Part II, Chapter eight, says: "All Ensigns seem universally twenty two years old,...", thus implying that Checkov is still 22 and The Making of Star Trek is less than one year after "Who Mourns for Adonais?"
In "Journey to Babel" between stardates 3842.4 and 3843.4, Sarek says his age is:

SAREK: One hundred two point four three seven precisely, Doctor, measured in your years. I had other concerns.

The Making of Star Trek  Stephen E. Whitfield, 1968, Part II, Chapter five, says:  "Sarek is 102 years old, or about middle-age in terms of Vulcan years (the Vulcan life span is about 250 years)."
If the same type of years is used both times, the chronological data in The Making of Star Trek should be valid less than 0.563 years after the scene in "Journey to Babel".
The Making of Star Trek  Stephen E. Whitfield, 1968, Part II, Chapter five, also says: "Mr. Spock has served aboard the U.S.S. Enterprise for 13 years, the first nine of them under Captain Christopher Pike, the last four of them under Captain James Kirk."  Thus Spock should have begun serving under Kirk on the Enterprise when Sarek was about 98.437 to 99.
The Making of Star Trek  Stephen E. Whitfield, 1968, Part II, Chapter four, also says: "Kirk has been in command of the Enterprise for more than four years, and was the youngest Academy graduate ever to have been assigned as a Starship Command Captain."
If Kirk became commander of the Enterprise about 4.0 to 5.0 years earlier, that would have been when Sarek was aged 97.437 to 99.
If Spock began serving with Pike on the Enterprise 13.0 to 14.0 years before The Making of Star Trek, and if the voyage to Talos IV in the "The Cage" was 13.0 to 14.0 years before "Menagerie", then "Menagerie" should be minus 1.0 to positive 1.0 years before The Making of Star Trek, and that seems too short a time span.
But if Spock served with Pike for 11.3415 years, and on the Enterprise with Pike for nine years, that leaves a period of about 2.3415 years when Spock would have served with Pike while not on the Enterprise.  And if some or all of those 2.3415 years was spent sometime after the voyage to Talos IV in "The Cage", while the Enterprise was being refitted and repaired perhaps, then "Menagerie" might be between 2.3415 and 3.3415 years before The Making of Star Trek.
So people who consider The Making of Star Trek canon or semi canon have another clue to the possible date when Kirk took command of the Enterprise.
Short Answer, no definite statements.  Long answer, a few possible hints or clues

A: According to Memory Alpha, The Original Series episode "Where No Man Has Gone Before" takes place in 2265 and The Animated Series episode "The Counter-Clock Incident" takes place in 2270, which is exactly 5 years. This makes sense if you where to watch both The Original Series and The Animated Series in production order and ignore all stardates and air dates.
Then Star Trek: Discovery began advertising that their show will take place 10 years before Kirk, Spock and the Enterprise, so I figured it would be set in 2255, which I was wrong. I bet the discovery writers did not take the second pilot into account because it was the 3rd episode aired, instead, "The Man Trap" was aired, which took place in 2266. Gene Roddenberry never wanted to give up specific date information as when the show took place, that is why we have the stardate system.
I bet you the only reason whey the series takes place between 2266 and 2269 is because the series ran from 1966 to 1969. They just add the animated series to 2270. You might have noticed that Star Trek IV's Earth date is 2386, which was released in 1986 and of course they go back in time to 1986 (and I always thought it was funny that they used Christopher Lloyd's ship as the time machine). There are some significant crew and uniform changes between "Where No Man Has Gone Before" and "The Corbomite Maneuver" witch would explain why the two episodes take place one year apart.
Also according to Memory Alpha, Captain's Pike last five-year mission ends in 2266, so spoiler alert: at the end of season 1 of Discovery, we might be seeing Captain Pike and his crew on their last mission on the Enterprise. According to "The Menagerie", "The Cage" takes place 13 years earler, which would put that episode in 2253, which would put that episode in 2253.
According to Star Trek: Online, in the mission "City on the Edge of Never" (spoiler alert), you are sent back in time within the game to the year 2270, when Kirk and the Enterprise are still in service. If you have not played the game, I recommend watching the breakthroughs because it explains how Klingons get their old cranial ridges back.

redpajama_set_name: RedPajamaStackExchange


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If not cleaned correctly, the smoke damage will leave residue throughout your home. An estimator from Highland Construction and Restoration will come to your home to get an accurate understanding of the project. The estimator will take pictures to document the damage and create an inventory list of all property, including damaged and undamaged. The estimator from Highland will communicate with your insurance adjuster to streamline the process.
As soon as the homeowner and insurance company approve the scope of work, certified and experienced technicians begin the restoration process. Salvageable items are taken to the Highland Construction and Restoration warehouse. Once there, the items are then cleaned, inventoried, and stored until the project is complete. The soot and smoke residue is then removed from the air and surfaces with specialized equipment. The structural damage in the home is repaired.
After an incident involving fire in your home, dealing with smoke and soot damage is stressful. Highland Construction and Restoration is a family owned company built on compassion and understanding. They are exactly the professionals to trust with restoring your home and getting your life back to normal as soon as possible.


text:
class Article < ActiveRecord::Base

  include RademadeAdmin::Hideable

  belongs_to :active_gallery
  belongs_to :author

  translates :name

  def to_s
    name
  end

end

redpajama_set_name: RedPajamaGithub


text:
Q: Send button from c# to client side I have a helper created as follows:-
public static MvcHtmlString FileDomElement(this HtmlHelper helper, FileUpload fileUpload)
{
    string strOutput = string.Empty;

    if (fileUpload.MultipleFile)
    {
        if (string.IsNullOrEmpty(fileUpload.MimeType))
        {
            strOutput = "<input type=\"file\" multiple=\"multiple\" id=\"chooseFiles\" value=\"Choose File(s)\" onchange=\"ehDisplayFileNames();\" style=\"opacity: 0\" >";
        }
        else
        {
            strOutput = "<input type=\"file\" multiple=\"multiple\" id=\"chooseFiles\" value=\"Choose File(s)\" accept=\"" + fileUpload.MimeType + "\" onchange=\"ehDisplayFileNames();\" style=\"opacity: 0;\">";
        }
    }
    else
    {
        if (string.IsNullOrEmpty(fileUpload.MimeType))
        {
            strOutput = "<input type=\"file\" id=\"chooseFiles\" value=\"Choose File(s)\" onchange=\"ehDisplayFileNames();\" style=\"opacity: 0;\">";
        }
        else
        {
            strOutput = "<input type=\"file\" id=\"chooseFiles\" value=\"Choose File(s)\" accept=\"" + fileUpload.MimeType + "\" onchange=\"ehDisplayFileNames();\" style=\"opacity: 0;\">";
        }
    }

    return MvcHtmlString.Create(strOutput);
}

Now in the onchange event of the file I want to send the button to the backend. Because I have multiple file input elements on a single page. So I want to detect which one was clicked.
Please help me with this.
Thanks
Abhishek

A: You could try catching the event info and using that to get a reference to the sender:
function ehDisplayFileNames(event) {
    console.log(event.target.id + ' has fired the event!');
}

Your markup would need to send the event:
    onchange="ehDisplayFileNames(event);"
Mark's comment above is very true, once you have a client side id you will still need some type of Ajax call to get this info back to the server.


libsrs2 is the next generation SRS library from the original designer of SRS. It implements the Sender Rewriting Scheme, a part of the SPF/SRS protocol pair. Libsrs2 has been written from an entirely clean codebase with compliance, speed and versatility in mind. It is platform independent and has no external dependencies. It is thread-safe and heap-safe, and is suitable for large scale applications and embedded systems and can operate without many standard system facilities.
SPF is one of a family of systems which aim to prevent email address forgery. A good SPF implementation is available from http://www.libspf2.org/.
[From spf.pobox.com]: The present SMTP standard for email allows anyone to forge anyone else's email address. This means I could send anyone a message claiming to be from you, and only an email expert would be able to tell the difference. SPF allows your mail servers to easily distinguish forgeries from real mail. Importantly, SPF works before the message body is transmitted, saving you the bandwidth cost of downloading the message and the CPU cost of filtering it.
SPF (and related systems) present a challenge to forwarders, since the envelope sender address might be seen by the destination as a forgery by the forwarding host. Forwarding services must rewrite the envelope sender address, while encapsulating the original sender and preventing relay attacks by spammers. The Sender Rewriting Scheme, or SRS, provides a standard for this rewriting which makes forwarding compatible with these address verification schemes, preserves bounce functionality and is not vulnerable to attacks by spammers. A graphical explanation is available here or here and a technical document describing the protocol and its strengths is at here.
It works with all address verification schemes..
SRS is an exceptionally clean and simple design which brings compatibility not just with SPF, but with many other ad-hoc address verification schemes, some of which are used by major ISPs.
It is well defined and secure.
The paper on SRS describes in detail how the protocol works and why it is invulnerable to attack.
SRS does not require changes to protocol or semantics.
SRS exists as a rewriting system entirely within current SMTP practices. Many alternative systems require changes to either SMTP syntax, semantics or both. SRS requires neither.
While anyone can implement SRS, only forwarders need to implement SRS. No modifications are required by any other party to MSAs, MTAs, MUAs or other software.
Write or maintain patches against MTAs?
Take responsibility for a distribution?
Write bindings for a language?
Web design "Sinorca" from OSWD.


Rock Hill, SC – Kenneth Roy Ferrell, 71, passed away on Thursday, March 28, 2019 at Atrium Health, Pineville, NC.
Funeral service for Mr. Ferrell will be held at 3:30 p.m. Sunday, March 31, 2019 at Fort Mill Church of God, 225 Academy Street, Fort Mill, SC 29715. Visitation will be at the church from 2:00 to 3:30 pm prior to the service. Burial will be private.
Mr. Ferrell was born December 23, 1947 in York County to the late Charlie Edward Ferrell, Sr. and the late Margaret Bechtler Ferrell. He was retired from Duke Power as a purchaser. He was a member of Fort Mill Church of God where he served as a layman for the Prison Ministry.
Mr. Ferrell is survived by his wife, Jean Cairco Ferrell; his son, Kenneth Ryan Ferrell and his wife, Teresa Ferrell of Six Mile, SC; his daughter, Heather Ferrell Stroud of Rock Hill; his grandchildren, Brandon Ferrell, Isaac Ferrell, Chase Stroud, Lincoln Stroud, Levin Stroud; and his brother, Charlie E. Ferrell, Jr. and his wife, Pat of Rock Hill.
In lieu of flowers, the family would like memorials to be made in Mr. Ferrell's name to Fort Mill Church of God, Freedom Ministry, 225 Academy Street, Fort Mill, SC 29715.


Village of Hope Inc. is a non-profit, 501 c (3) organization providing academic support services for girls in grades 6-12. It is our mission to help girls achieve academic success, create positive self images and build beneficial relationships among their peers. We encourage girls to pursue higher education as an option. Through encouragement and positive role-modeling these young women will be able to self-advocate, and effectively transition into adulthood. Girls gain opportunities to experience a stable foundation on which to build a wholesome and healthy life.
Founded in 2009, Village of Hope was created from a sincere desire to help increase the number of girls who go on to pursue post-secondary education, reduce teen pregnancy rates and positively deal with the many impacts poverty has on girls. Growing up in poverty founder, Wansheba Lawal knows what strain poverty can have on girls. She also understands how important education is in escaping poverty. She believes that having access to higher education can make all the difference in what girls will be able to do with their lives.


Part of the KYC Middle Office team, in coordination with the Front Office, Financial Security and the KYC controllers, the Analyst is in charge of analyzing and updating the KYC files.To seek public information (on internet, subscription databases etc.) to complete or update the KYC files.
In coordination with the Front Office, to complete or update any required non-public information.
To complete the search of all names for adverse information, PEP and OFAC/EU lists.
To ensure the global quality of the KYC files according to the compliance rules.
To carefully and pro-actively follow-up his/her own files.
Comply with all applicable legal, regulatory and internal Compliance requirements, including, but not limited to, the Singapore Compliance manual and Compliance policies and procedures as issued from time to time; Financial Security requirements, including, but not limited to, the prevention of Financial Crime and Fraud including reporting obligations to the Money Laundering Reporting Officer.
Maintain appropriate knowledge to ensure to be fully qualified to undertake the role.
- Good knowledge in corporation law and/or financial analysis.
- Self-motivated and excellent team player.
•Knowledge of finance and banking system.
•Experienced in research, knowledge of banking regulations.
•Capable to effectively interact with other teams.


text:
RAGOZIN, Zenaide Alekseevna
25. April 2017 7. August 2018
RAGOZIN, Zenaide Alekseevna. 1835 — 1924. Russian Traveller and Authoress of Popular Histories. Travelled much in Europe, in 1874 emigrated to the U.S.A. in (naturalized 1894).
Publications: The story of Vedic India as embodied principally in the Rig-Veda. 457 p. N.Y. 1895, 5th ed. 1923.
– Also in the series Story of Nations: The story of Chaldea. From the Earliest Times to the Rise of Assyria. 363 p. N.Y. 1887, 4th ed. 1893; The story of Assyria. From the Rise of the Empire to the Fall of Ninive. 1891; The story of Media, Babylon and Persia. 1888
– History of the World. N.Y. 1899; other books and translations.
Sources: briefly in Buckland, Dictionary; briefly Wikipedia.
RAFFLES, Thomas Stamford
RAHDER, Johannes
Buddhism University of Munich Indian studies University of Paris Western explorers and traders islam Historical studies British colonial time Language studies University of Bonn University of Leipzig University of Vienna Jainism Colonial India University of Berlin Tibet


Lung cancer kills more Americans than any other cancer: it is a statistic we are at the forefront of changing.
We also offer the coordination and support of a thoracic nurse navigator, ensuring seamless, convenient care—no matter what level of care is needed.
A pulmonary nodule is any "spot" in the lung that is less than 3 centimeters in diameter; most are discovered by accident during imaging for another problem. Our clinic is designed to thoroughly evaluate each nodule and determine what, if any, follow-up is necessary.
Treatment often involves "watchful waiting" with periodic imaging to assess for changes in shape or size. While the vast majority of nodules are benign, caused by old infections or scar tissue, some are early signs of lung cancer, requiring prompt testing and treatment. We are leaders in using evidence-based protocols to minimize unneeded testing or invasive procedures, offering patients exactly the right care at the right time.
For nodules that appear suspicious, we are one of the few centers in California to offer electromagnetic navigational bronchoscopy, a minimally-invasive diagnostic technology that uses GPS-like technology to locate and biopsy a pulmonary nodule without performing surgery or a needle biopsy.
If you have been diagnosed with a lung nodule, please contact Shannon Lindop, RN, OCN, Thoracic Nurse Navigator and Program Coordinator, at (714) 446-5355. Your diagnostic images will be evaluated by our panel of experts and, if needed, an individualized plan of care developed—including appropriate monitoring, additional testing, or treatment. A physician referral is required; if you don't have a referring physician, please contact the nurse navigator to make an appointment with one of our pulmonologists.
Current or former smokers, ages 55-77, who have smoked the equivalent of one pack of cigarettes a day for 30 years—or those who have smoked a pack a day for 20 years and have an additional risk factor— face a significantly higher risk of developing cancer or other lung diseases.
Low-dose CT screening is considered the gold standard in detecting lung cancer early. In addition to the latest generation CT imaging, we also offer the experience and expertise to accurately interpret and evaluate the images—ensuring the correct level of care.
If you are considered a heavy smoker with an additional risk factor (a family history of lung cancer, exposure to secondhand smoke, or exposure to asbestos, radon or arsenic), please contact (714) 446-5830 to schedule a lung screening. Most insurance plans including Medicare and Medi-Cal consider lung screening a covered benefit, as do all Covered California insurance plans.


United States of America has so much to offer for truly unique holidays such as The Rocky Mountains of Colorado, shopping and tourist attractions of New York, the glitz and entertainment of Hollywood and the unforgettable playground of Walt Disney World in Florida. Why not visit the city that never sleeps, Las Vegas with entertainment round the clock, it really is an incredible place.
The USA is an enormous country with a staggeringly diverse climate across its 52 states. Whether it's basking in the sun, surfing the waves or skiing in the mountains, the United States has the weather, the right timing is all you need.
America has so much to offer with a state to cover all your holiday needs, whether you're looking for a sunshine holiday in winter, an activity packed experience, to splash some cash on a shopping break or try your luck in a casino, a fun-filled holiday for all the family or a romantic break for two, you will be sure to find something that suits you in the United States of America.
Probably the first foods that spring to mind when thinking of American cuisine are the hamburger and hot dog, though the concepts were created here, the country also offers regional variations and local specialities such as Southern fried chicken and New England clam chowder, Boston baked beans and Texan chilli.
One thing to remember when dining out in the USA, portions are large and that includes the sides. Also tipping is common practise with 15-20% of your bill the expected amount.
Flight Time - Flight times vary from your UK departure point to the State you are going to, and some destinations may have stop overs.
Best Time to Visit: The USA is an all year round destination.
Here's a list of our top accommodations in the USA, ranked by our customer ratings and reviews.


Member for Lingiari, Warren Snowdon, said families, low incomes workers and seniors in rural and remote parts of the Northern Territory received their third round of tax cuts in three years in the Rudd Government's third budget.
"This round of tax cuts will put money in the pay packets of working families in the electorate of Lingiari every week, helping our families to balance the family budget," Mr Snowdon said.
"The Rudd government has delivered these tax cuts despite the pressures placed on the budget by the global crisis," he said.
In addition, from 1 July 2010 low incomes workers in Lingiari will be able to earn up to $16,000 and not have to pay income tax (up from $11,000 in 2007-08).
"This is a real benefit for low income and part time workers, in particular mums working part time in Lingiari, who will now be able to earn more income without having to pay income tax," Mr Snowdon said today.
This year's budget will also increase the amount of income a senior Australian eligible for the Senior Australian Tax Offset (SATO) can earn before they pay income tax or the Medicare Levy.
"These budget changes will increase the income threshold that eligible senior Australians can earn tax free from $29,867 to $30,685 for singles, and from $25,680 to $26,680 for each member of a couple," he said.
In previous budgets the Government has also delivered an historic increase in the base rate of the pension despite the budget impacts of the global recession, as well as stimulus payments to help older Australians with cost of living pressures.
The following table shows the weekly increases in disposable income from income tax cuts since 2008-09. The benefits are compared to 2007-08 and are cumulative from year-to-year.


text:
Natural Flood Defences and Biodiversity
by Nick Mann March 2, 2016
I pottered up to parliament yesterday to listen to the All Party Parliamentary Group meeting on natural flood defences, and very interesting it was too. I was there to listen to, among others, Jeremy Biggs of the Freshwater Habitats Trust, an excellent charity which we support. I'm the bald suit in the photo. I've been to several parliamentary group and committee hearings and don't understand why more folk don't bother to (including, in this instance, any MPs from Somerset!). It's a great opportunity to hear experts and meet interesting people, not least politicians, who, unfashionably, I generally find impressive. It also gives you a sense of the difficulty of establishing and implementing joined up policy in the environment.* Natural flood defences, it turns out, are unsurprisingly complicated. It's not just a question of paying hill farmers to plant trees in catchment areas. There were three themes running through yesterday's presentations and debate which were familiar from other environmental issues. Most obviously, there is no money available to action any new initiative, which will somehow have to be paid by the private sector. Secondly, there is limited available data to promote new schemes or help design them. Thirdly, policy has to be not just well informed but co-ordinated. I'm not convinced these things won't happen in this instance, and the gains for biodiversity and the voters could be immense. *This is one of the reasons I don't subscribe to organizations like Avaaz and Change.org, which reduce issues to 140 characters and spew out tens of thousands of emails into MPs' inboxes. Ironically, they increase a sense of alienation from the process of government which Americans are currently paying a heavy price for.
Subscribe for blog updates from Habitat Aid


Originally recorded in many spellings including Mac Scanlon, O' Scanlon, O' Scannill, O' Scanlan, but now more usually as Scanlon, and more rarely Scannell, Scanlan, Scanlin, and possibly others, this is an ancient Irish surname. It derives from the Gaelic pre 10th century personal 'Scanlain' which translates literally as 'The contentious one'. There were originally three quite distinct septs or clans, the first having the prefix 'Mac' and they held territory in County Louth and gave their name to Ballymascanlon near Dundalk.
Leave a message for SCANLON family members.
I believe I am a descendant of Mac Scanlon from the territory in County Louth. My Dad recently deceased had info tracing his family back 1849 born in Ireland coming to US in late 1860's to 1870. He also had a pamphlet about Cill C Lara which was beleved to be the tomb of Mac Scanlon chieftain of the 10th century. Any info would be great.
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text:
An Ode to Millie & Steve
Amusement Parks, Fairs, & Carnivals, Obsessions
Hello and welcome to my love story about Millie and Steve, two rollercoasters that the general public may know as Millennium Force and Steel Vengeance. I will try to keep this PG, only because I couldn't get Fabio to pose for the cover.
I've always been super into amusement parks (and county fairs until I almost died at one) but even though I like roller coasters, I never really considered myself an ENTHUSIAST. Then I rode T-Express at Korea's Everland and, for a wooden coaster, that thing impressed me more than any crazy-ass steel multi-inversion Jojo-rolled contraption ever has. It made me want to seek out other crazy wooden coasters, because up until then I equated wooden coasters with rickety old back-breakers. I started binge-watching rollercoaster videos on YouTube, becoming more and more obsessed. That's how I started learning about the different manufacturers, but the one that stuck out the most to me was Rocky Mountain Construction (RMC). I was fascinated by the way they take old, rough woodies and refurbish them into these head-spinning feats of engineering magic. The first one I got to ride was Lightning Rod at Dollywood and it was a game changer for me, I have been on a mission ever since to ride all of the RMCs. I am an RMC fan girl all the way.
Look. When I find something that I like, I don't just LIKE IT: I LIVE IT, I BREATHE IT, I DREAM IT. And, I watch YouTube videos until my Roku crashes, I read Wikipedia and personal blogs, I search Instagram hashtags, I adjust our budget so we can do weekend amusement park road trips (STOP GOING TO THE GROCERY STORE, HENRY – WE DON'T NEED FOOD WE NEED COASTER CREDITS. THIS IS THE GREATEST DIET EVER). It becomes my driving force, the thing that gets me through the work week and gives me something to anticipate.
I never had much of a burning desire to go to Cedar Point until last year, when RMC refurbished the old Mean Streak into Steel Vengeance. So when we finally went a few weeks ago, I was practically salivating on myself at the thought of riding this bad boy.
Because this coaster is still new-ish and world class to those who know some things about the coaster scene, the line for this was between 90-250 minutes all weekend. We kept putting it off and putting it off, but finally, around 8:00pm on our first day there, I told Chooch, "Look, if we want a night ride on this bad boy, it's now or never, bud."
I think it said it was an 80 minute wait when we got in line, because by that point, all the haunt attractions had opened so most of the people in the park were in line for those things. Well, 80 minutes was a lie. They must have changed the sign to 120 minutes as soon as we walked past, because we stood in that queue—which winds around underneath the tracks so you're like, majorly cut off from the rest of civilization when you're in that line—for so long that I started to forget what Henry looked like (he opted out) which is either good or bad depending on what kind of day I'm having when you ask me.
Luckily, they have TVs in the line, so we got to watch clips of horror movies, fight with each other over trivia, and watch random music videos while eavesdropping on people around us playing Heads Up (this one girl was SO LOUD and also extremely stupid—some of the things she couldn't figure out were maddening to us bystanders). There was a mom in front of us with her elementary school-age son and an older boy who I think was in college and also may be have been a Spanish exchange student? Look, we had a lot of time to spy on people.
So yeah, after standing in line for…I lost track but I want to say it was about 90 minutes, then it happened.
I will never forget it because we had made it to the last part of the serpentine path, all the twisty-windy parts of the queue were behind us, and we were finally on the lone path to the station. (Granted, that lone path still had some turns, a metal detector, and steps, but still!) We were standing right next to the giant billboard that had the Steel Vengeance character on it along with all the record-breaking stats.
This is where we were standing when the dreaded THIS RIDE IS CLOSED FOR MAINTENANCE announcement came out of some hidden hell-speakers and we could barely hear it over the loud roar of idle conversation around us, but it didn't matter because we KNEW.
"Huh, that train didn't have anyone in it," the guy behind us noted as an empty train soared past us. We were in the part of the line now that had an excellent view of the first drop.
Yeah no shit asshole, if you would stop talking about dumb video games for a second, you'd have heard the announcement! As people finally started to figure out what was going on, a small exodus happened and because of this, we kept moving further up in line.
Chooch and I kept waiting for the other to cry uncle and suggest to leave the line, because neither of us wanted to make that call. All I kept thinking was that it would be our luck that it would start running again after we got out of line.
Long story short (LOL yeah like my stories are ever short), we opted to stay in line and after about 30-45 minutes, the damn thing started running again and everyone cheered and fuck if it didn't feel like we were REALLY A PART OF SOMETHING, you know? Like, the Donner Party.
No, not the Donner Party.
Maybe a hostage situation, though. But then the hostage guy ends up having a squirt gun so we can all laugh about it later as the popo haul his soiled ass away.
Anyway, we rationalized that we had moved up in line just as far as we would have if the ride hadn't broken down, so it was all the same, really.
The ride attendant at the top of the steps was assigning people seats but we thought, Look, we stood in line for this long, what would it hurt to just ask if we could snag the back row?
So we asked.
And the broad was all, "Eh, sure go ahead."
And we had the most epic, glorious, whirlwind night ride on what is now my TOP ROLLERCOASTER BAE OF ALL TIME. I'm not even going to try and describe it other than it whips you around with ungodly force and everything happens so fast that you can't even wrap your mind around the logistics of it and then when you think you've gotten your bearings, you're suddenly being lurched through an inversion that makes you feel like dish water being sucked down a drain, and then suddenly you're back in the station, fingerbrushing knots out of your hair and looking at your riding companion like, "IS MY FACE IN ONE PIECE!?"
Chooch actually ran his hands through his hair and made this wild-eyed I'VE HAD AN EPIPHANY expression like he was about to convert to the Kabbalah or some shit, and honestly, I personally nearly wept.
It was that good.
I felt like Steel Vengeance had actually rescued me from a burning building or something and then, oh god, oh no, was I getting a crush on Steve?!
Meanwhile, Henry was sleeping on a bench like a regular old back-alley wino, just kidding, he was actually awake and not at all concerned even though we had been missing in action for two and a half hours and he had our phones so we had no way of telling him what was happening, but since this was Cedar Point, he assumed that the ride had broken down because that's what rides do at Cedar Point. So I guess he probably just ate a bunch of soft pretzels and scrolled through Reddit on his phone, because somewhere along the way I didn't pay enough attention to him and he turned into the type of person who loses himself in asinine threads of Internet memes. Coo-coo-cool.
We were just like, "O-M-G YOU HAVE NO IDEA HOW FUCKING SICKENING THIS RIDE IS, HENRY! YOU MISSED OUT BIG TIME! GOOD LUCK SLEEPING IN THAT BED OF REGRET TODAY!"
The next day, Henry wanted to see if Steve was all that we made it out to be, but as you might be aware, our plans of getting in a morning ride on this bad boy before the crowds rushed in was dashed when Steve was closed during early entry.
And then he proceeded to be down for most of the day it seemed. We kept tracking him through the Cedar Point app and sprinted over to him as soon as the status changed to "open." It said that the standby time was 45 minutes.
LIES. LIES LIES LIES.
We 100% stood in line for nearly as long as Chooch and I did the night before. This time, right as we shuffled past the Steve billboard thing, an announcement came on.
WE FROZE.
But it turned out that it was just a "slight delay" while they added another train.
We exhaled.
Not more than 10 minutes later, another announcement cut through the gaggles of groups engrossed in Heads Up and the weird mom and son who were arguing with each other the entire time they were in line.
IT WAS FUCKING DOWN AGAIN.
"You have to be fucking kidding me!" I cried dramatically. I could actually feel the synapses firing inside me and I imagined peeling my skin off and shooting into the air using nothing but the sheer force of my anger.
"You guys can leave, you've already ridden it," Henry said calmly, but I noted a twitch in his 'stache. "I'm invested at this point."
Well, I wasn't leaving! I had major FOMO just thinking about Henry riding Steve without me. So we all opted to stay in line. People started exiting in small waves. A ride attendant walked past us, en route to the entrance where a CLOSED sign needed to be erected (lol). "Just so you know, this isn't just a small problem," he monotoned to everyone within earshot. "It's probably going to be at least an hour."
We all exchanged looks. Even more people left, so we moved up significantly in line and shrugged.
Henry and Chooch argued about every single thing.
Arguing.
But then, less than 30 minutes later, they sent a test train. Everyone cheered. Then, they sent a fully-loaded train, and everyone REALLY CHEERED. The line started moving for real. Of course, they opened the Fast Lane right so loads of people filed through on that side and I was SO PISSED because the rest of us had demonstrated extreme levels of patience and endurance by waiting this out and Cedar Point could have rewarded us by at least keeping that line closed off for a few more minutes, goddammit.
I have never been the type of person who would stand in line for THAT LONG for a RIDE so I must really be thirsty for Steve and his wood. Henry and I snagged the back seat and as we buckled ourselves in, I yelled, "YOU'RE GOING TO LOVE THIS!" and he was just like, "WE WILL SEE ABOUT THAT" because Henry does not know how to have fun or display any sort of emotion aside from exhaustion, irritation, and disgust. Maybe envy whenever he sees someone wearing a better beverage t-shirt that him.
But when we hit the midcourse break run, he looked at me and mouthed, "WHAT THE FUCK" and I was like, "RIGHT?!!!?" Holy shit, this ride. It's everything. I was so fucking annoyed every other hour during our weekend at Cedar Point, but THIS RIDE was redemption. I would go back to Cedar Point every fucking weekend and be jerked around by the shitty operations if I knew I could ride this at least once each time.
It's that good.
It's world class.
There is a reason why so many of the experts and enthusiasts rank this as #1.
I AM TEARING UP JUST THINKING ABOUT YOU, STEVE. WAIT FOR ME! I'LL BE BACK.
Meanwhile, Chooch's favorite ride that weekend was actually Millie, a/k/a Millennium Force, though he said it was difficult to choose between the two. The first time we rode Millie, on our first day at Cedar Point, we had the back row and I experienced a pretty good greyout, to the point where I ran down the exit ramp in search of Henry, who had ridden on the train before ours, just to excitedly scream, "I GREYED OUT, DUDE!!!"
I have never greyed out on a ride before, and it was so awesome!
Millie is pretty fucking beautiful. My favorite things about her, aside from that wicked first drop, was the cool 1970s space-y soundtrack that plays in the station, and the killer views of Lake Erie that the lift hill offers. Honestly, for as many times as that park let me down that weekend, I can't deny that the atmosphere is unbeatable. I mean, it's not at DisneySea's level of beauty, but it is pretty fucking close. Oh, and the ride operators on Millie were phenomenal. They were entertaining (on our second ride, one of the operators asked, "Who knows the manufacturer of this ride?" and Chooch and I screamed, "INTAMIN!" before anyone else could answer even though we knew there wasn't a prize but we were born to be first, OK; the ride operator was like, "FRONT ROW GOT IT!" and we rode that wave for a good 45 minutes) and super efficient. They had three train ops down to a science.
Chooch and I had a good ride in the front row on our second day, but I think I preferred the back. We are definitely backseat riders for the most part, on most coasters, but I do really love front row at night. We unfortunately didn't get any night rides on Millie, so clearly we have to go back at some point. Like, this weekend. OK, probably not this weekend. I think Henry will murder me with his eyeballs if I even ask, lol.
I think I actually might be in love with Steve. Sorry, Henry.
Chooch just ran by and I said, "Chooch is there anything you want to say about Millie?" and he said, "Uh yes!" in a way that I expected a saccharine sonnet to come wisping out of his mouth, but instead all he said was, "It was good."
WOW. Just so you know, he teared up looking at it from the car window when we drove away, so.
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Cedar Point Weekend: Broken Down (Day 2) King's Dominion Weekend!


The retractable Cargo Cover keeps items in the back out of view, which provides peace-of-mind when your RDX is parked or otherwise unattended. Mounts behind the 2nd-row seats for maximum coverage. Tested to ensure it will not deform, crack, or blister when subjected to extreme temperatures. Installs easily by removing the factory covers in the side lining. Available in Ebony and Parchment.
Our 2017 Acura RDX Cargo Cover ships directly from an authorized Acura dealer. Since 2000 we have offered secure online ordering and fast affordable shipping. All our genuine Acura parts and Acura accessories are backed by Acura's warranty, which is covered by any Acura dealer in the US. Exactly like your local dealer only lower prices and shipped directly to your door. If you have questions about Acura accessories or Acura performance parts or need assistance, please don't hesitate to contact our friendly Acura parts and accessories experts online.