David's Blog

Living a quiet life in Coquitlam, B.C.

Name:
Location: Coquitlam, British Columbia, Canada

Sunday, October 28, 2012

Keplers Equation: Eccentric Anomaly Values at the Quarter-period


In this post I present some information about the solution to Kepler’s Equation of Elliptical Motion (KE) at the point t = T/4 (where T is the period of the motion). I will refer to E at t = T/4 as ET/4.

If e is 0, meaning a circle, the solution reduces to the usual value for sinusoidal motion at the quarter period: π/2. However, for elliptical motion, E is greater than π/2 at this time.

According to KE,
E = M + e sin(E)

Since the magnitude of sin(E) is between 0 and 1, that means ET/4 is restricted to the range M < E < (M + e):

π/2 < ET/4 < π/2 + 1

In fact, over the range e = 0 to e = 1, sin(ET/4) ranges from 1 to cos(D),
or D, where D is the "Dottie Number", the solution to the equation

cos(D) = D

(D is an irrational number; it does not have an exact value, but must be calculated numerically to as many digits as required: 0.73908513321516064...)

In other words, the range of ET/4 is further restricted:

π/2 < ET/4 < π/2 + D

Now recall the sine addition formula:

sin(θ1 + θ2) = sin(θ1) cos(θ2) + cos(θ1) sin(θ2)

Therefore, sin(π/2 + D) = cos(D).
So, sin(ET/4) is restricted to the range

cos(D) < sin(ET/4) < 1

This suggests a function that could bracket sin(ET/4) over the range e = 0 to e = 1. In fact, for the first graph below, the following functions were included to bracket the curve of sin(ET/4):
i) a simple straight-line function through the endpoints: 1 - m*e, where m = 1 - D = 0.260914866785, and
ii) cos(D*e).
Both these functions are equal to sin(ET/4) at the endpoints.
In other words,
(1 - m*e) <= sin(ET/4) <= cos(D*e)

Some notes for the following image:
i) sin(ET/4) is labeled sQ (the red line);
ii) y1 = 1 - m*e (the green line);
iii) cDe = cos(D*e) (the blue line).



Alternately, consider cos(ET/4).
cos(ET/4) goes from 0 for e = 0 to -sin(D) at e = 1. Note that the following graph depicts NEGATIVE cos(ET/4) (it is my personal preference to work with positive numbers.)

Notes for this image:
i) -cos(ET/4) is labeled ncQ (the red line).
ii) y1 = m*e (the green line). This time m = sin(D) = 0.673612029183.
iii) sDe = sin(D*e) (the blue line).



Some numerical data for e = 0.786151377748:
ET/4 = 2.20436934578188 rad,
cos(ET/4) = -0.592028088783,
sin(ET/4) = 0.805917329564,
tan(ET/4) = -1.36128225136,
√(1 - e²) sin(ET/4) = 0.498084301804
 
 

Labels: ,

Sunday, October 21, 2012

Kepler’s Equation: A Summary of Exact Points

I enjoy investigating Kepler’s Equation of Elliptical Motion (KE) and plan to present some information over the course of several posts. I would like to start by posting a tabulation of some exact points for motion on an ellipse. A diagram of the relevant characteristics of an ellipse are depicted at the following link: Kepler's Equation of Elliptical Motion

In the table below, data points are the result of the following procedure:

i) A value for E or sin(E) was specified.

ii) Then KE was used to work backwards to find the time at which this value is reached: ωt = M = E - e sin(E).
In fact, the first column in this table, t (algebraic), simply repeats this expression instead of writing out each point explicitly.

Note that several of the points included in the table are significant in the literature about KE, for example, the point at which E = π/2 radians. Knowing this point, it is easy to work backwards to find the time at which this value was reached:

E - e sin(E) = M
π/2 - e * sin(π/2) = ω*t
π/2 - e = t

In other words, t = π/2 - e

Unfortunately, this does not do us much good in terms of finding a general solution of KE in terms of time. This result isn't very useful by itself--unless we happen to want E at exactly this time. In fact, we could compute as many exact solutions to KE as we wanted in the same way (by working backwards). Again, these results aren't very useful by themselves unless we wanted the values of E at exactly those times. In any case, I have compiled several data sets in the table below. Perhaps this information will be useful for future work.

To better visualize the progress of the motion, some numerical results are also included in the table. These numbers were computed using the following parameters:
eccentricity: e = 0.786151377748
radial velocity of the mean anomaly, M: ω = 1.
(In other words ω = 2π/T = 1, where T is the period of the motion.)

t
(algebraic)
    t
(numeric)
    E
(algebraic)
    sin(E)
(algebraic)
    sin(E)
(numeric)

E - esin(E)     0.0916     E     (1 - e²)1     0.3820
E - esin(E)     0.1254     E     (1 - e²)¾     0.4859
E - esin(E)     0.1305     π/6     1/2     0.5
E - esin(E)     0.1804     E     (1 - e²)½     0.6180
E - esin(E)     0.2292     E     cos(e)     0.7066
E - esin(E)     0.2295     π/4     √2/2     0.7071
E - esin(E)     0.2865     E     (1 - e²)¼     0.7862
E - esin(E)     0.3664     π/3     √3/2     0.8660
π/2 – e     0.7846     π/2     (1 - e²)0     1
E - esin(E)     1.2039     E     sin(EsE=sM)     0.9334
T/4     1.5708     E     sin(Et=T/4)     0.8059

 

Labels: ,