Keplers Equation: Eccentric Anomaly Values when sin(E) = sin(M)

In this post I present some information about the solution to Kepler’s Equation of Elliptical Motion (KE) at the point at which sin(E) = sin(M). This point is significant because it is the only point at which the value of sin(E) might be found which is a linear expression in time. Plots of E and M intersect only at t = 0, π, and 2π. However, plots of sin(E) and sin(M) intersect at one more point, and because sin(M) operates on an operand that is linear in time, the value for sin(E) at this time would also be known. I will refer to M and E at this point as M_sE=sM and E_sE=sM, respectively.

According to KE, E = M + e sin(E)
Since sin(E_sE=sM) = sin(M_sE=sM), this equation can be re-written as follows:

E_sE=sM = M_sE=sM + e sin(M_sE=sM)

The only time at which this condition holds is when E and M are symmetric around π/2. Also at this time, M + E = π.

STATEMENT 1

E_sE=sM - π/2 = π/2 – M_sE=sM
= (e/2) sin(M_sE=sM) = (e/2) sin(E_sE=sM)

From STATEMENT 1, expressions can be made in terms of either E_sE=sM or M_sE=sM:

E_sE=sM - (e/2) sin(E_sE=sM) = π/2         EQUATION 1
M_sE=sM + (e/2) sin(M_sE=sM) = π/2       EQUATION 2

EQUATION 1 is very similar to KE itself, and can be solved numerically. A plot of sin(E_sE=sM) is shown in the graph below.

From a diagram of this geometry, note that

cos((e/2) sin(M_sE=sM)) = sin(M_sE=sM)
sin((e/2) sin(M_sE=sM)) = cos(M_sE=sM)

The half-angle identities could now be used:

cos(B/2) = √((1 + cos(B))/2)
sin(B/2) = √((1 - cos(B))/2)

Therefore,

cos((e/2) sin(M_sE=sM)) = √((1 + cos(e sin(M_sE=sM)))/2)
sin((e/2) sin(M_sE=sM)) = √((1 - cos(e sin(M_sE=sM)))/2)
respectively.

These forms don’t really help, because they are even more complicated equations to solve.
At this point, the most concise expression for sin(E_sE=sM) is as the solution to the equation

cos[(e/2) θ] = θ, where θ = sin(E_sE=sM).

This equation is analogous to the expression for sin(E_T/4), which is as the solution to the equation

cos[e θ] = θ, where θ = sin(E_T/4)

Some numerical data for e = 0.786151377748:
E_sE=sM = 1.9377086781046062 rad,
M_sE=sM = 1.20388397548 rad,
η_sE=sM = 2.290150905621918 rad,
r_η = 0.792451718158,
sin(E_sE=sM) = 0.933439441046,
cos(M_sE=sM) = 0.358735013494,
√(1 - e²) sin(E_sE=sM) = 0.576897301019

Also, on the graph below, note that a plot of cos(0.45e) looks very close to a plot of sin(E_sE=sM).

Some notes for the following image:
i) sin(E_T/4) is labeled sET4;
ii) sin(E_sE=sM) is labeled sqEM;
iii) c1e = cos(0.45e);
iv) as e goes to 1, the equation goes to cos(θ/2) = θ,
    where θ = sin(M_sE=sM). This equation is solved numerically:
    sin(M_sE=sM) = 0.900367222589,
    cos(M_sE=sM) = 0.435130859037,
    M_sE=sM = 1.1206127155 rad,
    E_sE=sM = 2.02097993809 rad.

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 E_T/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 E_T/4 is restricted to the range M < E < (M + e):

π/2 < E_T/4 < π/2 + 1

In fact, over the range e = 0 to e = 1, sin(E_T/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 E_T/4 is further restricted:

π/2 < E_T/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(E_T/4) is restricted to the range

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

This suggests a function that could bracket sin(E_T/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(E_T/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(E_T/4) at the endpoints.
In other words,

(1 - m*e) <= sin(E_T/4) <= cos(D*e)

Some notes for the following image:
i) sin(E_T/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(E_T/4).
cos(E_T/4) goes from 0 for e = 0 to -sin(D) at e = 1. Note that the following graph depicts NEGATIVE cos(E_T/4) (it is my personal preference to work with positive numbers.)

Notes for this image:
i) -cos(E_T/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:
E_T/4 = 2.20436934578188 rad,
cos(E_T/4) = -0.592028088783,
sin(E_T/4) = 0.805917329564,
tan(E_T/4) = -1.36128225136,
√(1 - e²) sin(E_T/4) = 0.498084301804