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