arcsinc( 1-y ), for small y

[Albert Chan]

On the railroad warp problem, same formulas still work for pi/2 < x < pi:
In other words, circle center and warp ok on the *same* side:

Prove sin(x) / x = m / (m + d):

m/2 = R sin(pi - x) -- chord length
(m + d) / 2 = R x -- arc length

sin(pi - x)/x = sin(x)/x = m / (m + d) -- QED

Prove e = (m/2) tan( x/2 ):

(m/2) / e = tan((pi - x) / 2) -- triangle: chord + e (contained circle center)

e = (m/2) / tan(pi/2 - x/2) = (m/2) tan( x/2 ) -- QED

My weighted e is not designed for big angle, but still work ok
For x < pi/2 (thus, warp < pi/2), weighted e has max error of 0.001%

For HUGE warp, weighted e formula run into problem:
Example: let m=1, 6.5X warp (d=5.5), calculate e:

sin(x) / x = 1 / (1+d) = 0.1538461538
x = 2.71131291 -- my Casio solver, x ~ 155 degrees
R = (1+d) / (2x) = 1.198681269
e = 1/2 tan(x/2) = 2.288101656 -- almost diameter of circle !

Assumed we don't know x, use 4 weighted e method:
z = 3/8 d = 2.0625
e(min) = sqrt(z) = 1.436140662
e(max) = sqrt(z + z z) = 2.513246158
e ~ (4 e(max) + e(min)) / 5 = 2.297825059 -- error = +0.425%

With warp that high, "improved" weight does not help:
y = 1 - 0.1538461538 = 0.8461538462
w = 4 - 27/56 y = 3.5920322967
e ~ (w e(max) + e(min)) / (w+1) = 2.27868654 -- error = -0.411%

For HUGE warp, weighted e method is not suited, below fit is better:
For pi/2 < x < pi, fitted e (below) has max error of 0.013%

e (warp 2.4X or more)
~ (((-0.690948 y + 1.95561) y - 2.47546) y + 1.52907) d

Redo above case (6.5X warp):

e ~ 0.4160261229 * 5.5 = 2.288143676 -- error = +0.002%