Perimeter of Ellipse

[Albert Chan]

(01-21-2020 05:16 PM)Albert Chan Wrote:  \(\large p(a,b) = 2× \left( p({a+b \over 2}, \sqrt{ab}) - {\pi a b \over AGM(a,b)}\right)\)

In other words, calculate ellipse perimeter from a much less eccentric ellipse.

We could even do it in a novel way, from the *more* eccentric ellipse.
Below, LHS is the more eccentric ellipse, with \({a_0 + b_0 \over 2} = a, \sqrt{a_0 b_0} = b\)

\( p(a_0,b_0) = 2× \left( p(a,b) - {\pi b^2 \over AGM(a,b)}\right)\)

Using my Casio FX-115MS, AGM2 method for Halley's comet, numbers from here

A = 2667950000
B = 678281900
C = A²
D = B²
E = 0.5

F=B-A : C=C-EF² : B=√AB : A=A+F/2 : E=2E

Quote:\(p(a,b) = {p(a_0,b_0) \over 2} + {\pi b^2 \over AGM(a,b)}\)

Keep pressing "=" until F=0 (so that A,B,C all converged)

\(\large\pi\)(C+D)/B       → ellipse perimeter = 11,464,318,984.1 km

This was similar to my lua code, with C=(A²+B²)/2, E=0.25, so that converged C/B = "radius".