Perimeter of Ellipse

[Albert Chan]

(01-19-2020 11:00 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.

Amazingly, the less eccentric ellipse have the eccentricity of h

lua> sqrt = math.sqrt
lua> a, b = 2667950000, 678281900 -- numbers from here
lua> = sqrt(1-(b/a)^2)     → e = 0.967142904276921
lua> = (a-b)/(a+b)           → h = 0.5945995852827773 = e'
lua>
lua> a, b = (a+b)/2, sqrt(a*b)
lua> = sqrt(1-(b/a)^2)     → e' = 0.5945995852827773
lua> = (a-b)/(a+b)           → h' = 0.10863394673347321 = e''

Prove: Assume a ≥ b, let a', b' = (a+b)/2, √(ab)

e' = √(1-(b'/a')²) = √(1 - 4ab/(a+b)²) = (a-b)/(a+b) = h