Perimeter of Ellipse

[Albert Chan]

(01-21-2020 05:16 PM)Albert Chan Wrote:  

(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

If we are at focus F2, and we let distance a0 = AF2, b0 = BF2, eccentricity of ellipse is also h0

ellipse major-axis a = OB = (a0+b0)/2
ellipse minor-axis b = OC = √((CF2)² - (OF2)²) = √(a² - (a-b0)²) = √((2a-b0)*b0) = √(a0*b0)

This matched ellipse perimeter recursive AGM formula!

Example, lets try the earth-moon orbit, average orbital distance 384748 km

a0 = 406731 km
b0 = 364397 km

e = h0 = (a0-b0)/(a0+b0) = 42334/771128 ≈ 0.0549

Update, Jan 26, 2023
In 2 years, California State University, San Bernardino picture link is dead.
It is ironic that CSUSB motto is "We Define The Future".
Above picture were from WayBack Machine, Thanks!