Perimeter of Ellipse

[Albert Chan]

(12-07-2019 07:57 PM)Gerson W. Barbosa Wrote:  Just a small refinement – more are still possible – and the overall error in the length of the orbit of Pluto is only 19.34 fm (femtometers!).

p ~ 2π{agm(a,b)[77h² - 768(h - 1)] - [54h² - 512(h - 1)]√(ab)}/[23h² - 256(h - 1)]

where h = [(a - b)/(a + b)]²

Example:

a = 5906376272 km

b = 5720637952.8 km

p ~ 36529672878.01583840603514191296851 km

Exact:

p = 36529672878.01583840603514193230844 km

If you have AGM, this recursive formula is better (based on AGM2 method):

\(\large p(a,b) = 2× \left( p({a+b \over 2}, \sqrt{ab}) - {\pi a b \over AGM(a,b)}\right)\)

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

You can do this recursively, but then you might as well use AGM2.
Note: AGM only needed to calculate once, since AGM(a,b) = AGM((a+b)/2, √(ab)) = ...

Code:

from gmpy2 import *
get_context().precision = 160
pi = const_pi()

def pade22(a,b):
    h = ((a-b)/(a+b))**2
    return pi*(a+b) * ((-21*h-48)*h+256) / ((3*h-112)*h+256)

>>> a=mpfr('5906376272')
>>> b=mpfr('5720637952.8')
>>> print pade22(a,b)
36529672878.015838406030163739762066543303728710139

>>> p = pade22((a+b)/2, sqrt(a*b))
>>> print 2*(p - pi*a*b / agm(a,b))
36529672878.015838406035141932308423167594568357066

AGM improved pade22 doubled accuracy, from 22 to 45 digits.