Perimeter of Ellipse

[Albert Chan]

(12-29-2019 07:58 PM)Gerson W. Barbosa Wrote:  We can do better:

\(p\approx \frac{\pi \left [ \left ( 15h\sqrt{1+\frac{3h}{8-3h\sqrt{2}}}-80\right )\left ( a^{2}+\frac{6}{5}ab+b^{2} \right ) +4h\left ( a^{2}+2ab+b^{2} \right ) \right ]}{\left ( 12h\sqrt{1+\frac{3h}{8-3h\sqrt{2}}}+4h-64\right )\left ( a+b \right )}\)


where

\(h = \left (\frac{a-b}{a+b} \right )^{2 }\)

This one wasn't particularly difficult to find (might explain later).

We can simplify this a bit.

Let \(\large c = \sqrt{1+\frac{3h}{8-3h\sqrt{2}}}\quad → p ≈ \pi(a+b)\Large\left({3ch^2 + 12(c-1)h - 64 \over 4(3c+1)h-64}\,\right) \)

If \(\large c = 1\), above simplified to: \(\quad\large p ≈ \pi(a+b)\Large\left({3h^2 - 64 \over 16h-64}\,\right) \)

This matched ellipse perimeter Pade [2,1] approximation.