Perimeter of Ellipse

[Gerson W. Barbosa]

(12-07-2019 07:57 PM)Gerson W. Barbosa Wrote:  

(12-04-2019 10:27 PM)Gerson W. Barbosa Wrote:  ———-

This first Ramanujan approximation brings the error down to under one meter:

π[3(a + b) - √((3a + b)(a + 3b))]

His second approximation is even better, with an error less than one micrometer:

https://www.johndcook.com/blog/2013/05/0...-ellipse/

Instead of

p ~ 2π[3*agm(a,b) - 2√(a*b)]

I will suggest a better approximation involving the arithmetic-geometric mean and the geometric mean:

p ~ 2π{a + b - a*b/agm(a,b) - 2[agm(a,b) - √(a*b)]}

Let’s use it to compute the length of the orbit of Pluto¹:

.
.
.



36529672878.01583848170946635744574

Exact result:

36529672878.01583840603514193230844 km

Difference:

0.000000075674324425137 km,

or

75.674 μm

——


¹ Assuming the parameters are exact and the orbit is perfectly elliptical.

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).

p ~ 36529672878.01583840603557428740268 km

p = 36529672878.01583840603514193230844 km

Difference: 0.432 nm

Contrary to what is stated above, the difference obtained with Ramanujan's second approximation is less than one nanometer, not one micrometer (actually 0.905 nm), about the double of the above result, but his formula is way more simple.