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May 14

Perimeter estimation

Do someone know a very accurate estimation for the perimeter of an ellipse? For example evaluation for (a=5;b=1). And what is the error % of this estimation? How do you make the evaluation? Wich formula?TASDELEN (talk) 18:51, 14 May 2010 (UTC)[reply]

See circumference or ellipse. Dmcq (talk) 20:37, 14 May 2010 (UTC)[reply]

The circumference on an ellipse with semiaxes a, b is

\int _{0}^{2\pi }{\sqrt {a^{2}\cos ^{2}t+b^{2}\sin ^{2}t}}\ dt

. This can be evaluated to any desired accuracy with numerical integration techniques. For

a=5,\ b=1

this is

21.010044539689000944699164588473738912894812339134152623096835657214254472447144946328558302691971167999951485...

, correct to the number of digits given. I could just as easily give 100,000 accurate digits. -- Meni Rosenfeld (talk) 17:53, 15 May 2010 (UTC)[reply]

The ellipse article gives this as a good approximation:

C\approx \pi \left(a+b\right)\left(1+{\frac {3\left({\frac {a-b}{a+b}}\right)^{2}}{10+{\sqrt {4-3\left({\frac {a-b}{a+b}}\right)^{2}}}}}\right);\!\,

For

a=5,\ b=1

this is 21.010027, so you can judge how good the approximation is from Meni's figure.→86.132.162.141 (talk) 19:26, 15 May 2010 (UTC)[reply]

Thanks a lot.By estimation I understand Meni's evaluation and not mathematica tool's estimation.He used Ramanujan's formula and find C=21,010026651559300 (at 15 digit).

So,the error %=(Lexact-Lestimated)/Lexact=0,000000851482682.May you be interested whith a new formula which gives error%=-000000417870650,a %50 less error than Ramanujan's estimation?TASDELEN (talk) 23:18, 15 May 2010 (UTC)[reply]

In situations where accuracy is important, you wouldn't use such formulas but rather advanced techniques which are scalable to any accuracy. If you must have an elementary closed form, you can find expressions which are arbitrarily accurate, but the more accurate expressions are also more complicated. So finding an expression which is marginally more accurate than another is not interesting. If the new expression is more accurate and also simpler than the commonly known estimates, it might be somewhat interesting. -- Meni Rosenfeld (talk) 08:58, 16 May 2010 (UTC)[reply]

Meni, I am impressed. How did you do that? Bo Jacoby (talk) 04:31, 16 May 2010 (UTC).[reply]

Mathematica. Specifically,

NIntegrate[(25 Cos[t]^2 + Sin[t]^2)^(1/2), {t, 0, 2 Pi}, WorkingPrecision->150]

For higher precision, the following may be more robust:

Integrate[(25 Cos[t]^2 + Sin[t]^2)^(1/2), t]
(% /. {t -> 2Pi}) - (% /. {t -> 0})
N[%, 100000];

-- Meni Rosenfeld (talk) 08:33, 16 May 2010 (UTC)[reply]

Gentlemen,I am using Thales theorem to find the perimeter of an ellipse.An exact formula but implicit and cracking Thales theorem I may reach to a nice and practically high accuracy for astroids (x/a)^r+(y/b)^r=1.(r=2 is an ellipse).Not "simpler",but more accurate than all known approximations.TASDELEN (talk) 09:37, 16 May 2010 (UTC)[reply]