# A Formula for the Perimeter of an Ellipse

Unlike for circles, there isn’t a simple exact closed formula for the perimeter of an ellipse. We compare several well-known approximations, and conclude that a formula discovered by Ramanujan is our favourite, due to its simplicity and extreme accuracy.

Published: 11th February 2019 Last updated: 22th November 2020

An ellipse satisfies that equation:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$$

For $a=b$, we have a circle, but if $a \neq b$ the result is like a circle that has been elongated in either the horizontal or vertical direction.

Now we know that the area of a circle is $A =\pi r^2$, and some of you might also know that the area of an ellipse with semi-minor and semi-major axes $a,b$ is simply given by $A = \pi a b$.

The fact that these two formulae are very similar is probably not surprising.

Therefore, one might reasonably expect that if the formula for the perimeter of a circle is $P = 2\pi r$, then there would be a similar, and relatively simple formula for the for ellipse.

Surprisingly, and very unfortunately, this is not the case. There is no simple exact closed formula for the perimeter of an ellipse.

Now it turns out that there are some expressions that give the result exactly, but they are typically in terms of advanced functions that are not studied until the later years of undergraduate mathematics. Specifically, the exact expression for the perimeter of an ellipse is can be written as :

\begin{array}{l} P_{\textrm{exact}}(a,b) \; &=   4 a \; E(1-\frac{b^2}{a^2}) & =  2\pi a \;\;  _2F_1 \left( \frac{1}{2}, -\frac{1}{2}; 1;1-\frac{b^2}{a^2} \right) &=  2\pi \sqrt{ab} \; P_{1/2} \left( \frac{a^2+b^2}{2ab} \right) \nonumber \end{array}

where $E(\cdot)$ is the complete elliptic integral of the second kind, $_2F_1(\cdot)$ is the Gaussian hypergeometric function, and $P_{\nu}(z)$ is a Legendre Function.

Therefore through the use of specialist mathematical software, we can calculate the perimeter of an ellipse to arbitrary precision, and the compare this value to various simpler approximations.

The following expressions have been proposed over the centuries, with the last one discovered by Ramanujan.

\begin{array}{rl} P_1(a,b) &= \pi (a+b)  \nonumber P_2(a,b) & = \pi \sqrt{2 (a^2+b^2)}  \nonumber   P_3(a,b) &= \pi \sqrt{2(a^2+b^2)- (a-b)^2/2} \nonumber P_4(a,b) &= \pi \left( 3(a+b) – \sqrt{(3a+b)(a+3b)} \right) \nonumber \end{array}

Firstly, we note that for $a=b$, these all turn out to be exactly equivalent to $P=2 \pi a$, which is a good start. Therefore, these approximations only differ from the exact value as they become more elongated.

Then for the really elongated ones, such as when it is 8x wider than it is high, the errors for $P_1, P_2, P_3,P_4$ are around 2

This last equation is traditionally called Ramanujan’s first approximation.

There are several elegant formulae which are all far are more accurate than all of the above-mentioned formulae.

They are typically expressed in terms of the particular ratio, $h = \left( \frac{a-b}{a+b} \right)^2$.

An elegant one derived from Pade approximations

$$P_5(h) = \pi (a+b) \frac{64-3h^2}{64-16h} .$$

And an even more accurate one (Jacobsen, 1985)

$$P_6(h) = \frac{256-48h^2-21h^4}{256-112h^2+3h^4}$$

And more accurate again, here is Ramanujan’s (more famous) second approximation

$$P_7(h) = \pi (a+b) \left( 1+ \frac{3h}{10+\sqrt{4-3h}} \right)$$

And finally the one that is used by many computer programs is [Rackauckas]

$$P_8(h)= \pi (a+b) \frac{135168-85760h-5568h^2+3867h^3}{135168-119552h+22208h^2-345h^3}$$

Note that although $P_5$ is the least accurate amongst these latter functions, it is still more accurate than the all the approximating functions in the earlier section of this post.

Summary of Very Accurate Approximations

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My name is Dr Martin Roberts, and I’m a freelance Principal Data Science consultant, who loves working at the intersection of maths and computing. “I transform and modernize organizations through innovative data strategies solutions.”

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## 3 Replies to “A Formula for the Perimeter of an Ellipse”

1. Maher Ezzideen Aldaher says:

Dear sir
I’ve worked to find firstly to find the perimeter of ellipse in a simple way then for lame curves and general astroid with partial arcs and have also simple eq. for area also and to find roots ,these put in my research which proceeded at ICMS 2012 Nagpur ,India.
you can find it at : Linkedin , academia.edu, slideshare.net, nanopdf.com
Thank you
With best regards

2. Ian Lindsay says:

Great article, very informative – who would have thought it is so complex! Just one thing I spotted – I suspect that the ‘6’ and the ‘8’ might be swapped in the text above in the Rackauckas approximation for the third term on the numerator. Rackauckas’ paper has 3867h^3, although he doesn’t really go into where the numbers come from so it is difficult to verify.

1. Hi Ian,
Thanks so much for picking up on this mistake. Now fixed. You must have an incredible eye for details to notice this discrepancy.
I definitely owe you a coffee next time you are ever in my passing through my home town. 😉
Glad the article was helpful!

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