Ellipse Perimeter (Ramanujan)

Perimeter of an ellipse — Ramanujan's approximation

The circle has a tidy perimeter formula. The ellipse does not. There's no closed elementary formula for it, and the exact value comes from the complete elliptic integral of the second kind. In practice the formula people reach for is Ramanujan's first approximation from 1914: P ≈ π·[3(a+b) - √((3a+b)(a+3b))], which is good to better than 0.01% for nearly every ellipse you meet in engineering. There's also a simpler but rougher estimate from Euler, P ≈ π·√(2(a²+b²)), but it overestimates noticeably once a and b drift far apart. Take a = 5 and b = 3: Ramanujan gives P ≈ 25.527, while Euler comes in at ~25.946, roughly 1.6% high.

Applications

Athletics tracks are built on oval geometries, so lane lengths have to be worked out precisely. Planetary and satellite orbits are ellipses too, and the path length feeds into fuel and timing calculations. In industrial design, oval tanks, gaskets, picture frames and aircraft fuselage cross-sections all need a good perimeter estimate before material gets cut.

FAQ

How accurate is Ramanujan's formula? Better than 0.01% for eccentricities below 0.95. The relative error only starts to grow near the degenerate flat-ellipse limit.

What if a = b? Then the ellipse is just a circle, and the formula collapses exactly to P = 2π·a.

Why no elementary closed form? The arc length integral involves √(1 - e²·sin²θ), and that just can't be written with finitely many elementary functions. It defines a new kind of function, the "elliptic" one.