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Ellipse Perimeter Ramanujan II Calculator

Computes ellipse perimeter from semi-axes a and b using Ramanujan second approximation formula with high accuracy.

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Ellipse perimeter โ€” Ramanujan's second approximation

In 1914 Ramanujan also published a second, more refined approximation for the ellipse perimeter: P โ‰ˆ ฯ€(a + b)ยท(1 + 3hยฒ / (10 + โˆš(4 โˆ’ 3hยฒ))), where h = ((a โˆ’ b)/(a + b))ยฒ. Its precision is remarkable. The relative error stays under 0.0001% for just about any ellipse you'll run into, and it only starts to slip near the degenerate flat-ellipse limit. That makes it considerably sharper than Ramanujan's first formula, and it's what most people reach for when they need high precision but would rather not evaluate the elliptic integral numerically. With a = 7 and b = 4, h works out to about 0.0735 and P โ‰ˆ 35.622.

Applications

Orbital ballistics uses it to find the path length of satellites and interplanetary probes, where the timing budget can come down to milliseconds and depends on how accurate the perimeter is. Stadium and racetrack design leans on high-precision elliptical perimeters to stagger lane lengths correctly. And in oval and ovoid industrial design, think pressure vessels, automotive headlights, optical mirrors, the formula is what lets you cut material to length without wasting much.

FAQ

How does it compare to the first Ramanujan formula? The first lands around ~0.01%. This one cuts that by two orders of magnitude, getting to roughly ~0.0001% for moderate eccentricities.

What does h represent? It's a dimensionless squared ratio between the difference and the sum of the semi-axes. When h = 0 you have a perfect circle; as h โ†’ 1 the ellipse degenerates.

When is it worth using the exact elliptic integral? Really only when eccentricity gets very high (above ~0.99) or you need precision below 1 ppm. For ordinary engineering work, Ramanujan II has you covered.

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