Ellipse Perimeter Calculator (Ramanujan)

Unlike a circle, an ellipse has no simple closed-form perimeter formula. The exact answer needs an elliptic integral, which has to be evaluated numerically. Ramanujan’s 1914 approximation gets you there with arithmetic you can do on paper:

P ≈ π × [3(a + b) − √((3a + b)(a + 3b))]

Where a is the semi-major axis (half the longer diameter) and b is the semi-minor axis (half the shorter diameter).

For a circle (a = b = r), this reduces to 2πr, the familiar circumference formula.

How accurate is it, really?

Better than most people expect, and it depends entirely on how stretched the ellipse is:

For any ellipse you would actually draw, order material for, or dig a pond in the shape of, the answer is exact to more decimal places than your tape measure has. The error only becomes visible on shapes so elongated they look like a line, and even then it stays under half a percent.

Where ellipse perimeters matter in real life:

• Race tracks. A 400 m running track is two straights joined by two curves, not a true ellipse. The standard geometry is 84.39 m straights and 115.61 m curves, measured in lane 1.
• Whispering galleries. An ellipse has the property that sound leaving one focus converges on the other, which is why you can hear a whisper across Statuary Hall in the US Capitol. Worth knowing that St Paul’s in London, the gallery everyone cites, is not an example: it is circular, and its effect comes from sound creeping around the curved wall rather than crossing between two foci. Different geometry, different physics, same party trick.
• Oval gardens and ponds. A 12 ft × 8 ft oval pond perimeter ≈ π × [30 − √(22 × 18)] ≈ 31.7 ft of edging stones needed.
• Decorative oval mirrors. A 24 × 18 in oval mirror (a = 12, b = 9) needs π × [63 − √(45 × 39)] ≈ 66.3 in of frame molding.
• Astronomy. Planetary orbital paths are ellipses with the sun at one focus. Earth’s orbital perimeter is approximately 940 million km.
• Stadium and amphitheater designs for sightline calculations.

Worked example: an oval pond

A backyard pond, ellipse 16 ft × 10 ft. Semi-axes: a = 8 ft, b = 5 ft. P ≈ π × [3(13) − √(29 × 23)] ≈ π × [39 − √667] ≈ π × [39 − 25.83] ≈ π × 13.17 ≈ 41.4 ft.

For edging stones: order 45 linear ft to allow for fitting around the curve.

Worked example: binding an oval rug

Oval bedside rug, 5 × 3 ft. Semi-axes: a = 2.5 ft, b = 1.5 ft. P ≈ π × [3(4) − √(9 × 7)] ≈ π × [12 − √63] ≈ π × [12 − 7.94] ≈ π × 4.06 ≈ 12.75 ft of edge binding tape.

Eccentricity (how oval, vs how circular):

e = √(1 − b²/a²)

Eccentricity is not very intuitive, because it climbs fast while the shape still looks fairly round. The axis ratio b/a is easier to picture, so here are both:

Note how e = 0.745, which sounds dramatic, is just a 3:2 oval. That is why this calculator reports the axis ratio alongside it.

At e = 1 the ellipse has collapsed to a line segment, because b has gone to zero. You will also see e = 1 described as a parabola; that is true of conic sections in general, but a parabola is a different curve, not the limit of this formula.

Earth’s orbital eccentricity is 0.0167. Draw it accurately and nobody would spot that it is not a circle.

Why Ramanujan’s formula is so loved: it is fast, it uses nothing beyond π and a square root, and over the range of shapes anyone actually builds it is indistinguishable from the exact answer. Ramanujan published several approximations of increasing accuracy; this is the simplest one that is good enough, which is usually the one that survives.