Ellipse Perimeter Calculator
From the two semi-axes, get the distance once around any oval — using Ramanujan's famously accurate approximation.
What is an ellipse perimeter calculator?
Two semi-axes in, perimeter out
An ellipse perimeter calculator turns two measurements — the semi-major axis a and the semi-minor axis b — into the distance once around an oval's edge. The semi-axes are half the longest and shortest diameters, measured from the centre out to the edge. That makes them the two inputs you need for oval tables and mirrors, running tracks, elliptical gardens and ponds, and the orbits of planets, which trace ellipses with the Sun at one focus. Unlike a circle, an ellipse has no simple exact perimeter formula, so this tool uses Ramanujan's renowned approximation.
Enter the two semi-axes in any length unit to get the ellipse perimeter instantly.
The perimeter has no simple exact formula, so it uses a famous approximation.
P ≈ π × [3(a + b) − √((3a + b)(a + 3b))]The exact perimeter of an ellipse needs an elliptic integral, which has no elementary closed form. So this tool uses Ramanujan's second approximation, π × [3(a + b) − √((3a + b)(a + 3b))]: add the semi-axes, scale and subtract a square-root correction term, then multiply by π (about 3.14159). It is accurate to better than one part per million for ovals of moderate eccentricity.
Suppose you have an ellipse with a semi-major axis of 5 and a semi-minor axis of 3.
Add and scale the semi-axes
3 × (5 + 3) = 3 × 8 = 24 — three times the sum of the axes.
Compute the correction term
√((3·5 + 3)(5 + 3·3)) = √(18 × 14) = √252 ≈ 15.874508 — the square-root term.
Subtract and multiply by π
π × (24 − 15.874508) = 25.526986 units — the perimeter once around the edge.
The perimeter (about 25.526986 units for a = 5, b = 3) is the distance once around the edge — the trim you would run around an oval mirror, the fencing around an elliptical bed, or the length of one lap on an oval track. The single most common mistake is using the full diameters instead of the semi-axes: the axes are halves, so an oval measuring 10 across and 6 tall has a = 5 and b = 3, not 10 and 6, and feeding in the full widths doubles the perimeter. A second insight is the link to the circle: when a equals b, the ellipse becomes a circle of radius r, and Ramanujan's formula collapses to the familiar 2 × π × r. So a circle is just a perfectly symmetric ellipse, and reasoning about an oval's perimeter is reasoning about a stretched circle. The accuracy also depends on shape: for round-ish ovals the estimate is essentially exact, while very long, thin ellipses introduce a slightly larger error.
The perimeter is an approximation, and your inputs deserve a second look.
Approximate perimeter and consistent units
This is Ramanujan's approximation, not an exact value — an exact ellipse perimeter needs an elliptic integral. The estimate is accurate to better than one part per million for moderate eccentricities, but the error grows for very long, thin ellipses where one axis dwarfs the other. The semi-axes are also unit-agnostic, so the answer is only meaningful if you keep one unit across both inputs: semi-axes in centimetres give a perimeter in centimetres, never a mix.