How this instrument works
An ellipse's circumference has no elementary closed-form formula at all — no arrangement of algebra, roots, logarithms, or trig functions gives the exact value, because the integral describing it belongs to a class mathematicians literally named 'elliptic integrals' after this very curve. A circle is the one special case where the integral collapses to something tidy, 2πr; separate the two axes even slightly and the exact answer becomes transcendental. The formula on this page, C ≈ π[3(a+b) − √((3a+b)(a+3b))], sidesteps that by trading the integral for a square root that lands extremely close.
The correction term does real work rather than sitting there for show. Without it, the obvious guess — average the two axes and treat the shape like a circle, C ≈ π(a+b) — already gets the right answer when a equals b, but drifts low as the ellipse elongates, because it ignores how much more tightly the boundary curves near the ends of the long axis than near the ends of the short one. The square root term measures exactly that extra curvature and adds it back in.
Set a = b and the whole expression folds back into the circle formula on the nose: 3(a+b) becomes 6a, the product under the root becomes 16a², its square root is 4a, and π(6a − 4a) = 2πa. At the opposite extreme, flattening one axis toward zero, the formula's relative error tops out at roughly four tenths of one percent against the true degenerate perimeter — worse than the near-perfect fit near a circle, but still tighter than most tape measures.
- Enter the longer of the two radii — center to the far edge along the long axis — into Semi-major axis, a.
- Enter the shorter radius, center to edge along the short axis, into Semi-minor axis, b.
- Read the result in Circumference (Ramanujan approximation), the estimated distance once around the boundary.
- Keep both axes in the same unit; the answer returns in that same unit, since this is a length, not an area.
- As a sanity check, swap the two numbers between the fields — the answer should not move, since the formula treats a and b symmetrically.
Worked example — coping for a 5-by-3 reflecting pool
A garden pool is dug as an ellipse with a semi-major axis of 5 metres and a semi-minor axis of 3 metres. First the sum and its triple: 3(a + b) = 3(5 + 3) = 24. Then the product under the root: (3a + b)(a + 3b) = (15 + 3)(5 + 9) = 18 × 14 = 252, and √252 ≈ 15.874508. Subtracting gives 24 − 15.874508 = 8.125492, and multiplying by π returns 25.526986 — matching the sheet's Circumference field to the last displayed digit.
It is worth seeing what the lazy shortcut would have missed: averaging the axes gives π(a + b) = π × 8 = 25.132741, about 1.5% short of the real 25.526986. Over a 25-metre pool edge that gap is roughly 39 centimetres of coping stone — not a rounding error, a missing course of bricks — which is exactly the gap the square-root term in Ramanujan's formula exists to close.
Questions
Why doesn't an ellipse have an exact circumference formula?
Because its circumference is an elliptic integral — the historical origin of that name — and elliptic integrals cannot be written as any finite combination of algebra, roots, logarithms, or trig functions. A circle is the one special case, a = b, where the integral collapses to 2πr; separate the axes and the exact value genuinely has no closed form, which is why this calculator falls back on a fast, close approximation instead of an identity.
How accurate is the Ramanujan formula used here?
Very, outside of extreme shapes: the relative error is exactly zero for a circle and grows smoothly as the ellipse elongates, topping out around 0.4% only in the limit where one axis shrinks toward zero. Across the proportions used in ordinary engineering drawings, orbits, or garden layouts, that error is smaller than a tape measure can register.
What's the common mistake people make estimating this by hand?
Averaging the two axes and reusing the circle formula, C ≈ π(a+b), on the assumption circumference scales the way area does. It doesn't: for a = 5, b = 3 that shortcut gives 25.132741 against the correct 25.526986, a shortfall of about 1.5%, because it drops exactly the curvature term the √((3a+b)(a+3b)) correction supplies.
How does this relate to the ellipse's area formula?
Area, πab, is exact and elementary because it scales as a simple product of the two axis lengths — Archimedes essentially had it in antiquity. Circumference has no such shortcut, since it depends on how tightly the boundary curves at every point around the loop, not merely on the axis lengths multiplied together, which is the reason one formula is ancient and the other still has no exact closed form.
Does it matter which axis I call the semi-major one?
Only for the labels to read sensibly, not for the arithmetic: the formula treats a and b symmetrically, so entering 5 and 3 or 3 and 5 into Semi-major axis, a and Semi-minor axis, b returns the identical 25.526986. Convention just calls the longer radius 'major', but swapping the two values leaves the result unchanged.
Who was Ramanujan, and why is his name on this formula?
Srinivasa Ramanujan published this approximation in a 1914 paper on modular equations and approximations to π, one of many results he produced with almost no formal training in elliptic function theory. It is still the standard quick approximation in engineering handbooks and software whenever computing the exact elliptic integral would be more machinery than the job needs.