How this instrument works
Perimeter is the general geometric word for the length of a shape's boundary — it applies to triangles, hexagons, and ellipses alike, whereas circumference is the older term kept specifically for round curves such as circles. For an ellipse the two words point at exactly the same question and the same number: how long the curve is if you walk all the way around it once. Unlike a circle, though, that boundary length has no elementary formula. It is a complete elliptic integral of the second kind, L = 4aE(k), a quantity mathematics cannot write down using algebra, roots, and trig functions alone — NIST's Digital Library of Mathematical Functions gives the exact identity in section 19.30.
Srinivasa Ramanujan published this formula in 1914 as the second of three approximations to the ellipse's boundary, refining a cruder first guess and later outdone in accuracy by a more elaborate third one. Its structure blends the two axes twice — once as a plain sum 3(a+b), and once inside a square root as the product of two weighted combinations, (3a+b) and (a+3b). The two blends are built so the square-root term cancels exactly the right amount of the sum whenever the ellipse happens to be a circle, leaving no error at all in that case and only a very small one elsewhere.
Set the two axes equal and every trace of approximation disappears: 3(a+b) becomes 6a, the square root becomes 4a exactly, and the formula reduces algebraically to P = 2πa, the familiar circle circumference recovered with zero error rather than by coincidence. The opposite pitfall is the shortcut of averaging the axes, P ≈ π(a+b), which treats the ellipse as a circle of radius (a+b)/2. That guess is always a little low, and increasingly so the more one axis grows relative to the other.
- Enter the ellipse's longer half-length into Semi-major axis, a.
- Enter the shorter half-length into Semi-minor axis, b, using the same unit as a.
- Read Perimeter (Ramanujan approximation) for the boundary length, reported in that same unit.
- Set a equal to b to check the circle case — the field should return exactly 2πa, with no rounding drift.
- Swapping the two numbers between a and b changes nothing: the formula is symmetric, so either axis can go in either field.
Worked example — semi-axes 5 and 3
An elliptical garden bed measures 5 m from center to the far end of its long axis and 3 m to the edge of its short axis, so a = 5 and b = 3. The sum 3(a+b) is 3 × 8 = 24. The two weighted blends are 3a+b = 18 and a+3b = 14, and their product is 252, so the square root term is √252 ≈ 15.8745. Subtracting gives 24 − 15.8745 ≈ 8.1255, and multiplying by π returns a perimeter of 25.526986 m — the exact figure the golden vector for this instrument returns is 25.526986393758545.
It is worth seeing why the shortcut fails here: averaging the axes gives P ≈ π(5+3) = 25.132741 m, about 1.6% short of the real figure. Buying edging trim by that naive number would leave the last stretch of the bed unbordered — the extra 0.4 m comes entirely from how sharply the boundary curves near the pointed ends of an elongated oval, a bend a simple average has no way to account for.
Questions
Why isn't there an exact formula for an ellipse's perimeter?
Because the arc-length integral for an ellipse does not reduce to elementary functions the way a circle's does — it is a complete elliptic integral of the second kind, L = 4aE(k), documented in NIST's Digital Library of Mathematical Functions, section 19.30. E(k) itself has no closed form in ordinary algebra, so every practical method, including this one, uses an accurate approximation rather than an exact expression.
How accurate is Ramanujan's approximation used here?
It is exact in the circular limit, where a = b gives precisely 2πa, and it stays extremely close to the true elliptic-integral value for ordinary, only moderately elongated ellipses; the error grows as the shape becomes very stretched. Ramanujan published it in 1914 alongside a cruder first estimate and a still sharper third one — this is the middle formula, trading a little accuracy for a shorter expression.
Is perimeter the same thing as circumference for an ellipse?
Yes — both name the identical boundary length. Circumference is the traditional term reserved for round curves like circles; perimeter is the general geometry term used for the boundary of any closed shape, ellipses included. Searching either word for this shape lands on the same question and the same Ramanujan approximation, since an ellipse's boundary has no simpler exact formula than a circle's C = 2πr does.
What mistake do people make estimating an ellipse's perimeter by hand?
The common shortcut is averaging the two axes as if the ellipse were a circle: P ≈ π(a+b). For semi-axes 5 and 3 that guess gives about 25.132741, roughly 1.6% under the true 25.526986 from Ramanujan's formula — a gap that widens the more elongated the ellipse becomes, since a plain average ignores how sharply the boundary curves near the pointed ends.
Does it matter which axis I enter as a and which as b?
Not for the arithmetic: swap the two values and the formula returns the same perimeter, because 3(a+b) and (3a+b)(a+3b) are both unchanged by exchanging a and b. By convention a is labeled the semi-major, or longer, axis and b the semi-minor, or shorter, one, which keeps the fields consistent with how an ellipse's shape is normally described.
What happens when the two axes are equal?
The ellipse becomes a circle of radius a, and the formula collapses exactly: 3(a+b) turns into 6a, and √((3a+b)(a+3b)) turns into √(16a²) = 4a, leaving P = π(6a − 4a) = 2πa, the ordinary circumference formula, reached with no approximation error at all.