Ellipse Circumference Calculator

What this calculator does

This works out the distance around the edge of an ellipse, an oval, from its two half-widths: the semi-major axis (a) the long way, and the semi-minor axis (b) the short way.

One thing to say plainly up front: this is an estimate, not an exact answer. The reason why is interesting, and it is worth a moment before the formula.

Using the calculator

1. Type the semi-major axis (a).
2. Type the semi-minor axis (b).
3. Press Calculate.

Both values need to be positive. The result is in the same unit you used for the axes, since the circumference is a length.

Why an ellipse has no exact circumference formula

A circle is easy. Its circumference is exactly 2 × π × radius, full stop. You would expect an ellipse to be just as tidy, but it is not, and this trips up a lot of people.

The trouble is that an ellipse curves by a different amount at every point along its edge, tight at the ends, gentle along the sides. Adding up all that changing curvature leads to something called a complete elliptic integral of the second kind. The exact circumference is 4 × a × E(e), where E is that integral and e is the ellipse's eccentricity. The snag is that this integral cannot be written out with ordinary algebra, powers and roots. There is simply no neat closed formula for it.

So every practical tool, this one included, reaches for an approximation instead. The honest thing is to tell you which one, and how good it is.

The estimate this tool uses

This calculator uses one of the simpler approximations:

circumference ≈ 2 × π × √((a² + b²) ÷ 2)

It takes the two half-axes, squares them, averages the squares, takes the square root of that, and multiplies by the same 2π that wraps a circle. That middle step is the root mean square of a and b.

It has one property going for it that matters: it is exact for a circle. When a and b are equal, √((a² + a²) ÷ 2) comes back to a, and the whole thing becomes 2 × π × a, the plain circle circumference. So the more circular your oval, the better this estimate behaves.

How close it gets, and a sharper formula

For ovals that are nearly round, the estimate is good. As the ellipse gets more stretched, it drifts, and it drifts upward: it tends to read a little high. On an ellipse where the long axis is twice the short one, it comes out about 2.5 percent over. On gentler ovals the gap is much smaller.

If you are working with a strongly stretched ellipse and need a tighter answer, the formula to reach for is Ramanujan's:

circumference ≈ π × [3(a + b) - √((3a + b)(a + 3b))]

That one stays within roughly 0.04 percent even for quite elongated ovals. This tool keeps to the simpler estimate above, which is plenty for most everyday shapes, but it is worth knowing the sharper formula is there when precision counts.

Units and rounding

There are no unit menus, so work in whatever unit you like, and read the answer in that same unit. The result is shown to four decimal places. Keep in mind that those trailing digits reflect the approximation, not the true oval, so they show what the formula gives rather than promising that much accuracy.

A worked example | a = 5, b = 3

Say the semi-major axis is 5 and the semi-minor axis is 3.

1. Square and average: (5² + 3²) ÷ 2 = (25 + 9) ÷ 2 = 17.
2. Square root: √17 = 4.1231.
3. Multiply by 2π: 2 × π × 4.1231 = 25.9062.

So the estimate is about 25.91. Ramanujan's sharper formula gives about 25.53 for the same oval, so the simple estimate here runs roughly one and a half percent high, just as you would expect for a stretched shape.

Questions people ask

Why is there no exact formula for an ellipse's circumference?

Because the edge curves by a different amount at every point, and adding that up leads to a complete elliptic integral of the second kind, which cannot be written with ordinary algebra. So tools use approximations instead.

How accurate is this estimate?

Very good for nearly round ovals, and it reads a little high as the ellipse stretches, about 2.5 percent over on a 2-to-1 ellipse. For a tighter answer on stretched ovals, use Ramanujan's formula.

Does it work for a circle?

Yes, exactly. When the two axes are equal the formula reduces to 2 × π × radius, the circle's circumference, with no error at all.

What are the semi-major and semi-minor axes?

The two radii of the oval, measured from the centre. The semi-major axis (a) is the longer one, the semi-minor axis (b) the shorter. If you have the full widths, halve each first.

What is a more accurate formula?

Ramanujan's approximation, π × [3(a + b) - √((3a + b)(a + 3b))], which stays within about 0.04 percent even for quite elongated ellipses.

References

A note on where this comes from. The exact circumference of an ellipse is 4 × a × E(e), a complete elliptic integral of the second kind, as set out in the NIST Digital Library of Mathematical Functions, which is why no elementary formula exists and approximations are used in practice. The sharper formula mentioned above is one of those given by Srinivasa Ramanujan in his 1914 paper on approximations to π. If you want the area inside the oval instead, the area of an ellipse calculator uses the exact π × a × b. For further reading, see Ellipse.

1. National Institute of Standards and Technology (NIST), Digital Library of Mathematical Functions, perimeter of an ellipse as a complete elliptic integral of the second kind. https://dlmf.nist.gov/
2. S. Ramanujan (1914), "Modular equations and approximations to π," Quarterly Journal of Mathematics, the source of the closer ellipse perimeter approximations.