An oval is like an elongated circle, and mathematicians often call it an ellipse. These shapes are found everywhere, from racetracks to windows to planetary orbits. The difficulty in measuring the circumference of an oval is that the calculation is based on two vertexes known as the foci, instead of one central point, as with a circle. There are no exact formulas to solve for the exact circumference of an oval, but some are more accurate than others.

## Major and Minor Axis

The most noticeable part of an oval is what differentiates it from a circle: It lacks a constant radius. Instead, ovals can be broken down into two key measurements -- the major axis and the minor axis. The major axis spans the length of the oval, running through the center and connecting the two farthest points. The minor axis sits perpendicular to the major and connects the two closest points. For purposes of calculation, the distance from the center of the oval to the farthest point across the major axis is represented as "A," and the distance from the center to the closest point across the minor axis is represented as "B."

## The Basic Formula

One way to measure the perimeter or circumference of an oval involves finding the average radius of the oval and using it in the same way the circumference of a circle is discovered. The formula to find the circumference of a circle is P = 2πr, where "P" represents the perimeter, or length of the circumference. "R" represents the radius, the distance from the center of the circle to the outside. π is the symbol for pi, a universal constant that, when multiplied by the diameter of the circle (the distance across through the center, also represented by 2 times the radius), calculates the distance around a circle.

Finding the average radius of an oval uses the same principles as looking for the hypotenuse -- the longest side -- of a right triangle: the Pythagorean theorem. This states that a single side, A, plus the other side, B, is equal to the hypotenuse, C. As a formula, it appears as A^2 + B^2 = C^2.

To solve, both A and B are multiplied by themselves or "squared"; the sum of those two numbers is equal to the third side squared. To find the actual value of C, the square root of the sum is found. This is symbolized in formulas as "√" and simply calculates what number must be multiplied by itself to attain a certain product.

Using the above information, the major and minor axes are used to represent two sides of a triangle and solve for their hypotenuse using the Pythagorean theorem by squaring them and adding them together.

Because the radius only represents the distance from the center to the edge and the major and minor axes span the length of the oval, you must divide the sum of their squares by 2 and then find the square root, √, of the dividend. When these calculation are substituted into the formula for finding the area of a circle, the results should look like this:

- P = 2π√((A^2 + B^2)/2)

While it creates a decent approximation in most circumstances, this method becomes highly inaccurate when dealing with flatter ovals, those with a major axis at least four times longer than the minor axis.

## Ramanujan's Formula 1

The second formula is a bit longer and slightly more accurate:

P = π[3(A + B) - √((3A + B) * (A + 3B))]

The two polynomials presented expand to 3A^2 + 10AB + 3B^2. Finding the square root of this is similar in approach to finding the average length in the previous formula, but it's then subtracted from three times the sum of the major and minor axes before multiplying by pi. The purpose for this formula was to deal with the "eccentricities" of an oval, or how far it can stretch in either direction.

## Ramanujan's Formula 2

The next formula formula uses an additional variable, "h," which replaces earlier calculations on how to find the average of the major and minor axes and compensate for the margin of error.

h = (a - b)^2/(a + b)^2

It is then implemented in the following formula to calculate perimeter:

P = π(a + b)[1 + (3h/(10 + √(4 - 3h))]

This formula comes substantially closer than the others and is comparable to using the approximation of 7/22 in place of pi, leaving the margin of error less than 0.05 percent.