The Math Forum

Ask Dr. Math: FAQ

   Circumference of an Ellipse   

Geometric Formulas: Contents || Ask Dr. Math || Dr. Math FAQ || Search Dr. Math

[Back to Ellipse & Parabola Formulas]


The circumference C of an ellipse must be computed using calculus. To do this, we set up a Cartesian coordinate system. We put the origin at the center of the ellipse, the x-axis along the major axis, whose length is 2a, and the y-axis along the minor axis, whose length is 2b. The eccentricity e is defined by
0 <= e = sqrt(a2-b2)/a < 1.
Then the equation of the ellipse is
x2/a2 + y2/b2 = 1,  a >= b > 0.
Now the formula for computing the arc length of any curve given by the parametric equations x = f(t), y = g(t), over the range c <= t <= d is
s = INTEGRAL  sqrt[(dx/dt)2+(dy/dt)2] dt.


For the above ellipse, we can use the parametric equations
x = a sin(t),
y = b cos(t),
0 <= t <= 2,
C = s,
dx/dt = a cos(t),
dy/dt = -b sin(t).
C = INTEGRAL  sqrt(a2cos2[t]+b2sin2[t]) dt.
We can use the 4-fold symmetry of the ellipse to rewrite this as
C = 4 INTEGRAL sqrt(a2cos2[t]+b2sin2[t]) dt.
This can be further rewritten, using cos2(t) = 1 - sin2(t), and the definition of the eccentricity e, as
C = 4a INTEGRAL   sqrt(1-e2sin2[t]) dt.
Now this integral is a famous one. It is called a "complete elliptic integral of the second kind." It is one of those integrals that cannot be expressed in closed form in terms of the familiar functions of calculus, except if e = 0, when we have a circle. That means that this is the simplest formula possible for the circumference of a general ellipse.

Another way to write this is

C = 4aE(/2,e).


This integral can be evaluated numerically, of course. Another way to compute its value is using an infinite series. Set
x = (a-b)/(a+b).
C = (a+b)(1 +  SUM [(2n-2)!/(n![n-1]!22n-1)]2x2n),

C = (a+b)(1 + x2/4 
             + x4/64 
             + x6/256 
             + 25x8/16384 + ...).
This series converges pretty rapidly, especially when x is small, that is, when a and b are close together, that is, when e is small.


If a = 15, b = 6, e = sqrt(21)/5, and x = 3/7, you find
C = (15+6)(1 + 9/196 
              + 81/153664 
              + 729/30118144 
              + 164025/94450499548 
              + 2893401/18512297918464 + ...),
  = 21(1 + 0.045918367 
          + 0.000527124 
          + 0.000024205
          + 0.000001737 
          + 0.000000156 + ...),
  = 21(3.1415926536)(1.046471589),
  = 69.039336580,
approximately. With six terms, we get 7 significant figures of accuracy for this value of x (the correct answer being 69.03933778699452855...), even though a and b are not close together (and e = 0.916515... is not very small).


The approximate formula

C = 2 sqrt((a2+b2)/2)
can be found in Mathematical Tables from the Handbook of Chemistry and Physics, 10th ed. (1954), p. 315.

The idea seems to be to use 2r, but for r use the root-mean-square of the semi-major and semi-minor axes. Then a > r > b, so one shouldn't be too far off.

In fact,

2 sqrt[(a2+b2)/2] = (a+b)(1 + x2/2 - x4/8 + ...),
so the first order term is right, and the second order term is double what it should be, so this is not too awful an approximation, but not very good, either. It's about as good as using r = (a+b)/2.

Better would be to use

C ~~ (sqrt[(a2+b2)/2] + [a+b]/2),
   = (a+b)(1 + x2/4 - x4/16 + ...),     
which agrees in both first and second order terms. If x4/16 is negligibly small, then this gives right answers, whereas the original approximation does only if x2/4 is negligible.

A short search of the Dr. Math archives turned up the following approximation due to Ramanujan:

C ~~ (3a + 3b - sqrt[(a+3b)(b+3a)])

This is even better, because

(3a + 3b - sqrt[(a+3b)(b+3a)]) = (a+b)(1 + x2/4 + x4/64 + x6/512 + ...),
which agrees even in the third-order term!

However, in the same paper [Ramanujan, S., "Modular Equations and Approximations to ," Quart. J. Pure. Appl. Math., vol. 45 (1913-1914), pp. 350-372], he gives another, even better approximation:

C ~~ (a+b)(1+3x2/[10+sqrt(4-3x2)])
which has a relative error of about (3/217)x10 for small values of x, since this function has series expansion
(a+b)(1 + x2/4 
         + x4/64 
         + x6/256 
         + 25x8/16384 
         + 95x10/131072 + ...),
agreeing with the actual series for C through terms of the fifth order, and even having sixth order term close to right!

For a relatively compact formula, this is clearly the winner, although it's rarely clear how Ramanujan derived his formulas. Anyone interested enough to read this far would probably greatly enjoy taking a detour to read about his fascinating life and contributions to mathematics in the MacTutor History of Mathematics archive.

Contributed by "Dr. Rob," Robert L. Ward

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. Math ®
© 1994- The Math Forum at NCTM. All rights reserved.