What is the formula for the perimeter of an ellipse?

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    Ellipse Perimeter
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SUMMARY

The perimeter of an ellipse cannot be expressed with an exact algebraic formula; instead, it requires numerical integration or approximations. The commonly referenced approximation is given by the formula 2π√((1/2)(a²+b²)), where a and b are the semi-major and semi-minor axes, respectively. For high eccentricities (0.9-1.0), the accuracy of this approximation may vary. Numerical integration can be performed using the formula 4a∫₀^(π/2)√(1-e²sin²t)dt, where e is the eccentricity.

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  • Understanding of ellipse properties, specifically semi-major and semi-minor axes
  • Familiarity with calculus, particularly integration techniques
  • Knowledge of eccentricity and its implications on ellipse shape
  • Basic understanding of numerical methods for approximating integrals
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  • Research the Elliptic Integral and its applications in calculating ellipse perimeter
  • Learn about numerical integration techniques for approximating complex integrals
  • Explore more accurate approximation formulas for ellipse perimeter
  • Investigate the impact of eccentricity on ellipse properties and perimeter calculations
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Mathematicians, physics students, engineers, and anyone involved in geometric calculations or ellipse-related studies will benefit from this discussion.

joecoss
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Could anyone direct me to an analytically correct algebraic formula for the Perimeter of an Ellipse based on either the eccentricity or the Semi-Major and Semiminor Axes other than the Elliptic Integral ? If so, how accurate will it be for relatively high eccentricities such as 0.9-1.0 ? Thanks.
 
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Given those properties of a given ellipse, you should be able to define a function, f(x) which describes the top half of the ellipse. The perimeter of the ellipse would be twice the length of f(x) on the interval on which it exists. The length of f(x) on that interval, let's call it [-a,a], is:

L = \int _{-a} ^a \sqrt{1 + [f'(x)]^2}dx
 
To see why this works, think of f'(x) as dy/dx. Now, put the "dx" under the square root, and you'll get:

L = \int _{-a} ^a \sqrt{{dx}^2 + {dy}^2}

Now, if you consider an infinitessimal piece of the function, you can treat it as a straight line segment. If you think of this segment as the hypoteneuse of a triangle with sides dx and dy, then clearly, the length of this hypoteneuse is the integrand. Sum the lengths of these tiny segments over the desired interval, and you get the length of the function on that interval.
 
I have a handbook that lists the perimeter of an ellipse as approximately:

2\pi\sqrt{\frac{1}{2}(a^2+b^2)}

a and b are the semi-major and semi-minor axes, respectively. No idea on the accuracy.
 
OK I am with you on the Arclength Formula, do you know the function if given a (S-maj), b (S-minor), or c (Focus to center) ? Thanks a lot.
 
Won't the arclength formula lead to the elliptic integral? You already said you don't want that.

Edit: There is no exact solution for the perimeter of an ellipse. You either have to numerically integrate this:

4a\int_0^{\pi/2}\sqrt{1-e^2\sin^2 t}\,dt}

(where e is the eccentricity)

Or use an approximation like the one I gave in my earlier post.

Edit Edit: This page seems to have some better approximations listed at the bottom:

http://mathforum.org/dr.math/faq/formulas/faq.ellipse.circumference.html
 
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