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We all know that the curvature of a circle is defined by the equation $$k=\frac{1}{r}$$

What about ellipses?

In terms of major axis $a$, minor axis $b$, $x$ and $y$, what is the curvature of an ellipse?

Thanks lots!

soupynoodles
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3 Answers3

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$$x=a\cos(t),y=b\sin(t)$$ $$\dot x=-a\sin(t),\dot y=b\cos(t)$$ $$\ddot x=-a\cos(t),\ddot y=-b\sin(t)$$

$$\kappa=\frac{\dot x\ddot y-\ddot x\dot y}{(\dot x^2+\dot y^2)^{3/2}}=\frac{ab}{(a^2\sin^2(t)+b^2\cos^2(t))^{3/2}}=\frac{ab}{((\frac ab y)^2+(\frac bax)^2)^{3/2}}$$

  • Thanks @YvesDaoust. So if I were to plub in x and y coordinates along with the major and minor axes, I get the curvature of the ellipse at that coordinate, yes? – soupynoodles Sep 26 '15 at 10:57
  • Hem, what else ? –  Sep 26 '15 at 11:07
  • Righto... Please don't get too mad at a mere high schooler. Just one last question - where do you get that formula for k?
    Actually one more thing - can you show to generalize this for a circle?     – soupynoodles Sep 26 '15 at 11:09
  • Have you heard of Google ? –  Sep 26 '15 at 12:05
  • @soupynoodles just put $a=b$ and $ x^2+y^2 =a^2$ in Yves Daoust's formula. – Narasimham Sep 26 '15 at 13:33
  • Just as a derivative gives you the “instantaneous slope” of a curve by finding the line that best approximates the curve at a point, the above gives you “instantaneous curvature” by finding the circle that best approximates the curve at a given point. – amd Sep 26 '15 at 21:23
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You can find the curvature in parametric form here, tag 11.

Narasimham
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The second answer is the maximum principal curvature, as explained on pages 71-76 of "Vector and Tensor Analysis" by Harry Lass, McGraw-Hill (1950). The denominator is G^{3/2}, where G is the third term in the first fundamental form for the Ellipse. The minimal principal curvature is c / ( a \sqrt{G}).

Van Snyder
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