Polar form of a superellipse?

What is the polar form for a superellipse with semidiameters $a$ and $b$, centered at a point $(r_0, θ_0)$, with the $a$ semidiameter at an angle $\varphi$ relative to the polar axis?

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1 Answer

For a regular ellipse oriented with the major axis along $x$, Wikipedia gives $r=\frac{ab}{\sqrt{(b\cos \theta)^2+(a \sin \theta)^2}}$. To rotate this by $\theta_0$, just subtract from $\theta$ giving $r=\frac{ab}{\sqrt{(b\cos (\theta-\theta_0))^2+(a \sin (\theta-\theta_0))^2}}$. To make it a superellipse, just change the exponent to $n$ giving $r=\frac{ab}{\sqrt[n]{|b\cos (\theta-\theta_0)|^n+|a \sin (\theta-\theta_0)|^n}}$. Translations are hard in polar coordinates, so I would give up and go to Cartesian.

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Insert absolute value signs where needed... :) –  Ｊ. Ｍ. Oct 26 '11 at 17:09
@J.M.: Good point. Inserted –  Ross Millikan Oct 26 '11 at 17:22