Polar form of a superellipse?

Question

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?

Answer 1

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.

Answer 2

For the ones still interested, two forms of a super ellipse in Cartesian and polar form:

Cartesian non-continuous:

$y=W\cdot \left (1-\dfrac{1}{L^{n}}\cdot \left |x-xextra\right |^{n}\right )^{\dfrac{1}{n}}$

Cartesian continuous:

$y=W\cdot \left |1-\dfrac{1}{L^{n}}\cdot \left |x-xextra\right |^{n}\right |^{\dfrac{1}{n}}$

Polar

$r = L\cdot W\cdot \dfrac{1}{\left (\left |L\cdot sin(\theta)\right |^{n}+\left |W\cdot cos(\theta)\right |^{n}\right )^{\dfrac{1}{n}}}$

Of which:

W = width along y axis

L = length along x axis

xextra = shift the center of the ellipse along the x-axis