I have an application where I need to determine the curvature of a superellipse at various points along the curve, where the Cartesian form of the superellipse is defined as: $$\frac{x^n}{a^n}+\frac{y^n}{b^n}=1$$ For an ellipse, the Cartesian equation for the curvature $\kappa$ is easy to find:
$$\kappa= \frac{1}{a^2b^2}\left(\frac{x^2}{a^4}+\frac{y^2}{b^4}\right)^{-{\frac32}}$$ but I have searched and searched and cannot find the Cartesian equation for superellipse curvature.
Could someone point me to a site that has this information?
Using Hessian matrix for implicit function $F(x,y)=0$,
\begin{align} F(x,y) &= \frac{x^n}{a^n}+\frac{y^n}{b^n}-1 \\ F_x &= \frac{nx^{n-1}}{a^n} \\ F_y &= \frac{ny^{n-1}}{b^n} \\ \nabla F &= \begin{pmatrix} F_{x} \\ F_{y} \end{pmatrix} \\ F_{xx} &= \frac{n(n-1)x^{n-2}}{a^n} \\ F_{yy} &= \frac{n(n-1)y^{n-2}}{b^n} \\ F_{xy} &= 0 \\ \mathbb{H}(F) &= \begin{pmatrix} F_{xx} & F_{xy} \\ F_{xy} & F_{yy} \end{pmatrix} \\ \kappa &= \frac{ \begin{vmatrix} \mathbb{H}(F) & \nabla F \\ (\nabla F)^T & 0 \end{vmatrix}}{|\nabla F|^3} \\ &= \frac{-n^3(n-1) \dfrac{x^{n-2} y^{n-2}}{a^n b^n} \left( \dfrac{x^n}{a^n}+\dfrac{y^n}{b^n} \right)} {n^3\left( \dfrac{x^{2n-2}}{a^{2n}}+\dfrac{y^{2n-2}}{b^{2n} }\right)^{3/2}} \\ &= -\frac{(n-1) x^{n-2} y^{n-2}} {a^n b^n\left( \dfrac{x^{2n-2}}{a^{2n}}+\dfrac{y^{2n-2}}{b^{2n} }\right)^{3/2}} \\ \end{align}
For $n>1$, $\kappa<0$ implies the centre of curvature being on another side of the outward normal.