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The squircle is given by the equation $x^4+y^4=r^4$. Apparently, its circumference or arc length $c$ is given by

$$c=-\frac{\sqrt[4]{3} r G_{5,5}^{5,5}\left(1\left| \begin{array}{c} \frac{1}{3},\frac{2}{3},\frac{5}{6},1,\frac{4}{3} \\ \frac{1}{12},\frac{5}{12},\frac{7}{12},\frac{3}{4},\frac{13}{12} \\ \end{array} \right.\right)}{16 \sqrt{2} \pi ^{7/2} \Gamma \left(\frac{5}{4}\right)}$$

Where $G$ is the Meijer $G$ function. Where can I find the derivation of this result? Searching for any combination of squircle and arc length or circumference has led to nowhere.

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    $\begingroup$ There's an eightfold symmetry you can exploit. Have you tried explicitly writing down the arc-length integral and seeing if you get the line integral definition of the Meijer G-function? (c.f. en.wikipedia.org/wiki/…) $\endgroup$
    – Neal
    Commented Dec 15, 2015 at 1:07
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    $\begingroup$ The squircle is a special case of a super ellipse. The arc-length intergral of a super-ellipse has been calculated here in detail: fractional-calculus.com/super_ellipse.pdf They get some series, which if you are lucky coincides with that particular Meijer G-function. $\endgroup$
    – Mankind
    Commented Dec 15, 2015 at 1:10
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    $\begingroup$ @Mankind if you can distill the relevant parts of that pdf to an answer is gladly award you the bounty! $\endgroup$ Commented Jun 11, 2022 at 13:45

2 Answers 2

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You can do this using this empirical formula to find perimeter of general super-ellipse

$$L=a+b\times\left(\frac{2.5}{n+0.5}\right)^\frac{1}{n}\times \left( b+a\times(n-1)\times\frac{\frac{0.566}{n^2}}{b+a\times\left(\frac{4.5}{0.5+n^2}\right)}\right).$$

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By your definition, $\mathcal{C} = \{(x,y) \in \mathbb{R}^{2}: x^4 + y^4 = r^4\}$. Which can be parametrized as \begin{align} \mathcal{C} = \begin{cases} \left(+\sqrt{\cos (\theta )},+\sqrt{\sin (\theta )} \right)r\\ \left(+\sqrt{\cos (\theta )},-\sqrt{\sin (\theta )} \right)r\\ \left(-\sqrt{\cos (\theta )},+\sqrt{\sin (\theta )} \right)r\\ \left(-\sqrt{\cos (\theta )},-\sqrt{\sin (\theta )} \right)r \end{cases} , \qquad 0 \leq \theta \leq \frac{\pi}{2}, \, 0<r \end{align}

enter image description here

Now, look at this curve in $\mathbb{R}^{2}_{+}$ as $y = \sqrt[4]{r^4-x^4}$, then observe that symmetry with both axis. It yields the arc length is just: $$c = 4 \int_{0}^{r} \sqrt{1+\left(\dfrac{d}{dx}\sqrt[4]{r^4-x^4}\right)^2} \,dx = 4 \int_{0}^{r} \sqrt{1+\frac{x^6}{\left(r^4-x^4\right)^{3/2}}} \,dx$$

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  • $\begingroup$ In the image, $r=1$ $\endgroup$ Commented Dec 15, 2015 at 10:13
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    $\begingroup$ Actually, the curve has eight-fold symmetry along the lines $x^2 = y^2$. We can therefore also take an arc sweep of $\frac{\pi}{4}$, and multiply that value by $8$. $\endgroup$ Commented Dec 15, 2015 at 12:12
  • $\begingroup$ Perfect addition, @JackLam. $\endgroup$ Commented Dec 15, 2015 at 12:47
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    $\begingroup$ I have done this, but I still don't see how this is equivalent to the stated result. $\endgroup$
    – Ben Longo
    Commented Dec 16, 2015 at 1:37

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