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I know a simple equation for a squircle, $x^4+y^4=a^4$

a squircle

What would be the equation for an hexagon with rounded corners?

In the equation, how can I take in account for the angle of rotation of the hexagon about its center?

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The following equation gives a rounded hexagon where $\theta$ can be changed in order to give any rotation of the hexagon about its centre and $r$ can be changed to affect the roundness of the hexagon itself. I found that $r=15$ works well. The value $10$ at the end of the equation can be increased to increase the size of the hexagon as well. $$\sum_{n=1}^6 \Big|x\cos{\Big(\frac{\pi n}{3}+\theta\Big)}+y\sin{\Big(\frac{\pi n}{3}+\theta\Big)}-\frac{1}{6}\Big|^r=10$$

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polar $$ r = 5 + \frac{1}{5} \cos (6 \theta) $$ looks very good. You can solve as rectangular using $x = r \cos \theta$ and $y = r \sin \theta$

The 6 is fixed, but you may adjust the ratio of coefficients as you like, for visual effect.

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