If you take a $sin$ function and use it as the "x" coordinates in a cartesian plane, and you use an out-of-phase $sin$ function for "y" coordinates, you will obviously get the unit circle as shape, if you use $\pi/2$ as the phase for the y-coordinate.
I was playing around with the "wave" function that maps to the shape of a circle, rather than a $sin$ wave (meaning, it has a discontinuous first derivative), and looked what came out of that, instead of the circle coming out of the $sin$ wave.
I have three questions:
- Is this a squircle (superellipse)? As per first definition of http://mathworld.wolfram.com/Squircle.html
- How is this way of constructing 2D shapes called so I can learn more about them?
- If you are interested, I am mostly interested in doing this recursively. Meaning, in this case I would take the squircle shape, create a wave from it, and again create a shape, and keep doing this.. Is there any knowledge about this field?
Let us focus on the first quadrant, in which the analogue of $\cos$ is $x=\sqrt{1-t^2}$ for $t\in[0,1]$, and the analogue of $\sin$ is $y=\sqrt{1-(1-t)^2}$. For it to be a squircle, we would need $x^4+y^4=\mathrm{const}$, but one can check that this does not hold for the given curve.
However, expressing $t$ in terms of $x$ and $y$, we have $$ t=\sqrt{1-x^2}, \\ 1-t=\sqrt{1-y^2}, $$ so the implicit equation of the curve, at least in the first quadrant, is $$\sqrt{1-x^2}+\sqrt{1-y^2}=1.$$ It turns out that this equation also works in the other quadrants too.
The implicit equation can be expanded to the equivalent form $$x^4 + y^4 - 2(1 - x^2)(1 - y^2) = 1.$$ Here the difference with the squircle equation $x^4+y^4=1$ is clear.