Uniform grid on a disc Do there exist any known methods of drawing a uniform grid on a disk ? I am looking for a map that converts a grid on a square to a grid on a disk.
 A: There are many possibilities to map a square on a disk. For example one possibility is:
$$
\phi(x,y) = \frac{(x,y)}{\sqrt{1+\min\{x^2,y^2\}}}
$$
which moves the points along the line trough the origin.
If you also want the map to mantain the infinitesimal area, it's a little bit more complicated. One possibility is to look for a map $\phi(x,y)$ which sends the circle to a rectangle by keeping vertical the vertical lines. On each vertical band you can subdivide the strip in equal parts.
This means that you impose $\phi(x,y) = (f(x),y/\sqrt{1-x^2})$. The condition that the map preserves the area becomes:
$$
\frac{f'(x)}{\sqrt{1-x^2}} = 1
$$
i.e. $f'(x) = \sqrt{1-x^2}$ which, by integration, gives
$$
f(x) = \frac 1 2 (\arcsin x + x\sqrt{1-x^2}).
$$
A: Fix $M, N\in \mathbb N$.
The area of the annulus between $r=\sqrt {\frac mM}$ and $r=\sqrt {\frac {m+1}M}$ is $\frac \pi M$ for $m=0,\ldots, M-1$. Now divide everything into $N$ sectors like a pie. Then each part has area $\frac{\pi}{MN}$ and the diameter is $\sim \frac1{\sqrt M}$ for parts near the center and $\sim\frac{2\pi}N$ for parts near the boundary. Choosing $M,N$ such that $M\approx \frac{N^2}{4\pi^2}$ somewhat balances this. Unfortunately area by diamter squared is like $\frac \pi N$, so the shapes become far from square-like, so even though the diameter tends to $0$ this may not be what you really want.
