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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.

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  • $\begingroup$ I don't understand you question: uniformity is a property of the grid itself, not of the surface to which it is applied. Or do you mean a conformal transformation like between the complex z- and w- planes? $\endgroup$
    – Lucozade
    Commented Jul 8, 2013 at 20:32
  • $\begingroup$ What I am asking is: How would you go about dividing the disk into small "boxes" of equal area, where the "diameter" of each box is small (i.e. you can't just use arbitrarily thin arcs, since the diameter is not small in that case). $\endgroup$ Commented Jul 8, 2013 at 20:34

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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 $$ f(x) = \frac 1 2 (\arcsin x + x\sqrt{1-x^2}). $$

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  • $\begingroup$ Would this result in a uniform grid on circle, if starting with uniform grid on square ? $\endgroup$ Commented Jul 8, 2013 at 20:36
  • $\begingroup$ No. I don't understand what your mean with "uniform" $\endgroup$ Commented Jul 8, 2013 at 20:38
  • $\begingroup$ It means each subdivision has equal area. $\endgroup$ Commented Jul 8, 2013 at 20:38
  • $\begingroup$ Do you mean that the "infinitesimal" grid has equal area? Otherwise the map will depend on the size of the grid... $\endgroup$ Commented Jul 8, 2013 at 21:09
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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.

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