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I have googled this and not come up with an answer yet, but basically, I'm trying to find out the distance between each point or vertice on a sphere (all points are evenly spaced). I already know this information, but I want to figure it out using math.

Radius = 1
Number of points = 382
Distance between points: 0.156918

I have tried calculating the surface area, then dividing that by the number of points, is that correct?

(4 * pi * sqrt(radius)) / 382

but no luck, assuming that my idea is correct, should finding the surface area, divided by the number of points give me the distance between each?

Shannon

Willie Wong
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Shannon Hochkins
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    What do you mean by "all points are evenly spaced"? This is usually hard to achieve on a sphere (Which tessellation of the sphere yields a constant density of vertices?, How to tile a sphere with points at an even density?). –  Feb 28 '13 at 02:20
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    No. That would give you the average surface area per point. Additionally, your formula for surface area is wrong. It should be $4\pi r^2$, not $4\pi\sqrt{r}$ – SSumner Feb 28 '13 at 02:20
  • That second question link he gave you should point out the issue: "There is also a common confusion between uniform random distribution and evenly-spaced distribution, which are very different things" – SSumner Feb 28 '13 at 02:22
  • @RahulNarain okay, basically I'm trying to get my calculation to arrive at "0.156918" (the distance between points), I know the number of points in the sphere, I know the radius, is there any way of calculating this? – Shannon Hochkins Feb 28 '13 at 02:32
  • How did you get the number $0.156918$ in the first place? And where does your distribution of points come from? –  Feb 28 '13 at 02:50
  • @RahulNarain Because I can measure the distance manually in maya by selecting the two points, it then gives me the value of 0.156918. – Shannon Hochkins Feb 28 '13 at 03:00
  • Do you still get exactly the same value if you select a different pair of points? And where does your distribution of points come from? –  Feb 28 '13 at 03:05
  • it would be the distance from any 2 points immediately adjacent to each other in any direction, not for example one point on one side, to another on the other side. What I'm trying to calculate is the distance between to points, right next to each other so that I can space all the points evenly. – Shannon Hochkins Feb 28 '13 at 03:11
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    There's no such thing as equally or evenly spaced points on a sphere in general; there are only some special configurations like the vertices of Platonic solids that could be considered equally/evenly spaced. – joriki Feb 28 '13 at 23:37

2 Answers2

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For a large number of points, the difference between a sphere and a plane will not be important, and you can equidistribute the points on a plane. What pattern are you using? Equilateral triangles, squares, and hexagons are all candidates. For a hexagonal pattern, there will be half as many hexagons as points, so the area of a hexagon is $A=\frac {8 \pi r^2}{n}$ where $n$ is the number of points. The side will then be $s=\sqrt{\frac {2A}{3\sqrt 3}}=\sqrt{\frac {16 \pi r^2}{3n\sqrt 3}}$ per Wikipedia. You can do a similar calculation for the other lattices. The distance will again decrease as the square root of $n$, but the constant will be somewhat different.

Ross Millikan
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As a very rough approximation to pairwise distances, the following can be used: imagine the sphere were perfectly packed by $N$ disks of radius $r$. The area of each such disk is approximately $\pi r^2$ (approximately because they are not really flat, but this is a small source of error). The total area of the sphere is $4\pi R$ where $R$ is the radius. Therefore, $N\pi r^2 \approx 4\pi R^2$, and we get a ballpark figure for pairwise distances by doubling the radius $r$: $$ \text{pairwise distances}\approx 2r \approx 4R/\sqrt{N} \approx 0.2 \tag1$$ In reality, $4R/\sqrt{N}$ is an overestimate. We should at least try to account for the fact that disks don't pack perfectly. The densest packing of equal disks in the plane has density $\pi\sqrt{3}/6$. If we multiply the area $4\pi R^2$ by this factor, the new estimate is $$ \text{pairwise distances}\approx 4\cdot 3^{1/4}R\sqrt{\frac{\pi}{6N}} \approx 0.195 \tag2$$ Well, that did not help much, but at least we now have a sophisticated-looking formula (2).