Calculate Spherical Distance between points

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

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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?). – Rahul Feb 28 '13 at 2:20
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 2:20
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 2:22
@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 2:32
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

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.
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).