Saturday, October 29, 2011

Math Problem: Evenly spaced points on a sphere - Part 1

If place points around the circumference of a circle, it is not difficult to position them so that they're evenly spaced.

regular polygons

32π / 3120°triangle
4π / 290°square
52π / 572°pentagon
n2π / n360 / nn-gon

But suppose you want to position points around the surface of a sphere so that they're evenly spaced.

What is the angle between adjacent points? Is it possible to position the points so that the angle is the same between all adjacent points on the sphere?

Let's look at some examples. For n=4, you have a tetrahedron.

According to wikipedia, the angle between adjacent points is arccos(-1/3) or about 109.4712°.

For n=6, you have an octahedron. The angle between adjacent points is π / 2 or 90°.


To summarize and expand:

nfacestri. facesedgestri. edgesangleanglename
6881212π / 290°octahedron
86121218multiple anglescube
12202030302 arctan(2/(1+√5))63.435°icosahedron

So what's the answer? I don't know, but one thing we have learned is that a cube's points are not all evenly spaced. The angles between them follow a regular pattern. There are twelve edges with an angle between the points of 2 arcsin(1/√3) ≈ 70.528°, and if you include all triangle edges, there are six edges with an angle between the points of 2 arcsin(√2/√3) ≈ 109.4712°.

Continue reading Part 2