Given x points that must be on the surface of a sphere, how should they be placed so they are a maximum distance from each other?

Question

Given x points that must be on the surface of a sphere, how should I place them so they are the maximum distance from each other?

I'm attempting to create a script that will place any given number of objects on the surface of a sphere. I think the simplest way to represent the position of each point would be two degrees, the first representing horizontal position and the second the vertical. (0, 0) would the highest point, (90, 90) would the leftmost point, (270, 90) the rightmost point, etc.

(x, y) would be like:

Is there a way to get the positions for each point on the sphere given the number of total points? For example, if I had 2 points, the furthest they could be is (0, 0) and (180, 180). 6 would have a point on each of the furthest points (top, bottom, front, back, left and right)

2 points example:

Is this possible? Also is there a better way to do the coordinate system?

Answer 1

This is a very active area of mathematics. This is one of the recent papers:

Hardin, D. P.(1-VDB-CAX); Michaels, T. J.(1-VDB-CCA); Saff, E. B.(1-VDB-CAX) Asymptotic linear programming lower bounds for the energy of minimizing Riesz and Gauss configurations. (English summary) Mathematika 65 (2019), no. 1, 157–180.

Answer 2

This is known as Fejes Tóth's Problem. The general problem is unsolved; configurations are known for some values of $x$.

Known examples of such configurations:

• 2 points at the opposite poles

• 3 points at the vertices of an equilateral triangle on any great circle

• 4 points at the vertices of a regular tetrahedron

• 6 points centered on each face of a cube

• 8 points at the vertices of a square antiprism with equal edges (not a cube!)

• 12 points centered on each face of a regular dodecahedron

• 20 points at the vertices of a regular dodecahedron, or centered on each face of a regular icosahedron

If you know of any further examples, please add them to this list.