Geodesic distance between equidistant points on a sphere

Question

On the unit sphere equidistant points can be found for $1, 2, 3, 4, 6, 8, 12, 20$. The geodesic distance between the points are $\pi$ for $2$, $2\pi\over 3$ for $3$, $\pi\over 2$ for $6$, etc... Is there a natural formula that kind of yields the distance for every $n$, even if the points can't be equidistantly placed?

Answer 1

you want the article by Saff and Kuijlaars. This problem is traditionally called the Tammes problem. https://perswww.kuleuven.be/~u0017946/publications/Papers97/art97a-Saff-Kuijlaars-MI/Saff-Kuijlaars-MathIntel97.pdf

Asymptotically, for $n$ vertices on the sphere, one expects geodesic distance to the closest neighbors at about $4/\sqrt n.$

By hand calculation, we cannot expect a packing to have a larger proportion of the area of the sphere than circles in a hexagonal packing in the plane. This observation suggests a geodesic radius of $1.9046 / \sqrt n,$ or pairwise distance of nearest neighbors $3.809 / \sqrt n.$