6
$\begingroup$

My goal is to put $n$ points on a sphere in $\mathbb{R}^3$ to divide it in $n$ parts, so that their disposition would be as "equivalent" as possible. I don't exactly know what "equivalent" mathematically means, perhaps that the min distance between two points is maximal.

Anyway in $2$ dimensions it is simple to divide a circle in $n$ parts. In $3$ dimensions I can figure out some good re-partitions for particular values of $n$ but I lack a more general approach.

  • 3
    Related: http://math.stackexchange.com/questions/9846/which-tessellation-of-the-sphere-yields-a-constant-density-of-vertices/, http://math.stackexchange.com/questions/31619/well-separated-points-on-sphere, http://math.stackexchange.com/questions/51009/optimal-number-of-points-for-integration-over-the-surface-of-a-sphere2011-08-17
  • 0
    Given $n$ points on a circle $S^1$ these points divide $S^1$ in $n$ parts in an obvious way, and it is easy to describe a configuration where these points are "equivalent". On a $2$-sphere this is another matter. If your $n$ points are the vertices of a regular $n$-gon on the equator they are certainly "equivalent" insofar as there is a group of isometries of $S^2$ permuting these points transitively. But perhaps you have something else in mind. In any case, $n$ points on $S^2$ do not divide $S^2$ into $n$ parts without further ado.2011-08-17
  • 0
    The ones cited by Rahul Narain may be better, but you could also look at http://math.stackexchange.com/questions/11499/possible-to-imitate-a-sphere-with-1000-congruent-polygons/11512#115122012-04-14

2 Answers 2