Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can one generate a distribution of N points over the surface of a sphere so that the all N voronoi cells have the same area? Which is the best algorithm for this?

share|cite|improve this question
You need to describe this rather more precisely. Do you want the points randomly distributed or as far apart as possible? For example, if you had two points you might want the second to be exactly opposite the first, with no randomness in that relationship. And what do you mean by distance between each pair of points to be as long as possible? For example the shortest distance between any two points, or the total sum of the squares of the distances between all pairs of points? – Henry Sep 21 '11 at 13:54
thanks, now I guess it is more clear – flow Sep 21 '11 at 14:10
I doubt that this measures what you want it to measure. For instance, if you have three points, the maximal sum of squared distances is achieved if two points are the same and the third is opposite to them (for a sum of $2(\pi r)^2$), whereas what I suspect you'd want is for the three points to form an equilateral triangle on a great circle (for a sum of $(4/3)(\pi r)^2$). – joriki Sep 21 '11 at 14:31
Related to… – lhf Sep 21 '11 at 14:32
ok, I reedited it again – flow Sep 21 '11 at 14:43

The following is a classical paper on distributing points on a sphere:

Edward Saff, Arno Kuijlaars, Distributing Many Points on a Sphere, The Mathematical Intelligencer, Volume 19, Number 1, 1997, pages 5-11.

share|cite|improve this answer
is it there any software implementation for this? – flow Sep 22 '11 at 6:03
Yes, if you search on the names of the authors and the title of the paper I gave, you'll find some implementations. – Raskolnikov Sep 22 '11 at 16:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.