# How to get a larger triangle when we discretize a sphere?

In order to solve this problem,I want to get larger triangles (Of course less amount). But I don't know how to do this. I have tried MaxCellMeasure and MeshRefinementFunction, but this doesn't work

DiscretizeRegion[Sphere[], MaxCellMeasure -> 10]


and

DiscretizeRegion[Sphere[],
MeshRefinementFunction -> Function[{vertices, area}, area > 10]]


Always gives

Any suggestions ?

• There's a limit, of course: the tetrahedron. – J. M. will be back soon Mar 31 '17 at 5:51
• @J.M. You mean we cannot get another mount triangle? – yode Mar 31 '17 at 5:55
• Think about it: can you cover a sphere with three triangles? – J. M. will be back soon Mar 31 '17 at 5:56
• @J.M. Fun,It's seem this answer response your question..:) – yode Mar 31 '17 at 5:58
• That's an icosahedron (twenty triangles), so still far away from the limit of four. ;) – J. M. will be back soon Mar 31 '17 at 6:00

a = BoundaryDiscretizeRegion[Ball[{0, 0, 0}, 1],
MaxCellMeasure -> {"Length" -> 3}, PrecisionGoal -> 0.01]


Length determines the size of the triangulation

which gives:

• Amazing,I don't know we can use MaxCellMeasure like this.. – yode Mar 31 '17 at 5:59
• I still haven't figured out how to force it to show an octahedron, much less a tetrahedron. – J. M. will be back soon Mar 31 '17 at 5:59
• Why not DiscretizeRegion[Sphere[]...] :) – yode Mar 31 '17 at 6:01
• True ;) i.e. DiscretizeRegion[Sphere[{0, 0, 0}, 1], MaxCellMeasure -> {"Length" -> 3}, PrecisionGoal -> 0.01] would also work – Dunlop Mar 31 '17 at 6:04
• or DiscretizeRegion[Sphere[], MaxCellMeasure -> 1, PrecisionGoal -> 1] – halmir Mar 31 '17 at 15:14