4
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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 ?

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

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

Length determines the size of the triangulation

which gives:

discretised surface

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

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