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For a given 3D domain, two unstructured tetrahedral meshes M_1 and M_2 are generated. The average edge length in M_1 is double the average edge length of M_2.

Are there estimates on the ratio of the number of tetrahedra in both meshes?

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    $\begingroup$ Wouldn't it simply be ~8x? $\endgroup$ – stali Jun 30 '13 at 23:26
  • $\begingroup$ @stali Adaptive meshes could have a factor less than that. $\endgroup$ – Shuhao Cao Jul 1 '13 at 4:05
  • $\begingroup$ @stali That would be true for hexahedral meshes. $\endgroup$ – Nico Schlömer Jul 1 '13 at 7:58
  • $\begingroup$ Why dont you quickly test in Cubit/Gmsh? $\endgroup$ – stali Jul 1 '13 at 13:13
  • $\begingroup$ @stali I played around a little with Gmsh, and found that when the average edge length is scaled by a factor of 0.5, the number of tetrahedra increases by a factor of about 6. I don't know under which conditions this holds though. $\endgroup$ – Nico Schlömer Jul 1 '13 at 18:38
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If the edge lengths decrease by a factor $k$, the number of tetrahedra will typically increase by between $k^2$ and $k^3$ depending on the adaptivity of the mesh. A purely uniform mesh will see $k^3$, just as a uniform grid of length $1/n$ has $O(n^3)$ cells. A highly adaptive mesh with full resolution near a surface will increase by roughly $k^2$, since most of the elements will occur in a band near the surface. It is also possible to see only a $k$ factor increase if the refinement is near a curve, but this is less common.

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  • $\begingroup$ What do you mean by "a purely uniform mesh"? The unit cube subdivided into hexahedra subdivided into tetrahedra? $\endgroup$ – Nico Schlömer Jul 1 '13 at 8:25
  • $\begingroup$ "Purely uniform" is perhaps an unfortunate word choice. By "uniform" I mean a mesh where the edge lengths (and ideally altitudes) of most tetrahedra are within a small constant factor of each other. $\endgroup$ – Geoffrey Irving Jul 1 '13 at 16:27

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