create an (almost) hexagonal mesh on an ellipsoid

EDIT I edited the question in order to take into @Kuba's comment.

I want to create this figure with Mathematica (in particular an almost hexagonal mesh on an ellipsoid; thanks to @Kuba I know this is not 100% possible).

I use the function hexTile defined by @R.M. as his reply in 39879.

hexTile[n_, m_] :=
With[{hex = Polygon[Table[{Cos[2 Pi k/6] + #, Sin[2 Pi k/6] + #2},
{k, 6}]] &},
Table[hex[3 i + 3 ((-1)^j + 1)/4, Sqrt[3]/2 j], {i, n}, {j, m}] /.
{x_?NumericQ, y_?NumericQ} :>
2 \[Pi] {x/(3 m), 2 y/(n Sqrt[3])}]


E.g.

ht = With[{ell = {7 Cos[#1] Sin[#2], 5 Sin[#1] Sin[#2], 3 Cos[#2]} &},
Graphics3D[
hexTile[20, 20] /. Polygon[l_List] :> Polygon[ell @@@ l],
Boxed -> False]]


How can we modify the function so that the distribution of hexagons and pentagons resembles closely that of the first image?

Thanks.

• @Kuba. I accept the dublicate nature of the question, but still how we can get the same output as the first image with Mathematica? Dec 4, 2015 at 11:03
• @Kuba I have edited the question. I think now it is not a duplicate anymore. Dec 4, 2015 at 11:12
• voted to reopen :) let me leave the link to 100% hexagonal mesh on sphere problem
– Kuba
Dec 4, 2015 at 11:12
• I apologise for the incovience! Thanks a lot. Dec 4, 2015 at 11:17
• No worries, we learn each day, it's not obvious it is not possible.
– Kuba
Dec 4, 2015 at 11:18

If we compute the dual polyhedron of an appropriate triangularization of a surface we can get another polygonal mesh. This is pretty much the same as Kuba's approach except the code below computes the dual polyhedron more efficiently.

The basic function iDual computes the dual of a polyhedron given by a list of coordinates and lists of faces (by the indices of their vertices in the coordinate list). (Technically, the function assumes some approximate regularity of the polyhedron and that it can be considered centered at the origin. The mean of the vertices of a face serve as the "midpoint" of the face and form the vertices of the dual. Polyhedra with folds in them are probably not going to work.) The user-level function dual translates graphics and regions into input for iDual. While there is combinatorial data for determining the ordering of vertices about a face of the dual, doing it numerically with sortvertices is both easier and faster.

ClearAll[dual, iDual, sortvertices];

sortvertices[coords_, normal_, face_] :=
With[{proj = DeleteCases[
Orthogonalize[
Join[{normal}, N@IdentityMatrix[3]]
], {0., 0., 0.}][[2 ;; 3]]},
SortBy[face, ArcTan @@ (proj.coords[[#]]) &]
];

iDual[coords_?MatrixQ, faces : {{__Integer} ..}] :=
With[{nvertices = Max@faces, nfaces = Length@faces},
With[{mat = SparseArray@ Flatten@Table[{v, f} -> 1, {f, nfaces}, {v, faces[[f]]}],
dualcoords = Mean[coords[[#]]] & /@ faces},
Graphics3D@ GraphicsComplex[
dualcoords,
Polygon[
Table[
sortvertices[dualcoords, coords[[v]], dualfaces[[v]]],
{v, Length@dualfaces}]]
]
]]];

(* user-level functions *)
dual[polyhedron : Graphics3D@GraphicsComplex[coords_, Polygon[faces_]]] :=
iDual[coords, faces];
dual[polyhedron_?MeshRegionQ /;
RegionDimension[polyhedron] == 2 && RegionEmbeddingDimension[polyhedron] == 3] :=
iDual[MeshCoordinates[polyhedron], MeshCells[polyhedron, 2] /. Polygon -> Sequence];
dual[polyhedron_?BoundaryMeshRegionQ /; RegionDimension[polyhedron] == 3] :=
iDual[MeshCoordinates[polyhedron], MeshCells[polyhedron, 2] /. Polygon -> Sequence];


Here is Kuba's example, with the dual vertices projected back onto the ellipsoid:

dual@DiscretizeRegion[Sphere[], MaxCellMeasure -> .02] /.
GraphicsComplex[pts_, stuff___] :>
GraphicsComplex[(Normalize /@ pts).DiagonalMatrix[{1, 2, 3}], stuff]


Note region functions do not always produce an appropriate triangularization:

dual@ DiscretizeRegion[Sphere[]]
dual@ DiscretizeRegion[Sphere[], MaxCellMeasure -> {"Area" -> 0.01}]


% /. GraphicsComplex[pts_, stuff___] :>
GraphicsComplex[(Normalize /@ pts).DiagonalMatrix[{1, 2, 3}], stuff]


It seems impossible to use region functions directly on Ellipsoid:

dual@ BoundaryDiscretizeRegion[Ellipsoid[{0, 0, 0}, {1, 2, 3}], MaxCellMeasure -> .5]
dual@ BoundaryDiscretizeRegion[Ellipsoid[{0, 0, 0}, {1, 2, 3}],
MaxCellMeasure -> {"Area" -> 0.03}]


It works on other polyhedra, too.

GraphicsRow[{#, dual@#}] &@ PolyhedronData@ "TruncatedDodecahedron"


• Thank you, Michael. I've used this for triangulated spherical tilings and it works well. +1 Aug 26, 2018 at 17:44

Very inefficient but short:

I assumed that DiscretizeRegion of a Sphere gives us a mesh that have 5 or 6 triangles at each vertex.

 ms = DiscretizeRegion[Sphere[], MaxCellMeasure -> .01];

(*groups of polygons with one common vertex*)
data = Sow[#, #[[1]]] & /@ MeshCells[ms, 2] // Reap // #[[-1, All, All, 1]] &;

data0 = MeshCoordinates[ms];

(*this function replaces polygons which have common vertex
with polygon composed of those polygons centroids.*)
reshape[list_] := #[[FindShortestTour[#][[2]]]] &@ Map[Mean[data0[[#]]] &, list]

Polygon[reshape /@ data] // Scale[#, {1, 2, 3}] & // Graphics3D