Skip to main content

graphics - 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).


enter image description here


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]]

enter image description here


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


Thanks.




Answer



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},
With[{dualfaces = mat["AdjacencyLists"]},
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]


Mathematica graphics


Note region functions do not always produce an appropriate triangularization:


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

Mathematica graphics Mathematica graphics


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


Mathematica graphics


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}]

Mathematica graphics Mathematica graphics


It works on other polyhedra, too.


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


Mathematica graphics


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

Is there a way to do conditional matrix loop using 'continue'

I have the following: n = 3; m = 5; ww = RandomReal[{0, 0.1}, {n, n}]; uu = RandomReal[{0, 1}, {m, n}]; pp = RandomReal[{0, 1}, {n, n}]; ss = RandomInteger[{0, 5}, {m, n}]; Grid[{{"ww", "uu", "pp", "ss"}, {ww // TableForm, uu // TableForm, pp // TableForm, ss // TableForm}}, Spacings -> {5, 2}, Dividers -> All] where I would like to look at every element of matrix ss and produce a matrix tt , with zeroes at the locations in ss which have zeroes, and in all other positions do the following: tt = (-1/Subscript[ww, m]) Log[(1 - uu)/(Subscript[pp, m - 1])], where Subscript[ww, m] is the value at index of ww matrix and where Subscript[pp, m - 1] is the value at index-1 of pp matrix. So for example if the first value ever read from matrix ss happens to be 2, then value taken from matrix ww would be from the row 2, but from pp would be from row 1. Also how to tell difference between a 0 as a valid value from within the matrix elemen...