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

Post a Comment

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
Post a Comment
Read more

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....
Post a Comment
Read more