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[GIF] Cantellation (Cross sections of the 16-cell)

Cross sections of the 16-cell

Cantellation

Another GIF given by taking cross sections of some convex shape (previously here and here). This time, we're looking at cross sections of the 16-cell, so the cross sections are 3-dimensional polyhedra. In fact, this turns out to be a nice illustration of the fact that the cuboctahedron is the cantellated tetrahedron and, moreover, that if you continue the cantellation as far as it can go, you end up with the tetrahedron dual to the one you started with.

As usual, I'm using my general slicing function:

slices[edges_, vec_, plotrange_] := 
 Module[{projector, pedges, n, times, positions, v},
   projector = Orthogonalize[NullSpace[{vec}]];
   pedges[t_] := (1 - t) #[[1]] + t #[[2]] & /@ edges;
   n = Length[pedges[.5]];
   times = 
    Table[NSolve[{pedges[t][[i]].vec == #, 0 <= t <= 1}, t], {i, 1, 
      n}];
   positions = Flatten[Position[times, a_ /; a != {}, 1]];
   v = Table[
     pedges[t][[positions[[i]]]] /. 
      Flatten[times[[positions[[i]]]], 1], {i, 1, 
      Length[positions]}];
   ConvexHullMesh[projector.# & /@ v, PlotRange -> plotrange, 
    PlotTheme -> "Polygons"]
   ] &

I also need the vertices and edges of the 16-cell:

sixteencellvertices = 
  Normalize /@ 
   Flatten[Permutations[{-1, 0, 0, 0}]^# & /@ Range[1, 2], 1];
sixteencelledges = 
  Select[Subsets[sixteencellvertices, {2}], #[[1]] != -#[[2]] &];

And then, finally, here's the Manipulate (notice that it struggles for values of s near $0$ and $\pi$, I believe because the BoundaryMesh consists of a huge number of simplices for these values of s):

DynamicModule[{cols},
 cols = Append[RGBColor /@ {"#88BEF5", "#F469A9"}, GrayLevel[.2]];
 Manipulate[
  Graphics3D[{EdgeForm[None], 
    Blend[{cols[[1]], cols[[2]]}, 1/2 - 1/2 Cos[s]], 
    GraphicsComplex[MeshCoordinates[#], MeshCells[#, 2]] &[
     TriangulateMesh[
      slices[sixteencelledges, Normalize[{1, 1, 1, 1}], 1][
       1/2 Cos[s]], MaxCellMeasure -> 100]]}, PlotRange -> 1, 
   ImageSize -> 540, Boxed -> False, ViewPoint -> 10 {0, 0, 1}, 
   ViewVertical -> {0, -1, 0}, ViewAngle -> ?/30, 
   Lighting -> "Neutral", Background -> cols[[3]]], {s, 0, ?}]
 ]
7 Replies

This is very nice, thanks. There is a good MathWorld article with some references to built in data about 16-cell. For example.

The vertices of the 16-cell with circumradius 1 and edge length $\sqrt2$ are the permutations of $(\pm1, 0, 0, 0)$. There are 2 distinct nonzero distances between vertices of the 16-cell in 4-space. The skeleton of the 16-cell is implemented in the Wolfram Language as GraphData["SixteenCellGraph"]. When embedded in three-space, the 16-cell skeleton is a cube with an "X" connecting diagonally opposite vertices on each face.

GraphData["SixteenCellGraph", "Graphs"]

enter image description here

GraphData["SixteenCellGraph", "Graphs3D"]

enter image description here

POSTED BY: Sam Carrettie

@Sam Carrettie Good point. In fact, skeletons of five of the six regular convex 4-polytopes are given in the RegularPolychoron class of GraphData. Strangely, when you run GraphData["RegularPolychoron"] the output is

{"HundredTwentyCellGraph", "SixHundredCellGraph", "SixteenCellGraph", "TesseractGraph", "TwentyFourCellGraph"}

meaning that PentatopeGraph (which is a valid named graph and is indeed the skeleton of the 5-cell) isn't in the RegularPolychoron class. Does anybody know why?

enter image description here - another post of yours has been selected for the Staff Picks group, congratulations !

We are happy to see you at the tops of the "Featured Contributor" board. Thank you for your wonderful contributions, and please keep them coming!

POSTED BY: Moderation Team

Not sure, but maybe @Ed Pegg knows.

POSTED BY: Sam Carrettie

We posted this from the official Wolfram Twitter and Facebook. Thank you!

POSTED BY: Moderation Team

@Moderation Team Very cool.

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