Skip to main content

graphics3d - How to draw a dodecahedron with each face modified to a pentagram?



I'd like to draw a dodecahedron with each face carved on the sides so it becomes a pentagram. I wonder how to start to do this kind of task in the Wolfram Language?


Edit:


The result should still be a completely enclosed polyhedron; i.e., the carved out parts should be connected by newly added faces. I don't want the result to have holes.



Answer



Solution from @chuy looks really nice. Although I think that it was a little bit of work around because it's a visualization only, but the defined structure doesn't really represent the carved dodecahedron. Here is my approach of carving a dodecahedron pumpkin into pentagrams.


First we define a function that makes a pentagram from a pentagon.


tau = (2 Sqrt[5])/(5 + Sqrt[5]);
pentagram[pts_] :=
Riffle[pts, #] &@(pts[[# + 1]]*tau + (1 - tau)*
pts[[1 + Mod[# + 2, 5]]] & /@ Range[0, 4, 1]);


Then we apply this function to all faces of dodecahedron.


ind = PolyhedronData["Dodecahedron", "FaceIndices"];
vert = PolyhedronData["Dodecahedron", "VertexCoordinates"];
polyVerts = Reverse@*pentagram /@ (vert[[#]] & /@ ind);

Note the Reverse, it doesn't have to be there, since it just changes the orientation of the pentagram, but it's required to avoid weird artifacts while rendering, see more discussion here.


Now we need to create inner faces of our pumpkin.


pairs = Partition[#, 2] &@Riffle[#, #*85/100] &@polyVerts; 


pairs contain the outer face and inner face. The last thing to do is create wedges that will connect inner faces with outer faces.


wedges[face_] := (Permute[#, Cycles[{{4, 3}}]] &@Flatten[#, 1] &@
face[[1 ;; -1, #, 1 ;; -1]]) & /@
Partition[#, 2, 1] &@(Range[1, 10]~Join~{1});

Now we need to draw all our polygons: faces and wedges:


Graphics3D[
Join[{EdgeForm[{Black, Thick}], Orange},
Polygon /@ Join[wedges[#], #] & /@ pairs], Boxed -> False]


Carved pumpkin


Edit: It has been requested to have no holes in the resulting polyhedron. So no more pumpkin carving.


Let's make a list of all added vertices and include the original pentagon vertex indices that produced these additional concave vertices.


pairList[l_, r_] := Partition[#, 2] &@Riffle[#, RotateLeft[#, r]] &@ l;

concVerts[vert_, face_] :=
Partition[#, 2] &@
Riffle[Sort /@ pairList[face, 1],
vert[[First[#]]]*tau + (1 - tau)*vert[[Last[#]]] & /@
pairList[face, 2]];

concave = Flatten[#, 1] &@(concVerts[vert, #] & /@ ind);

Now we will fill holes with triangles, every triangle has two concave vertices and one original pentagon vertex.


triang[vert_, up_, edge_, 
conc_] := {vert[[#[[1, up]]]], #[[2]], #[[4]]} &@Flatten[#, 1] &@
Select[concave, #[[1]] == edge &];

tri = Flatten[#, 1] &@
Table[triang[vert, i, edges[[j]], concave], {i, 1, 2}, {j, 1,
Length@edges}];

Graphics3D[
Join[Polygon /@ tri, {EdgeForm[{Black, Thick}], Orange},
Polygon /@ polyVerts], Boxed -> False]

Carved dodecahedron


Comments

Popular posts from this blog

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}]

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}]

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.