I made it out of white cartridge paper back in the 60s.
I'm thinking of making it again, but this time in colour. Presumably six colours for the coplanar pairs of pentagrams, ten colours for the coplanar pairs of triangles, but how many colours will I need for the coplanar pairs of squares?
Miller's Monster
According to Wenninger, there are 30 pairs of squares. They are diametral or "hemi" faces, so there is no parallel pair. To colour each pair differently would need 30 colours.
That makes 6+10+30=46 colours in all. Ouch!
Wenninger describes a simpler colour scheme. He uses a simplified set of 5 colours for the icosahedral (triangular) faces, based on the observation that the 20 face planes are those of 5 tetrahedra: one colour for each tetrahedron. He also uses a single colour for all squares. This gives 6+5+1=12 colours which is a bit more practical.
Although the pentagons cannot be divided into equal subgroups in the way that the triangles can, I am not sure about the squares. I suspect that they might be divisible in various ways, depending on your preference. One way to find out would be to study the Archimedean duals having 60 faces, pick (the?) one with equivalent face planes, and look for sub-symmetries there.
That makes 6+10+30=46 colours in all. Ouch!
Wenninger describes a simpler colour scheme. He uses a simplified set of 5 colours for the icosahedral (triangular) faces, based on the observation that the 20 face planes are those of 5 tetrahedra: one colour for each tetrahedron. He also uses a single colour for all squares. This gives 6+5+1=12 colours which is a bit more practical.
Although the pentagons cannot be divided into equal subgroups in the way that the triangles can, I am not sure about the squares. I suspect that they might be divisible in various ways, depending on your preference. One way to find out would be to study the Archimedean duals having 60 faces, pick (the?) one with equivalent face planes, and look for sub-symmetries there.
Cheers,
Guy. Guy's polyhedra pages
Guy. Guy's polyhedra pages
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Peter Kane
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I tried it in the following way:
Take the compound of 5 tetrahedra and put it into memory 1. Then take Millers Monster, select a pentagram (they all should be red in the auto colour option), you hit c and you slightly shift the red colour a bit lighter or darker to avoid colour conflicts in the following steps. You set the radius of the model to the same like that of the compound in memory 1. Select a triangle and hit shift-h to hide all of them, and you do the same with the coplanar one. Then hit shift-ctrl-t to toggle between shown and hidden faces. Now you see all triangles only. You reduce the symmetry to tetrahedral and you add the model from memory 1. If you select a triangle of a tetrahedron now and you look at that face (view->orientation->look at selected face), you see two coplanar triangles parallel above the selected face. You give them the colour of the selected one, using ctrl-c. You repeat this with all faces of the 5 tetrahedra compound, then you delete the tetrahedra one by one. Now you see an arrangement of 5 groups of 4 pairs of coplanar triangles, each group in tetrahedral orientation. (You best save it now in a memory, so that you can retrieve it, when needed.)
You hit shift-ctrl-h to show all faces, select a pentagram, hit h to hide all of them and shift-ctrl-t to toggle between shown and hidden faces. You see twelve pairs of coplanar pentagrams in a dodecahedral arrangement. You colour them like 6 groups of parallel faces, using ctrl-c.
After showing all faces again, you select a square and give all of them the same colour using shift-c.
Al colours can easily be modified afterwards.
Ulrich
Take the compound of 5 tetrahedra and put it into memory 1. Then take Millers Monster, select a pentagram (they all should be red in the auto colour option), you hit c and you slightly shift the red colour a bit lighter or darker to avoid colour conflicts in the following steps. You set the radius of the model to the same like that of the compound in memory 1. Select a triangle and hit shift-h to hide all of them, and you do the same with the coplanar one. Then hit shift-ctrl-t to toggle between shown and hidden faces. Now you see all triangles only. You reduce the symmetry to tetrahedral and you add the model from memory 1. If you select a triangle of a tetrahedron now and you look at that face (view->orientation->look at selected face), you see two coplanar triangles parallel above the selected face. You give them the colour of the selected one, using ctrl-c. You repeat this with all faces of the 5 tetrahedra compound, then you delete the tetrahedra one by one. Now you see an arrangement of 5 groups of 4 pairs of coplanar triangles, each group in tetrahedral orientation. (You best save it now in a memory, so that you can retrieve it, when needed.)
You hit shift-ctrl-h to show all faces, select a pentagram, hit h to hide all of them and shift-ctrl-t to toggle between shown and hidden faces. You see twelve pairs of coplanar pentagrams in a dodecahedral arrangement. You colour them like 6 groups of parallel faces, using ctrl-c.
After showing all faces again, you select a square and give all of them the same colour using shift-c.
Al colours can easily be modified afterwards.
Ulrich
Last edited by Ulrich on Sun Mar 29, 2015 8:39 am, edited 1 time in total.
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Peter Kane
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robertw
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There's a much easier way. Stella already has various colour arrangements built in. From the menu, select these:
- Color->Basic Color Scheme->Auto Color (probably already set).
- Color->Special Color Arrangements->Dodecahedral Arrangement 2 (6 colors)
- Color->Special Color Arrangements->Icosahedral Arrangement 1 (5 colors)
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Peter Kane
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