Steven DollinsIf you walk triangle strips on this 67 vertex sphere triangulation with 5-fold dihedral symmetry, it makes one complete circuit.
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In contrast, the 12-vertex icosahedron has 6 triangle strips.
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16 vertices gives a triangulation with tetrahedral symmetry. It has 7 triangle strips.
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22 vertices can also have tetrahedral symmetry, but now only have four strips that each wrap the sphere twice.
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40 vertices in tetrahedral symmetry gives a mix of the two with 4 strips that wrap twice and three that only wrap once.
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22 vertices can also arrange with 2-fold dihedral symmetry with two strips of six pentagons each separated by a single loop of 10 hexagons. The triangulation has two long triangle strips and two short ones.
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And 22 vertices can also arrange with 2-fold cylindrical symmetry that runs all the pentagons together into one long strip. It produces one long triangle strip and three short ones.
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23- and 29-vertex spheres both arrange in D3 symmetry with a single triangle strip.
50 vertices arranged in D6 symmetry is interesting in that it forms two different but close in length triangle strips -- one following the longitudes and the other the latitudes.
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Tetrahedral symmetry requires that a general point be in a set of 12 -- on each of the 4 faces in each of 3 orientations. You can also add 4 points at the vertices, 4 at each face center, or 6 at each edge center. Combined, any even number of points >= 4 can be arranged with tetrahedral symmetry, albeit not always evenly.
Here is 50 points in tetrahedral symmetry which requires that some of them have valence 7.
Here is an 80-vertex sphere in tetrahedral symmetry with 24 valence-7 vertices.
We can also get 80-vertex tetrahedral symmetry with a more "traditional" arrangement of 12 pentagons and the rest hexagons.
80 vertices in 2-fold dihedral symmetry has triangle strips of 4 different lengths.
Jocelyn Etienne – research@scdollins
Very nice.
How close are those to geodesic curves? (That is, if we put them under tension, does it have some chance to be at least a marginally stable arrangement when laid around a slippery sphere?)
@jocelyn_etienne Approximating geodesics was part of my goal. I wanted to find paths that could enclose a sphere uniformly and symmetrically.
Segments that cross hexagons are the closest to geodesics, but across a pentagon they curve towards its center and across a heptagon, they bend away.
If you subdivide an icosahedron, you can get several strips in parallel, but, of course, only the middle will be a geodesic as in this 92-vertex 3,0 Goldberg polyhedron.