[math-fun] Rolling polyhedra
I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees. For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is). Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)? Jim Propp
If the angle sums are 2pi/k1, 2pi/k2, … for integers k1, k2, … greater than 1, then the curvature sum-rule implies 4pi =2pi (1-1/k1)+2pi (1-1/k2)+ … But the only solution is the case where there are just four terms, and k1=k2=k3=k4=2. If you allowed for general rational multiples of 2pi there would be more solutions (cube, regular icosahedron, …). -Veit
On Jul 16, 2015, at 11:35 PM, James Propp <jamespropp@gmail.com> wrote:
I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees.
For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is).
Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)?
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
What about stellated polyhedra? Jim Propp On Friday, July 17, 2015, Veit Elser <ve10@cornell.edu> wrote:
If the angle sums are 2pi/k1, 2pi/k2, … for integers k1, k2, … greater than 1, then the curvature sum-rule implies
4pi =2pi (1-1/k1)+2pi (1-1/k2)+ …
But the only solution is the case where there are just four terms, and k1=k2=k3=k4=2.
If you allowed for general rational multiples of 2pi there would be more solutions (cube, regular icosahedron, …).
-Veit
On Jul 16, 2015, at 11:35 PM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees.
For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is).
Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)?
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Thanks, Veit. What if we use re-entrant polyhedra (so that the curvature sum-rule requires modification)? I think the small stellated dodecahedron (viewed as a re-entrant polyhedron with just 12 vertices) has the property I stated: its faces are pentagrams with angles of pi/5 meeting in groups of 5, so that angle-sum at each of the 12 vertices is pi. The way I want to "roll" a re-entrant polyhedron on a plane is somewhat peculiar: the faces that are currently rolling along the plane are required to stay on or above it, but other faces may penetrate the plane. Analogy: We can "roll" pentagram ABCDE along a line so that (for instance) as it rests on segment BC, vertices A and D are above the line but vertex E is below the line. I'd like to do something similar with polyhedra, but for simplicity I'd like to start with the "monodromy-free" case. Note that when you roll a cube, there's nontrivial monodromy: when the cube returns to where it started, in need not be in the same orientation. That's because the three 90-degree angles at each corner add up to 3 pi / 2, which is not a submultiple of 2 pi. But there's more to niceness than trivial monodromy, because if you start rolling a stellated dodecahedron around, its "footprints" will form a re-entrant tiling of the plane by pentagrams, and the set of footprints won't be discrete. What I'd like is something that I can "re-entrantly roll" on a plane so that the set of footprints is discrete, AND the monodromy is trivial. Are there any, besides the tetrahedra I mentioned? Jim Propp Are there other examples? Jim Propp On Friday, July 17, 2015, Veit Elser <ve10@cornell.edu> wrote:
If the angle sums are 2pi/k1, 2pi/k2, … for integers k1, k2, … greater than 1, then the curvature sum-rule implies
4pi =2pi (1-1/k1)+2pi (1-1/k2)+ …
But the only solution is the case where there are just four terms, and k1=k2=k3=k4=2.
If you allowed for general rational multiples of 2pi there would be more solutions (cube, regular icosahedron, …).
-Veit
On Jul 16, 2015, at 11:35 PM, James Propp <jamespropp@gmail.com> wrote:
I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees.
For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is).
Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)?
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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You could pucker the tiling out-of-plane. For example, there’s a nice tiling of the plane by 60-degree rhombi, where either 3 obtuse angles meet at every vertex or 6 acute. Now push the vertices out of the plane so the rhombi become squares. Rolling a cube on that polyhedral surface (of squares) has trivial monodromy. You get other examples like this one by using Voronoi cells of lattices (e.g. the rhombic dodecahedron). -Veit
On Jul 17, 2015, at 10:55 AM, James Propp <jamespropp@gmail.com> wrote:
Thanks, Veit.
What if we use re-entrant polyhedra (so that the curvature sum-rule requires modification)?
I think the small stellated dodecahedron (viewed as a re-entrant polyhedron with just 12 vertices) has the property I stated: its faces are pentagrams with angles of pi/5 meeting in groups of 5, so that angle-sum at each of the 12 vertices is pi.
The way I want to "roll" a re-entrant polyhedron on a plane is somewhat peculiar: the faces that are currently rolling along the plane are required to stay on or above it, but other faces may penetrate the plane. Analogy: We can "roll" pentagram ABCDE along a line so that (for instance) as it rests on segment BC, vertices A and D are above the line but vertex E is below the line. I'd like to do something similar with polyhedra, but for simplicity I'd like to start with the "monodromy-free" case. Note that when you roll a cube, there's nontrivial monodromy: when the cube returns to where it started, in need not be in the same orientation. That's because the three 90-degree angles at each corner add up to 3 pi / 2, which is not a submultiple of 2 pi.
But there's more to niceness than trivial monodromy, because if you start rolling a stellated dodecahedron around, its "footprints" will form a re-entrant tiling of the plane by pentagrams, and the set of footprints won't be discrete.
What I'd like is something that I can "re-entrantly roll" on a plane so that the set of footprints is discrete, AND the monodromy is trivial.
Are there any, besides the tetrahedra I mentioned?
Jim Propp
Are there other examples?
Jim Propp
On Friday, July 17, 2015, Veit Elser <ve10@cornell.edu> wrote:
If the angle sums are 2pi/k1, 2pi/k2, … for integers k1, k2, … greater than 1, then the curvature sum-rule implies
4pi =2pi (1-1/k1)+2pi (1-1/k2)+ …
But the only solution is the case where there are just four terms, and k1=k2=k3=k4=2.
If you allowed for general rational multiples of 2pi there would be more solutions (cube, regular icosahedron, …).
-Veit
On Jul 16, 2015, at 11:35 PM, James Propp <jamespropp@gmail.com> wrote:
I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees.
For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is).
Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)?
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I like it! But I'd like an example with an unpuckered plane even better, since I want to think about 3-D analogues of polygonal cycloids. Are there any re-entrant polyhedra of the sort I need? Jim Propp On Fri, Jul 17, 2015 at 12:07 PM, Veit Elser <ve10@cornell.edu> wrote:
You could pucker the tiling out-of-plane. For example, there’s a nice tiling of the plane by 60-degree rhombi, where either 3 obtuse angles meet at every vertex or 6 acute. Now push the vertices out of the plane so the rhombi become squares. Rolling a cube on that polyhedral surface (of squares) has trivial monodromy. You get other examples like this one by using Voronoi cells of lattices (e.g. the rhombic dodecahedron).
-Veit
On Jul 17, 2015, at 10:55 AM, James Propp <jamespropp@gmail.com> wrote:
Thanks, Veit.
What if we use re-entrant polyhedra (so that the curvature sum-rule requires modification)?
I think the small stellated dodecahedron (viewed as a re-entrant polyhedron with just 12 vertices) has the property I stated: its faces are pentagrams with angles of pi/5 meeting in groups of 5, so that angle-sum at each of the 12 vertices is pi.
The way I want to "roll" a re-entrant polyhedron on a plane is somewhat peculiar: the faces that are currently rolling along the plane are required to stay on or above it, but other faces may penetrate the plane. Analogy: We can "roll" pentagram ABCDE along a line so that (for instance) as it rests on segment BC, vertices A and D are above the line but vertex E is below the line. I'd like to do something similar with polyhedra, but for simplicity I'd like to start with the "monodromy-free" case. Note that when you roll a cube, there's nontrivial monodromy: when the cube returns to where it started, in need not be in the same orientation. That's because the three 90-degree angles at each corner add up to 3 pi / 2, which is not a submultiple of 2 pi.
But there's more to niceness than trivial monodromy, because if you start rolling a stellated dodecahedron around, its "footprints" will form a re-entrant tiling of the plane by pentagrams, and the set of footprints won't be discrete.
What I'd like is something that I can "re-entrantly roll" on a plane so that the set of footprints is discrete, AND the monodromy is trivial.
Are there any, besides the tetrahedra I mentioned?
Jim Propp
Are there other examples?
Jim Propp
On Friday, July 17, 2015, Veit Elser <ve10@cornell.edu> wrote:
If the angle sums are 2pi/k1, 2pi/k2, … for integers k1, k2, … greater than 1, then the curvature sum-rule implies
4pi =2pi (1-1/k1)+2pi (1-1/k2)+ …
But the only solution is the case where there are just four terms, and k1=k2=k3=k4=2.
If you allowed for general rational multiples of 2pi there would be more solutions (cube, regular icosahedron, …).
-Veit
On Jul 16, 2015, at 11:35 PM, James Propp <jamespropp@gmail.com> wrote:
I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees.
For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is).
Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)?
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Picking up an old thread and taking it in a different direction: Does a cube rolling around the outside of another cube have trivial monodromy? Is this true for every convex polyhedron rolling around its mirror-image? I *think* so, and I even have an idea for how to prove it (use a moving coordinate system in which the two polyhedra are mirror-images of each other in a fixed plane at all times). But then it would be true for spheres as well, and this smells wrong to me. Jim On Fri, Jul 17, 2015 at 2:21 PM James Propp <jamespropp@gmail.com> wrote:
I like it! But I'd like an example with an unpuckered plane even better, since I want to think about 3-D analogues of polygonal cycloids.
Are there any re-entrant polyhedra of the sort I need?
Jim Propp
On Fri, Jul 17, 2015 at 12:07 PM, Veit Elser <ve10@cornell.edu> wrote:
You could pucker the tiling out-of-plane. For example, there’s a nice tiling of the plane by 60-degree rhombi, where either 3 obtuse angles meet at every vertex or 6 acute. Now push the vertices out of the plane so the rhombi become squares. Rolling a cube on that polyhedral surface (of squares) has trivial monodromy. You get other examples like this one by using Voronoi cells of lattices (e.g. the rhombic dodecahedron).
-Veit
On Jul 17, 2015, at 10:55 AM, James Propp <jamespropp@gmail.com> wrote:
Thanks, Veit.
What if we use re-entrant polyhedra (so that the curvature sum-rule requires modification)?
I think the small stellated dodecahedron (viewed as a re-entrant polyhedron with just 12 vertices) has the property I stated: its faces are pentagrams with angles of pi/5 meeting in groups of 5, so that angle-sum at each of the 12 vertices is pi.
The way I want to "roll" a re-entrant polyhedron on a plane is somewhat peculiar: the faces that are currently rolling along the plane are required to stay on or above it, but other faces may penetrate the plane. Analogy: We can "roll" pentagram ABCDE along a line so that (for instance) as it rests on segment BC, vertices A and D are above the line but vertex E is below the line. I'd like to do something similar with polyhedra, but for simplicity I'd like to start with the "monodromy-free" case. Note that when you roll a cube, there's nontrivial monodromy: when the cube returns to where it started, in need not be in the same orientation. That's because the three 90-degree angles at each corner add up to 3 pi / 2, which is not a submultiple of 2 pi.
But there's more to niceness than trivial monodromy, because if you start rolling a stellated dodecahedron around, its "footprints" will form a re-entrant tiling of the plane by pentagrams, and the set of footprints won't be discrete.
What I'd like is something that I can "re-entrantly roll" on a plane so that the set of footprints is discrete, AND the monodromy is trivial.
Are there any, besides the tetrahedra I mentioned?
Jim Propp
Are there other examples?
Jim Propp
On Friday, July 17, 2015, Veit Elser <ve10@cornell.edu> wrote:
If the angle sums are 2pi/k1, 2pi/k2, … for integers k1, k2, … greater than 1, then the curvature sum-rule implies
4pi =2pi (1-1/k1)+2pi (1-1/k2)+ …
But the only solution is the case where there are just four terms, and k1=k2=k3=k4=2.
If you allowed for general rational multiples of 2pi there would be more solutions (cube, regular icosahedron, …).
-Veit
On Jul 16, 2015, at 11:35 PM, James Propp <jamespropp@gmail.com> wrote:
I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees.
For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is).
Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)?
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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This is a very interesting question. Yes, I think this is a proof that any convex polyhedron rolling on its mirror image has trivial holonomy. (Meaning: If they both start by sharing a common face in 3-space, a reflection in whose plane identifies each polyhedron with its mirror image.) Sketch of proof: Then rotating one of them by angle theta about an edge of the latest shared face is equivalent to rotating both of them in opposite directions by an angle theta/2 about the same edge. As for rolling one sphere on an identical one (with neither slipping nor twirling about the line connecting their centers): Indeed, essentially the same proof sketch as above shows that once each point of a sphere is identified (once and for all) with its mirror image, then (after some rolling) reflection about the tangent plane separating the spheres will always carry a point of one sphere to its (original) mirror image on the other sphere. * * * Here's a related problem I worked on for a while but only got as far as a good approximation to the answer: Given a sphere rolling on a horizontal plane (without slipping or twirling): Suppose it begins tangent to the origin (0,0) of the plane, with its point of tangency equal to the south pole S of the sphere. Suppose it's now rolled along a simple closed curve C of the plane so that when it returns to (0,0), its final point of tangency is the north pole N. Then, among the simple closed curves C for which this occurs: 1) Which C is the shortest? 2) Which C surrounds the smallest area? —Dan
On Friday/25December/2020, at 2:31 PM, James Propp <jamespropp@gmail.com> wrote:
Does a cube rolling around the outside of another cube have trivial monodromy? Is this true for every convex polyhedron rolling around its mirror-image? I *think* so, and I even have an idea for how to prove it (use a moving coordinate system in which the two polyhedra are mirror-images of each other in a fixed plane at all times). But then it would be true for spheres as well, and this smells wrong to me.
The paragraph beginning with "Yes" got inadvertently truncated in previous e-mail. The following is, I hope, fixed. —Dan
On Friday/25December/2020, at 4:40 PM, Dan Asimov <dasimov@earthlink.net> wrote:
This is a very interesting question.
Yes, I think this is a proof that any convex polyhedron rolling on its mirror image has trivial holonomy. (Meaning: If they both start by sharing a common face in 3-space, a reflection in whose plane identifies each polyhedron with its mirror image ... then after rolling to any number of adjacent faces so the two polyhedra end up sharing the same face they did initially — reflection in that face's plane associates the same pairs of point, one in each polyhedron, as were associated initially.)
Sketch of proof: Rotating one of them by angle theta about an edge of the latest shared face is equivalent to rotating both of them in opposite directions by an angle theta/2 about the same edge.
As for rolling one sphere on an identical one (with neither slipping nor twirling about the line connecting their centers): Indeed, essentially the same proof sketch as above shows that once each point of a sphere is identified (once and for all) with its mirror image, then (after some rolling) reflection about the tangent plane separating the spheres will always carry a point of one sphere to its (original) mirror image on the other sphere.
https://www.simonsfoundation.org/event/parking-cars-rolling-balls-and-fallin... This lecture was memorable to me. The four point tetrahedral symmetry of the triangular tiling has some important role in Socolar’s tiling theory, but who knows what. The “viererbaum” is the closest I ever got to figuring out the matching rule game, but we still haven’t measured it, ha ha. —Brad
On Dec 25, 2020, at 4:40 PM, James Propp <jamespropp@gmail.com> wrote:
Picking up an old thread and taking it in a different direction:
Does a cube rolling around the outside of another cube have trivial monodromy? Is this true for every convex polyhedron rolling around its mirror-image? I *think* so, and I even have an idea for how to prove it (use a moving coordinate system in which the two polyhedra are mirror-images of each other in a fixed plane at all times). But then it would be true for spheres as well, and this smells wrong to me.
Jim
On Fri, Jul 17, 2015 at 2:21 PM James Propp <jamespropp@gmail.com> wrote:
I like it! But I'd like an example with an unpuckered plane even better, since I want to think about 3-D analogues of polygonal cycloids.
Are there any re-entrant polyhedra of the sort I need?
Jim Propp
On Fri, Jul 17, 2015 at 12:07 PM, Veit Elser <ve10@cornell.edu> wrote:
You could pucker the tiling out-of-plane. For example, there’s a nice tiling of the plane by 60-degree rhombi, where either 3 obtuse angles meet at every vertex or 6 acute. Now push the vertices out of the plane so the rhombi become squares. Rolling a cube on that polyhedral surface (of squares) has trivial monodromy. You get other examples like this one by using Voronoi cells of lattices (e.g. the rhombic dodecahedron).
-Veit
On Jul 17, 2015, at 10:55 AM, James Propp <jamespropp@gmail.com> wrote:
Thanks, Veit.
What if we use re-entrant polyhedra (so that the curvature sum-rule requires modification)?
I think the small stellated dodecahedron (viewed as a re-entrant polyhedron with just 12 vertices) has the property I stated: its faces are pentagrams with angles of pi/5 meeting in groups of 5, so that angle-sum at each of the 12 vertices is pi.
The way I want to "roll" a re-entrant polyhedron on a plane is somewhat peculiar: the faces that are currently rolling along the plane are required to stay on or above it, but other faces may penetrate the plane. Analogy: We can "roll" pentagram ABCDE along a line so that (for instance) as it rests on segment BC, vertices A and D are above the line but vertex E is below the line. I'd like to do something similar with polyhedra, but for simplicity I'd like to start with the "monodromy-free" case. Note that when you roll a cube, there's nontrivial monodromy: when the cube returns to where it started, in need not be in the same orientation. That's because the three 90-degree angles at each corner add up to 3 pi / 2, which is not a submultiple of 2 pi.
But there's more to niceness than trivial monodromy, because if you start rolling a stellated dodecahedron around, its "footprints" will form a re-entrant tiling of the plane by pentagrams, and the set of footprints won't be discrete.
What I'd like is something that I can "re-entrantly roll" on a plane so that the set of footprints is discrete, AND the monodromy is trivial.
Are there any, besides the tetrahedra I mentioned?
Jim Propp
Are there other examples?
Jim Propp
On Friday, July 17, 2015, Veit Elser <ve10@cornell.edu> wrote:
If the angle sums are 2pi/k1, 2pi/k2, … for integers k1, k2, … greater than 1, then the curvature sum-rule implies
4pi =2pi (1-1/k1)+2pi (1-1/k2)+ …
But the only solution is the case where there are just four terms, and k1=k2=k3=k4=2.
If you allowed for general rational multiples of 2pi there would be more solutions (cube, regular icosahedron, …).
-Veit
On Jul 16, 2015, at 11:35 PM, James Propp <jamespropp@gmail.com> wrote:
I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees.
For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is).
Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)?
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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https://m.youtube.com/watch?v=9qMv2exawsA
On Dec 25, 2020, at 9:27 PM, Brad Klee <bradklee@gmail.com> wrote:
https://www.simonsfoundation.org/event/parking-cars-rolling-balls-and-fallin...
This lecture was memorable to me. The four point tetrahedral symmetry of the triangular tiling has some important role in Socolar’s tiling theory, but who knows what.
The “viererbaum” is the closest I ever got to figuring out the matching rule game, but we still haven’t measured it, ha ha.
—Brad
On Dec 25, 2020, at 4:40 PM, James Propp <jamespropp@gmail.com> wrote:
Picking up an old thread and taking it in a different direction:
Does a cube rolling around the outside of another cube have trivial monodromy? Is this true for every convex polyhedron rolling around its mirror-image? I *think* so, and I even have an idea for how to prove it (use a moving coordinate system in which the two polyhedra are mirror-images of each other in a fixed plane at all times). But then it would be true for spheres as well, and this smells wrong to me.
Jim
On Fri, Jul 17, 2015 at 2:21 PM James Propp <jamespropp@gmail.com> wrote:
I like it! But I'd like an example with an unpuckered plane even better, since I want to think about 3-D analogues of polygonal cycloids.
Are there any re-entrant polyhedra of the sort I need?
Jim Propp
On Fri, Jul 17, 2015 at 12:07 PM, Veit Elser <ve10@cornell.edu> wrote:
You could pucker the tiling out-of-plane. For example, there’s a nice tiling of the plane by 60-degree rhombi, where either 3 obtuse angles meet at every vertex or 6 acute. Now push the vertices out of the plane so the rhombi become squares. Rolling a cube on that polyhedral surface (of squares) has trivial monodromy. You get other examples like this one by using Voronoi cells of lattices (e.g. the rhombic dodecahedron).
-Veit
On Jul 17, 2015, at 10:55 AM, James Propp <jamespropp@gmail.com> wrote:
Thanks, Veit.
What if we use re-entrant polyhedra (so that the curvature sum-rule requires modification)?
I think the small stellated dodecahedron (viewed as a re-entrant polyhedron with just 12 vertices) has the property I stated: its faces are pentagrams with angles of pi/5 meeting in groups of 5, so that angle-sum at each of the 12 vertices is pi.
The way I want to "roll" a re-entrant polyhedron on a plane is somewhat peculiar: the faces that are currently rolling along the plane are required to stay on or above it, but other faces may penetrate the plane. Analogy: We can "roll" pentagram ABCDE along a line so that (for instance) as it rests on segment BC, vertices A and D are above the line but vertex E is below the line. I'd like to do something similar with polyhedra, but for simplicity I'd like to start with the "monodromy-free" case. Note that when you roll a cube, there's nontrivial monodromy: when the cube returns to where it started, in need not be in the same orientation. That's because the three 90-degree angles at each corner add up to 3 pi / 2, which is not a submultiple of 2 pi.
But there's more to niceness than trivial monodromy, because if you start rolling a stellated dodecahedron around, its "footprints" will form a re-entrant tiling of the plane by pentagrams, and the set of footprints won't be discrete.
What I'd like is something that I can "re-entrantly roll" on a plane so that the set of footprints is discrete, AND the monodromy is trivial.
Are there any, besides the tetrahedra I mentioned?
Jim Propp
Are there other examples?
Jim Propp
On Friday, July 17, 2015, Veit Elser <ve10@cornell.edu> wrote:
If the angle sums are 2pi/k1, 2pi/k2, … for integers k1, k2, … greater than 1, then the curvature sum-rule implies
4pi =2pi (1-1/k1)+2pi (1-1/k2)+ …
But the only solution is the case where there are just four terms, and k1=k2=k3=k4=2.
If you allowed for general rational multiples of 2pi there would be more solutions (cube, regular icosahedron, …).
-Veit
> On Jul 16, 2015, at 11:35 PM, James Propp <jamespropp@gmail.com> wrote: > > I want to think about rolling polyhedra, and the polyhedra I want to think > about rolling are polyhedra in which the sum of the angles at each vertex > is a submultiple of 360 degrees. > > For instance, given any triangle T, we can create a tetrahedron whose four > faces are all congruent to T, with angle-sum 180 degrees at each vertex > (there's a name for such tetrahedra but I forget what it is). > > Are there other polyhedra in which the angles at each vertex sum to a > submultiple of 360 degrees (not necessarily the same one at each vertex)? > > Jim Propp > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (4)
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Brad Klee -
Dan Asimov -
James Propp -
Veit Elser