Timeline for Wrapping a convex polyhedron with string
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Jun 10, 2012 at 15:57 | comment | added | Misha | @Will: Not me, but my 3-year old. You are welcome. | |
| Jun 10, 2012 at 15:47 | comment | added | Will Jagy | @Misha, thanks. You wake up really early. | |
| Jun 10, 2012 at 15:12 | comment | added | Misha | @Will: The fact that a geodesic cannot pass through a cone point with angle $\alpha<2\pi$ is standard and easy to check. If such geodesic $\gamma$ exists, cut the cone by a geodesic disjoint from $\gamma$ to a corner $C$. Then the vertex angle of $C$ is $\alpha<2\pi$ and one immediately sees the contradiction with minimality of $\gamma$. | |
| Jun 10, 2012 at 14:46 | comment | added | Joseph O'Rourke | @Misha: whoops, I was inadvertently assuming a simple closed geodesic. My mistake . | |
| Jun 10, 2012 at 11:43 | comment | added | Misha | ...since in the case of Riemannian metrics there are infinitely many closed geodesics, see your question mathoverflow.net/questions/75652/… | |
| Jun 10, 2012 at 11:37 | comment | added | Misha | @Joseph: Suppose that the total angle at every vertex is a rational multiple of $\pi$. (This situation is "rare" but dense among convex polyhedra.) Then for every $\delta>0$ there exists a closed (i.e., periodic) geodesic in $S$ which is $\delta$-dense. Here $S$ is the boundary of your solid. The proof is the same as for the billiards. Note that a closed geodesic need not be simple (most of them will not be simple). As I said above, I expect the existence problem for periodic geodesics to be quite hard and I do not think that there is an easy way to disprove their existence... | |
| Jun 10, 2012 at 10:53 | comment | added | Joseph O'Rourke | @Misha: If the closed geodesic closes smoothly, then it must partition the curvature into $2\pi + 2 \pi$, which is generically rare. So most convex polyhedra have no closed geodesic. | |
| Jun 10, 2012 at 1:20 | comment | added | Will Jagy | @Misha, you win. I thought there might be circumstances when a string through a vertex would still be a local minimum of length, but apparently not. | |
| Jun 10, 2012 at 1:18 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Jun 10, 2012 at 0:59 | comment | added | Misha | ...entering a vertex. In view of this, I expect the following problem to be also difficult or to have negative solution: Given the boundary $S$ of a convex solid, is there a closed geodesic in $S$? (Again, I do not allow geodesics which pass through vertices here.) | |
| Jun 10, 2012 at 0:56 | comment | added | Misha | @Will and @Joseph: The setup here is very similar to the one you would find in the theory of billiards, see e.g. Tabachnikov's book "Geometry and Billiards". An outstanding open problem there is existence of a periodic billiard trajectory for an arbitrary convex polygon. If you have a convex 2-dimensional polygon $P$, glue two copies of $P$ along the boundary and think of this surface $S$ as a degenerate case of the boundary $S$ of a convex solid. Studying geodesics on $S$ is equivalent to study of billiard trajectories in $P$. The usual convention is that billiard trajectory stops after ... | |
| Jun 10, 2012 at 0:38 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Jun 10, 2012 at 0:34 | comment | added | Will Jagy | @Misha, I think you may have a point. If the desire is to have physical realism...a string hitting a vertex has a choice of two (or more) neighbors to continue to, and the direction change is violently unstable depending on that...if you have string with thickness $\delta$ you can get away with being pretty near a corner. | |
| Jun 10, 2012 at 0:26 | comment | added | Will Jagy | @Joseph, I just edited in something about corners. Give me a couple of minutes...I see, if there are only three polygons at a vertex, you unfold just those three, there is a wedge with positive angle on the current polygon where the string is not permitted. Right, upon unfolding, the string must be straight, and there must be an unfolded neighbor polygon to receive the continued path. | |
| Jun 10, 2012 at 0:25 | comment | added | Misha | @Will: I would not worry about the corners and just assume that you have a periodic geodesic on the boundary $S$ of the polyhedron, disjoint from vertices and $\delta$-dense in $S$. | |
| Jun 10, 2012 at 0:21 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Jun 10, 2012 at 0:19 | comment | added | Joseph O'Rourke | @Will: Right, what is wanted is a geodesic, but then every once in a while it turns around a vertex in such a way that it kinda catches there and so wouldn't slip over it. Perhaps as in Figure (b.i). Although that is actually a geodesic right up to the vertex... | |
| Jun 9, 2012 at 23:50 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Jun 9, 2012 at 23:44 | history | answered | Will Jagy | CC BY-SA 3.0 |