# Wrapping a convex polyhedron with string

This is a meta-question, rather than a specific mathematical question. I am seeking a mathematical definition that captures the following physical idea.

Suppose you have a convex polyhedron $P \subset \mathbb{R}^3$. I would like to "wrap" $P$ with string so that (a) the entire surface area is covered with string, and (b) the string does not tend to slip off, even with minimal friction. The closest related literature I've found is this very current paper:

"Wrapping the cube and other polyhedra," T. Tarnai1, F. Kovács, P. W. Fowler and S. D. Guest, Proceedings of the Royal Society, 2012.

I am seeking a mathematical definition of what it might mean to "wrap without slippage." Then the natural follow-on question is to minimize the total string length, for a given $P$, and given string thickness $\delta$. But first I need a clear defintion! Any help would be appreciated—Thanks!

PS. Yes, I am familiar with the closely related work by Demaine, Demaine, and Mitchell.

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I'm not sure about the corners, but the discrete analogue of geodesic for a convex polyhedron is a sort of Snell's law, the angle approaching an edge must equal the angle of the string leaving the edge. Consider that the string has no idea it is in three dimensions, so if you cut out those two adjacent faces and fold them flat, you see that the path of the string can be shortened unless it is a straight line in the flat unfolded plane.

Note that when you fold it back up, this condition may be tricky to interpret.

Well, that's a start. Maybe they say it in one of your articles.

Yes, in your illustration c(ii), different choices of how to unfold the cube might make it possible to show one of the bands in full length as a bunch of parallel straight lines. Worth trying to draw that, label the faces and edges and all and see how well that goes. I'm not saying some arbitrarily complicated polyhedron can be unfolded in such a fortunate way.

Alright, I misunderstood the illustration a bit, (c.ii) is a single band of strings. So there is no way to unfold the cube to get everything resembling straight lines. Meanwhile, I think the rule for corners should begin with this: no string is permitted to go along an edge. There must be a way to decide, essentially by continuity, which polygon comes next. Need to think about other ways to say that...For the cube, it is impossible to implement this rule. Any string hitting a vertex, in the unfolded triple of squares, has only open air ahead of it, so the choice of where to go next in the cube is arbitrary and physically unstable. Not looking good for head on collision with a vertex.

EDITTTTT: It seems Misha wins. I did a little experiment with a regular (pentagonal) dodecahedron. Given a short segment that passes through a vertex and is orthogonal to the joint edge of two neighboring faces, and is given equal length on both faces, it is still shorter to slip the string off the vertex and take the "geodesic" that passes through the third pentagon at that vertex. So my one scheme, involving large angles, is still hopeless.

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@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 $S$. –  Misha Jun 10 '12 at 0:25
@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 ... –  Misha Jun 10 '12 at 0:56
...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.) –  Misha Jun 10 '12 at 0:59
@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 $\gamma$. –  Misha Jun 10 '12 at 15:12
@Will: Not me, but my 3-year old. You are welcome. –  Misha Jun 10 '12 at 15:57