Fred: (1) I'd like to see the movie (2) Convert it to a video and put it up on YouTube! Or maybe tinypic.com, which hosts images, so no conversion to video required. --Michael On Tue, Aug 11, 2009 at 2:02 AM, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 8/9/09, Allan Wechsler <acwacw@gmail.com> wrote:
It's fairly clear that this problem will fall soon; somebody will give an exact construction of an example. I'm pretty sure I've seen origami models with zigzag pleats that could probably be extended and bent into a torus with little effort.
Jim Propp asked for a polyhedral origami torus, that is one properly embedded in 3-space, and having vertices where the face angles always add to 2\pi. The following construction gives a pencil of examples with 20 faces (4 square + 16 triangular), 12 vertices, 32 edges.
The vertices with their Cartesian coordinates comprise 8 corners of a cube: U1 at (+1,+1,+1), U2 at (+1,-1,+1), U3 at (-1,-1,+1), U4 at (-1,+1,+1), V1 at (+1,+1,+1), V2 at (+1,-1,-1), V3 at (-1,-1,-1), V4 at (-1,+1,-1), together with 4 points forming an interior skew quadrilateral: W12 at (+q,0,+h), W23 at (0,-q,-h), W34 at (-q,0,+h), W41 at (0,+q,-h), where q,h are related parameters.
The faces comprise a band of 4 square faces around the cube: {U1,U2,V2,V1}, {U2,U3,V3,V2}, {U3,U4,V4,V3}, {U4,U1,V1,V4}, 8 triangles joining top and bottom edges to the nearest interior points: {W12,U2,U1}, {W23,U3,U2}, {W34,U4,U3}, {W41,U4,U1}, {U1,U2,W12}, {U2,U3,W23}, {U3,U4,W34}, {U4,U1,W41}, 8 triangles joining corners to nearest quad edges: {U2,W12,W23}, {U3,W23,W34}, {U4,W34,W41}, {U1,W41,W12}, {V2,W23,W12}, {V3,W34,W23}, {V4,W41,W34}, {V1,W12,W41}.
The internal "radius" q must satisfy 0 < q < 1 for proper embedding, but is otherwise arbitrary; the "height" h = h(q) is then fixed apart from sign, taking the (same!) value h = 1.224744871 at q = 0 and 1, with maximum h = 1.287188506 near q = 0.58579.
This behaviour is rather surprising: there is no obvious reason why the angle sums at the U/V and W vertices should be related, as q and h vary independently; yet whenever one takes the value 2\pi, so does the other simultaneously!
I prepared a movie showing this polyhedron as a function of q, but at the moment I can't think of anywhere to put it --- if anybody wants to see it, I'll happily email directly to them (.GIF movie --- any browser should be able to cope).
Fred Lunnon [11/08/09]
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