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]

On 8/11/09, Fred lunnon <fred.lunnon@gmail.com> wrote:

... 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. ...

Maybe it should be emphasised that (for 0 < q < 1) these polyhedra are =properly= embedded in 3-space --- there are no touching, overlapping or self-intersecting faces, edges or vertices --- the interior is open and connected. The only regrettable feature is that the corners and some edges are re-entrant: easily seen to be unavoidable if they are to be "flat" (developable). A rather attractive robotics and graphics project would be to animate the folding of the max h(q) case, starting from a flat sheet: the "map" is easy to construct. I'm trying not to think about it ... Regarding the relation between height h and radius q of the interior vertices: by symmetry, it's obvious that the implicit constraint f(h, q) = 0 say must involve only h^2. In fact, for every instance I've examined, if q is rational, then h^2 is also apparently rational. Surely this must mean that h^2 = g(q), where it seems reasonable to conjecture that g(q) might be polynomial. I haven't tried to deduce this from the problem; however, solving equations for its coefficients doesn't seem to work. Any ideas, anybody? The values I have (guessed) so far are h(-sqrt2) = h(+sqrt2) = 0; h^2(0) = h^2(1) = 3/2, h^2(1/2) = h^2(2/3) = 119/72, h^2(-1) = 11/18, h^2(-1/2) = 231/200, h^2(1/3) = 731/450, h^2(-4/3) = 59/450, h^2(-2/3) = 287/288, h^2(-1/3) = 1139/882, h^2(3/4) = 1311/800, h^2(1/4) = 2511/1568. The plot of f(h, q) = 0, or h^2 = g(q) if such it be, is a quartic-like oval, symmetric about the h-axis, and flattened on the positive side of the q-axis. It's noteworthy that on the other side, the extremum (q, h) = (-sqrt2, 0) occurs just at the place where the (by now) self-intersecting polyhedron has everted so far that, as q increases further, it is just about to become properly embedded once more --- but cannot, as h becomes imaginary! Incidentally, I tried using Plouffe's inverter for these computations --- are you there, Simon? On the home page http://pi.lacim.uqam.ca/ for "Histoiry" read "History"; for "Tables de constantes" read "Tables of Constants". It didn't cope too well with some cases, I have to say: for example, it failed to recognise that 0.9965277777777785 ~ 287/288; and it claimed 1.285604051711792 ~ 1/12*(-238)^(1/2), nearly correct apart from a mysterious intrusive factor of \iota! Fred Lunnon