The original problem was to construct a polyhedron, homeomorphic to a torus, with angular defect zero at every corner (and everywhere else): properly embedded in 3-space, so that the interiors are homeomorphic. I found a 2-parameter family of "origami polytore" solutions with 20 faces (4 square + 16 triangular), 12 vertices, 32 edges, each comprising: a square cuboid of depth c with corners (+/-1,+/-1,+/-c), deprived of its top and bottom faces, with 8 edges and 8 corners connected by triangle faces to the skew quadrangle with corners (+q,0,+h), (0,-q,-h), (-q,0,+h), (0,+q,-h). Given the depth c, what radii q and heights h are possible? Theorem: when the polytore is origami, either h^2 = -(q^2-2)*( 1/2 + c^2/(q-2)^2 ) --- the "oval" branch, or h^2 = -(q^2)*( 1/2 + c^2/((q-2)^2-2) ) --- the "cusp" branch. There are just two congruency classes of corner: cuboid, and quadrangle. By the Gauss-Bonnet theorem, the sum of the angular defects vanishes; so we need ensure only that they vanish at cuboid corners. The latter having two square faces, it remains to ensure that the other three face angles A,B,C sum to \pi.
From the coordinates of the corners, the squares of the edge lengths can be found by Pythagoras; then in turn by the cosine rule, the squared sines SA,SB,SC of the face angles are expressible as _rational_ functions of c,q,h.
The squared sine SABC of the sum angle satisfies a quartic polynomial equation with coefficients polynomial in SA,SB,SC [of which I am inordinately proud --- see "Horrid trig identity"], which we need to have root zero, so constant coefficient zero. The latter turns out to factorise as the square of the product of the polynomials implicit in the theorem; so we have a necessary condition for origami-city. For sufficiency, in practice, all we need to do is check that the sum is correct at some convenient point, such as q = 1, then rely on continuity of the root as a function of q. In theory, I'm not entirely happy about this: just exactly _how_ can we be sure that some other root of the quartic does not intrude? That caveat aside, the interesting upshot of this frivolous investigation was that there is another branch of solutions (cusp) besides those originally found (oval) by trial-and-lots-of-error. Both branches are plotted for c = 1 at <hq_plot.jpg>; those solutions giving proper embeddings (rather than mere immersions) lie between 0 < q < 1. The new branch is less well house-trained than the old: at q = 2-sqrt2, h roars off to oo. At one point the branches actually cross: although somehow I keep feeling this solution should have some interesting property, in fact it looks just like most of the others <polytore.jpg>. There is available a Maple worksheet <polytore.mw> on this topic, incorporating the above still in rotatable form, and an animation of the polyhedron as q varies. Further projects suggested by this investigation include the extension to regular base polygons with any even number edges (rather than just squares, as above); and an animation of the polyhedron as folding up from a planar map. The latter poses a nontrivial robotics question: is it possible to fold the polyhedron from a sheet on which opposite edges have already been joined [forcing it to be folded flat, but four layers thick]? My attempts to find somewhere public to put all this stuff having been so far cruelly stymied, I must again offer to send the files to anybody expressing an interest. Fred Lunnon [18/08/09]
On 8/18/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
The original problem was to construct a polyhedron, homeomorphic to a torus, with angular defect zero at every corner (and everywhere else): properly embedded in 3-space, so that the interiors are homeomorphic. ... My attempts to find somewhere public to put all this stuff having been so far cruelly stymied, I must again offer to send the files to anybody expressing an interest.
The situation is improving. There is now a still shot of a particular polytore (cubical crossover q = 0.674013, h = 1.285291, c = 1.0) at http://www.mapleprimes.com/files/8970_polytore2.pdf with an A4-sized planar net for the same at http://www.mapleprimes.com/files/8970_flattore2.pdf and the plots of h against q (both branches) for c = 1 at http://www.mapleprimes.com/files/8970_hq_plot2.pdf There should also be a Maple worksheet with algebra and movies, just as soon as we can work out how to upload an updated file (!), and can persuade MapleNet to actually run the graphics over the internet as advertised ...
The latter poses a nontrivial robotics question: is it possible to fold the polyhedron from a sheet on which opposite edges have already been joined [forcing it to be folded flat, but four layers thick]?
I dunno about that --- but it's pretty tricky trying to build the wretched thing from a conventional net anyhow! Dividing into (at least) two pieces is recommended, to anyone reckless enough to attempt the feat. Fred Lunnon
The situation is improving. There is now a still shot of a particular polytore (cubical crossover q = 0.674013, h = 1.285291, c = 1.0) at http://www.mapleprimes.com/files/8970_polytore2.pdf with an A4-sized planar net for the same at http://www.mapleprimes.com/files/8970_flattore2.pdf
[...]
it's pretty tricky trying to build the wretched thing from a conventional net anyhow! Dividing into (at least) two pieces is recommended, to anyone reckless enough to attempt the feat.
Nonsense! I have one in my hands now, printed from the planar net above, and it folds and unfolds with almost no effort. The key is that you need to start by creasing all the edges in the correct direction: most of them should be mountain folds, but the two long edges of the four big triangles (touching the squares on their third edge) should be valleys. (Maybe the lines should be drawn differently, as is origami-standard.) Fred's polytorus has two types of vertices -- (a) eight that are corners of the original cube, where meet two squares and three triangles, and (b) four more where six triangles come together. In the pdf above, the (a)s are all interior vertices of the net, which means you get a really visceral feeling that they're flat: you watch them go from planar to part of the polyhedron. The planarity of the four (b)s, on the other hand, is not at all obvious here -- the six incident triangles are divided up three and three. Perhaps the model would be improved if the two mostly-interior type-(b) vertices were augmented with a little bit of webbing -- not a full extra copy of the three other triangles that meet there, but just a small stub with (two valley) fold lines, that the counterpart faces can snuggle into when the whole thing is assembled. --Michael -- Forewarned is worth an octopus in the bush.
On 8/19/09, Michael Kleber <michael.kleber@gmail.com> wrote:
... Nonsense! I have one in my hands now, printed from the planar net above, and it folds and unfolds with almost no effort. The key is that you need to start by creasing all the edges in the correct direction: most of them should be mountain folds, but the two long edges of the four big triangles (touching the squares on their third edge) should be valleys. (Maybe the lines should be drawn differently, as is origami-standard.)
Congratulations! [Guess I must be out of practice.]
... Perhaps the model would be improved if the two mostly-interior type-(b) vertices were augmented with a little bit of webbing -- not a full extra copy of the three other triangles that meet there, but just a small stub with (two valley) fold lines, that the counterpart faces can snuggle into when the whole thing is assembled.
Agreed. Unfortunately, apparently straightforward mods like this are rendered all but impossible by Maple's byzantine plotting architecture, which was evidently never actually designed but just growed like Topsy. Somehow, Maple has to be fooled into thinking that the net is either a function plot in 2-D, or a solid shape in 3-D ... I'll work on it. WFL
On 8/19/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
... The situation is improving. There is now a still shot of a particular polytore (cubical crossover q = 0.674013, h = 1.285291, c = 1.0) at http://www.mapleprimes.com/files/8970_polytore2.pdf with an A4-sized planar net for the same at http://www.mapleprimes.com/files/8970_flattore2.pdf and the plots of h against q (both branches) for c = 1 at http://www.mapleprimes.com/files/8970_hq_plot2.pdf
Since I can't update these, I've changed all the filenames from ...2 to ...3 in the current versions.
There should also be a Maple worksheet with algebra and movies, just as soon as we can work out how to upload an updated file (!), and can persuade MapleNet to actually run the graphics over the internet as advertised ...
I've become brassed-off waiting for this to happen, so the current program is (hopefully) downloadable from http://www.mapleprimes.com/files/8970_polytore3.mw --- if anybody manages to make any kind of use of these, I'd be pleased to hear from them. On 8/19/09, Michael Kleber <michael.kleber@gmail.com> wrote:
... Fred's polytorus has two types of vertices -- (a) eight that are corners of the original cube, where meet two squares and three triangles, and (b) four more where six triangles come together. In the pdf above, the (a)s are all interior vertices of the net, which means you get a really visceral feeling that they're flat: you watch them go from planar to part of the polyhedron. The planarity of the four (b)s, on the other hand, is not at all obvious here -- the six incident triangles are divided up three and three.
Perhaps the model would be improved if the two mostly-interior type-(b) vertices were augmented with a little bit of webbing -- not a full extra copy of the three other triangles that meet there, but just a small stub with (two valley) fold lines, that the counterpart faces can snuggle into when the whole thing is assembled.
After a grotesque amount of finagling, I managed to put some rather crude dashed lines along the re-entrant edges. The "webbing" sounds a nice idea, but I'm not convinced it's worthwhile here: it is actually pretty obvious from the diagram that the whole net constitutes a tile which covers the plane in chessboard fashion. Or the triangular faces could all be transferred onto the same side of the squares, yielding bounding straight lines to left and right --- but spoiling the symmetry. WFL
On 8/25/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
... Thus f_5 in 3 sec on my ancient Mac G4; though f_6 with an estimated 10^6 terms remains out of reach, leaving (you will be desolate to hear) the direct verification of planarity for the inner corners of any general polytore family likewise unattainable.
More twaddle from the apparently inexhaustible fountain of elementary incompetence which this fine problem has afforded me! The "W" corners of the polytore (i.e. not corners of the initial cuboid) are already feasible using f_4, since two pairs of the 6 angles there are equal; so we just substitute a_i -> 4 a_i (1-a_i), by the addition law for sines, and get on with it. After 5 minutes out pops a numerator of total degree 44 in q,h, which factors into --- among other things --- (oval branch) (cusp branch)^2 --- phew! Intriguingly, besides c^2 h^2 there are 3 other factors, of degree > 4, which of course must have no (new) real roots in 0 < q < 1, otherwise total angular defect from both U/V and W corners would be nonzero. Is this the first time the Gauss-Bonnet theorem has been employed to find roots of algebraic equations, I wonder? Fred Lunnon
Having clambered out of one elephant trap, I tumble straight into another. The story so far --- We have a family of "polytores", polyhedra of genus 1 having the symmetry of a square, parameterised by inner quadrangle radius q, inner quadrangle height h, outer cuboid height g (formerly c). Provided 0 < q < 1 and h,g > 0 these solids are properly embedded in Euclidean 3-space; and provided q,h,g satisfy certain other constraints, they are "planar" --- isometrically embedded, with face-angles summing to 2\pi at the corners, and unfoldable into a flat net which tiles the plane. Necessary for planarity is the polynomial equation f_k(a, b, c, ...) = 0 [see "Horrid trig identity"], where A,B,C,... denote face-angles, a = sin^2(A) etc, and k varies depending on the number of faces at the corner under inspection. When f_k = 0, we are assured that, for some choice of the signs, A (+/-) B (+/-) C (+/-) ... = 0 mod \pi . There are two classes of polytore corner, "U/V" with 5 faces and "W" with 6. For either class, we find f_k has two distinct quartic factors, christened "oval" and "cusp" after the shapes of their plane curves when g = 1 is fixed. The first class is characterised by the factor (due originally to Fred Helenius) oval(q,h,g) = 2(q - 2)^2 h^2 + (q^2 - 2)(q^2 - 4q + 2g^2 + 4) ; the second by the factor (whose author now prefers to remain anonymous) cusp(q,h,g) = 2(q^2 - 4q + 2)h^2 + q^2(q^2 - 4q + 2g^2 + 2) . The constraint oval(q,h,g) = 0 appears to be sufficient, as has been verified numerically, and by building models of examples. Just exactly how might these be proved sufficient? For g fixed, oval(q,h) = 0 defines a plane curve. If at some point along this curve --- q = 1 is a promising candidate --- we can verify that all the signs are positive --- more precisely A + B + C + ... = 2\pi --- then by continuity of the sum and of the curve, this will remain true as q varies along the remainder of the curve. I earlier remarked
In theory, I'm not entirely happy about this: just exactly _how_ can we be sure that some other root of the quartic does not intrude? Well, it seemed to work alright, so it must be OK, mustn't it ...
The model built (thanks to Michael Kleber) was of a particular example:
The two branches cross where q = 0.674013, h = 1.285291, g = 1; Somehow it seems that this case should have an interesting property, though it's not clear what. This wasn't very observant: besides the two obvious right-angles at a U/V corner, there is a third belonging to the UWW face; and at a W corner, in the flat net there are two less obvious right-angles between UW and VW edges.
Hmmm --- awful lot of (1/2)\pi angles there --- this "cross-over" case really deserves to be renamed "right-angled" ... Anyway pressing on, since it satisfies both constraints, we have for free a point on the second curve corresponding to a planar polytore; now by continuity, there should be an entire second family of planar polytors satisfying cusp = 0. But 'ang 'abaht --- as q approaches 2-sqrt2 along the cusp curve, h approaches oo. And while the outer U/V cuboid remains fixed, as each W corner disappears into the distance, the sum of the face-angles around its ever-sharpening spike quite plainly approaches zero --- and cannot possibly be planar! So what has gone wrong? Seeing that things seem a bit slow on the list at the moment, I'll leave this for now as an exercise for the student. Fred Lunnon
Fred summarized: Necessary for planarity is the polynomial equation f_k(a, b, c, ...) = 0
[see "Horrid trig identity"], where A,B,C,... denote face-angles, a = sin^2(A) etc, and k varies depending on the number of faces at the corner under inspection. When f_k = 0, we are assured that, for some choice of the signs, A (+/-) B (+/-) C (+/-) ... = 0 mod \pi .
[...]
The two branches cross where q = 0.674013, h = 1.285291, g = 1; Somehow it seems that this case should have an interesting property, though it's not clear what. This wasn't very observant: besides the two obvious right-angles at a U/V corner, there is a third belonging to the UWW face; and at a W corner, in the flat net there are two less obvious right-angles between UW and VW edges.
Hmmm --- awful lot of (1/2)\pi angles there --- this "cross-over" case really deserves to be renamed "right-angled" ...
Anyway pressing on, since it satisfies both constraints, we have for free a point on the second curve corresponding to a planar polytore; now by continuity, there should be an entire second family of planar polytors satisfying cusp = 0.
But unfortunately the "cusp" branch doesn't actually keep working; the signs turn out to be wrong. Those right angles Fred mentions are exactly the reason: at that point the angles around the degree-6 vertex can be partitioned into two parts (2 and 4 angles) that each sum to pi. So while the intersection point happens to satisfy the angular constraint with all + signs, continuity gets you bupkis. With Fred's exercise out of the way, what I really wanted to ask was about this other fact: There are two classes of polytore corner, "U/V" with 5 faces and "W" with 6.
For either class, we find f_k has two distinct quartic factors, christened "oval" and "cusp" after the shapes of their plane curves when g = 1 is fixed. The first class is characterised by the factor (due originally to Fred Helenius) oval(q,h,g) = 2(q - 2)^2 h^2 + (q^2 - 2)(q^2 - 4q + 2g^2 + 4) ; the second by the factor (whose author now prefers to remain anonymous) cusp(q,h,g) = 2(q^2 - 4q + 2)h^2 + q^2(q^2 - 4q + 2g^2 + 2) . The constraint oval(q,h,g) = 0 appears to be sufficient, as has been verified numerically, and by building models of examples.
The fact that f_5 and f_6 both turn out to be 0 at the same time is at first glance not surprising: Gauss-Bonnet says the total angular defect is 0, so if either type of vertex is flat, the other must be too. But the "cusp" branch *isn't* guaranteeing flatness, we now know; it's instead guaranteeing a 0 of the angle sum with some other combination of signs. So why on earth does this cusp branch show up as a factor of both f_5 and f_6? What else is forcing these to be simultaneously zero? (Pity it happens, whatever it is -- otherwise we could eliminate stray branches by taking the gcd of f_5 and f_6!) --Michael -- Forewarned is worth an octopus in the bush.
On 9/1/09, Michael Kleber <michael.kleber@gmail.com> wrote:
... But unfortunately the "cusp" branch doesn't actually keep working; the signs turn out to be wrong. Those right angles Fred mentions are exactly the reason: at that point the angles around the degree-6 vertex can be partitioned into two parts (2 and 4 angles) that each sum to pi. So while the intersection point happens to satisfy the angular constraint with all + signs, continuity gets you bupkis.
With Fred's exercise out of the way, ...
Well, yes, that certainly is what happens in detail to the polyhedron! But it doesn't actually answer the question I had in mind --- why doesn't the continuity argument work? Assuming g fixed say, both plane curves defined by the vanishing of the individual quartic factors are analytic around the right-angled / crossover point in the (q,h) plane. But the big polynomial (f_3 or f_4) has a double point there: so it's _not_ analytic, and we cannot rely on its continuity there! If we separate out the quartics and argue each one individually, continuity works just fine. In fact, it works so well that I now realise we didn't actually need those great big sin^2 polynomials at all, or indeed any angles --- plain old Pythagoras does the trick! New drastic rewrites loom ... can this saga get any more embarrassing?
what I really wanted to ask was about this other fact:
There are two classes of polytore corner, "U/V" with 5 faces and "W" with 6. ... The fact that f_5 and f_6 both turn out to be 0 at the same time is at first glance not surprising: Gauss-Bonnet says the total angular defect is 0, so if either type of vertex is flat, the other must be too. But the "cusp" branch *isn't* guaranteeing flatness, we now know; it's instead guaranteeing a 0 of the angle sum with some other combination of signs.
So why on earth does this cusp branch show up as a factor of both f_5 and f_6? What else is forcing these to be simultaneously zero?
(Pity it happens, whatever it is -- otherwise we could eliminate stray branches by taking the gcd of f_5 and f_6!)
I've puzzled over this a good deal. By angle chasing through the triangular faces (without assuming isometric embedding) --- in the same way that Gauss- Bonnet is proved --- it's easy to show that a number of signed angle-sums share the property of holding simultaneously at both types of corner, which sort of accounts for the phenomenon. Otherwise, it almost looks as if there might be another angular invariant lurking in there somewhere, analogous to the Gauss-Bonnet / Euler characteristic. Computing these polynomials for some other polyhedra might be interesting --- if somebody can propose any promising candidates! By the way, I didn'tt actually compute f_5 and f_6 at these corners, but used symmetry and right-angles to get away with f_3 and f_4. The bigger polynomials might well have more common factors --- but we will probably never know! Fred Lunnon
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Michael Kleber