Origami & Folding
what one sheet can do
Series

Self-contact — the series

8 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 20° a crease0.50 turns of papernothing touching anything34° a crease0.85 turns of papernothing touching anything36° a crease0.90 turns of paper1 pair through one another50° a crease1.25 turns of paper5 pairs through one anotherone strip of 10 panels, seen end-onit laps itself at 36.0° a crease, which is where its cross-section closesevery panel is the same length in every frame; the only thing changed is how far each crease is turned

    Paper through paper

    Every test the subject has for rigid folding is a statement about a neighbourhood, and a neighbourhood cannot see the far side of the sheet. So a pattern can satisfy all of them while driving one panel straight through another, and the sharpest witness has no interior vertex in it at all.

    part 1 · rigid
  2. 5 of 6 solved meshes are solid at every angle sampledthe bar is the deepest interpenetration found anywhere in the motion, in panel widthsmesh 11closes to 9e-14solid at every anglemesh 17closes to 1e-121.18 — panels 3:0 and 3:2, 2 steps apartmesh 19closes to 6e-14solid at every anglemesh 23closes to 5e-14solid at every anglemesh 27closes to 2e-12solid at every anglemesh 71closes to 4e-12solid at every angle

    Closing is not building

    A quadrilateral mesh solved so that every loop closes to within a millionth of a radian is a mesh whose fold angles are consistent. It is not necessarily an object. One of the six solved here drives a panel through another at every angle of its motion — there is no part of the fold at which it could be made of solid panels — and the pair that crosses is two steps apart in the sheet, where nothing evaluated at a vertex could see it.

    part 2 · rigid
  3. the bar is the share of the population with a folded stateevery pattern in all four passes every condition at every interior vertexthe printed patterns4 of 80 cannot be placed · 0 cannot be ordered · 4 undecidedtwist tessellations2 of 125 cannot be placed · 2 cannot be ordered · 3 undecidedquadrilateral meshes2 of 60 cannot be placed · 4 cannot be ordered · 0 undecidedfold-and-cut patterns5 of 70 cannot be placed · 0 cannot be ordered · 2 undecidedundecided is a real answer here and is not rounded toward either side

    A collision is an order

    Paper passing through paper is treated here as a thing that happens during a motion and is caught by watching for it. At the flat state it is not an event at all: it is the absence of an ordering, and it can be proved rather than observed. Four of the six quadrilateral meshes this site solves for rigid folding place perfectly and admit no ordering of their nine panels — so every one of them must pass through itself, and none of them was ever driven to find out.

    part 3 · rigid
  4. the bar is the nodes the ordering search visitedthe letters are consistent on every one of these, so the one-pass test says nothing about any of themmesh 37,4739 panels · 7,473 nodes · no order existsmesh 58,0079 panels · 8,007 nodes · no order existsmesh 89,3469 panels · 9,346 nodes · no order existsmesh 111,0159 panels · 1,015 nodes · an order existsmesh 141449 panels · 144 nodes · an order existsmesh 199,0629 panels · 9,062 nodes · no order existsa red bar is a pattern with no folded state, found only by visiting every ordering it might have had

    Two refusals that refuse differently

    Four of the six developable quadrilateral meshes this collection solves have no ordering of their nine panels — they must pass through themselves, and a search over every ordering proves it. On all four, the letters agree with themselves perfectly. The linear proof and the exponential search are not a fast test and a slow one: they answer different questions, and neither contains the other.

    part 4 · rigid
  5. the bar is how many letterings of the mesh can have their panels stackedout of every labelling of its twelve creases, enumeratedmesh 3016 pass every vertex · 16 agree with themselves · arrived refusedmesh 5032 pass every vertex · 32 agree with themselves · arrived refusedmesh 8832 pass every vertex · 32 agree with themselves · arrived refusedmesh 11832 pass every vertex · 32 agree with themselves · arrived foldablemesh 141416 pass every vertex · 14 agree with themselves · arrived foldablemesh 19416 pass every vertex · 16 agree with themselves · arrived refusedtwo of the meshes have none at all, and two more were refused only at the lettering they came with

    Refused at one lettering

    Four of six quadrilateral meshes here have no arrangement of their nine panels — established by searching every ordering, at the labelling each mesh arrived with. Enumerate every labelling instead and two of the four fold perfectly well at a different one. What was reported as a fact about four meshes is a fact about two meshes and two labellings.

    part 5 · rigid
  6. the bar is how many letterings of the mesh can have their panels stackedout of every labelling of its twelve creases, enumeratedmesh 3016 pass every vertex · 16 agree with themselves · arrived refusedmesh 5032 pass every vertex · 32 agree with themselves · arrived refusedmesh 8832 pass every vertex · 32 agree with themselves · arrived refusedmesh 11832 pass every vertex · 32 agree with themselves · arrived foldablemesh 141416 pass every vertex · 14 agree with themselves · arrived foldablemesh 19416 pass every vertex · 16 agree with themselves · arrived refusedtwo of the meshes have none at all, and two more were refused only at the lettering they came with

    A search with nothing to reorder

    One search on a crease pattern costs eighty steps or fifteen thousand depending on the order it takes its decisions in. The other search on the same crease pattern costs 1,188,571 steps whatever order it is given — twelve permutations of the panels, twelve identical counts. The difference between them is one line of code that neither has and one has.

    part 6 · rigid
  7. panels with nothing below them, and where they aresquare ×1125 panels, 16 of them touching the edge · all 1 at the edgesquare ×2281 panels, 32 of them touching the edge · all 2 at the edgesquare ×33169 panels, 48 of them touching the edge · all 3 at the edgetriangular ×1369 panels, 39 of them touching the edge · all 3 at the edgetriangular ×25233 panels, 79 of them touching the edge · all 5 at the edgehexagonal ×1469 panels, 39 of them touching the edge · all 4 at the edgehexagonal ×27233 panels, 79 of them touching the edge · all 7 at the edgehexagonal ×310493 panels, 119 of them touching the edge · all 10 at the edgeelongated ×12105 panels, 48 of them touching the edge · all 2 at the edgeelongated ×23369 panels, 96 of them touching the edge · all 3 at the edgethe sheet these letters belong to has no such panel at all

    An order with no least element

    Enumerating every way a folded pattern can be stacked works by building upward from a panel with nothing below it. The smallest square twist patch has exactly one such stacking and takes eleven thousand steps to find it. The pattern that patch was cut from has no panel with nothing below it at all, so the enumeration has nothing to start from — and the sheet is perfectly well stacked anyway.

    part 7 · rigid
  8. one period of the square grid's twist tessellationa ring is where a crease leaves and returns on the far side40 crease pieces → 32 creases25 drawn panels → 16 panels16 vertices, every one interiorV − E + F = 0mountainvalleyraw edge

    Two panels that are one panel

    Paper cannot pass through paper, and every test for it compares pairs of panels. On a glued sheet two pieces of the drawing can be the same piece of paper — so a test that does not know the identification either reports a collision between a panel and itself, or misses one where the sheet meets itself round the loop.

    part 8 · rigid

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