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The figure library — page 5

Every picture here is generated from code at build time. This is every generator, ordered by how many essays call it, each rendered at its defaults.
A boundary vertex is a stripA vertex where creases meet the edge of the paper, drawn as the fan of paper it has and again as the one-dimensional crease pattern that fan is. The sectors come in a line rather than in a ring, so the four conditions the subject states at an interior vertex are not weakened there — they are about a different object, and the object this is has a decidable condition of its own.the edge of the paperMVM40°60°20°60°the same sectors, in a lineMVM40°60°20°60°this lettering folds4 of 8 letterings foldVMV MMV VVM MVMno vertex theorem applies here at all— the sectors do not close, and there is no cycle to alternate round
A shrink is two numbersEvery pattern placed by how much it draws the sheet in along each axis. A corrugation sits on the line where one direction is untouched; a twist sits on the diagonal, drawing in equally both ways; the Miura sits off both, which is the whole of what makes it a Miura rather than a pleat.24682468times shorter across the sheettimes shorter down itaccordionMiuraleaf corrugationYoshimurasquare twistwhat each one doesaccordion8.00× across, 1.00× downMiura2.73× across, 1.75× downleaf corrugation3.57× across, 1.22× downYoshimura4.00× across, 4.00× downsquare twist1.52× across, 1.52× downthe dashed line is a directionthe fold never touchesone direction untouched: 1 of 5 · equal both ways: 2 · neither: 2the single number a corrugation is usually quoted by is these two multiplied together
Nothing in a body folds on a lineA hinge that cannot go below a radius consumes a fixed length of surface every time it is used, so the share of the sheet spent on hinges is proportional to the number of folds. Past some count the pattern is mostly hinge and a finer fold buys nothing, and where that happens is arithmetic.010203040506000.20.40.60.81foldsshare of the sheet lost to hinges50% of the sheet64 foldshinge radius 0.08 on a 10 unit sheet · (π − 2)ρ = 0.0913 lost per fold
Where the cheap paper is, with a hole and withoutThe paper a flap of a given length can claim at each point of the sheet, dark where it claims least. A hole in the middle makes the paper around it as cheap as the paper at the sheet's own edge, and the two sheets hold the same amount of paper.a flap of 0.12 of the sheet's side, on two sheets of the same areadarker is cheaper: less of the flap's disc is paper that has to be paid forwith a holesolid, same areamean claim 0.8646mean claim 0.8959
A wedge out, and the cone that closesA disc of paper with a 60° wedge marked for removal, and the cone the rest of it closes into when the two cut edges are brought together. Nothing is stretched: the paper that is left is exactly the paper that was there. What changed is how much of it surrounds the centre, and that fixes the cone's half-angle at 56.44° with nothing left to choose.the sheeta wedge of 60° marked for removal60°56.4°what it closes intoa cone of half-angle 56.44°83.3% of the turn is left, and the sine of the half-angle is that same fractionthe circles of latitude are the disc's own, arriving shorter than a flat sheet would needno fold can do this: folding moves paper about and never alters how much of it surrounds a point
Twos and threes, and nothing elseEach number with the degree of the simplest rational equation it satisfies. A fold reaches a number exactly when that degree is a product of twos and threes — a compass supplies the twos and one fold supplies the threes — so a fifth root is out of reach at any number of folds, and saying so needs no new argument once the reachable set is known to be a field.the degree of the equation, and what it is made ofnumberdegreemade ofwhere it comes from½11a fold in half√222^1the diagonal of the squareφ22^1the silver rectangle's cousin∛233^1doubling the cube2 cos(2π/7)33^1the regular heptagon∜242^2a square root of a square root∛2 · √262^1 · 3^1a product of two of them2^(1/5)5not twos and threesa fifth root2 cos(2π/11)5not twos and threesthe regular hendecagonchecked by exhaustion: no number here satisfies a rational equation of lower degree with coefficients up to 6
Every flap along one lineThe tree method does not produce an arbitrary shape; it produces a base whose flaps all lie along a single axis, with each flap as long as its edge of the skeleton. That restriction is what makes the circle argument work, because the base's shadow square to the axis is the tree itself.the treeleg0.62leg0.62body0.34arm0.50head0.78the basethe axis0.620.620.340.500.78the flaps are the tree's edges, at the tree's lengths, all square to one lineso the base's shadow along the axis is the tree, and nothing else can be designed this waywhich is the restriction the circle argument quietly depends on
A crease is not a lineA fold carries the paper round a small radius rather than through a point, and the arc uses more of the sheet than the stack advances by. One crease loses a fraction of a millimetre. A grid with hundreds of them loses that on every line at once, which is why an ambitious tessellation is folded from thin paper and why a large grid comes out short.ρ = 0.12 mmarc 0.377 mm, stack advances 0.240 mmlost per crease (π − 2)ρ = 0.1370 mmone crease, at its real radiuson a 150 mm sheet8 × 81.0 mm — 0.6%16 × 162.1 mm — 1.4%24 × 243.2 mm — 2.1%32 × 324.2 mm — 2.8%48 × 486.4 mm — 4.3%lost along every line of the grid, in both directionswhich is why an ambitious grid is folded from thin paper
A strip folded by the all-layers machineThe pile of paper after each fold, drawn from the simulator's own states rather than from a description of them. The dashed line marks where the next fold happens. Each layer is one run of the strip that has not yet been folded anywhere along its length.creases at 0.20, 0.40, 0.60, 0.80 — assignment MVMVflat1 layerafter fold 12 layersafter fold 23 layersafter fold 34 layersafter fold 45 layers4 folds, and the finished pile satisfies the assignment — checked against the layer-ordering rules
Why the difference is twoThe cross-section of a flat-folded vertex is a closed path that turns through exactly one full circle. Every mountain turns it one way by half a turn and every valley the other, so the counts must differ by exactly two — which is Maekawa's theorem, and it is a statement about winding rather than about paper.MVMwalk the folded edge and count the turns:each mountain turns +180°, each valley −180°the walk closes, so the total is ±360° — which forces |M − V| = 2the sheet must come back to where it started
Twenty lengths, four conditionsThe same mesh before and after. Nothing about the flat-folding conditions has changed — both are exact at every interior vertex — and only the lengths of the creases are different. The one on the right folds rigidly and the one on the left does not.mismatch 0.066 radiansmismatch 8.5e-14 radiansevery vertex of both is developable and Kawasaki-exact to the last bit a double holds
How much of the sheet the flaps claimThe fraction of a square filled by n equal discs, for the best arrangement a seeded search could find. The dashed line is the density of the hexagonal packing of the whole plane, which is proved and which no packing inside a square reaches, because the boundary wastes a strip. For most n the true optimum is unknown.2 discs53.9%r = 0.29293 discs61.0%r = 0.25434 discs78.5%r = 0.25005 discs67.3%r = 0.20716 discs66.3%r = 0.18767 discs66.9%r = 0.17448 discs72.8%r = 0.17029 discs78.5%r = 0.1667hexagonal density 90.69%every bar is the best a seeded search found, not a proved optimum —which is the honest state of the problem for all but the first few values of n