Concept

Constraint — where it appears

A requirement that narrows what is allowed. In folding, constraints remove answers and shrink the space a search must cover, and which of those two effects dominates is rarely obvious in advance.

Named by 23 essays across 8 fields — each of them below, with the objects they name alongside it.

geometrypacks tocorrugation8 panels at 0.42 rad40.8% — 2.5× smallerMiura6 × 4, 15 interior vertices16.6% — 6.0× smallerfan8 sectors about one point25.0% — 4.0× smallerroll8 turns12.5% — 8.0× smallerpacked area as a fraction of deployed, computed from each geometry — not measured from any animal

The same corrugation in four places

A leaf, a wing, a crushed cylinder and a solar array arrive at nearly the same fold, and none of them copied any of the others. Convergence stories are cheap; this one is checkable, because the constraint that forces it can be computed rather than admired.

biology · Convergence
30°60°90°0.250.400.550.700.85how much of the room between two vertices the twists takeno paper leftno assignment existstwist angleboth curves are measured rather than plotted from a formula

Fenced at both ends

The twist angle of a tessellation looks like a free dial, and it is fenced twice. Turn too far and the pleats have no paper left. Turn too little and something stranger happens: every angle condition in the subject goes on holding and the pattern loses its mountain-valley assignment entirely.

tessellation · Twists
what the vertex conditions settle once one crease is chosenpatternsettled, against what is therepreliminary base1 of 81 vertices still choosingmiura 6×41 of 3815 vertices still choosingwaterbomb 4×41 of 7625 vertices still choosingyoshimura 6×51 of 8422 vertices still choosingsquare twist grid3 of 14464 vertices still choosingtriangular twist grid2 of 236104 vertices still choosingKawasaki was settled by the angles before a letter was written; the letters are what is left, and they are nearly all left

How little the conditions decide

Local is not global is a statement about sufficiency: every vertex can pass and the sheet still fail. There is a sharper complaint available, and it is about strength. Fix one crease of a tessellation and propagate every condition the subject has to a fixed point: three creases out of a hundred and fifty-eight follow, and sixty-six vertices are still holding more than one answer.

flat-folding · Flat-foldability
23456789101105101520flapshow much the requirement costs, per cent14.61.70.0-0.05.60.510.810.44.72.8

The shapes the optimum has

Requiring a circle packing to be its own mirror image halves the number of coordinates a search has to find, so the same effort covers a much smaller space. Whether that helps depends on something the search cannot know in advance — whether the best packing was symmetric — and measured flap count by flap count the answer alternates without a pattern anybody could use.

design · Symmetry
what a cut adds, in letterssquare ×148 creases become 12 · 4 vertices either waysquare ×2832 creases become 40 · 16 vertices either waysquare ×31272 creases become 84 · 36 vertices either waytriangular ×11024 creases become 34 · 12 vertices either waytriangular ×22096 creases become 116 · 48 vertices either waytriangular ×330216 creases become 246 · 108 vertices either wayhexagonal ×11024 creases become 34 · 12 vertices either wayhexagonal ×22096 creases become 116 · 48 vertices either wayhexagonal ×330216 creases become 246 · 108 vertices either wayelongated ×11240 creases become 52 · 20 vertices either wayelongated ×224160 creases become 184 · 80 vertices either wayelongated ×336360 creases become 396 · 180 vertices either wayrhombille ×11248 creases become 60 · 24 vertices either wayrhombille ×224192 creases become 216 · 96 vertices either wayrhombille ×336432 creases become 468 · 216 vertices either waythe bar is how many creases the cut divides; nothing else about the two sheets differs

The rim is four letters a cell

Cut a rectangle out of a tessellation and it asks exactly the vertices the tessellation asks, exactly the same questions. What it adds is four free letters for every period of edge — the creases the cut divides, which become two independently answerable creases instead of one. Eight letters on a two-period square, sixteen on a four-period one, and nothing else about the two objects differs at all.

flat-folding · Boundary
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

The bottom layer is at the rim

A hundred and sixty-nine panels of folded tessellation, and three of them have nothing underneath. All three touch the paper's edge, and the same is true on every tiling at every size measured. Which panel is at the bottom of a stack turns out to be a fact about where the sheet was cut rather than about the pattern, and the pattern itself has no bottom at all.

flat-folding · Forced order
the Yoshimura at 6 by 5, at nine proportionsrow height 1.257 nodes30 labellings a vertex · 0.88 nodes a panelrow height 1.557 nodes30 labellings a vertex · 0.88 nodes a panelrow height 1.757 nodes30 labellings a vertex · 0.88 nodes a panelrow height 1.732050857 nodes30 labellings a vertex · 0.88 nodes a panelrow height 1.732050919 nodes8 labellings a vertex · 0.29 nodes a panelrow height 1.7419 nodes8 labellings a vertex · 0.29 nodes a panelrow height 1.819 nodes8 labellings a vertex · 0.29 nodes a panelrow height 219 nodes8 labellings a vertex · 0.29 nodes a panelrow height 2.519 nodes8 labellings a vertex · 0.29 nodes a panelthe equilateral Yoshimura is drawn at √3 = 1.732050808, on the dear side

A knife edge nine decimals wide

Draw the Yoshimura with its rows 1.7320508 half-columns tall and each vertex admits thirty labellings and the pattern costs fifty-seven steps. Draw it at 1.7320509 and each admits eight and it costs nineteen. The number between them is √3, which is the proportion everybody draws — and below it the sectors are unequal and the lemma is still silent, because the small ones sit next to each other.

flat-folding · Vertex degree
proving the glued square cell has no lettering1×1, 4 panels3proved there is none · the other test found one in 32×2, 16 panels35proved there is none · the other test found one in 93×3, 36 panels3,455proved there is none · the other test found one in 6254×4, 64 panels200,000still running at the budgeta bar at the budget is a search still running, not a proof

Pruning on proofs alone

A search that discards a branch it cannot prove wrong is not a search. Deciding whether a periodic pattern's layer relations really contradict themselves is far dearer than the disc's one-pass test, so the cheap test is asked first — it is sufficient, so it settles almost everything — and the expensive one runs only on what the cheap one rejects. Five of nine steps on a small cell, fifty thousand of fifty-seven on a large one.

complexity · Search order
proving the glued square cell has no lettering1×1, 4 panels3proved there is none · the other test found one in 32×2, 16 panels35proved there is none · the other test found one in 93×3, 36 panels3,455proved there is none · the other test found one in 6254×4, 64 panels200,000still running at the budgeta bar at the budget is a search still running, not a proof

The cost of proving something false

A search closing its whole tree is the strongest result this collection can produce, and on a glued tessellation it produces one that is wrong. What it costs to reach is three steps at one period, thirty-five at four, three thousand four hundred and fifty-five at nine, and more than two hundred thousand at sixteen — growing far faster than the cost of finding the lettering it says does not exist.

complexity · Hardness of folding
sliding the cut across one period of the square tessellation36 vertices at every position, and a different set of creases divided at each0102030cut at the start of a periodone period alongnodes; the axis starts at zero, and the whole spread is inside a factor of 1.32

Where you cut hardly matters

Slide the same rectangle across one whole period of the same tessellation and every position gives a different patch: different creases divided, different half-panels round the edge, panel counts from forty-nine to sixty-one. The cost of lettering them runs from twenty-five steps to thirty-three. Whether a cut is made changes the answer by three orders of magnitude; where it falls changes it by a third.

complexity · Typical instances
the twist patches: nodes against panels050100150one a panel0 panels157every vertex of this family keeps 4 labellings

The most decided vertex here

Sixteen ways to letter four creases; Maekawa allows eight; the big-little-big lemma allows four. A twist polygon's corner is one of the few vertices in this collection where the second cut applies, so it keeps four labellings where a grid, a leaf, a Miura and a crumple all keep eight — and the family the collection long called difficult turns out to be the one whose conditions decide the most.

tessellation · Twists
clipped tessellation patches, nodes per panel0.000.250.500.751.00one node a panelthe square gridthe triangular gridthe honeycombthe elongated triangular tiling0 panels413 panelsthe family the collection called hard is the one below the line

The edge was not what made it hard

Five families of pattern searched at one step per panel and a tessellation patch did not, and the property left standing after four alternatives were killed was having a rim. Measured under a fixed letter order the patches cost between a half and two-thirds of a step per panel, at every tiling and every size — below the line rather than above it, and the rim is why.

design · Sheet shape
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.

rigid · Self-contact
the same 2×2 glued cell, searched under two rulesa cycle is a contradictiona cycle whose steps add to zero isand what the loops dothe square gridnothing, in 359 nodesevery loop travels (2 directions)the triangular gridnothing, in 12,143455 nodesevery loop travels (2 directions)the honeycombnothing, in 9,6191,043 nodesevery loop travels (3 directions)the elongated triangular tilingnothing, in 9,123162 nodesevery loop travels (5 directions)the rhombille tilingunfinished at 200,000unfinished at 200,000“nothing, in n” is an exhausted search: a proof that the pattern has no consistent lettering, which is false

A test imported without its hypothesis

The rule that a loop in a folded sheet's layer relations proves the pattern cannot fold arrives from the layer-ordering literature, where the sheet is a disc and the panels are finitely many. This collection took the rule and not the sentence that says which sheets it is about, then applied it for years to patterns whose whole interest is that they repeat.

history · Rediscovery
the period cell of the Miuraone period, with its neighbours round it2 interior vertices in the cell7 crease pieces drawnperiod 1.000 × 2.000one column wide and two rows high, because the zigzag returns after twothe cell is a rectangle of ordinary paper until somebody says its edges are one edge

A mechanism that closes on itself

A rigid-foldable pattern is a mechanism: panels as rigid plates, creases as hinges, and a motion counted by degrees of freedom at each vertex. Close the sheet into a tube and the mechanism has to come back to itself after a circuit — a constraint that is not at any vertex and that the degree-of-freedom count does not see.

rigid · Rigid folding
02468101214010203040506070clearance above the basesurface, as a multiple of the basewalls stop at 40.0sheet 0.01, channel 0.05plies keep risingthey cross at 9.40the comb saturates at twice the reciprocal of its channel's share, and the stack does not saturate at all

The channel grows with what it feeds

A comb of standing walls beats a stack of plies by eight because its members do not share the clearance that pays them. Supply takes that back, and asymmetrically: a wall's channel has to be sized for the surface the wall carries, so it grows with the wall's height and is charged against the pitch, while a ply's channel is a constant charged against the clearance. The comb then saturates at twice the reciprocal of the channel's share, the stack does not saturate at all, and the two cross at a clearance the model gives in closed form.

biology · Surface in a volume
a body with four legs and a tail — one internal edge of 0.8circles of radius m·ℓ, and every pair's requirement drawn — the ones through the body ask for more than the two circles do

Every pair, not every circle

A uniaxial base is designed by packing a circle for each flap, and the circles are not the condition. The condition is that every pair of the subject's extremities be separated on the sheet by the distance between them through the tree — which for two flaps meeting at one point is the sum of their lengths, and for two flaps across a body is more. Circles are the case with no body in it, so a design read off circles alone is promised a base sixteen to thirty per cent larger than the sheet can give.

design · Uniaxial bases
a body of three segments — two internal edges, 0.6 and 0.9the segments are the pairs at their limit — 4 of 8 of them measured through the body rather than around it

What the condition does not decide

A tree of seven leaves imposes twenty-one separations and eight of them bind. The rest are slack, the eight pin six of the seven leaves against the sheet's own edges, and the seventh can be moved half a per cent of the sheet for nothing. The requirement that looks quadratic is doing linear work, and what it leaves undecided is the part a designer is actually choosing.

design · Uniaxial bases
what lengthening each edge of a body with wings, legs, a head and a tail coststhe scale falls from 0.2651 by this much per unit of extra length, measured by re-solving the arrangementchest–head0.0786a flap, 5.9% of the scale per 0.2rump–ll0.0471a flap, 3.6% of the scale per 0.2chest–rump0.0429the body, 3.2% of the scale per 0.2chest–wl0.0361a flap, 2.7% of the scale per 0.2chest–wr0.0264a flap, 2.0% of the scale per 0.2rump–tail0.0123a flap, 0.9% of the scale per 0.2rump–lr0.0120a flap, 0.9% of the scale per 0.2an edge no tight pair passes through is an edge the design can spend freely, and the condition says which

The price of a limb is not its length

Lengthening an edge of a subject's tree costs the design some of its scale, and the amount can be measured by re-solving the arrangement. It is not proportional to the edge, and it is not the body that is dearest. On a bird whose wings are twice its legs, the head — nine tenths of a unit against the wings' one and six — costs three times what a wing costs, and two edges of a lizard cost nothing at all.

design · Uniaxial bases
00.20.40.60.811.21.41.600.10.20.30.4arc from the pole (radians)excess, as a share of the circle7 rings5.0% stretchfirst at 0.35last at 0.98of the way to the rima ring goes in wherever the residual excess would otherwise pass what the material takes

Where a ring of divisions belongs

A pattern that divides the circle everywhere as finely as its rim requires is over-divided for most of its radius, because the excess grows from nothing. Putting a ring of new divisions in wherever the residual would otherwise pass what the material takes gives seven rings on a hemisphere at five per cent of stretch, at 0.35, 0.50, 0.62, 0.72, 0.81, 0.90 and 0.98 of the way out — and the first of those sits where a completely different criterion put its first tuck start.

material · Developability
the reach, the member length and the member count, against the anglesurface counted as a multiple of the base, lengths as multiples of the clearance90°60°30°10°the angle the members stand atthe reach: 200flat, at every anglehow long each member ishow many of them there areclearance 1, sheet 0.01 · the reach is 200 at every angle; the two factors move by 57

The angle the eight does not know

A comb's members are always drawn standing square to the base, and nothing has asked why. Lean one to an angle and it must be longer to reach the same clearance, which is more surface; it also takes more of the base to stand on, which is fewer members. The two are reciprocal and cancel exactly — the surface a comb holds is the same number from a right angle down to one degree, where each member is fifty-seven times the clearance long and there are two of them where there were a hundred.

biology · Surface in a volume
two linings of one tube, and the ceiling neither passessurface per unit length of tube, from a sheet 0.01 thickradiuslayersfins at bestthe ceilinglayers over fins0.516078.51572.040016353146282.020022526125725132.01004100785027100532.0050the ceiling is the tube's cross-section over the sheet thickness, twice over, and no arrangement of flat sheet passes it

In a tube the standing members lose

Members standing across a clearance beat layers lying along it by eight, and every drawing of that argument has a flat base under it. Curve the base into a tube and the ranking inverts: radial fins converge, so the room they need is the room at their tips, and their best arrangement fills exactly half the cross-section. Concentric layers fill all of it. The eight becomes a half, and the half is exact.

biology · Surface in a volume
123456780.40.50.60.70.80.91how many lengths of finshare of the ceilingtips equally spacedheights halvingtwo thirdsradius 1, sheet thickness 0.01 · the share of the ceiling 2πR²/τ that fins of m lengths hold

The wedge belongs to one length

Radial fins inside a tube reach at best half of what any lining of sheet could hold, because converging fins leave empty wedges behind their tips. Tapering the fins cannot help: the tip already sets the count, and a fin cannot be thinner there than the sheet it is made of. Fins of several lengths can. Counted along the radius they are a staircase under a straight line, and the staircase with m steps is best with its steps equally spaced, where it holds exactly m ⁄ (m + 1) of the ceiling. The factor of two belonged to fins of one length, not to fins.

biology · Surface in a volume

Named alongside it

The objects these essays reach for when they reach for this one.

TessellationAssignmentPanelPeriodicityBoundarySearch costLayer orderTrade-offThe big-little-big lemmaConstraint propagationDegrees of freedomDesign

All concepts