Origami & Folding
what one sheet can do
Concept

Self-folding — where it appears

A sheet that folds without being handled, driven by something built into it. It needs one input if the pattern has one freedom, and it needs the folded state to be determined by that input, which is not automatic.

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

-10010000.20.40.60.81how far the vertex is drivenstored energybranch oneMVMMbranch twoMVVVboth run downhillfrom the flat state,and end at zeroso the energy does notprefer either branch —the noise decides

Paper that folds itself

A self-folding sheet has to supply the fold and then choose what to fold into. The second half is where these things fail, and no amount of torque helps, because the two outcomes are equally downhill.

rigid · Self-folding one crease decided, and how much of the sheet followsthe flat-folding conditions, propagated2 of 12the rigid-folding conditions, propagated12 of 12and the rigid propagation leaves 1 consistent set of fold angles

One crease decides the sheet

Fix one crease of a flat-folding problem, propagate every condition the subject has, and three creases out of a hundred and fifty-eight follow. Fix one fold angle of a rigid one and every crease on the sheet follows, with a single consistent answer. The same experiment, two questions, opposite answers — and it is why a self-folding sheet needs one biased vertex rather than one per vertex.

rigid · Self-folding creaseworst amplification of an error in itc:0:11.76c:0:21.77c:0:31.74c:1:11.76c:1:21.77c:1:31.74c:2:11.76c:2:21.77c:2:31.74c:3:11.76c:3:21.77c:3:31.74r:1:01.00r:1:11.00r:1:21.00r:1:31.00r:2:01.58r:2:11.58r:2:21.58r:2:31.58r:3:01.65r:3:11.65r:3:21.65r:3:31.65the best crease is 1.77 times better than the worst, and it is on the sheet's edge

Which crease to push

Deciding one fold angle settles every other one on a quadrilateral mesh, which is what makes a self-folding sheet buildable with a single actuator. It leaves a question that sounds like an afterthought: which crease. Driving each of a mesh's twenty-four in turn gives twenty-four different answers to how far an error in it travels — and on the sheet that repeats one vertex, it gives several answers to what shape the sheet takes.

rigid · Self-folding the height is the worst amplification anywhere on the sheetone line per crease; the horizontal axis is the driven crease's own fold angle3.010.32.5fold angle of the driven creasea mesh with no two vertices alike6 of 6 creases are worst near the flat sheet0 steps refused as branch changes

The hardest instant

Driving one crease of a quadrilateral mesh settles every other one, and an error in the driven crease arrives elsewhere multiplied. That multiplier was measured once, at one fold angle. Followed along the whole motion it is worst at the flat sheet on twenty of twenty-four creases — and on the Miura the measurement has to refuse to answer.

rigid · Self-folding how many folded states each crease of a four-by-four Miura leavesdriven to 0.6 radians, with every consistent assignment enumerated rather than the first eight1 state4an actuator belongs on one of these2 states82 states, so the sheet has a choice4 states44 states, so the sheet has a choice8 states88 states, so the sheet has a choicethe four that leave one are c:3:2, c:3:3, r:3:2, r:3:3 — all of them at the same corner of the sheet

Only four creases decide a Miura

Driving one crease of a rigid quadrilateral mesh settles every other one — except that on the pattern everybody builds it often does not. Enumerated properly, four of a four-by-four Miura's twenty-four creases leave exactly one folded state and the other twenty leave two, four or eight. A mesh whose vertices all differ leaves one from every crease. The ambiguity is not a property of quadrilateral meshes; it belongs to the symmetry.

rigid · Self-folding 0123400.010.020.030.040.05gearing between the two creasesradians of errormoved at the firstleft at the secondworst at a gearing of onetwo actuators disagreeing by 0.05 radians, equal stiffness · the sheet settles where the stored energy is least

Two drivers and one freedom

Two actuators on a sheet with one degree of freedom are two commands for one number, and if they disagree by a hundredth of a radian the sheet cannot satisfy both. Where it settles is decided by the gearing between the two creases: a strongly geared pair absorbs the disagreement and leaves a quarter of it standing, while a weakly geared pair keeps ninety per cent. The loosest coupling is the expensive one, which is the opposite of what coupling usually means.

rigid · Self-folding the energy two actuators store in a disagreement, pair by pairenergy k₂δ² ⁄ (1 + k₂g² ⁄ k₁); the last column is the second crease's holding stiffness, k₂ + k₁⁄g², kept when k₂ is cut to a tenthgearingbalancing k₂ ⁄ k₁equal stiffnesssecond ten times stifferten times softerstiffness kept0.31610.010.915.000.09992%0.5393.440.772.560.09780%0.7841.630.621.400.09466%0.9341.150.531.030.09258%1.0001.000.500.910.09155%1.7610.320.240.310.07632%a mesh with no two vertices alike, driven at c:0:1 · energy stored in the fight, in units of δ² times the first actuator's stiffness

A gearing reflects stiffness squared

Two actuators on a sheet with one freedom disagree, and the sheet settles where their stored energy is least. With unequal stiffnesses the answer depends on them only through k₂g² ⁄ k₁ — the second actuator, seen from the first crease, is a spring of stiffness k₂g², the gearing entering squared as a gear train reflects any stiffness. That settles which actuator to make compliant. On a rigid mesh's loosest pair a second actuator ten times stiffer than the first stores fifty times the fighting energy of one ten times softer, and softening it gives up only 8 per cent of how firmly that crease is held, because the first actuator already holds it ten times over through the gearing. On the tightest pair softening saves four times the energy and gives up 68 per cent of the hold. Compliance is cheap exactly where the fight is expensive.

rigid · Self-folding a 4-by-4 Miura, every crease driven in turn, at 10 angles along the motionevery consistent assignment enumerated at each crease, with both configurations found at every vertexanglecreases × states they leavethe creases that decide it0.24×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:30.44×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:30.64×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:30.84×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:31.04×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:31.24×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:31.64×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:32.04×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:32.44×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:32.84×1 8×2 4×4 8×8c:3:2 c:3:3 r:3:2 r:3:3every row is the row above it: the census is a property of the pattern, not of how far it has folded

The deciding set does not move

A driven Miura leaves several folded states from most of its creases and exactly one from a few, and those few are where an actuator belongs. It was reported that the few change along the motion — four of twenty-four at 0.6 radians, fourteen at 0.8 — and that a five-by-five sheet had a crease leaving fifteen states where every other count was a power of two. Mapped at twenty angles from 0.1 to 3.0 radians on three sizes of sheet, neither survives. Every crease leaves the same number of states at every angle, every number is a power of two, and the same creases decide the sheet throughout. The changes were the vertex solver losing one of a vertex's two configurations on 138 of 8,640 solves, and the configurations it lost can be carried exactly from an angle where it finds both.

rigid · Self-folding

Named alongside it

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

ActuationRigid foldingBifurcationQuadrilateral meshActuatorBranch selectionGearingDegrees of freedomFold anglePropagationSymmetry breakingMiura

All concepts