The amplification along the whole motion, crease by crease
gearing-motion is one function. Everything below came out of it during this
build, at arguments taken from the essays rather than invented for this page — so a figure
here is the same figure a reader meets in an essay, and when the generator changes, this
page changes with it.
At its defaults
sub: "states", show: "split", delta: 0.05
sub: "states", show: "stiffness-curve", g: 0.3
sub: "states", show: "stiffness-curve", g: 2
What it checked while it drew
Collected by running this generator with a listener on the assertions, not written here. The count is how many separate times this build put that claim to the test.
- 76 steps refused on the Miura and 0 on the solved mesh, at the same threshold ×2
- and the worst crease leaves more states on a larger sheet, reaching 16 at 5×5 ×2
- on a 4-by-4 Miura driven to each of 10 angles from 0.2 to 2.8 radians, every crease leaves the same number of folded states at every angle, every number is a power of two, and the same 4 creases decide the sheet throughout ×2
- 4 of the 24 creases of a four-by-four Miura leave exactly one folded state when driven — c:3:2, c:3:3, r:3:2, r:3:3 — and the rest leave two, four or eight ×1
- a step is refused when a crease's angle moves by more than a fiftieth of a radian while the driven one moves by a ten-thousandth, which is not a derivative ×1
- and a strongly geared pair leaves almost none of the disagreement at the second — 5.9 per cent of it at a gearing of 4.0 against 100 at 0.05 ×1
- and every count is a power of two — 1, 2, 4, 8 — because each vertex reached with only one crease known has two configurations and the choices are independent ×1
- and the share of a disagreement left standing at the second actuator runs from 91 per cent down to 24 — the weakly geared pairs are the expensive ones ×1
- and two pairs with the same k₂g² leave the same disagreement standing and move the other crease by the same share — the stiffness a gearing reflects is multiplied by its square ×1
- at 0.8 radians some creases of each sheet decide it and some do not — 4 of 12 at 3×3, 4 of 24 at 4×4 ×1
- at 1.2 radians some creases of each sheet decide it and some do not — 4 of 12 at 3×3, 4 of 24 at 4×4, 6 of 40 at 5×5 ×1
- each crease is driven in turn and the whole sheet re-propagated on each side of the angle, so the amplification is measured rather than differentiated symbolically ×1
- on a mesh with no two vertices alike every one of its 24 creases determines the folded state uniquely — the ambiguity is not a property of quadrilateral meshes ×1
- one crease of this mesh reaches the others at 6 distinct gearings, from 0.32 to 1.76 ×1
- one crease of this mesh reaches the others at 6 gearings; a second actuator ten times stiffer than the first stores 5.0δ² of fighting on the loosest pair and 0.31δ² on the tightest, while ten times softer it stores under 0.10δ² on every pair ×1
- over 8,640 vertex solves on a 4-by-4 Miura, the fixed lattice of starting points finds only one configuration 138 times, and carrying the configurations from a well-conditioned angle recovers the second on every one of them ×1
- the settled position found by minimising the stored energy directly agrees with k₂gδ ⁄ (k₁ + k₂g²) to 1e-9 at every gearing and stiffness ratio tried ×1
- the sheet moves furthest from the first actuator's command when the gearing between the two is one, and less at every other gearing ×1
- the two branches of the motion give different assignments, so the vertex genuinely has a choice to make and the figure has something to show ×1
- while on the Miura, which is one vertex repeated, only 4 of 24 do — so it is the symmetry that produces the choice ×1
Where it is called
Changing this generator changes every figure on this list, which is what makes the list worth publishing rather than keeping in a check script.
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.
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.
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.
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.
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.
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.
Every generator · The rigid folding field · The patterns a reader can fold