Which crease to push

One crease decides the sheet ran the propagation experiment on rigid folding and got the opposite answer from the flat-folding version. Fixing one crease of a flat-folding problem settles one crease in eighty; fixing one fold angle of a rigid one settles the entire sheet, with exactly one consistent answer.

That is the result a self-folding sheet is built on. One actuator, not one per vertex, and the geometry does the rest.

It leaves a question that sounds like an afterthought and is not. Which crease? If any of them determines the rest, then an engineer has twenty-four equally good options and there is nothing more to say. There is more to say, and it comes in two parts.

Determining is not determining well #

A fold angle set at one crease arrives at a distant crease through a chain of vertices, and each vertex has a gearing — the fold-angle multiplier a degree-four vertex applies, measured off a solved motion at two hundred points. So a small error in the driven angle arrives elsewhere multiplied by the product of the gearings along the way.

That product is a property of a pair of creases, and the number an engineer wants is the worst of it: how far the largest amplification anywhere on the sheet is from one, for each possible choice of driven crease.

It is measured directly. Drive the mesh at ρ − h and at ρ + h, propagate both, and divide the change in every crease by 2h. No closed form is used anywhere, because a closed form would have to assume something about the vertex and the whole point of these meshes is that nothing is assumed.

The site has measured that gearing directly at two hundred points of a solved motion and checked it against a closed form the solver never saw. What is new here is not the factor at one vertex but the product across a sheet, and the fact that the product is what an actuator’s placement is a choice between.

What the survey says #

On a mesh with no two vertices alike, at a fold angle of 0.6 radians, the worst amplification runs from 1.00 to 1.77 depending on which crease is driven. At 0.9 radians it runs from 1.00 to 1.28.

Two things about those numbers deserve saying immediately. The first is that they are modest: choosing the worst crease rather than the best costs a factor of about two, not of ten. An error of a degree at the actuator arrives somewhere else as a degree and three quarters at worst, which is a real difference and not a catastrophe. Put beside the tolerance a solved mesh demands of its cutting — a third of a millimetre on a 150 mm sheet — the actuator’s placement is the smaller of the two problems by a wide margin, and it is the one that costs nothing to get right.

The second is that the number is never below one. A crease reports a gearing of exactly one about itself, so the worst amplification over the whole sheet cannot be smaller than that, and a driven crease whose worst is 1.00 is one whose error reaches everywhere else reduced — every other crease moves less than the driven one does.

Where the good creases are #

The survey has a structure that is more interesting than its range. Read the rows of that figure and the amplifications fall into groups: every crease in one row of the mesh reports the same number, and so does every crease in one column.

That is a strong statement about what the answer depends on. It does not depend on where along a row a crease sits, on whether it is near the middle or the edge, or on the size of the panels beside it. It depends on the row’s index and on nothing else.

The best crease on this mesh is a row crease in the first interior row — on the edge of the sheet rather than in the middle. That is worth stating because the intuition runs the other way: a driver at the centre has the shortest paths to everywhere, so it ought to accumulate the least. It does not, because the amplification is not a function of path length. The chains from a crease to everywhere else pass through the same vertices whatever the starting crease is, and what changes is which gearing the ratio is taken relative to.

Why the amplification depends on so little #

The grouping is worth taking apart, because a quantity that ignores position along a row is a quantity with a structure behind it.

The amplification from crease X to crease Y is a product of gearings along a chain from one to the other. On a mesh that folds, that product is independent of the chain taken — if it were not, the mesh would be asking two different things of the same crease and would not fold. So each crease carries a single number relative to any fixed reference, and the ratio of two of those numbers is the amplification between them.

That is why the worst amplification for a driven crease X is the largest of those numbers divided by X’s own. Choosing a driven crease is choosing a denominator, and the best choice is the crease with the largest number of its own — because then every other crease’s error is a fraction of the actuator’s rather than a multiple.

The grouping by row follows: on these meshes the number is constant along a row, so the choice is really a choice of row, and there are four of them.

Which makes the survey much cheaper than it was run #

The chain-independence has a consequence for the measurement itself, and it is worth stating because it turns a survey into a single reading.

If every crease carries one number $g$ and the amplification from $X$ to $Y$ is $g_Y / g_X$, then the whole table is fixed by the twenty-four values of $g$ — and those are obtained from one propagation. Drive any crease at $\rho \pm h$, propagate both, and the finite differences are $g_Y/g_{X_0}$ for every $Y$ at once. The common factor $g_{X_0}$ cancels out of every ratio the survey reports.

So the survey as described costs two solves per driven crease and forty-eight in total, and the same table is available from two. Nothing is approximated in the collapse: the values are the same values, obtained once instead of twenty-four times.

That also settles the two facts the essay states separately. The worst amplification for a driven crease is $\max_Y g_Y / g_X$, which is at least one because $Y = X$ is in the maximum — the crease’s gearing about itself. And the best crease is simply the one with the largest $g$, so ranking the actuators requires no comparison of surveys at all: it is one propagation, sorted.

The spread the table reports is then a single number, $\max g / \min g$: 1.77 at $\rho = 0.6$ and 1.19 at $\rho = 0.9$. Four rows, four values, one ratio.

The ambiguity counts are powers of two, and that says what is loose #

The other column admits the same kind of reading. Driving a Miura leaves one, two, four or eight consistent foldings — and every one of those is a power of two, with nothing in between.

A count of six or three would mean the residual freedom was a tangled thing. A run of powers of two means it is a set of independent binary choices, each of which can be made either way without affecting the others. Eight is $2^3$, so there are three of them on this sheet, and driving a crease removes as many as the crease happens to touch.

That turns the open question at the end of this rung into an arithmetic one. If the three choices are three lines of the pattern that may each flip, then a crease determines the sheet exactly when it lies on all three, and leaves eight when it lies on none — and the two intermediate counts say the crease meets one or two. The prediction is checkable on the same run that produced the counts, and it would explain them rather than list them.

The part that is not about error at all #

The second half is sharper, and it is a fact about the sheet rather than about the actuator.

Driving one crease of the general mesh leaves exactly one consistent assignment of fold angles to all twenty-four. Driving one crease of a Miura leaves one, two, four or eight, depending on which crease is driven.

So the sentence “one crease decides the sheet” is true of the mesh nobody could have drawn and false of the mesh everybody builds. A Miura driven at a badly chosen crease is a sheet that has been told its fold angle and not told its shape, and a self-folding version of it can settle into the wrong one.

The reason is the repetition. Every vertex of a Miura is the same vertex, so the two configurations available at each of them are the same two, and a choice made at one vertex is compatible with either choice at its neighbour more often than it would be on a sheet whose vertices differ. The property that makes a Miura manufacturable is the property that makes it ambiguous, and the two folded states are the subject of a rung of their own.

What that means for a sheet that folds itself #

Put the two halves together and the design rule has an awkward shape.

A self-folding sheet needs one driven crease. On a general mesh any crease will do and the choice is worth a factor of about two in accuracy. On a Miura — which is the pattern anybody would actually build, because its reason for folding survives being cut badly — the choice is worth a factor of about two in accuracy and the difference between a sheet that knows its shape and one that does not.

So the two properties an engineer wants pull apart. Robustness to manufacture wants the repeating sheet; determinacy of shape wants the sheet with no two vertices alike; and the mesh that is easy to make is the one that has to be told, by something other than its own geometry, which of its folded states to enter.

In practice that something is a bias: a crease pre-folded a little further than the others, or an actuator that pushes rather than merely holds. A self-folding sheet needs one biased vertex rather than one per vertex is the earlier rung’s statement of the same fact, and this rung says which vertex — the one whose crease leaves a single consistent folding.

Which theorem was checked, and how #

Three anchors, and the first is the one that makes the rest measurable.

A crease’s gearing about itself must be exactly one. It is, to a part in a thousand, and if it were not the whole survey would be reporting the arithmetic of the finite difference rather than a property of the mesh.

The propagation itself is the one the previous rung built, driven from an arbitrary crease rather than from the row crease at one corner — which is the whole extension this rung needed, since the earlier question only ever asked about one. A crease that changes branch between the two propagations is skipped and counted, not averaged. A branch change is not a derivative, and a survey that silently included one would report an amplification of several thousand for a reason that has nothing to do with the mesh.

And a mesh that does not fold must leave the survey nothing to measure. Run on a mesh drawn at random, every row reports no consistent propagation and the spread is computed over an empty list — so the check requires that case to come back empty rather than to produce a number.

The measurement in one table #

It is worth setting the four numbers down together, because the essay’s whole content is a comparison and the comparison is small.

The left column is the same on both sheets and the right column is not, which is the shape of the result: the accuracy question does not distinguish the two meshes and the determinacy question does. An engineer choosing between the patterns on the strength of the left column would find nothing to choose, and would be choosing between a mechanism that knows its shape and one that does not.

Where the model stops #

The gearing is a geometric ratio: how much one fold angle changes when another does. It is not a force, a torque or a mechanical advantage, and nothing here computes what an actuator would have to be able to push. A sheet whose worst amplification is 1.8 needs its actuator to be 1.8 times more accurate than the tightest crease requires, and how strong it has to be is a different question with a material in it.

The survey is also of a mesh with one degree of freedom, which is what a developable quadrilateral mesh that folds rigidly has. A sheet with several would need several actuators and the question would change shape entirely — which crease becomes which set of creases, and the counting is a different subject.

What the picture cannot show #

A bar chart of amplifications cannot show where on the sheet a crease is, and the interesting structure — that the answer depends on the row index — is visible in the figure only as a run of equal bars. The crease names carry the position and a reader has to decode them.

The mode figures have the opposite difficulty: they draw two folded states side by side and the reader has to compare them, when what matters is that a single sheet at a single fold angle is both of them until something decides. A picture of an ambiguity has to show the alternatives, and showing them makes them look like two objects rather than one undecided one. Nothing shows an error travelling. The quantity measured is a derivative, which is a statement about two configurations infinitesimally apart, and the picture of a wrong fold angle propagating through a sheet would be a picture of two nearly identical folded objects.

The generalisation #

The useful form of this is that a mechanism with one freedom has a choice of coordinate, and the choice is not free. Any of its variables can be used to parameterise the motion, and they are equivalent as descriptions and not as inputs: driving the mechanism through one of them applies a different distribution of sensitivity than driving it through another.

That is a general fact about single-degree-of-freedom mechanisms and it is why the question is worth asking of any of them. What is specific to folding is the answer’s shape: the amplifications group by row, the range is a factor of about two, and the best coordinate is on the boundary.

The ambiguity result generalises differently and more sharply. A mechanism whose components are identical has more symmetry than one whose components differ, and symmetry in a mechanism means multiple configurations at the same input. Manufacturing prefers identical components for obvious reasons; control prefers distinguishable ones. The Miura sits exactly on that trade and nothing about it can be adjusted to avoid it.

Who found it, and when #

Self-folding as an engineering subject dates from the 2010s and the single-actuator argument is standard in it: a rigidly foldable pattern with one degree of freedom needs one input, and the literature is largely about how to supply the input — heat, light, magnetics, pre-stressed layers.

The placement question does not appear to be asked in geometric terms. That is understandable: an engineer chooses an actuator’s position for reasons that are mostly not about closure sensitivity, and the factor of two this essay measures would be lost under any of them. What survives the objection is the ambiguity half, which is not a matter of degree — a sheet that can settle into the wrong shape is a control problem however the actuator is chosen.

Where the ladder goes next #

The immediate continuation is the ambiguity: for a Miura, which creases leave one folding and which leave eight, and whether the pattern of that is as structured as the amplification’s. If some crease of a Miura determines its shape uniquely, that crease is the one an actuator belongs on, and the argument would be complete.

A third question is whether the amplification’s row structure survives on larger meshes. Four rows is few enough that the pattern could be a coincidence of size, and a six-by-six mesh would settle it in one run — which is the sort of continuation that costs nothing and either confirms a structure or removes a paragraph.

The other direction is what happens with two actuators on a sheet with one freedom, which is over-determination rather than redundancy: two drivers that disagree by a hundredth of a radian are a sheet being asked to be in two places, and how it resolves that is a materials question with a geometric core.