The freedom that survives repetition

The Miura sheet is the cleanest object in this field, and the reason is that it can be made as large as anybody likes without becoming a different mechanism.

Fold one crease and the whole sheet follows. That is true of four panels and of a hundred and forty-four; it is what makes the pattern useful, and it is the thing to measure, because a mechanism whose behaviour does not change with its size is exactly where a count that does change with size can be caught doing it. The count in question is the one this site has been checking against a rank since its first field, and on a chain the two have never differed by more than a couple.

What the sheet is #

An $n$ by $n$ Miura pattern is a grid of parallelogram panels with the vertical crease lines zigzagging left and right by $b\cos\alpha$ from row to row, so that every interior vertex has sectors $\alpha$, $\pi - \alpha$, $\alpha$, $\pi - \alpha$. It is developable, like every flat crease pattern — a property that distinguishes nothing, since a vertex drawn on a flat sheet has no choice about it — and it satisfies the alternating-sum condition, which some vertices do and most do not.

Counting up: there are $n(n-1)$ horizontal creases between two panels and $(n-1)n$ zigzag ones, so $2n(n-1)$ unknowns. There are $(n-1)^2$ interior vertices, each contributing a closure that costs three scalars, so $3(n-1)^2$ equations.

The count is the difference:

$$2n(n-1) - 3(n-1)^2 = (n-1)(3-n).$$

Positive at two, nought at three, and negative for ever after. At six by six it is $-15$; at twelve by twelve, a hundred and forty-four panels, it is $-99$.

How the sheet is folded in the first place #

The flat state cannot be used as a starting point, and that is worth a paragraph because it is the first thing anyone tries.

At the flat state every crease direction lies in the plane, so each vertex’s three rows span two dimensions and the Jacobian is rank-deficient by one per interior vertex. A Newton step taken there goes into a subspace that leaves the sheet flat; the solve converges instantly to the configuration it started in and reports success. A rigidly foldable pattern has to be pushed off its flat state by hand, and which way it is pushed decides which of its branches it lands on.

The push used here is the pattern’s own mountain-and-valley assignment written as signs — a horizontal crease takes the sign of its row and a zigzag crease the sign of its row plus its column — with a magnitude of a tenth of a radian. Newton does the rest in six or seven iterations.

Seeded differently the solver finds a different answer, and the different answer is instructive rather than wrong. Given all four creases of a vertex at the same magnitude it converges to a configuration in which some creases come to rest at exactly nought and the sheet folds along the rest: a genuine rigid folding, with mobility one like the other, and not the Miura mode. A pattern’s branches are components of its configuration space exactly as a linkage’s are, and a solver lands on whichever one it was aimed at.

What the rank says instead #

One freedom, on every row.

The mechanism is folded first — Newton on the vertex closures, seeded with the pattern’s mountain-and-valley assignment written as signs, converging in six or seven iterations to a residual of $10^{-14}$ or better. Then the Jacobian is written at that folded state and its rank taken. At twelve by twelve that is a 363 by 264 matrix; its rank is 263, its nullity is 1, and the decision is made across a gap of $4.8 \times 10^{12}$ between the smallest singular value kept and the largest discarded.

The redundant column is the difference between the two answers, and it is a perfect square:

$$s = 3(n-1)^2 - \big(2n(n-1) - 1\big) = (n-2)^2.$$

Nought at two, one at three, four at four, nine at five, sixteen at six, thirty-six at eight, a hundred at twelve. That is not fitted; it falls out of subtracting the measured rank from the constraint count, and the measured rank is $2n(n-1) - 1$ on every row.

So the sheet at a hundred and forty-four panels has one freedom and a hundred constraints that repeat what the others have already said, and $1 - 100 = -99$ is a perfectly true sentence about a mechanism that folds.

The check that the fold is real #

A rank taken at a configuration is worth exactly what the configuration is worth, so it is worth saying how the folded state is established and what confirms it.

The unknowns are the fold angles. The residual is the logarithm of the rotation left over after going once round each interior vertex, three numbers per vertex, and it is computed by multiplying rotations — a purely algebraic route that never places a panel anywhere.

The check is geometric and independent of that. Every panel is walked out from the first one across a spanning tree of the panel graph, which gives each panel a rigid transform in space; then every vertex’s position is computed from each of the panels that meets it, one reading per panel. Those readings have to agree, and they agree only if the fold angles close every vertex loop.

The worst disagreement across the whole family is $4.5 \times 10^{-14}$ of a panel’s length, at twelve by twelve. That is not the same number as the closure residual and it is not computed the same way; it is the closure read in millimetres, on the drawing, at the place a reader would look.

There is a third route and it is the one that would have caught a wrong sign. The Jacobian above is written down from geometry — the crease directions in space, with a traversal sign and an edge orientation neither of which is guessable — while the residual is a product of rotations. Comparing the two by finite differences at a converged fold agrees to $3 \times 10^{-11}$, and the first version of the Jacobian, which had the frames right and the traversal wrong, disagreed by 0.35. Nothing else would have noticed: the solver still converged, on its damping, in twenty-seven iterations instead of three, and every figure would have drawn perfectly.

What grows and what does not #

Reading down the table by columns rather than by rows is where the field’s own point is.

Panels, creases, vertices and constraints all grow as the square of the side. Rank grows as the square of the side. Mobility does not grow at all, and redundancy grows as the square of the side less two. So of the six quantities on the table, five are quadratic in $n$ and the one a reader cares about is constant — which is the whole reason a count that mixes the quadratic ones cannot be trusted to report the constant one.

The identity from the field’s first essay holds on every row and is worth applying here to see what it does and does not buy. It says $m - s$ equals the count, exactly, so knowing the count and either nullity gives the other. It does not say which of the two is small. On a four-bar $s$ is nought and the count is the answer; on a twelve-by-twelve sheet $s$ is a hundred and the count is a hundred less than the answer, and there is no way to tell those two situations apart without taking a rank.

That is also why the redundancy column is the one worth reading. It is the count’s error, it is a property of the geometry rather than of the graph, and it is the quantity that decides how hard the pattern was to draw.

Where the repeated constraints are #

It is tempting to look for the hundred redundant constraints, and the temptation is worth resisting in a specific way.

There is no set of a hundred equations that could be deleted, leaving the rest independent, and no natural choice of which they would be. What the rank says is that the row space of a 363-row matrix is 263-dimensional. The hundred-dimensional space of dependencies is a subspace of $\mathbb{R}^{363}$: each of its vectors is a combination of the vertex closures that comes to nothing identically, for every configuration of the sheet, and it has a basis but no canonical one.

What the dependencies do have is a second interpretation, and it is the one that makes the number mean something. Because a dependency is a direction in residual space that no change of fold angles can reach, it is also a direction in which a change to the pattern cannot be absorbed. So the number of conditions on where the vertices may be drawn is at most the number of dependencies — and on this family it is exactly that. A hundred repeated constraints and a hundred conditions on the drawing are the same hundred, seen from either end, which is the subject of its own essay.

Why the sheet is not a special case of itself #

A reasonable objection at this point is that the Miura is one pattern, that its scaling is a property of that pattern, and that nothing general has been shown.

Half of that is right. The $(n-2)^2$ is the Miura family’s own number, and a different foldable tessellation gives a different one.

The other half is not. What is general is the shape of the result: a network’s mobility does not follow its count, the discrepancy is the redundancy, and the redundancy grows with the assembly because the thing being repeated is a constraint that repeats. The deployable ring shows it in a different representation, with a count of nought at every size and a mobility of four at every size and four dependencies wherever it is measured.

And the lazy tong shows the opposite: an assembly built from a repeated unit whose count is right at every size and whose redundancy is nought at every size. Repetition on its own does not produce dependencies. What produces them is repetition of a unit whose geometry has been chosen so that its constraints overlap — which is what a Miura’s parallelograms are and what an angulated element’s kink angle is.

Where the numbers came from, and what would have broken them #

Three things had to be right before any row of that table meant anything, and each of them was found by a check rather than by reading.

The rank has to be taken with an instrument that can see the null space. The routine this site had used everywhere until now accumulates the Gram matrix and takes its eigenbasis, which squares the condition number and cannot resolve a singular value below $\sqrt{\varepsilon}$ of the largest. Asked at the tolerance this table needs, it over-states the rank of every constraint matrix in the field.

The Jacobian’s two signs have to be right, and only a finite-difference comparison against the residual says so. On a symmetric pattern a wrong traversal sign is itself a symmetry and hides completely.

And the fold has to be a fold, which is what the independent vertex-position check is for. A configuration whose closure residual is small but whose panels do not meet is not a folded sheet, and it is the failure that would leave every number on the table plausible and wrong.

None of the three would have announced itself. Each was found by asking for a second answer to a question that already had one — which is the habit the whole site runs on, applied at a size where a wrong answer is no longer inspectable by eye.

What one freedom means at this size #

Three things follow from the measurement that are worth having explicitly, because they are what the number is for.

One number decides the whole sheet. Drive any crease to any angle in range and every other fold angle is determined — 263 of them at twelve by twelve, solved, not chosen. That is what a deployable structure needs and it is not a property most assemblies of panels have.

There is no partial fold. A mechanism with one freedom has a one-dimensional configuration space, so the sheet cannot be half-folded in one corner and flat in another. Every panel’s attitude is a function of the same parameter. A real sheet of card can of course be creased unevenly; what it cannot do is that while every panel stays flat and every crease stays straight.

And the mechanism is stiff in every direction but one. That is a rank statement of the kind a parallel platform makes about its own singularities, with the difference that here the count of constrained directions is in the hundreds. That is a statement about the rank rather than about a material: the constraint system leaves a single tangent direction, so any displacement off it violates a length or an angle somewhere. Turning that into a stiffness needs a material and is not this field’s; the geometric half is the count of directions, and it is $263$ against $1$.

One number to drive it and a hundred to draw it #

The two columns run opposite ways — one freedom at every size, and a redundancy growing as $(n-2)^2$ — and putting them together says something about the mechanism that neither says alone.

The freedom column is what makes the sheet easy to use. One number decides a hundred and forty-four panels; a single actuator deploys the whole thing; a single sensor knows where all of it is. That is the property deployable structures are built for, and it does not degrade with size at all.

The redundancy column is what makes the sheet hard to specify. Each dependency is a condition the drawing has to satisfy before any folded state exists, so a hundred dependencies is a hundred conditions on the pattern — and a pattern that misses them does not fold badly, it does not fold.

Those are the same number. The ease of driving and the difficulty of drawing are two readings of one square, and they grow together at the same rate. A larger Miura is more convenient to actuate and more demanding to draw, in exact proportion.

That explains a shape in the practice that would otherwise look like conservatism. Nobody draws a large origami mechanism freehand and then repairs it; every one is generated from a family — a parameterised construction that satisfies the conditions by symbolism rather than by fitting. The conditions are met once, at the family level, and then inherited by every member however large.

It also says exactly what a designer gives up by leaving the family. A freeform folded surface — panels all different, vertices fitted numerically — buys shape and pays the whole square: a hundred conditions to satisfy by solving rather than by construction, and no way to check the answer except by folding it. That is a real design activity and it is a different one from choosing a Miura’s angles.

Which is the honest summary of the field’s central object. The Miura sheet is not remarkable for having one freedom; a lazy tong has one freedom too. It is remarkable for having one freedom while carrying a hundred repeated constraints, and every property that makes it useful and every property that makes it fragile comes from that pairing.

The two ends of the table #

The row at the top is worth a sentence of its own, because it is the only row where the count is right.

At $n = 2$ there is one interior vertex, four creases and three equations: the count is $+1$ and the mobility is 1, with no dependency at all. The whole family’s disagreement grows out of a case where the arithmetic works perfectly, and it grows out of it smoothly — nought dependencies, one, four, nine — with no threshold anywhere and nothing that looks like an exception.

At the other end the limit is worth naming. Nothing here stops at twelve by twelve; the arithmetic runs to any size, and the honest way to take it to the end is to stop counting bodies altogether and measure one cell of a pattern that repeats for ever. That is the last rung of this field, and the surprise there is not that the count fails but that the answer depends on whether the pattern’s period is allowed to change — a freedom that belongs to no joint and that a finite sheet has no way to express.