The cell that repeats for ever

Every measurement in this field so far has had the assembly’s size in it. A Miura sheet’s redundancy is $(n-2)^2$; a tong’s error is $n$ times a unit’s; a ring’s kink angle is decided by how many pairs there are.

There is one thing left to do with a size, which is to take it away. The sheet’s own table runs from four panels to a hundred and forty-four and the trend in it is perfectly clear; this rung asks what is at the end of it.

A pattern that repeats for ever has no boundary, no total count of bodies, and no size at all. What it has instead is one cell, and the arithmetic of the whole thing is done on that.

How a cell is measured #

The construction is short. Give the cell a list of joints and a list of bars, and let each bar record which neighbouring cell its far end lies in: joint $i$ of this cell to joint $j$ of the cell $a$ lattice vectors along and $b$ up.

The rigidity matrix is then the finite one with the wrapped bars written against the translated position of their far end. Its columns are one cell’s coordinates. There is no boundary anywhere in it, and there is no approximation: the matrix is exact for the infinite pattern, and what it measures is the pattern’s behaviour under displacements that repeat with the same period.

The trivial motions are the two translations of the whole plane, so the count is

$$2n - b - 2.$$

The kagome cell has three joints and six bars, so it counts at $-2$: a structure with two redundant constraints.

What the two readings agree about #

Before the disagreement, the part that does not depend on the choice, because it is what makes the measurement worth making at all.

The cell is exact. Nothing about the periodic construction is an approximation of a large finite patch. The rigidity matrix written on one cell, with each bar’s far end taken in whatever neighbouring cell it lies in, describes the infinite pattern’s response to periodic displacements exactly. There is no truncation, no boundary condition to choose and no convergence to check.

And the identity holds under both readings. Freedoms minus dependencies equals unknowns minus constraints, with the unknowns and the trivial motions counted according to which reading is being made, and it is exact on all twelve rows of the table. The count is as right about the difference here as it is on a four-bar; what it is not is an answer.

A joint’s own bar contributes nothing when the period is held. A bar from a joint to its own image in the next cell is a constraint that constrains nothing: translating that joint moves both of its ends by the same amount, so the length is unchanged. A square grid’s two bars are both of that kind, which is why its rank under the held reading is nought and why both its constraints are dependencies. It is worth stating because it is the one entry in a periodic rigidity matrix that has no finite analogue — on a finite grid every bar does something, and it is only in the limit that a whole class of them stops.

Under the free reading the same bars do impose conditions, because stretching the lattice moves one end and not the other. Same bar, same graph, different matrix — which is the arithmetic behind the whole of this essay.

And then the question a finite network cannot ask #

Here is the thing that has no analogue in anything else in this field.

The construction above holds the lattice fixed. The joints may move within their cell, but the period may not change: the pattern is not allowed to stretch, shear or rotate as a whole.

That is a choice, and the other choice is available. Let the four entries of the lattice matrix be unknowns too. The count then gains four columns, and one more trivial motion — because rotating everything at once now rotates the lattice with it, which it could not do before — so it becomes

$$2n + 4 - b - 3 = 2n + 1 - b.$$

Two readings of the same infinite network, and they are not the same mechanism.

The graph of a pattern with no edge #

The loop arithmetic survives the translation and is worth stating, because it is where the infinite pattern stops being alarming.

What is left of the joint graph is the quotient graph: the cell’s own nodes, with an edge for every bar and a note of which neighbouring cell its far end lies in. The kagome cell’s quotient graph has three nodes and six edges, so its circuit rank is four — four independent cycles, on a graph of three nodes, because a bar may run from a node back to itself in another cell.

That is the honest generalisation of the loop count and it has one feature the finite case does not. A bar from a joint to its own image in another cell contributes a row of nothing at all under the held reading: translating that single joint moves both ends equally, so the bar’s length is unchanged and it imposes no condition. A square grid’s two bars are both of that kind, which is why its rank under the held reading is nought and its two constraints are both dependencies.

Under the free reading those same bars do impose conditions, because stretching the lattice moves one end and not the other. The same bar, the same graph, and a different matrix entirely — which is the arithmetic behind everything in this essay.

What changes #

A plain square grid — one joint per cell, bars to the right-hand and upward neighbours — is rigid with its period held. Two bars, two constraints, and the only motions are the translations: freedoms nought, dependencies two.

With the period free it has exactly one mechanism and no dependency at all. That mechanism is the shear everybody knows a square grid has.

The kagome does the same thing. Held, it is rigid with two dependencies; free, it has one mechanism and none — and the mechanism is the floppy mode the lattice is known for. Two dependencies among the constraints under one reading and none under the other, on one pattern.

The braced cases go the other way and are worth having as controls. A square grid with one diagonal per cell is determinate under the free reading: no mechanism and no dependency, count nought. Add the other diagonal and there is still no mechanism and now one dependency. A grid with five bars per cell has two dependencies and no mechanism.

So neither reading is a softer version of the other. Under one reading a bar can be redundant that is load-bearing under the other, and a mechanism can exist that does not exist.

Six lattices, read twice #

The table is worth walking, because each row says something different about what holding a period does.

The square grid is the simplest disagreement: rigid held, one mechanism free. The kagome is the same disagreement on a mechanism people care about — its free mode is the one that makes a kagome lattice notable.

The square grid with one diagonal is the row where both readings agree that there is no mechanism, and the free reading finds no dependency either: three bars, three constraints, and a determinate cell. That is worth having on the table because a determinate periodic cell is possible, and it is possible only under the free reading — held, the same cell has three dependencies and nothing to show for them.

Braced both ways adds a fourth bar and the free reading gains a dependency. The triangulated grid with five bars per cell has two dependencies free and five held, no mechanism either way, and is the row that shows the redundancy count climbing with the bracing exactly as it would on a finite truss.

And the last row — two joints, five bars — is the one that shows the two readings can agree about the mechanism count and disagree about everything else: no freedom under either, three dependencies held and none free.

Across the six, two change their mechanism count when the period is released and four do not. There is no rule visible in the table for which do, and none is claimed: it is a rank, taken twice, on two different matrices.

Where the freedom lives #

The square grid’s shear is worth looking at directly, because it is a freedom that belongs to nothing.

Every joint in the sheared grid is in a different place than it was. But relative to its own cell, no joint has moved at all: there is one joint per cell and it sits at the cell’s origin in both configurations. The whole of the motion is in the two lattice vectors.

The freedom belongs to the period, which is the one part of an infinite network that has nowhere to be counted. A finite patch of the same grid has this motion too, of course — a finite square grid shears — and there it appears as a motion of its joints, distributed over the patch, growing with distance from wherever is held. Take the patch to infinity and that description falls apart; the periodic one does not.

That is the honest reason both readings exist. Neither is right in general. The held reading is the right question when something outside the network fixes the period — a frame, a boundary, a substrate. The free reading is the right question when nothing does.

What a fixed period actually is #

It is worth being concrete about when each reading is the right question, because “does the period count as a body” sounds like a modelling preference and is not.

A repeating pattern set into a rigid frame has its period fixed by the frame: the number of cells across is an integer and the frame’s width is a number, so the cell’s width is decided from outside. That is the held reading, and it is the right one for a lattice bonded to a substrate, a grid welded into a border, or a repeating mechanism whose ends are bolted down.

A pattern free at its edges does not have its period fixed by anything. That is the free reading, and it is the right one for a deployable that has to change size, a sheet being folded in the air, or a lattice being tested for what it can do.

The difference is not a refinement. A square grid welded into a frame is a structure and the same grid loose is a mechanism, and a designer who computes one when they meant the other has an answer that is not slightly wrong.

There is a version of this on every mechanism in this field and it has been invisible because the assemblies were finite. Whether a lazy tong has four freedoms or one depends on whether its first unit is bolted down; whether a deployable ring has four or one depends on the same. The habit of subtracting the rigid motions hides the question because everybody agrees what a rigid motion is. On a periodic pattern the analogue of a rigid motion is the lattice itself, and nobody agrees, because the answer depends on what is outside the network.

What this does to the count #

The identity from the field’s first essay survives the translation intact. Freedoms minus dependencies equals unknowns minus constraints, on every row of the table and under both readings, with the unknowns and the trivial motions counted according to which reading is being made.

What does not survive is any expectation about which term is large. The square grid held has $m = 0$ and $s = 2$; free it has $m = 1$ and $s = 0$. The count changes from $-2$ to $+1$ and both changes are real.

Reading it back into the finite case #

The periodic measurement says something about finite sheets, and the something is a warning.

A large finite patch of a foldable or shearable pattern behaves like the free reading in its middle and like the held reading near anything that fixes it. So a measurement made on a small patch with its boundary pinned reports one answer — and pinning is what a count assumes when it subtracts three —, the same pattern with a free boundary reports another, and the difference does not go away as the patch grows — it moves to the edge and stays there.

For the Miura sheet in this field the boundary is free at every size, and the measured mobility of one is the free-boundary answer. Pinning a Miura sheet’s edge panel to a frame removes three of its coordinates and does not change its mobility, because the sheet’s one freedom is not a rigid motion. Pinning two edge panels does change it, and the sheet becomes a structure — which is how a deployable is held open, and which is a design act rather than a measurement.

What a cell cannot say #

Three limits, and the first is the one that would be easy to forget.

It only sees motions with the cell’s own period. A displacement that repeats every two cells, or every seventeen, or not at all, is invisible to this matrix. Such motions exist and are the subject of a whole apparatus — a wave vector, a band structure — that this field does not build. It is the same limitation a rank at one configuration has, moved from the configuration to the period. What is measured here is the mobility of the pattern under displacements as periodic as itself, and a mechanism could have others.

It says nothing about a real material. A kagome lattice of steel bars has stiffness, and the mode measured here is the direction in which it has least; converting one into the other needs a modulus and is not this field’s. The geometric statement is the count, and the count survives the bars being made of anything.

And it has no boundary, which is not a small idealisation. Every real repeating structure ends somewhere, and what happens at the end is often the whole of the engineering. What the periodic reading buys is the interior, exactly, with the edge removed rather than approximated.

Where the field ends up #

The route from the first rung to this one is a route through what a size can be.

A lazy tong’s size is a multiplier: everything about the unit, times $n$. A deployable ring’s size chooses a kink angle and then does nothing — four freedoms and four dependencies at every count of pairs. A Miura sheet’s size drives the count away from the answer quadratically while the answer stays at one. And a periodic pattern has no size, so the count is per cell and the only question left is whether the cell’s own period is a body.

That last question is the one this field could not have asked in any of its other rungs, and it is a fair place to stop: a network’s mobility depends on what is being held — as six things a network is not says from the other end, the count of bodies has stopped being a number, and the freedom that turns up belongs to the arrangement rather than to anything in it.

The question a finite network cannot ask is the real content of the rung and it is worth stating in its most general form. A quantity can depend on something that does not exist at finite size. A finite grid has no period — it has an extent, and its cells sit inside it — so the question is the period a body has no referent. Take the limit and the period becomes a real degree of freedom of the pattern, and whether it counts changes the answer. That is a genuinely unusual failure of the limit to be the limit of the answers: every finite grid is rigid, and the infinite one shears or does not depending on a convention that only exists in the limit. The lesson is worth carrying past lattices. When a limit introduces a new object, the limit of the answers and the answer about the limit are different questions, and the second is the one that needs its conventions stated. A reader taking a periodic result back to a finite assembly has to know which convention produced it, because on a finite assembly only one of the two readings is available at all.