The count says how many and not where

Every network in the field so far has been a mechanism somebody wanted: a tong that reaches, a ring that deploys, a sheet that folds, a grid that a brace makes rigid. The question each time was whether the thing moves and how many ways. The cell that repeats for ever asked it of a lattice with no edge at all, and found that the kagome lattice — a pattern of corner-sharing triangles — is rigid when its period is held and has one mechanism when it is not.

Nothing is built for ever. A lattice in a real material, a folded panel or a deployable surface is a piece of one, and a piece has an edge. This essay is about what the edge does, and it turns on a lattice that counts to exactly nothing.

Three joints and six bars in every cell #

A cell of the kagome lattice holds three joints and six bars: three bars of the triangle pointing up, and three of the triangle pointing down that joins it to its neighbours. In the plane a joint brings two freedoms and a bar removes one, so the count for one cell is 2 × 3 − 6 = 0. Every freedom a joint brings is taken by a bar. A lattice like that is the borderline between one that moves and one that is rigid, and in the bulk it has no freedom to spare in either direction.

Cut a rhombus of L cells a side out of it, keeping each cell’s three joints and every bar whose two ends the rhombus holds. The triangles along the edge are left incomplete: along one side of the rhombus every down-triangle has lost all three of its bars, along another it has lost two, and the joints on the remaining two sides have lost the down-triangles they would have shared with cells outside. Every joint still brings two freedoms. So the patch has $L^2$ cells’ worth of freedoms and $6L^2 - (5L - 2)$ bars — $5L - 2$ short of six a cell — and, less the three rigid motions of the whole patch, 5L − 5 mechanisms.

The rank of the rigidity matrix confirms it at every size from two cells to eight: 5, 10, 15, 20, 25, 30 and 35 mechanisms, and not one bar redundant. It is the count at its best. On the grid of squares the count could not say which braces made a difference, and on the three parallel bars it called a mechanism a structure. Here every bar removes a freedom no other bar removes, the count’s one assumption holds, and its answer is exact.

The figure plots a second lattice with the first, and the count is identical for it too. That second lattice is where the argument starts.

Turning the triangles #

Take every triangle that points up and turn it about its own centre by some angle, leaving the cells where they were. The triangles that point down are dragged into new shapes, but every joint is still in one up-triangle and one down-triangle, and every bar still joins the same two joints. The lattice is the twisted kagome. Its graph is unchanged, so Maxwell’s count per cell is still nought, and the patch still has 5L − 5 mechanisms.

The two patches differ in one thing the count cannot see. In the straight lattice, bars line up. Every bar of an up-triangle continues straight through a joint into a bar of a down-triangle, and those continue in turn, so the patch is crossed by straight lines of bars from one edge to the other in three directions. A line like that carries a mechanism of its own. Hold every joint of the patch except the sixteen on one row of bars eight cells long, and those sixteen can still move: each slides along the line by the same amount, and alternate joints step off the line to either side by tan 30° of that — 0.577 of it — which is exactly what keeps the bars leaving the line at sixty degrees from changing length to first order. The bars along the line only turn. It is a flex of the rigidity matrix that runs from one edge of the patch to the other, and each of the patch’s eight rows has exactly one.

Twist the triangles by any amount and every line is kinked at every joint. Nothing crosses the patch straight any more. Twisted by 17.2°, seven of the eight rows held that way have no mechanism at all; the eighth is the bottom row, where the incomplete triangles of the edge leave it one.

Where the lost bars are #

The count treats the patch’s 5L − 2 missing bars as a single total. They are not spread evenly, and the way they are spread is already a statement about where the mechanisms will be — though only a partial one.

On the side where the down-triangles have lost all three bars, every up-triangle still hangs from its neighbours by two joints, and a triangle held at two joints cannot turn. On the side whose joints have lost the down-triangles they share with the cells outside, each up-triangle hangs by a single joint and can swing about it. The bottom of the rhombus sits between the two: each down-triangle there keeps one bar of its three, a single strut where there should be a triangle. The same number of missing bars can therefore leave a stiff edge or a floppy one, depending on which bars they were.

That is still a statement about the graph, and it is the same for the straight and the twisted patch, because both have the same graph. It says which edge will carry more of the mechanisms. It says nothing about how far into the patch they reach, and that is where the two lattices part company. A count of lost bars per side cannot separate them; neither can the network’s graph, which each unit of a network hides its failure in as readily as its success.

Measuring where a mechanism is #

A patch with thirty-five mechanisms has a thirty-five-dimensional space of them, and any one mechanism drawn from it is a choice. The decomposition that computes the null space returns one basis; a different method returns another; and a picture of the first basis vector is a picture of how the method happened to arrange the space rather than of the space.

The measurement that does not depend on the choice is the projector onto the mechanism space. Its diagonal gives every coordinate a number between nought and one, and adding a joint’s two gives that joint’s weight: how much of the mechanism space moves it. A joint that no mechanism can move has weight nought. The weights over the whole patch add to the number of mechanisms, 35.000 to twelve decimal places, whatever basis the decomposition returned.

On the straight patch the outermost ring of cells has a mean weight of 0.250 a joint and the innermost 0.118. The edges are heavier because the lost bars are there, and they are not equally heavy: on the left-hand side of the rhombus each edge triangle hangs from its neighbours by a single joint and can turn about it, and that side carries 11.2 of the 35, while the other three sides carry about 5 each. But the middle is far from empty. A joint three cells from any edge still carries nearly half what an edge joint does, and that is the lines of bars reaching in.

The figure at the head of the essay is the same patch with its triangles turned by 0.3 radians, 17.2°. The count is the same and the pictures are not. The edge joints are heavier, 0.341 a joint on the outer ring, and the middle has almost nothing: 0.015 a joint, less than a twentieth of the edge.

The mechanism that reaches furthest in #

Mean weights say where the space is on average. A sharper question is whether any single mechanism of the patch puts most of its movement in the middle, and it has an exact answer that again involves no choice of basis.

Take the joints two or more cells in from every edge — a quarter of the patch. Of every combination of the patch’s mechanisms, find the one with the largest share of its squared movement at those joints. That is the top singular vector of the mechanism basis restricted to the middle joints, and its share is the square of the top singular value: the best that any mechanism of the patch can do.

On the straight patch that mechanism puts 59% of its movement in the middle quarter. Its largest movements are there rather than at the edge, and every bar keeps its length to first order, to about 10⁻¹⁴ of the largest displacement. The middle is a place the straight patch can move.

On the twisted patch the best any mechanism can do is 16%. That is less than the 25% the middle joints are of the patch, so no mechanism of the twisted patch even treats the middle fairly, let alone favours it. The one the computation finds still does its moving at the left edge, and every other mechanism does less in the middle than it does.

The two patches have the same joints, the same bars, the same count, the same rank and the same number of mechanisms. What the measurement separates is the one thing a count never had access to: the directions of the bars.

How fast the edge lets go #

The twist is a dial, and the obvious next measurement is how the answer changes with it.

Every curve starts high at the edge, where the lost bars are. The straight lattice’s levels off: 0.250, 0.135, 0.124, 0.118. With a small twist of 2.9° it barely changes. From about 6° the curves begin to fall away from the edge, and by 17° the innermost ring has a twentieth of the outermost’s weight. The more the triangles are turned, the more of the weight crowds into the outer ring — 0.363 a joint at 25.8° against 0.250 straight — because the same 35 mechanisms have to live somewhere, and the middle is the place they cannot.

The rate at which the weight falls from one ring to the next measures how far a mechanism reaches. Between the two innermost rings of the eight-cell patch it is 0.195 a cell at a twist of 0.1 radians, 0.479 at 0.2, and 0.735 at 0.3 — decay lengths of 5.1, 2.1 and 1.4 cells.

Over every twist measured the decay per cell is between 1.9 and 2.4 times the twist in radians. That is an empirical law of this patch and not a derivation, and the patch limits it in both directions. A decay length much longer than the patch cannot be measured on it: the straight lattice’s rate of 0.053 is a length of nineteen cells on a patch of eight, which says only that the weight has not fallen. And a decay length shorter than a cell leaves only the outer ring with any weight, so the rate between two inner rings saturates, as it begins to at 0.45 radians.

What the law says plainly is the order of magnitude. A twist of a few degrees gives mechanisms that reach tens of cells. A twist of fifteen or twenty confines them to the outer two or three.

Why a count cannot be refined into this #

It is tempting to hope that a cleverer count would see the difference: count lines of collinear bars, perhaps, and add a freedom for each. The measurement rules it out, and for a reason worth stating.

Both patches have exactly the same mechanisms by number. The straight lattice’s lines do not add mechanisms; they change which mechanisms there are. Any count, however refined, returns a number, and the number here is already right for both patches. What differs is a property of a subspace — where in the patch it sits — and no integer describes a subspace’s position. The quantity that does is continuous: the decay length, or the weight three cells in, and both move smoothly with the twist while the count does not move at all.

The same shape of answer turned up where a folded sheet meets its flat state. There, the rank reported freedoms that were not motions, and the correction was a second-order computation that no count could have replaced; here the rank is right and the correction is a projector. In both cases the extra information is geometric — which way a bar points, how two constraints curve — and in both it is invisible to anything that reads only which joints are joined.

This is the reverse of the loops argument. There, the count was wrong about a number and the rank corrected it. Here the count is right about the number, the rank agrees, and both are silent about the thing that matters to anyone who makes the patch: whether pushing on its edge moves its middle.

What the edge buys a designer #

A floppy boundary and a stiff interior, from one pattern. A patch of twisted kagome is a sheet whose every mechanism is near its edge. Held at the edge, it has no mechanism left that moves the middle by more than a small fraction, and the fraction falls with every cell of depth. The straight lattice held at the same edge still has line mechanisms crossing it. The two differ only in the shape of the down-triangles, which is a drawing decision.

A reason to hold one side and not another. The rhombus’s left-hand side, where each edge triangle hangs from a single joint, carries about a third of the weight on both lattices. Pinning those joints removes a larger share of the mechanisms than pinning any other side, which is the kind of choice holding the period of a lattice made for the infinite case.

A pattern condition that is not a condition on the pattern’s count. A crease pattern has to satisfy conditions before it folds at all, and each condition is a dependency among constraints that the count cannot see. The twist is a different kind of design variable: it changes nothing a count or a rank reports and everything about where the patch gives. A designer who checked a lattice patch by counting and by rank would pass both of these patches as identical, and one of them moves in the middle.

A dial with a known direction. A twist of 0.3 radians gives a decay length of about one and a half cells, and a twist of 0.1 about five. A designer who wants a surface that is soft only within a stated distance of its edge can read the twist off a curve like the one above for that pattern and size.

What these measurements do not settle #

They are first-order. A mechanism here is a velocity the rigidity matrix permits, and a first-order flex of a framework with no redundant bars does extend to a motion for small displacements, because the second-order obstruction needs a dependency among the bars and there is none. How far each mechanism goes before it runs into a collision or a flattened triangle is a different computation, and it is not made here.

They are on one patch shape. A rhombus has two kinds of edge, and its left-hand side is special. A hexagonal patch, or a rhombus cut at a different angle through the triangles, loses its bars differently, and its count and the share of the weight on each side both change. The decay into the interior should not, because it is a property of the lattice rather than of the cut, and that is a claim the measurement has not yet been asked to make.

They say nothing about a lattice that counts to more or less than nought. A lattice with spare bars in every cell is rigid inside whatever its edge does; one with too few moves inside regardless. The edge-versus-middle question is sharp exactly on the borderline, which is why the kagome was the lattice to ask it on.

Still open: an edge that knows which side it is on #

Both lattices here have edge mechanisms on every side of the rhombus, in roughly the proportions the cut dictates. There is a way of distorting the kagome lattice — shifting each up-triangle’s joints unevenly rather than turning them all by one angle — for which the count on each side of a patch stops being set by the cut alone: one side of a strip carries more mechanisms than the lost bars at that side account for, and the opposite side correspondingly fewer, while the total is still Maxwell’s count.

The distinct argument there would be to measure that imbalance on a strip rather than a rhombus: the mechanisms counted by weight on each long edge, as a function of the distortion, with the total held by the count. If the imbalance is a whole number that does not change as the distortion is varied smoothly, and jumps only when the lattice passes through a straight configuration like the one this essay started from, then it is a number the lattice carries in its cells and not in its edge — and the straight kagome, with its lines of bars, is the boundary between two lattices that cannot be told apart by any count and put their mechanisms on opposite sides.