Machine Kinematics
solved, then drawn
Field

Many of one thing

Every mechanism here so far is a chain: a handful of links, one or two loops, and a mobility somebody can check on the back of an envelope. A scissor lift, a folded sheet and a deployable ring are one small unit repeated, and three things change at once. The count of bodies is a parameter; mobility becomes the rank of a matrix rather than an arithmetic; and a unit that moves can be rigid the moment it is joined to another of itself.
Six assemblies, one routine, three disagreements and one accident. Every row is the same three steps: write down the constraint Jacobian, take its rank, and subtract it from the number of unknowns. The representations differ — bars between points, bodies joined by pins, panels joined by creases, one cell of a pattern that repeats for ever — and the routine does not. The counted column is the arithmetic on the numbers of bodies and joints; the measured column is the nullity of the matrix. They agree on the lazy tong and on the kagome cell and disagree on the other four, most sharply on the deployable ring, which the count declares immobile and which is sold as a mechanism that opens. The right-hand column is the reason: constraints that repeat what another constraint has already said, which the count has no way of seeing and the rank cannot help seeing. The fourth row is worth reading twice: the count says nothing can move and nothing can, so the two agree — and they agree for the wrong reason, because that pattern's flat state shows four freedoms and not one of them is a motion.

Many loops, one freedom

What changes when a mechanism is many copies of one thing: six assemblies put through one routine, and the four rows where counting and measuring disagree.

 The mechanism, and the graph that decides how many loops it has. A lazy tong of 5 scissor units drawn over its own joint graph: a node for every body — 10 of them — and an edge for every pin, 13 of those. The number of independent loops is e − v + 1 = 13 − 10 + 1 = 4, which is how many closure equations somebody writing this mechanism out by hand would have to find and is the one quantity in the field that can be read straight off a drawing. It is also all the count knows: Grübler's 4 is 3(n − 1) − 2j and contains no geometry at all, which is why it is right here and wrong four rows further down the ledger. positioned by solving, not by drawing.

The loops are in the graph

Where a mechanism's closure equations come from: the circuit rank of its joint graph, what the count is made of, and the loop no vertex can see.

 What a pattern that cannot fold leaves behind. The best a least-squares solve can do with the vertex closures, against the fold it is asked for. The Miura pattern closes at every angle, at the arithmetic's own floor — the line along the bottom is 10⁻¹⁵ and below. The same grid with its vertices moved by a tenth of a panel does not close at any angle at all: its residual starts at 5.9e-6 at the smallest fold and grows with it, and no seed and no number of iterations moves it. Both patterns have the same panels, the same creases, the same graph and the same developable vertices, and every one of those vertices folds perfectly well on its own.

Each one moves, and together they do not

The compatibility question a network raises and a chain cannot: units that each move, joined into an assembly that does not, and what actually decides which patterns fold.

 Six sizes, one freedom, and a count going the other way. Creases and constraints both grow as the square of the sheet's side, and they grow at different rates: two per panel against three per interior vertex. So the counted column runs (n − 1)(3 − n) and is positive at two, nought at three and increasingly negative after that, while the measured mobility is one on every row. The redundant column is the difference and it is exactly (n − 2)² — one at three, four at four, nine at five, thirty-six at eight. A twelve-by-twelve sheet of a hundred and forty-four panels is counted at minus ninety-nine and has a hundred repeated constraints, and it is the same mechanism as the smallest one on this table.

The freedom that survives repetition

What happens to a network's arithmetic as it grows: mobility measured at seven sizes, a count that diverges, and a redundancy that is a perfect square.

 A vertex is a spherical linkage, and the sectors are its link lengths. The four creases of one folded vertex, drawn as directions from the vertex itself, with the great-circle arcs between consecutive ones. Those arcs are the sector angles of the flat pattern — 80°, 60°, 100°, 120° — and they are those angles at every fold, to 8.9e-16 radians. That is the whole of the claim in the title: four axes through a point at fixed arcs from each other is a spherical four-bar, the object this site's spatial field built two phases ago, and a crease pattern's vertex is one of them with the arcs printed on the paper. The dihedral angle of the sheet at each crease is a half turn less that crease's fold angle, which here run 68.8°, 14.4°, 68.8°, 14.4°. positioned by solving, not by drawing.

Every vertex is a spherical linkage

What a crease pattern's vertex is, as a mechanism: the arcs that are its link lengths, the two branches it has, the relation that holds along one of them, and the degree that cannot fold.

 What the count is right about. Freedoms minus dependencies, against what the count predicts, on seven assemblies from three different representations. The difference is exact every time and it is exact for a reason that has nothing to do with mechanisms: the count is unknowns minus constraints, the rank is a number no larger than either, and the two nullities are what each of them has left over. So a count is not wrong in the way a mismeasurement is wrong. It is a statement about a difference being read as a statement about one of the terms — and on four of these seven rows both terms are large and the difference is nearly meaningless.

A constraint that has been said already

The second null space of a constraint Jacobian: what a dependency among constraints is, why it is not a force, and how it accounts for every disagreement in the field.

 The ring a count says cannot move. 8 pairs of angulated elements, each pair two mirror-image bent bars pinned at their kinks, with each pair joined to the next at two pins on a common radius. 16 bodies and 24 pins give Grübler's 3(n − 1) − 2j = 0: no freedom at all, a structure. The rank of the constraint Jacobian is 44 of 48, which leaves 4 — the three rigid motions of the whole ring and one deployment — and 4 constraints that repeat what the others have already said. The kink angle is not a style: it is 135.0000°, a half turn less the 45.0000° the ring subtends per pair, and any other value gives a ring that will not deploy. Inner radius 0.6876, outer 2.2688. positioned by solving, not by drawing.

The ring that closes at every size

Why an angulated element makes a deployable ring and a straight bar cannot: the invariant angle, the closure condition derived rather than quoted, and a deployment the count calls impossible.

 The network where counting works. 5 scissor units, each two bars pinned at their middles, with consecutive units sharing their end pins. Sixteen bodies and twenty-two pins at eight units, and Grübler's 3(n − 1) − 2j gives 4 — three rigid motions and one internal freedom — against a measured nullity of 4 and no redundant constraints at all. That holds at one unit and at sixteen. The tong is here to make the field's point in the direction nobody expects it: a network is not a place where counting fails, it is a place where counting stops being checkable by hand, and one of the two assemblies that spans this plate is counted perfectly. Span 8.1388 mm at this opening, which is 5 times one unit's 1.6278. positioned by solving, not by drawing.

One input at one end

The one network in this field whose count is right at every size, what its exactness buys, and where one unit's error ends up.

 The smallest mechanism that is not one. Two bars from a free joint to two pinned ones, with the three points in line. The constraint matrix has two rows and two columns, both rows are horizontal, and its rank is 1: one freedom left over, pointing straight up, and one dependency among the two bars. Move the joint up and neither bar changes length to first order, which is what the freedom says. To second order both bars get longer, by the same amount and in the same direction, and there is nothing to trade off against — which is what the dependency says. The obstruction is the dependency applied to the second-order stretch and comes to 1.414214; anything but nought there and the freedom is not the beginning of a motion. Lifted by 0.34 the bars are 0.05622 longer, which is the whole argument drawn to scale.

It moves to first order and not at all

A nullity is a candidate for a motion, not a motion: the second-order test that refuses one, the walk that decides, and a framework the count calls rigid that moves freely.

 Two branches, meeting where the sheet is flat. The same vertex's two folding modes, plotted as one crease's fold angle against another's. Both curves pass through the origin, which is the flat sheet, and they cross there and nowhere else. That crossing is what a rank cannot see: at the origin the tangent directions of both branches are available to the constraint matrix at once, so the nullity there counts them all and the mechanism has only one of them once it has left. Every branch argument made in these essays — a four-bar's assembly configurations, an arm's eight postures, the components a solve's paths turned out to run between — is this picture with different axes.

Where the branches meet

The flat state of a crease pattern as a bifurcation: why the rank drops there, what the extra nullity is, and why the same numbers come out for a pattern that folds and one that cannot.

 What the repeated constraints cost the drawing. Move an interior vertex of the flat pattern and the folded state generally stops existing. It survives if the change to the vertex closures can be absorbed by a change in the fold angles — and the part that cannot be absorbed is exactly the part that lies along a dependency, because a dependency is a direction in residual space the fold angles cannot reach. So the number of conditions a pattern's shape has to satisfy is at most the number of dependencies among its constraints, and on the Miura family it is exactly that: one at three by three, four at four, nine at five, measured by taking the rank of the obstruction. A twelve-by-twelve sheet has a hundred conditions on where its vertices may be. That is why a grid whose vertices are anywhere at all does not fold, and it is the same number, read the other way round, as the amount by which the count is wrong.

What a pattern has to satisfy

The number of conditions a crease pattern's shape must satisfy before it folds at all, measured as the rank of an obstruction map, and why it equals the redundancy of the constraint system.

 A network with no boundary at all. The kagome lattice, drawn out to 5 cells across and continuing for ever. The measurement is made on one cell: 3 joints, 6 bars, and a bar that leaves the cell comes back into it, written against the far end's position in the neighbouring cell. There is no boundary anywhere in the arithmetic, and the size of the network has gone from being a parameter to not existing. The highlighted triangle is the cell; every other line on the page is a copy of one of its six bars. positioned by solving, not by drawing.

The cell that repeats for ever

A network with no boundary, measured on one cell: what the count becomes, and how the answer depends on whether the pattern's period is held.

 A 4-stage scissor, open and as flat as it goes. On paper a scissor stack's height is n·L·sin φ and goes to nothing as the bars lie down. The bars are made of something: the two colours are the two planes the stack needs — two, at any number of stages, which is why a scissor folds at all — and the bars that share a plane are the ones two stages apart. They meet at φ = 0.0951, leaving each stage 0.0949 high, which is exactly the two bosses that touch: 2 × 0.0473. The stowed height of the whole stack is 0.380 rather than zero, and it grows with the stage count in the way a stowed lift's does.

A stack that has to fit

How flat a scissor stack folds once its bars have a width: two planes at any stage count, a fold angle bisected for, and a stowed height that is a boss diameter per stage.

 5 braces, and it is rigid. A 3×3 grid of squares with 5 of its cells braced by a diagonal, and no freedom left. The bipartite graph on the 3 columns and 3 rows, with one edge per braced cell, has 1 component — and the number of freedoms is one less than that, at every bracing there is. Nothing in the rank computation knows about columns, rows or graphs.

Which diagonal rigidifies a grid

Every bracing of a small square grid enumerated, with the rigidity matrix's nullity checked against the component count of the bipartite graph on columns and rows, and the minimal bracings counted against Cayley's formula.

 The same patch twisted 17.2°: 35 mechanisms, all at the edge. A rhombus of 8 × 8 kagome cells — 192 joints, 346 bars — with every up-pointing triangle turned by 17.2° about its own centre. Each joint is drawn with an area proportional to its weight: its share of the diagonal of the projector onto the patch's mechanisms, which does not depend on how the mechanisms are written down and adds up over the joints to the number of mechanisms, 35. That number is Maxwell's count, 2 × 192 − 346 − 3, and the rank agrees with no redundant bar. The mean weight is 0.341 on the outermost ring of cells and 0.015 on the innermost. The twist kinks every line of bars at every joint, and the mechanisms fall away from the edge 23-fold in 3 cells.

The count says how many and not where

Finite patches of the straight and twisted kagome lattice: the mechanism count as Maxwell's count, exact at every size from two cells to eight; where the mechanisms are, measured by the projector onto the mechanism space rather than by any one mode; the mechanism that reaches furthest in; and the decay per cell from the edge, between 1.9 and 2.4 times the twist.

 A strip of distorted kagome with its mechanisms drawn by weight: one edge, the other, or both. A strip of kagome lattice 10 cells wide, periodic along its length (6 of its 12 cells drawn), with each up-triangle's three joints moved along the triangle's sides by 0.1, 0.1 and −0.1. Each joint is drawn with an area proportional to its share of the strip's mechanisms, summed over every wavevector along the strip. The cut leaves each long edge the same number of missing bars, so the count says the two edges share the mechanisms equally. They do when the three moves have one sign. With the third reversed, the top half carries 19.99 of the 24 and the bottom 4.01. The bulk's windings, measured from the twisted kagome's, are (1, −1).

An edge that knows which side it is on

Topological polarization in a distorted kagome lattice: a strip periodic along its length split by wavevector, each wavevector's two mechanisms placed on one edge or the other; the windings of the bulk compatibility determinant round the Brillouin zone giving a whole-number polarization for each of eight sign patterns; the strip's edge shares converging exponentially on the prediction; and the polarization stepping only where the bulk gap closes, at configurations with a straight family of bars.

 A seam that holds two mechanisms. Ten rows of a strip of kagome lattice sixty cells wide, centred on the line where two differently distorted lattices meet. Below it every up-triangle's joints are moved along its sides by 0.3, 0.3, −0.3, polarization −1 in the component that counts along the seam; above it by −0.3, −0.3, 0.3, polarization +1. No bar is missing anywhere along the seam. Each joint is drawn with an area proportional to its share of the strip's two mechanisms at one wavevector. They sit on the seam and die away from it into both lattices, and the strip's outer edges, out of the picture, have none. The wavevector along the seam is 3/12 of a turn per cell.

A seam with no bar missing

Two kagome lattices with opposite topological polarization joined along a seam in a strip periodic along its length: mechanisms from the compatibility matrix's null space and states of self-stress from its conjugate transpose's, weighted across the strip's rows at each wavevector; the bottom edge, seam and top edge holding 1 + p, p′ − p and 1 − p′ of the two-per-wavevector count; the long-wavelength rows where seam and edges overlap; and the width dependence of the states that are exact only in a wide strip.

 Six cables hold a three-by-three grid only when they run round a cycle. A grid of three by three square cells whose bars are rigid and whose six braces are cables, each on one diagonal of a different cell. A cable resists stretching and not shortening, so it forbids its cell's shear in one sense only. Right, the graph that decides it: one node per column and one per row, and one arrow per cable, from the column to the row for a rising diagonal and back for a falling one. This grid is rigid: its arrows run round a single cycle through all six nodes, so every node can reach every other, and a self-stress pulls on all six cables. Of the 5,376 ways to place six cables in six of the nine cells, twelve are rigid.

A cable is half a brace

A square grid braced with cables that resist stretching only: a three-by-three grid with six cables and the directed graph on its columns and rows; every placement of five, six and seven cables tested by a self-stress and by strong connection; the twelve rigid six-cable grids; the fewest struts and cables for three grid sizes; a strip one cell high with crossed cables; and the self-stress that holds a rigid six-cable grid.

 Twelve cables, any one of which can go slack. A three-by-three grid of square cells braced with twelve cables: every row and every column of cells holds one cell with a rising cable, one with a falling cable and one crossed with both (RFX / FXR / XRF, reading R rising, F falling, X crossed). Right, its directed graph, one arrow per cable, and every column and row node has exactly two arrows in and two out. The dashed cable is the one gone slack, the rising cable of the cell in column 1, row 1. With it removed the graph is still strongly connected and a self-stress still pulls on all eleven remaining cables, so the grid is still rigid. The same is true whichever of the twelve goes slack.

Twelve cables survive a slack one

A square grid braced with cables that must stay rigid when any one goes slack: a twelve-cable Latin-square bracing with each cable made slack in turn; every filling of a three-by-three grid's cells with nothing, one cable either way or two crossed, by rigidity, by the count at each node and by survival; the thirty-six fewest survivors in three families; the fewest struts and cables, bare and with a spare, on seven grid sizes; three four-by-four bracings that meet the count, of which two fail; and the self-stress before and after a cable goes slack.

 A pretension chosen for the loss, and what a loss leaves. A three-by-three grid braced with twelve cables in a Latin square (XRF / FXR / RFX), pretensioned in the self-stress that keeps the most load capacity after any one cable goes slack, the largest cable pretension at one; line widths are in proportion to force. All twelve are taut: the cables carry from 0.81 to 1.00, and the smallest load, in any direction, that slackens another is 1.098, in units of the largest pretension. Intact it is 1.098; after its worst loss 0.404.

A spare is worth what its pretension leaves

Twelve-cable bracings of a three-by-three grid that stay rigid when any cable goes slack, treated as an elastic structure: the best Latin square pretensioned for the loss, with each cable made slack in turn; all 36 bracings by their capacity before and after their worst loss, pretensioned for the intact grid and for the loss; the best of each family after each of its twelve losses; the three classes of Latin square; and a ledger of the families before a loss, after one, re-tensioned, and how tight a cable is pulled.

 Two cables slack, and what the other sixteen keep. A three-by-three grid with both diagonals of every cell cabled, the only placement of eighteen that stays rigid when any two cables go slack, pretensioned in the self-stress that keeps the most load capacity after the worst pair, the largest cable pretension at one; line widths are in proportion to force. Here the worst pair has gone slack together and the rest have taken up their pull elastically: the cables carry from 0.18 to 1.38, and the smallest load, in any direction, that slackens another is 0.242, in units of the largest pretension. Intact it is 1.695; after its worst pair 0.242.

A grid that loses two crosses every cell

Cable-braced grids that stay rigid when any two cables go slack: the three-by-three grid with every cell crossed, pretensioned for losing two, with each of its six worst pairs slack; the count of arrows, fewest cables and survivors for one loss and two on three-by-three and four-by-four grids; every pair's leftover capacity on the crossed grid under two pretensions; the four-by-four bracings by how many pairs leave a cable slack; all of them pretensioned for their worst pair; the best of each family drawn; and a ledger of capacities.

All essays