Machine Kinematics
solved, then drawn
Series

Network — the series

21 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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

    A scissor lift, a folded sheet and a deployable ring are one small unit repeated thirty times, and three things change at once: the count of bodies becomes a parameter, mobility becomes the rank of a matrix, and a unit that moves can be rigid the moment it is joined to another of itself.

    part 1 · networks
  2. 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

    Before a network is a mechanism it is a graph, and the one quantity that can be read straight off a drawing is how many independent loops it has: edges less nodes plus one. Grübler's count is that arithmetic and nothing else — which is why it is right about a tong at every size and says a deployable ring cannot open.

    part 2 · networks
  3. 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

    Take the pattern a Miura sheet folds along and move every interior vertex by a tenth of a panel. Every vertex still folds on its own — each is a spherical four-bar with a freedom of its own — and the four of them together fold to no angle at all, with a residual that starts at six millionths and never falls.

    part 2 · networks
  4. 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

    A Miura sheet has one freedom at four panels and one at a hundred and forty-four, and the count runs the other way: plus one, then nought, then minus three, minus fifteen, minus ninety-nine. The gap between them is exactly (n − 2) squared, which is a hundred repeated constraints on a sheet with one degree of freedom.

    part 3 · networks
  5. 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

    Four creases through a point at fixed arcs from one another is a spherical four-bar — the object the spatial field is built on — with its link lengths printed on the paper as sector angles. The arcs hold to four parts in ten thousand million million at every fold, and on a flat-foldable vertex the half-angle tangents keep a ratio constant to nine figures.

    part 3 · networks
  6. 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

    Every constraint matrix leaves two null spaces, and a mechanism only lives in one of them. The other is the set of combinations of constraints that come to nothing — and its dimension is exactly the amount by which the count is wrong, on a deployable ring, a Miura sheet and a framework with twelve bars and six joints.

    part 4 · networks
  7. 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

    Two bent bars pinned at their kinks hold the angle between their connection lines at 135.000000° whatever you do to them, and two straight ones hold it at nothing. That is the whole difference between a scissor chain that grows in a line and a ring of eight that opens and shuts — and the count says the ring cannot move.

    part 4 · networks
  8. 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

    A lazy tong's reach is 2nL cos θ — exactly, to a part in a million million, over eight sizes and sixty openings — so it multiplies its input by the number of units. It multiplies everything else by the same number, including the part of the drawing nobody wanted multiplied: a unit cut a hundredth of a radian out puts a tong of thirty-two units 0.374 out at the far end.

    part 5 · networks
  9. 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

    Two bars from one joint to two pinned ones, all three in line: the rank leaves a freedom pointing straight up, and lifting the joint stretches both bars. The obstruction is 1.414214, the walk travels a millionth of what it is asked to, and how far it gets is a property of the tolerance rather than of the mechanism.

    part 5 · networks
  10. 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

    A flat sheet is the one configuration every folding of a pattern passes through, and it is the one configuration where the rank is wrong about all of them. Three of a three-by-three Miura sheet's four apparent freedoms are not motions — and a grid that folds to no angle at all reports exactly the same four.

    part 6 · networks
  11. 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

    Move an interior vertex of a crease pattern and the folded state generally stops existing. How many conditions the drawing has to meet is not a matter of taste: it is exactly the number of dependencies among the constraints, measured at one, four and nine on three sizes of sheet, and a hundred on a sheet of a hundred and forty-four panels.

    part 6 · networks
  12. 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

    Take the size of a network to infinity and it stops being a parameter. What is left is one cell, six bars, and a question nobody has to ask about a finite assembly: does the pattern's period count as a body? A square grid is rigid if it does not and shears if it does, and so does the kagome.

    part 7 · networks
  13. 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

    A scissor stack's height is n·L·sin φ and goes to nothing as the bars lie down — on paper. The bars are made of something, and what stops the fold is two bosses meeting: every stage keeps 0.0949 whatever the stack does, which is exactly twice the boss radius.

    part 8 · networks
  14. 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

    A three-by-three grid of squares needs five diagonals and eighty-one of the hundred and twenty-six ways of placing five will do. Which ones is not a rank question at all: it is whether a graph on the grid's columns and rows is connected, and eighty-one is the number of that graph's spanning trees.

    part 9 · networks
  15. 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

    A kagome lattice has three joints and six bars in every cell and counts to exactly nothing, so a patch cut from it has as many mechanisms as its edge has lost bars: 5L − 5 for a rhombus of L cells a side, which the rank confirms at every size with no bar redundant. Straight or twisted, the number is the same. Where the mechanisms are is not: a straight patch keeps nearly half its edge weight in the middle, and a patch whose triangles are turned by 17° keeps a twentieth.

    part 10 · networks
  16. 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

    Cut a strip from a kagome lattice and each long edge loses the same bars, so the count says the two edges share the strip's mechanisms equally. Move each triangle's three joints by the same signed amount and they do. Reverse one move and every mechanism that is not a translation moves to one edge. The shift is a whole number per cell, set by the winding of one six-by-six determinant round the lattice's zone. It changes only where one family of bars lines up straight.

    part 11 · networks
  17. 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

    Join two distorted kagome lattices of opposite polarization along a line. Every joint on the line keeps all four of its bars, so Maxwell's count says the line holds nothing. It holds two per wavevector: two mechanisms one way up, two states of self-stress the other way up, exactly the difference of the two polarizations. The self-stresses are tensions the lattice carries with no load on it, along a line where no bar is spare. In a strip sixty cells wide the prediction holds at every wavevector a quarter of the zone or more from its centre.

    part 12 · networks
  18. 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 rigid diagonals is rigid when a graph on its columns and rows is connected, so a three-by-three grid needs five braces and eighty-one sets of five will do. Brace it with cables instead, which pull and do not push, and each cable becomes one direction of an edge. The grid is rigid when that directed graph can get from every column and row to every other, which takes six cables, and only twelve of the 5,376 ways of placing six do it.

    part 13 · networks
  19. 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 three-by-three grid braced with cables is rigid on six and never survives the loss of one while each cell holds a single cable. Let a cell hold two crossed and every one of the 262,144 fillings can be tried: none of eleven cables survives a slack one, thirty-six of twelve do, and the rule for which is a count at each row and column, two cables in and two out. That count is the whole rule on every grid narrower than four cells. At four by four it stops being enough, and a grid can meet it everywhere and still hang on a single cable.

    part 14 · networks
  20. 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

    Thirty-six ways of placing twelve cables on a three-by-three grid survive a slack one, and the count at each node cannot tell them apart. Their loads can. Pretensioned at its best, every one of them takes a load of about 1.1 times its largest pretension before a cable goes slack. Lose a cable and the pull it carried is shared out elastically, and what survives is between a quarter and a four-hundredth of that. A pretension chosen for the loss instead of for the intact grid lifts the worst case: to 0.32–0.40 on the Latin squares, 0.22–0.27 when four cells are crossed, and never above 0.096 when six are. The family to build is the Latin square, and even among Latin squares three kinds of spare are not equal.

    part 15 · networks
  21. 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

    To stay rigid when any two of its cables go slack, a cable-braced grid needs three arrows into and out of every row and column of cells. On three by three that is eighteen cables, both diagonals of every cell, and no placement of seventeen survives. On four by four it leaves 576 placements of 24, and for once the count is the whole rule. Pretensioned for the intact grid, 570 of them lose a cable outright to some pair of losses. Pretensioned for the pairs, all of them keep something, and the bracings with no empty cell keep the most, up to 0.198 of the largest pretension.

    part 16 · networks

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