Machine Kinematics
solved, then drawn
Concept

Self-stress — where it appears

A set of forces in a framework's bars that balance at every joint with no load applied. It is the combination of constraints that says nothing new, and it decides both what a built framework's play must absorb and what pretension a cable structure needs.

Named by 7 essays across 2 fields — each of them below, with the objects they name alongside it.

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.

networks · Network 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.

networks · Network 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.

networks · Network Each family carries one self-stress, and the price is how hard the errors load it. Dixon's framework pushed 0.5 along B₂'s line, on the family it starts in. Pinned and pushed it has eight free coordinates against nine bars, so it carries exactly one self-stress: bar forces that balance at every joint. Each bar is drawn with a width in proportion to its share of that stress, tension solid and compression dashed: B₁W₁ -0.126, B₁W₂ 0.210, B₁W₃ -0.140, B₂W₁ 0.270, B₂W₂ -0.395, B₂W₃ 0.350, B₃W₁ -0.333, B₃W₂ 0.587, B₃W₃ -0.337. A framework as built with joint errors d pays, per unit ρ, |ω·e| / Σ|ωᵢ|, where e is the bar-length error its joints make. Framework 4 pays 0.0293 on the near family and 0.0267 on the far one; framework 7 pays 0.0131 and 0.0023.

The price is one self-stress

A Dixon framework built with its joints a little out of place has to be carried past its crossing by play, and the play it needs was measured by bisection, framework by framework. It has a closed form. Pinned and pushed, the framework carries exactly one self-stress on each family, and the least play is the errors' bar misfits weighted by that stress, over the stress's size. The formula matches every measured price to 0.6%, says why two frameworks of ten prefer the far family, and finds that one in twelve does.

constraint · Mobility 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.

networks · Network Either side of the crossing, only W₂ is somewhere else. Dixon's framework, pinned at B₁ with the bar B₁W₁ held in direction, a distance s = 0.3 along its motion before its first crossing (solid) and the same distance beyond it (dashed, W₂'s three bars only). The motion is Dixon's closed form with t = 1 − s²: every joint but W₂ is a function of s², so it is in the same place on both sides — here to the last digit — and W₂ is at s and −s, on opposite sides of the line through the black joints. The push on B₂ is the same at s and −s and is largest at the crossing, a fold; W₂'s side of the black line passes through nought.

A push turns back at the crossing

Dixon's framework as built has play enough to go on past its first crossing onto either family, and the question left open was which one a push on B₂ chooses: the family whose self-stress is cheaper, or something else. Neither. Dixon's own formula says that near the crossing every joint but W₂ moves as the square of the distance along the motion, so the push is at a fold there on both sides, and no push can carry the framework through. Ten frameworks as built, pushed to their turning point and pulled back, all retrace, including the two that are cheaper beyond. Driven instead by W₂'s side of the black line, all ten go through, needing exactly the worst price on their path, and the price is continuous through the crossing.

constraint · Mobility B₂ and W₂ go round like the ends of a trammel bar. Dixon's framework in its own frame: the black joints on the horizontal line, the white on the vertical, nine bars joining every black to every white. Round the whole motion, all four families joined at their crossings, B₂ is at √2 cos ψ on the black line and W₂ at √2 sin ψ on the white, so the bar between them (heavy) slides with its ends on two perpendicular lines and turns once as the framework goes once round. The corner of the rectangle on B₂ and W₂ (small dot) is at angle ψ on the dotted circle of radius √2, and goes round it once. Here ψ = 60°: B₂ at 0.707, W₂ at 1.225. The arc at W₂ is the motor's angle, from the bar W₂B₂ to the bar W₂B₃, 37.2°.

A drive that goes round is an angle

Dixon's framework moves round a loop through four crossings, and a push on B₂ folds at two of them while a drive on W₂'s side folds at the other two. A lever moving both was proposed as the drive that would go all the way round. It cannot, and no drive that is one number read off the joints can: round the whole loop B₂ and W₂ sit at √2 cos ψ and √2 sin ψ on their two lines, the ends of a trammel bar, so the loop is a circle and every such number has a top and a bottom on it. A lever only moves its two folds off the crossings. What goes round is an angle. Four of the framework's eighteen joint angles turn once as it goes once round, never stopping, and a motor at W₂ takes ten frameworks as built all the way round at the worst price on the loop — 1.00 to 1.46 times their worst crossing — after which every turn repeats the first.

constraint · Mobility

Named alongside it

The objects these essays reach for when they reach for this one.

Design ruleNetworkRankRedundant constraintClearanceOverconstraintShearChange pointEnumerationJoint graphSingular valueAssembly branch

All concepts