Machine Kinematics
solved, then drawn
Concept

Redundant constraint — where it appears

A constraint that removes nothing, because what it forbids is already forbidden by the others. It is the reason a hinge with two pins in line works at all, and the reason it stops working as soon as the two pins are not quite in line.

Named by 36 essays across 12 fields — each of them below, with the objects they name alongside it.

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.

networks · Network 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.

networks · Network Three parallel bars, and a formula that says this cannot move. Five links and six pin joints, so Grübler's criterion gives 3(5−1) − 2(6) = 0 and calls it a structure. The Jacobian has rank 5 against 6 free coordinates, so it measures one degree of freedom — and the sweep assembles 59 of 60 positions, which settles the matter. The third bar removes no freedom because its constraint is already implied by the other two, and a formula that counts joints cannot notice that they happen to be parallel. Mechanisms of exactly this kind carry drafting machines, anglepoise lamps and locomotive coupling rods, where the redundant bar is there for load sharing and for keeping the linkage out of its change point.

The mechanism Grübler says cannot move

Three parallel bars between two frames. Five links, six pins, and the criterion every engineering course teaches gives zero degrees of freedom — a structure. It is a mechanism, it is in drafting machines and locomotive coupling rods, and the formula cannot see why.

constraint · Mobility Three verdicts, and only one instrument can give all three. Every 10-link graph that satisfies Grübler's count, split by what is actually true of it. 230 are mechanisms with 10 links. 1,165 carry a subchain whose own count is exactly nought — and neither of the two standing routes can see them: the count returns one and the rank returns one, and both are right, because a rigid subchain removes exactly the freedoms it is supposed to. What is false is the description. 483 carry a subchain whose count is below nought, and those the rank does catch: the surplus pins repeat a constraint already imposed, the Jacobian loses rank, and the measured mobility comes out above the count. The third instrument — a count run over every subset of the links — is the only one that answers the question at all.

What a count cannot see

At ten links, 1,878 graphs satisfy Grübler's rule and 230 are mechanisms. The other 1,648 contain a subchain that is already a structure — and on 1,165 of them the count says one degree of freedom, the rank of the constraint Jacobian says one degree of freedom, and both are right about a mechanism that does not have ten links.

topology · Topology Bennett's four-bar at 40°. Four bars, four revolute joints, and axes that are not parallel — a spatial four-bar, which Kutzbach counts at -2 degrees of freedom. Bennett's condition, sin α / a = sin β / b, makes the screw system rank 3 instead of 4, so the mechanism has 1. Driving the first joint through a full turn, 48 of 48 positions assemble. Orthographic projection, viewed from 40° azimuth and 24° elevation; dashed stubs mark the joint axes.

Bennett, and the condition that moves it

A spatial four-bar is immobile by every count there is, and generically it cannot even be assembled at more than isolated configurations. Bennett found the one relation between four lengths and two twists that makes it turn through a full revolution — and break the relation by two parts in a thousand and most of the travel is gone.

spatial · Spatial 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.

networks · Network The contacts that fight each other — four-legs. With more contacts than freedoms, some weighted sum of the contact forces is zero: those contacts push against each other and not against the part. The weights are the left null space of the wrench matrix and they are drawn here. For four legs on a floor they come out as +−+− — the alternating sum — which is why a table rocks about a diagonal and never sideways, and why the gap under the fourth leg is δ₁ − δ₂ + δ₃ − δ₄ exactly.

The seventh contact

Add a contact to a part that is already exactly constrained and it adds no rank, so it constrains nothing — and it is the only contact in the set that can fail to touch. For four legs on a floor the combination that constrains nothing is the alternating sum of the four, which is why a table rocks about a diagonal and never sideways.

applied · Mobility Three bars, and no rotation left. Each of a delta robot's legs ends in a parallelogram, which keeps the bar on the platform parallel to the bar on the arm. That leg therefore permits the platform no turn about either direction perpendicular to its bar: it imposes two couple constraints, drawn here as rings about the directions they act on. Three legs impose 6; together they span only 3, so 3 are redundant. What is reciprocal to them is three couples — three pure translations, and the platform cannot turn at all.

Why the platform stays flat

A delta robot has three legs and three freedoms, and there is no obvious reason those freedoms should be the three translations rather than some mixture. The reason is a parallelogram in each leg, and the argument from there to "the platform cannot turn at all" is a constraint computation that takes six wrenches and a rank.

parallel · Parallel 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.

networks · Network A bevel differential, turning. cage in, hold left, with every member's speed taken from the train's null space and every angular position that speed integrated. The teeth are marked at the pitch points rather than cut as involutes — the flank is the teeth field's subject — but the count is the tooth count and the positions are the solved ones, so what turns and how fast is real. Drag it and watch which way each member goes: left 0.000 · right 2.000 · cage 1.000.

One wheel on ice

A differential with one wheel stopped turns the other at exactly twice the cage, and the relation it imposes is satisfied the whole time — nothing has failed, nothing is confused, and the reason the car does not move is not in this site. What is here is the other half: a locked axle is an overconstrained mechanism, and the sliding it produces is 2π times the track per circle driven, whatever the radius.

transmission · Transmission 5 contacts, and 1 of them free not to touch. A hexagon on five contacts. Each contact is removed in turn and the hold recomputed; the ones drawn in the warning colour are those whose removal leaves the part still held, which is to say the ones that are constraining nothing the others were not already constraining. There is one here, and the margin without it is 0.091 — unchanged, to every figure. That is the unilateral form of what a redundant constraint costs, and it costs something different from the bilateral form: a redundant bilateral constraint has to be satisfied and cannot be, so it leaves a gap somewhere; a redundant contact is simply free not to touch, and whether it does is decided by errors nobody controls. positioned by solving, not by drawing.

The contact that is free not to touch

A hexagon on five contacts holds, and taking one of the five away leaves the margin at 0.0914 — unchanged, to every figure. That contact constrains nothing the others were not already constraining, and what it actually does is become the one member of the set that is free not to touch, with the decision made by errors nobody controls.

holding · Restraint Kutzbach, plus the constraints counted twice. Each row is 6(n − j − 1) + Σf, then the redundant constraints ν measured from the two legs' constraint systems, then the mobility measured from the rank of the whole loop's screw system. The first column is wrong for 5 of 6 of these mechanisms; adding ν repairs every one. The catch is that ν is not a property of the joint graph — Bennett's linkage and a spatial four-bar with one twist changed have the same graph and different ν — so the corrected formula needs the measurement it was supposed to replace.

The formula is repaired by the thing it replaced

Kutzbach's count is wrong about most of the mechanisms worth building, and every textbook gives the same repair — add back the constraints that were imposed twice. The repair works on every loop this site has. It is also not a formula, because the number it adds cannot be read off the joint graph.

constraint · Spatial 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.

networks · Network The same error, twice, in two directions. A Sarrus linkage with one axis of one chain tilted off true, and the motion range that survives. Tilted within the plane the chain works in, it does not care: at 0.2 radians — eleven and a half degrees, which is not a manufacturing error by any standard — it still drives through a full turn. Tilted out of that plane, 0.001 radians stops it dead. Two hundred times the error, in the other direction, for no cost at all. What separates them is whether the perturbation lies in the screw system the mechanism leaves unconstrained — so "an overconstrained mechanism must be exact" is not merely crude, it is wrong about the case it is usually said of.

Fragility has a direction

Tilt one axis of a Sarrus linkage out of true by a thousandth of a radian and it stops dead. Tilt the same axis of the same mechanism by two hundred times as much, in the other direction, and it drives through a full turn with nothing measurably wrong. Three orders of magnitude between two errors of the same size — and the direction that matters is the one the reciprocal screw system names.

practice · Overconstraint 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.

networks · Network What each instrument returns, on each kind of graph. The 8-link census, three rows, and the same three questions asked of every graph in it. Grübler returns 1 in every row — it has to, because that is what the census selected on. The rank returns 1 in the first two rows and 2 in the third. Only the third column changes across all three rows, and it is the one this site did not have before this field: a mobility computed for every subset of the links rather than for the whole. Read down the middle two columns and the site's standing pair of routes is unanimous about 62 graphs, of which only 16 are what it says they are.

The count was right and the name was wrong

The constraint field has checked Grübler's count against a Jacobian rank since the foundation, and the two disagree only where the geometry is special. Here is an assembly where they agree, where both are correct, and where the mechanism does not have the number of links it is described as having.

constraint · Mobility What the clearance has to swallow. Bennett's linkage with its second length multiplied by 1 + δ, and the closure error the solver drives down to and then cannot improve on. The loop does not close at any δ tried, including one part in a million. But the gap is exactly proportional to δ — the ratio varies by 0.07% across four decades — with a measured constant of 0.507. Shared over 4 joints that is 0.127 δ of play per pin, so a linkage machined to one part in a thousand needs about 0.20 mm of clearance in a link of 1.6 m, or a hundredth of a millimetre in a link of 1.6 cm. That is an ordinary running fit, and it is why a mechanism that cannot be built is in every folding table.

Why a hinge works

A door hinge with three knuckles is overconstrained — three axes imposed where one would do, and exactly parallel is a condition no bored hole has ever met. It works because the misfit is 0.507 times the error and the play in each knuckle is larger than that. The mechanisms this site called unbuildable are built every day, and the thing that builds them is the clearance that was already there.

practice · Overconstraint A brace is one redundant equation, on purpose. The compiled machine, counted and measured, with and without 4 braces. The count says the braced machine has -3 degrees of freedom — it cannot move — and the rank of the constraint Jacobian says it has 1, the same as before. Every brace contributes exactly one equation the others already imply, which is what overconstraint is, and here it is being added deliberately: the redundancy is what removes the assemblies the count knows nothing about. This is Grübler being wrong for the useful reason rather than the embarrassing one.

A bar between two midpoints

In a parallelogram the midpoints of two opposite sides are exactly one side apart, and in the crossed assembly they are not. One bar between them admits the first and refuses the second — and it is one redundant equation per parallelogram, added on purpose, on a site whose constraint field is otherwise about overconstraint arriving by accident.

computing · Compute How much each set of wheels forbids. Every wheel contributes the same row, and whether it is a constraint or a drive is one factor of sin γ in it — γ being the angle the rollers make with the wheel's own axle. At γ = 0 the row says the body may not move across the wheel and the wheel's speed drops out of the statement; at γ = 45° the row says nothing about the body at all and fixes the wheel's speed instead. The whole difference between a machine that shuffles and one that slides sideways is in that factor.

The wheel that forbids nothing

Every wheel contributes the same row to the same matrix, and whether that row is a constraint on the vehicle or a statement about the wheel's own speed is decided by one factor of sin γ. At γ = 0 the vehicle may not move across the wheel; at 45° the row says nothing about the vehicle at all, and sideways costs exactly what forwards costs — to the last digit, and at no other angle.

rolling · Rolling The framework Maxwell's count calls a structure. Six joints and twelve bars in space. Three coordinates each gives eighteen unknowns, six rigid motions come off, and twelve bars is exactly twelve constraints — Maxwell's count is 6 against six rigid motions, which is the definition of isostatic: no mechanism, no redundancy, every bar carrying its own share and nothing spare. The rank is 11, not twelve. There is one dependency among the bars and one freedom left over, and the freedom is a genuine finite motion: walked here with every bar held to 4.4e-16 of its own length. The reason is a symmetry — three pairs of joints exchanged by a half turn about one line — and it is built into the coordinates rather than asserted about the result. positioned by solving, not by drawing.

Twelve bars and a symmetry

Six joints and twelve bars in space is Maxwell's count exactly: no mechanism, no redundancy, nothing spare. Place three pairs of the joints so that a half turn about one line exchanges them and it moves — a finite motion, walked with every bar held to five ten-thousand-billionths of its own length, on a framework the arithmetic calls a structure.

spatial · Spatial Every axis through one point. Four wheels on one rigid body, each rolling without sliding. Each turns about some point on its own axle line, and a rigid body has one such point, so every axle line has to pass through it. That is the whole of steering geometry, and it is a rank condition rather than a formula: here the four rows have rank 2 of 3, leaving a one-dimensional family of twists, and the centre they agree on is 12.000 m to the side. The scrub is 3.2e-17 m per metre — zero, to the last digit.

Every axis through one point

Bolt several rolling wheels to one rigid body and they impose one condition between them: every axle line must pass through a single point. The familiar steering formula falls out of it as a consequence rather than being quoted — cot δₒ − cot δᵢ = 0.574074 at a turn of six metres, of eight, of twelve and of twenty, on a track of 1.55 m and a wheelbase of 2.7.

rolling · Rolling One freedom, whatever the count says. Every one of these machines has exactly one degree of freedom, measured as the number of unknowns minus the rank of the constraint Jacobian. Unbraced, the count agrees. Braced, the count says the largest machine has -153 — that it cannot move, by a wide margin — and the rank says it still turns exactly as it did. The gap is one equation per brace and every one of those equations is implied by the others. This is the constraint field's oldest example, at a scale nobody would try by hand: a count that is wrong by a hundred and fifty-three about a mechanism that works.

One freedom and four hundred links

Braced, the machine compiled from a quintic has 1,249 equations in 1,096 unknowns and a Grübler count of minus a hundred and fifty-three. It turns. The rank of its constraint Jacobian is 1,095, so its mobility is one — and every one of the hundred and fifty-four surplus equations was added deliberately.

constraint · Mobility 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.

Six things a network is not

A count that is right about a difference and read as an answer, a nullity taken for a mobility, a flat state that cannot tell a mechanism from a structure, a scissor ring that closes nowhere, a vertex that folds while its sheet does not, and a null space computed with an instrument whose floor is above the answer. Six claims, each with the number that kills it.

wrong · Misconception 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.

networks · Network Three legs and four, at 0°. The same slice of positions at a platform angle of 0°, shaded by how well the platform is held — pale is near singular. Left, three legs: 3,312 reachable samples and a singular curve through them in 102 segments. Right, the same three legs and a fourth: 3,198 reachable, because the fourth leg must reach too, and no curve. What is left of the singular set in this slice is 1 isolated point, at (-1.449, -0.811), where all four lines meet. Positions held above 0.1 go from 2,989 to 3,178, and at no sampled position is the four-legged platform held less well than the three-legged one.

What a fourth leg buys

Three leg lines fail to hold a platform when they meet at a point, which is one condition, so in every slice of the workspace the failures form a curve. Four lines fail only when all four meet at a point, which is two conditions, so the curve becomes isolated points. The fourth leg buys that and more, and it costs a machine that can no longer be assembled from any four motor angles.

parallel · Parallel 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.

networks · Network A parallelogram at its flat position, exact and built slightly wrong. Left, a parallelogram with ground 3, cranks 2 and coupler 3 lying flat, where its two assemblies meet; faintly, the two ways it can go on, drawn at 30°. Middle, the same linkage with its coupler 0.05 too long, drawn at the closest it can come to the flat position: 7.49° away on either side, so the input's circle is thick where the input can go and red across the 15.0° it can never enter. Right, the input crank 0.05 too short, at the flat position: its two assemblies put the output crank 43.76° apart, and they do not meet at any input angle.

A parallelogram a micron wrong

A parallelogram linkage sits exactly where two kinds of four-bar meet, so a parallelogram that has actually been made is always one of four other machines. Make one bar a micron wrong on a 300 mm frame and the input stops a tenth of a degree short of lying flat, or the output turns round there with an acceleration that grows as one over the square root of the error. A third bar turns the square root back into a misfit of one micron.

constraint · Mobility Whether the part goes in is one inequality. Two of the four contacts are moved and the other two left where the drawing says; the horizontal and vertical axes are those two errors, inward positive. The shaded region is where the part still goes in and the unshaded region is where it does not fit at all — not fits badly, not is located wrongly: there is no position and no orientation the part can take. The boundary is the straight line 0.250·e₁ + 0.250·e₂ = 0, whose coefficients are the shares from the previous figure. Four probe points are marked, each checked twice — once by the inequality and once by a linear program that looks for a pose and reports the program infeasible when there is none — and the two agree at every one. A hold turns a set of tolerances into a single condition, and the weights in it are what say which contact is worth making accurately.

Which contact to make accurately

A hold turns a set of contact tolerances into one linear inequality, and the weights in it are the coefficients of the combination that cancels — a quarter each on a square held by four, and 0.144 to 0.424 on a hexagon held by five. Above that line the part goes in and below it there is no pose it can take at all: not badly located, not out of position, no fit.

holding · Tolerance Nine bars joining two sets of three joints, at three placements of one motion. Joints B₁, B₂ and B₃ lie on the horizontal line at -2, 1, 3, joints W₁, W₂ and W₃ on the vertical line through the same point at -1.5, 1, 2.5, and every joint of one set is barred to every joint of the other. Counted, nine bars on six joints leave no freedom. Drawn here at three placements, the joints have slid along their lines — B₂ at 0.632 and W₂ at 1.265; B₂ at 1.000 and W₂ at 1.000; B₂ at 1.265 and W₂ at 0.632 — and every one of the nine bars has the same length in all three, to 4.4e-16.

Nine bars that ought to be rigid

Join each of three joints to each of three others and the nine bars leave no freedom, by the count and by the rank, wherever the joints are. Put one set on a line and the other on a line at right angles and the framework moves, all the way round a loop, with no bar repeating any other: take away any one of the nine and the motion is unchanged, take away any two and it gains a freedom. Tilt the lines by a degree and it still has a freedom by rank and cannot move at all.

constraint · Mobility 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.

networks · Network The two circuits at a flat position, and the play that joins them. The input and output angles of a parallelogram whose input is short by 1e-3, near the flat position where the exact parallelogram's two motions would cross. With no play the linkage's configurations are two curves, an upper and a lower, 0.1095 radians apart in output angle at the flat input angle — the square root of the error, not the error. Each shaded band is what a radial play of a stated fraction of the error makes reachable, and the innermost boundary is the play-free pair. At a play equal to the error the bands meet and the linkage can pass from one circuit to the other.

A length error is undone by its own size

A parallelogram built a thousandth wrong loses its change point, and the two motions it could have chosen between end up a tenth of a radian apart — the square root of the error rather than the error. The radial play that joins them again is a thousandth exactly: not of that order, that number. It is the same number a third crank charges the same linkage in misfit, and no pin is worth more of it than any other.

constraint · Mobility 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.

networks · Network A framework built off square, and the play it asks for to move at all. The nine bars with their two lines 4° from perpendicular, as built and after the first joint has been pushed 0.2 along its line. The framework has a freedom by rank and no motion, so the push cannot be taken with the bars at their lengths. With every bar allowed to be wrong by 1.509e-4 — which is a radial clearance of 7.547e-5 at each end — a placement exists, and the worst bar in it is out by 1.509e-4. The same push on the perpendicular framework needs no allowance at all, because there it is a motion.

The right angle as a tolerance

Dixon's nine bars move only when their two lines are exactly perpendicular, and a framework built a degree off square has a freedom by rank and no motion at all. Give its joints clearance and it moves a bounded distance: the play each bar needs is proportional to the tilt and to the square of the travel, one constant serves every tilt, and all nine bars end up at that play exactly. Then the framework reaches its first crossing and the law is left three hundred times behind.

constraint · Mobility The same multiplier, applied to the error. A tong whose units are cut to an angle 0.01 radians away from the drawing. If one unit is out, the span is out by that unit's share and nothing more; if every unit is out the same way — which is what a machine setting or a worn tool produces — the error is multiplied by the unit count, exactly, to 7.6e-14. The third column is what would happen if the errors were independent and equally likely either way: the accumulation goes as the square root of the count instead, and the difference between the two columns at thirty-two units is a factor of 5.66. Which column applies is a question about how the parts were made, not about the mechanism.

The error that is repeated

Thirty-two units cut on one setting of one machine are thirty-two copies of one error, not thirty-two draws from a distribution — so a tong's span is out by thirty-two times a unit's, not by the square root of thirty-two times it. The two estimates differ by a factor of 5.66, and the second one is the comforting one.

practice · Tolerance Two circuits at an ordinary change point, and how far apart a length error leaves them. A four-bar on the Grashof boundary g + a = b + c — ground 4, crank 1, coupler 2.50, output 2.50 — with its crank short by 1e-3, near the one input angle at which its two assemblies would meet. Built exactly, the two curves would cross at the origin. Built with the error they pass each other 0.0400 radians apart at the flat input angle, against the law 2√(2bδ/c(g + a)) = 0.0400, and the pin clearance that rejoins them is 1.0000e-3: the error itself. Dragging the coupler's share of b + c moves the separation and leaves the clearance where it is.

Every change point lies flat

A parallelogram a thousandth wrong is rejoined by a pin clearance of exactly a thousandth, at any of its four bearings. The obvious guess is that an ordinary change point — a linkage on one Grashof boundary with no equal bars — would need a clearance with a constant in front and would reveal which bearing is loose. It does neither, because every change point has its four joints on one line. What does acquire a constant is the angle: 2√(2bδ/c(g + a)) when the circuits separate, and a stall constant with no coupler or output in it at all.

constraint · Mobility The bores, and the one line that has to pass through all of them. A hinge of 6 knuckles, its bores drawn at the distance each was made from the nominal axis in units of the bore tolerance. The leaf is a rigid body, so its pins are on one straight line — two parameters of position and two of direction — and it assembles when some line passes within the clearance of every bore. The line drawn is the one whose largest miss is smallest, and that miss is 0.875 of the tolerance. 2 of the 6 bores are at that distance and hold the fit; the rest are slack and could have been bored anywhere inside it without changing the answer.

A piano hinge is not forty door hinges

A three-knuckle hinge works because the misfit its bore errors create is smaller than the play already in its pins. A piano hinge has forty knuckles and thirty-nine of them are redundant, so the obvious reading is that it needs thirteen times the play. It needs two and a half times, and it can never need more than the bore tolerance itself — because a rigid leaf has one axis and a line through the middle of the errors misses every bore by at most the largest of them.

practice · Overconstraint

Named alongside it

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

MobilityOverconstraintRankNetworkToleranceGrübler's criterionConstraintClearanceConstraint jacobianDegrees of freedomNull spaceDeployable

All concepts