Machine Kinematics
solved, then drawn
Series

Mobility — the series

24 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. Three bars and four bars. On the left, two bars to a common point: three links, three joints, and Grübler gives 3(3−1) − 2(3) = 0. The Jacobian agrees — two free coordinates, rank 2, nothing left over — and the shape cannot change without a bar changing length. On the right, one more bar and one more joint gives mobility 1, and the whole of this site follows from that difference. The triangle is why bridges are triangulated and the quadrilateral is why machines are not.

    What decides whether it moves

    Before a mechanism does anything it has to be able to. Two bars pinned to a point cannot move; three can. The count that separates them is one subtraction, it is the first thing anybody computes about a machine, and it can be wrong in a way that no amount of care with the arithmetic will catch.

    part 1 · constraint
  2. Mobility, counted and measured. Grübler's criterion counts links and joints and knows nothing about the dimensions; the rank of the constraint Jacobian measures the dimensions and knows nothing about the topology. They agree for four of these five. The parallelogram with a redundant third bar is the exception: the formula declares it a structure with zero degrees of freedom, and it moves. The formula is the one that is wrong, because it cannot see that the third bar's constraint equations are already implied by the other two.

    Counting and measuring mobility

    Grübler's criterion counts links and joints and never asks how long anything is. The rank of the constraint Jacobian measures the lengths and never asks what a joint is. Two calculations with no inputs in common, producing one number — which is the only arrangement under which agreement is evidence.

    part 2 · constraint
  3. What each kind of joint takes away. Grübler's formula is M = 3(n − 1) − 2j₁ − j₂, and the 2 and the 1 in it are not conventions. A lower pair — a pin or a slide — holds two bodies together over a surface and leaves one relative freedom, so it costs 2. A higher pair — a cam against a follower, a wheel on a rail — touches at a point, the contact travels along both surfaces, and it costs 1. Five chains, each built and each measured from the rank of its constraint Jacobian, which has never heard of the formula. The last row is the one worth having: count that cam contact as a pin, as is very easily done, and the formula returns 0 where the mechanism has 1. The Jacobian does not move.

    What each joint takes away

    Grübler's formula has a 2 in it for pins and a 1 for cam contacts, and those numbers are not conventions to be memorised. They are the number of constraints each kind of joint imposes, they are measurable as the rank of a matrix, and miscounting one of them is the commonest way the formula is got wrong.

    part 2 · constraint
  4. 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.

    part 3 · constraint
  5. A 3-stage stack at 35°. Every bar is the same length and every stage is at the same angle, which is a consequence of the solve rather than an assumption in the drawing: the pins at the crossings and the roller on the ground leave one freedom, and it has been used. The platform is at 1547 mm and the ram — the heavy line from the ground pin — is 428 mm long. Moving the ram one millimetre from here raises the platform 5.0 mm; at the bottom of the travel the same millimetre is worth 17.6 mm.

    A roller is not a slider

    A scissor lift has one degree of freedom, at every height and for any number of stages. Grübler's criterion agrees — if the rollers under it are counted as pins in slots. Count them as slider blocks, which is how every textbook draws a slider, and the same formula declares a machine holding a car in the air to be a structure with minus one.

    part 4 · applied
  6. kelvin: 6 contacts, 6 of the six freedoms taken. A plan view of the arrangement, with each contact drawn where it acts and an arrow along the direction of the force it can carry. A ball's contact force passes through the ball's centre whatever surface it rests on, so the whole constraint system is these points and these directions and nothing else. Written as wrenches — a force along a line, which is a screw of zero pitch — their rank is 6. Six independent wrenches leave nothing free: the part has one place to be, and putting it down twice puts it in the same place.

    Six points and no more

    A ball resting on a surface is a joint: it takes one freedom away, and the force it can carry is a line through the ball's centre. Six of them, arranged well, take all six freedoms and leave a part with one place to be. Six arranged badly take five, and the sixth freedom is a screw with an axis that can be named.

    part 5 · applied
  7. 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.

    part 6 · applied
  8. Two numbers, and where they differ. Every mechanism in this field, with what its constraints leave and what its brackets fill. On every mechanism without a rolling contact the two columns are the same number, which is why nobody had to say which one mobility meant. Here only the rail agrees with itself — and the rail is the one mechanism in the table that cannot go anywhere new.

    The count that counts the wrong thing

    Mobility has meant one number for six fields, because until a wheel appeared no mechanism could tell two questions apart. A rolling wheel has two velocity freedoms and a three-dimensional reachable set, and the formula that gives 2 is not wrong — it is answering the question about instants when the question anybody asks is about intervals.

    part 7 · constraint
  9. Seven arrangements, one routine, and the two that hold. Every row is the same three steps: write down one row per contact — the moment of its normal about the origin, then the normal itself — take the convex hull of those rows, and ask whether the origin is inside it. The parts differ, the numbers of contacts differ, and the routine does not. Two of the seven hold. The other five leave the part something, and the interesting column is what: four rays of rotation for the pinwheel, a translation straight out of the vee, and for the last two a whole line rather than any number of rays, which is what a rank below three means and is the case a reader has to be warned about. Note that the four contacts of the second row are the four of the first row, on the same four edges of the same square, at the same distance along each. positioned by solving, not by drawing.

    Six hold nothing

    Every exact-constraint coupling in these essays — Kelvin, Maxwell, three-two-one, and a Kelvin clamp with a seventh pad added — has rank six and holds the part not at all. The escape a Maxwell coupling leaves is a pure vertical translation with nothing else in it, which is not a defect: it is what a coupling is, and gravity is the seventh contact nobody draws.

    part 8 · holding
  10. 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 the earlier counts did not have before this one: 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.

    part 9 · constraint
  11. What it takes to build each of the twelve. The same twelve, read as a bill of materials. Six of them are one joint, because a lower pair permits the whole symmetry group of its surface and those six groups are exactly the symmetry groups surfaces have. The other five with a dimension take a chain: two slides for planar translation, three for Cartesian motion, a thread and two slides for the screw-in-a-plane group, and three parallel pins with a slide along them for Schoenflies motion — which is a SCARA arm, and is why a pick-and-place machine has four joints and not one. The group each chain produces is measured from four hundred sampled poses rather than declared, and every row agrees.

    The freedom that is a set

    Grübler's rule has been in this collection since its first essay, and it adds up numbers. Each of those numbers is the dimension of a group of displacements, and the group has eleven siblings the number cannot distinguish. The count is not wrong; it is a projection, and this is what the projection discards.

    part 10 · constraint
  12. 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.

    part 11 · constraint
  13. A geared five-bar at 3 : 2, its coupler pin traced over 2 input turns. Two cranks of length 1 on pivots 3 apart, meshed through gears whose pitch circles, dashed, have radii 1.800 and 1.200, so the second crank turns 1.5000 times for every turn of the first and the other way. Coupler bars of 3 and 2.5 join them at the pin C, and the thin curve is where C goes over 2 turns of the input. The two crank pins are always between 1 and 5 apart and the bars can span 0.5 to 5.5, so the chain never meets a dead position. After 2 turns the machine is exactly where it began and the curve is closed.

    One freedom, and a motion that never repeats

    Mesh a gear on each crank of a five-bar and the count and the rank agree that one freedom is left, at every gear ratio. Whether the machine ever comes back to where it started is a different question, and neither instrument can see it: at 3 to 2 it is home after two turns, at 37 to 23 after twenty-three, and at the golden ratio never, with its nearest returns at the Fibonacci numbers.

    part 12 · constraint
  14. 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.

    part 13 · constraint
  15. 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.

    part 14 · constraint
  16. 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.

    part 14 · constraint
  17. 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 it reaches its first crossing, where the push is as large as the motion ever makes it, and a hundredth of a push further costs fifty-eight times the law — square or not.

    part 15 · constraint
  18. 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.

    part 15 · constraint
  19. Dixon's nine bars with every joint somewhere inside its clearance. Dixon's framework — joints at -2, 1, 3 on one line and -1.5, 1, 2.5 on a line square to it, every joint of one set barred to every joint of the other — with each joint moved to a seeded point inside a disc about its nominal place. The discs are drawn with a radius of 0.12 so that they can be seen; the framework measured has joint errors about a hundred times smaller. The nominal joints lie on two lines, which is one conic, and that is what lets nine bars on six joints flex. The joints as built lie on no conic: at an error of ρ, seed 2's smallest singular value is 0.1746ρ where the nominal framework's is nought, so by rank it is a structure.

    The clearance that makes the framework generic

    A clearance lets Dixon's nine bars move, and it also moves their joints off the two lines that let them move at all. Six joints off a conic make the framework rigid by rank, with a smallest singular value proportional to its distance from a conic. The same clearance that did the damage pays for it fourteen times over: a framework built anywhere inside its clearance discs is pushed to its first crossing using at most 7% of the allowance those discs give. The tilt of the lines decides how far it goes; its own joint errors change that by about 1%.

    part 16 · constraint
  20. Past the crossing, one push names two placements, one in each family. Dixon's framework, joints at −2, 1, 3 on one line and −1.5, 1, 2.5 on a line square to it, held by B₁ and the direction of B₁W₁, with B₂ pushed 0.5 along its line. Solid: the placement on the family the framework starts in (++), reached before the crossing. Dashed: the placement at the same push on the family beyond it (+−), reached after W₂ has passed the other line at a push of 0.591 — the faint framework — and the push has begun to fall again. Every bar has its drawn length in both. A framework with no errors moves from one to the other through the crossing; a framework as built has no motion and must be carried there by its play.

    The price of the other family

    At its crossing, Dixon's framework passes from one family of placements to another, and past it every push names two placements, one in each. A framework built with its joints anywhere inside a clearance ρ has no motion to follow, so its play has to carry it. The price of going on is proportional to ρ, from 0.002ρ to 0.20ρ across ten frameworks, exactly as a parallelogram's change point is — and reversing every joint error leaves it unchanged. Eight of the ten find the far family dearer than their own and two find it cheaper, and all of them find it far inside the 2ρ their clearance gives.

    part 17 · constraint
  21. 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.

    part 18 · constraint
  22. 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.

    part 19 · constraint
  23. 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.

    part 20 · constraint
  24. Moved apart, the nearest joints still go round a circle. A Dixon framework with B₂ starting 1 from the crossing point on the black line and W₂ starting 1.4 on the white, every other joint where Dixon put it: blacks at −2 and 3, whites at −1.5 and 2.5. Round its whole motion B₂ and W₂ stay on a circle of radius √(1 + 1.4²) = 1.720 (dotted), because the bar between them (heavy) keeps its length. Here ψ = 250°: B₂ at −0.588, W₂ at −1.617, W₁ at −1.704. The angle at B₂ between the bars to W₂ and W₁ turns once round the loop with its speed between 0.046 and 1.954 per unit of ψ, slowest as B₂ crosses the white line.

    Put the motor where the rocking bar is longest

    Dixon's framework goes round its loop with its two nearest joints on the ends of a trammel bar, and a motor at a joint angle with that bar for a side carries it round with no fold. With the nearest joints starting at different distances it looked as though the loop would become an ellipse and some angle would nearly stop. The loop stays a circle at every ratio, because the bar between the two joints is a bar. Each of the four turning angles runs at one radian per unit of the loop, plus or minus the rate of its other side, which rocks fastest at a crossing and there turns at the trammel bar's length over its own. That gives each angle's speed ratio in closed form, says none of them ever stops, and puts the motor at whichever joint has the longest rocking bar.

    part 21 · constraint

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