Machine Kinematics
solved, then drawn
Series

Mobility — the series

18 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 this site can name.

    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 on this site — 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 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.

    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 on this site 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 the framework reaches its first crossing and the law is left three hundred times behind.

    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

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