Machine Kinematics
solved, then drawn
Concept

Change point — where it appears

A configuration where a linkage's links line up and it can pass from one assembly to another. It is where the branches of a configuration space meet, and a mechanism designed to sit near one is a mechanism that cannot be relied on to stay in its branch.

Named by 13 essays across 4 fields — each of them below, with the objects they name alongside it.

Grashof's classification, predicted and then swept. Four sets of link lengths. For each, Grashof's condition predicts from the lengths alone whether the input can rotate a full turn, and the solver then attempts all 180 positions and reports how many assembled. The prediction and the measurement agree in every case, which is what licenses quoting the classification for a mechanism nobody has swept.

Grashof, predicted and then swept

Add the shortest link to the longest. If the total does not exceed the other two, some link can turn a full revolution. It is a sentence about four numbers, it was published in 1883, and it is the kind of claim these essays refuse to print without measuring — so every linkage here is also asked for all 360 positions and required to agree.

linkages · Fourbar 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 Where a tolerance stack-up stops meaning anything. The ratio between the first-order tolerance estimate and the band measured by building every corner linkage, for two four-bars at ±0.002 on each length. The crank-rocker's ratio is 1 at all 180 positions — a stack-up is exactly right for it, everywhere. The parallelogram is a change-point linkage, where all four bars can lie on one line, and at that position the estimate exceeds the measurement by a factor of 4.8e+5. The difference is not in the arithmetic, which is identical; it is that a derivative describes a map that can be inverted, and at a change point the map cannot.

Where a stack-up stops working

On a crank-rocker the first-order tolerance estimate matches the measured band at every one of 180 positions, to eight parts in ten thousand. On a parallelogram it exceeds it by a factor of 475,512. Same arithmetic, same tolerance, two linkages that differ only in their proportions — and nothing in the calculation says which one it is being run on.

practice · Tolerance A parallelogram's sextic is a circle and a quartic. The parallelogram with ground 4, crank 1.5, coupler 4, rocker 1.5, coupler point at u = 0.45, v = 0.50. Its eliminated sextic divided by the circle of radius 1.5000 about (1.800, 2.000) leaves a remainder of 6.3 × 10⁻¹⁶ of its largest coefficient, and the quotient is a quartic whose leading form is exactly a multiple of (x² + y²)². Of the configurations solved across a full turn on both assemblies, 720 are on the circle — the machine as a parallelogram, coupler parallel to the ground — and 720 on the quartic, the machine crossed. The two factors meet at four finite points: 2 are change points, marked solid, where one configuration belongs to both; the other 2 are places the two drawings merely cross.

A sextic that comes apart

A parallelogram chain's coupler curve is not one curve. Its sextic divides exactly by a circle, leaving a quartic, and each factor is one of the two things the machine can do. The division leaves rounding and nothing else, a coupler one millionth too long leaves a remainder a million times larger than rounding, and the two factors meet at the two places where the machine has to choose.

curves · Coupler Eight four-bars, one from each region the three signed sums cut. One linkage from each of the eight sign patterns of T₁ = g + c − a − b, T₂ = g + b − a − c and T₃ = b + c − a − g, each solved at a crank angle in the middle of its range. The thick arc round the left pivot is where the input's pin can go and the arc round the right pivot is where the output's can: crank-rocker + + +, input 100% of a turn and output 22%; double crank − − +, input 100% of a turn and output 100%; rocker-crank − + −, input 22% of a turn and output 100%; double rocker + − −, input 23% of a turn and output 21%; 0–π rocker + + −, input 76% of a turn and output 72%; π–π rocker + − +, input 84% of a turn and output 51%; π–0 rocker − + +, input 84% of a turn and output 85%; 0–0 rocker − − −, input 67% of a turn and output 91%. Grashof's condition names four of these and calls the other four one thing; the arcs show that the four triple rockers differ in which way along the ground line each rocker swings through.

Eight kinds of four-bar

Grashof's condition gives a four-bar one of four names and calls every linkage that fails it a triple rocker. Three signed sums of the lengths give eight, and a census of four thousand random linkages finds every one moving exactly as its signs say — because the planes where those sums vanish are the only places a four-bar's motion can change its kind.

linkages · Fourbar Four offset slider-cranks, one from each region the limit leaves. One slider-crank from each region of T₂ = b − a + e and T₃ = b − a − e, with the slide the dashed vertical line a distance e from the crank's pivot and the ground line dashed across. The thick arc round the pivot is where the crank's pin can go and the thick stretch of the slide is where the slider can: crank-rocker + +, crank 1, rod 3.5, offset 0.8, the crank reaching 100% of a turn and the slider between 2.37 and 4.43 on each side of it; double rocker − −, crank 3.5, rod 1.2, offset 0.6, the crank reaching 22% of a turn and the slider between 2.22 and 4.66 on each side of it; 0–π rocker + −, crank 2, rod 2.5, offset 1.6, the crank reaching 65% of a turn and the slider running from −4.21 to 4.21 through the ground line; π–π rocker − +, crank 2, rod 2.5, offset −1.6, the crank reaching 65% of a turn and the slider running from −4.21 to 4.21 through the ground line. These are the four of the eight four-bar kinds in which T₁ is positive.

Four kinds of slider-crank

An offset slider-crank is a four-bar whose output bar and ground have grown without bound, and in that limit the three signed sums that sort four-bars into eight kinds lose one of their signs. Four kinds survive. A census of four thousand finds every slider-crank moving as its region predicts, the textbook condition for a full crank turn turns out to be one region exactly, and each of the four kinds that vanish is carried, at a length that can be written down, into the survivor that shares its other two signs.

linkages · Fourbar 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 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 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 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.

constraint · Mobility 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 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.

constraint · Mobility

Named alongside it

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

Grashof's conditionAssembly branchClearanceDead centreOverconstraintParallelogramToleranceConfiguration spaceCrank-rockerRedundant constraintCircuitClassification

All concepts