Machine Kinematics
solved, then drawn
Concept

Repeatability — where it appears

How closely a machine returns to the same place when commanded the same joint values twice, which is a much smaller number than its accuracy. It is the number a robot is sold with, and it says nothing about whether the tool is where the program thinks it is, which is accuracy and is usually far worse.

Named by 5 essays across 4 fields — each of them below, with the objects they name alongside it.

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.

applied · Mobility A 4-sided bar, and the 13.6% it is out by. 3 flat jaws advancing together on a regular 4-sided bar of unit circumradius, with the bar turned 10.3° from square. The dashed circle is the axis the chuck is turning about and the marked point is where the bar's own centre has ended up: 0.13567 of a circumradius away. The arithmetic is one line — the jaws touch when c·u_k + h(u_k) = d, three unit vectors at 120° satisfy Σ u u ᵀ = 3/2 I, and so c = −⅔ Σ h(u_k) u_k — and it says that the offset vanishes exactly when the bar's own support function is unchanged by a 120° turn. Round, triangular, hexagonal, nine- and twelve-sided bars centre at any orientation; everything else does not, and by an amount that depends on how it happened to go in. Checked here against a linear program that closes the jaws without knowing the identity. positioned by solving, not by drawing.

Where the jaws put it

Three jaws closing on a bar put its axis at −⅔ Σ h(u_k) u_k, which vanishes exactly when the section's support function is unchanged by a 120° turn. So a three-jaw chuck centres round, triangular and hexagonal stock perfectly and a square bar by up to 17.3 per cent of its own circumradius — and the workshop rule about symmetry that predicts this is wrong, because a six-jaw chuck centres a square.

holding · Restraint The pose set is finite exactly when the part is held. Two arrangements, both with 0.06 of clearance on every contact, with the set of positions the part's centre may occupy drawn to scale. The one on the left holds: its pose set is a small bounded polyhedron, and every dimension of it is proportional to the clearance. The one on the right does not: its pose set runs off the page in the direction the part slides out of the vee, and giving the contacts a tighter tolerance narrows the box without ever closing that direction. A tolerance cannot buy a hold. The clearance decides how big a finite pose set is and the arrangement decides whether it is finite, and the second question has to be settled first because no amount of the first will settle it.

Held is not located

Back every obstacle off by a clearance and the permitted poses become a polyhedron — bounded exactly when the arrangement is a hold, since an unbounded direction of it would be a ray of the escape cone. So whether a part is held is whether its pose set is finite, the clearance is what gives that set a size, and the two questions have to be settled in that order because no tolerance settles the first.

holding · Restraint Every joint's lever arm. The thin lines run from the tool to each joint's axis, meeting it square. Their lengths are what a radian of error at each joint costs the tool in metres — not a rule of thumb but the Jacobian column, which is ω × r and therefore that perpendicular exactly. The shortest of them belongs to the joint nearest the work and the longest to the joint furthest from it, which is why an arm's accuracy is decided at the shoulder and its resolution at the wrist.

Where an error at the shoulder ends up

The same angular error at every joint of an arm, and the tool is out by 0.156 mm because of the shoulder, 0.017 mm because of the wrist roll and exactly nothing because of the last joint. The numbers are not properties of the joints. Each one is the distance from the tool to that joint's axis, measurable off the drawing with a ruler.

serial · Tolerance What preload takes away, and what it leaves. The same four-bar with 0.01 of clearance at each pin, driven so that the load through every joint keeps one sign. The upper line is the lost motion it had before: about 2.80° of crank rotation thrown away on every reversal, unrepeatable, and not removable by calibration. The lower curve is what is left once the pins are held against one side of their holes — a fixed offset of between -0.330° and 0.013°, which is a dimensional error rather than play, so its average of -0.136° comes out in a calibration and only the 0.343° of variation survives. A factor of 8.2, bought with a permanent parasitic load that this site does not model.

Taking up the play

Preload does not make a clearance smaller. It takes away the clearance vector's direction, which is the property that made the error unrepeatable — so 2.80° of lost motion becomes a 0.343° offset, of which 0.136° is a constant that calibrates out. A factor of eight, bought with a permanent parasitic load this site does not model.

practice · Clearance

Named alongside it

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

SensitivityToleranceClearanceFixtureForm closureLinear programPoseBacklashChuckClosure marginConstraintContact normal

All concepts