Machine Kinematics
solved, then drawn
Concept

Linear program — where it appears

The problem of maximising a linear function over a set of linear inequalities, answered by a simplex rather than by sampling. It reports three different facts — an optimum at a vertex, an unbounded objective, or an infeasible system — and the third is what decides whether a part with contact errors goes in at all.

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

How far inside the hull the origin actually is. The same seven arrangements with their margins drawn rather than tabulated, because the shape of this chart is the argument: the quantity is not a probability and not a percentage, it is a distance — how far the origin sits from the nearest face of the hull of the contact rows, with every row a unit vector so the number is comparable across arrangements. The two that hold come in at 0.211 and 0.091; the five that do not come in at exactly nought, and they are drawn at nought rather than left off. A margin that falls smoothly to nothing is what makes this a measurement: an arrangement approaching one that lets go says so before it does.

The test is a program, not a rank

Three independent routes to one yes-or-no: enumerate the escape cone's extreme rays by cross products, take the convex hull of the contact rows and ask where the origin is, or hand the whole thing to a simplex. They agree on every arrangement — and the first version of the third one reported a disc as held, which is the one part in the field that no number of contacts holds.

holding · Restraint 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 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

Named alongside it

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

Form closureFixtureClearanceClosure marginEscape conePosePositive spanRepeatabilitySensitivityToleranceChuckConditioning

All concepts