Machine Kinematics
solved, then drawn
Concept

Positive span — where it appears

The property that every vector of a space is a combination of a given set with non-negative coefficients. It needs at least one more vector than the dimension, which is why a body with six freedoms cannot be held by six frictionless contacts and why the classical minimum is seven.

Named by 9 essays across 2 fields — each of them below, with the objects they name alongside it.

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.

A constraint that only pushes

Every constraint in these essays so far has been an equation: a pin holds two points together, a bar holds two apart, a mesh holds a ratio. A part resting against another part says only *do not come closer* — so what it may do is a cone rather than a subspace, and whether it can move at all stops being a rank.

holding · Restraint 6 contacts, and the box lifts straight off. 3-2-1, six contacts. Each pad is a contact and each arrow the direction the box is free to move there — the inward normal, and the whole of what the contact contributes. The rank is 6, which is full: these 6 contacts are 6 independent constraints and a bilateral version of them would fix the box completely. The margin is nought, so the box is free to leave, and no amount of tightening the tolerances on where the pads are would change that. Rank and hold are different questions and this is the pair of pictures that separates them. positioned by solving, not by drawing.

Four in the plane and seven in space

Six independent constraints fix a body in space and six contacts fix nothing, because d vectors can span d dimensions and can never positively span them. The minimum is one more than the dimension — and it is a floor rather than an answer: six contacts on a box held it in none of four thousand random arrangements and seven held it in twenty-one.

holding · Restraint 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 5 contacts, and 1 of them free not to touch. A hexagon on five contacts. Each contact is removed in turn and the hold recomputed; the ones drawn in the warning colour are those whose removal leaves the part still held, which is to say the ones that are constraining nothing the others were not already constraining. There is one here, and the margin without it is 0.091 — unchanged, to every figure. That is the unilateral form of what a redundant constraint costs, and it costs something different from the bilateral form: a redundant bilateral constraint has to be satisfied and cannot be, so it leaves a gap somewhere; a redundant contact is simply free not to touch, and whether it does is decided by errors nobody controls. positioned by solving, not by drawing.

The contact that is free not to touch

A hexagon on five contacts holds, and taking one of the five away leaves the margin at 0.0914 — unchanged, to every figure. That contact constrains nothing the others were not already constraining, and what it actually does is become the one member of the set that is free not to touch, with the decision made by errors nobody controls.

holding · Restraint 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 things a hold is not

A rank read as a restraint, a count read as an answer, four contacts placed the wrong way round, a nullity taken for a spin, a part free in every direction and unable to leave, and a tolerance offered as a cure for an arrangement that was never a hold. Six claims, each with the number that kills it.

wrong · Misconception 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 The best four contacts on six regular polygons. The largest-margin placement of four frictionless contacts on regular polygons of 3, 4, 5, 6, 8, 12 sides, each found by exhaustive search over edge ends and refinement along the edges. 3 sides: margin 0.231 against a half-edge of 0.866; 4 sides: margin 0.333 against a half-edge of 0.707; 5 sides: margin 0.235 against a half-edge of 0.588; 6 sides: margin 0.293 against a half-edge of 0.500; 8 sides: margin 0.284 against a half-edge of 0.383; 12 sides: margin 0.223 against a half-edge of 0.259. From six sides up every contact sits at an end of its edge; on the triangle and the pentagon two of the four settle near the middles of edges instead. On the square and the even polygons the corner contacts take alternate ends of four edges a quarter-turn apart; an odd polygon has no edge exactly a quarter-turn round and holds less than either even neighbour.

The hold is in the corners

A disc cannot be held by frictionless contacts and a regular polygon can, so a polygon with more and more sides has to lose its hold somewhere. Searched exhaustively, the best four contacts sit at alternate ends of four edges a quarter-turn apart, and their margin is the half-edge sin(π/n) less a correction that falls as 1/n² — 74% of it at eight sides, 99.4% at sixty-four. The hold is lost as the side shrinks, not as its square, and it is carried entirely by how far a contact sits from its edge's middle.

holding · Restraint Two jaws on a polygon, and the cones they can push within. A regular polygon of 5 sides and circumradius 1, gripped by two contacts with a friction coefficient of 0.3. Each wedge is the cone of directions its contact can push in, half-angle atan 0.3 = 16.7° about its normal. Two contacts hold the part — resist every load, given a squeeze — exactly when the line between them lies inside both cones. An odd polygon has no parallel faces, and its most nearly opposed faces are 36.0° short of parallel, so the line can be inside both cones only if their half-angle is at least half that: a friction of tan(π/10) = 0.3249. Here the part is not held. Dragging changes the number of sides.

Two jaws and a coefficient

Every hold in this field has been frictionless, and two of its essays say what that leaves out. Put friction back and a part is held by two contacts in the plane instead of four and by three in space instead of seven; a disc, which no frictionless fixture can hold at all, is held by two jaws at any friction above nought; and an odd polygon needs a friction of exactly tan(π/2n). What friction takes away is the thing that made a frictionless hold a fact about geometry: a frictional hold is a hold per unit of squeeze.

holding · Restraint A cam cleat holds while its pivot leans inside the friction cone. A rope along the bottom, pulled to the right, pressed against a fixed backstop by a cam on a frictionless pivot. The cam can push on the rope only along the line from the contact to its pivot, because a body on one pin with one other force is a two-force member. That line leans 15° from the contact's normal; the shaded wedge is the friction cone, 21.80° either side of the normal for a friction coefficient of 0.4. The line is inside the cone, so the cam's force can resist the pull, and it squeezes the rope with 3.73 times the pull. Dragging moves the pivot.

A hold the load squeezes shut

A cam cleat bites harder the harder the rope pulls, and a wedge dragged by a bar into a narrowing gap squeezes the bar harder the harder it is drawn. Neither needs a screw: the load supplies its own squeeze, if the geometry lets it. A cam on a free pivot can push only along the line to its pivot, so it holds exactly while that line leans from the contact's normal by no more than the friction angle. A wedge's squeeze passes through two contacts, and it jams while its angle is no more than the sum of their friction angles — twice the figure usually quoted. And the set of loads either one holds does not shrink towards its threshold. It is lost there, all at once.

holding · Restraint

Named alongside it

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

Form closureFixtureUnilateral constraintClosure marginConvex hullLinear programRankClearanceContact normalEscape coneDegrees of freedomExact-constraint

All concepts