Restraint — the series · Machine Kinematics

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.

part 1 · holding

What one contact forbids

A rotation about a point is a twist, and a twist is affine in the point — so what a single contact permits is a half-plane of centres, with the boundary being the contact surface's own line. Reuleaux drew it in 1875 and it is exact rather than sampled, which is why every figure in this field is a picture of the plane rather than of a cone.

part 2 · holding

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.

part 2 · holding

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.

part 3 · holding

4 contacts, and the centres they still allow. The same four, placed pinwheel. The same square, the same four edges, the same distance along each — and taken the same way round rather than alternately. Every row's moment then has the same sign, so no positive combination can cancel it, and the part turns. Each contact contributes one half-plane of permitted centres per sense, and the shaded regions are what survives all 4 of them: the darker one is where an anticlockwise rotation is still permitted and the lighter one where a clockwise one is. What is left is the escape, and it is a region rather than a direction: any point inside it will do as a centre. The enumeration finds 4 extreme rays, of which 4 are rotations and the rest are translations — the corners of the region and its unbounded directions respectively. positioned by solving, not by drawing.

The escape is a place

A part that is not held escapes, and the useful thing is not that it escapes but where. The extreme rays of the cone are the corners of a region of the plane and its unbounded directions are translations — so the answer to 'this does not hold' is a picture with a shape, and the shape says where the next contact has to go.

part 3 · holding

Free to spin, and it cannot turn at all. An ellipse of semi-axes 1.4 and 0.9 inside four flat walls that touch it at the ends of its own axes. Every normal points at the centre, so every row's moment is nought and the four rows span two dimensions rather than three: the cone of permitted twists is the whole spin axis, a line through the origin, and the first-order answer is that the part is free to turn either way. Drag the angle and watch what happens. The ellipse's reach in the direction of the top and bottom walls is √(a²sin²θ + b²cos²θ), which is smallest at θ = 0 and grows from there, so any rotation whatever drives it into both of them — by 0.016 at this angle. A nullity is a candidate and not a motion, and this is the shape of case the fields before this one could not produce: not a mechanism at a singularity, but an ordinary part in an ordinary pocket. positioned by solving, not by drawing.

Free to turn and unable to

An ellipse in a pocket the size of its own bounding box has four contacts whose rows span two dimensions, so the cone of permitted twists is a whole line and the first-order answer is that it spins both ways. It cannot turn by any amount whatever: the penetration grows as the square of the angle, with a fitted exponent of 1.9944, and a circle in the same pocket turns for ever.

part 4 · holding

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.

part 4 · holding

Free at every instant and going nowhere

Three points on a circle of 1.1 radii around a unit disc leave it free in every direction at every configuration — rank two, margin nought, the whole plane of centres shaded — and it cannot get out. The threshold is 1/sin(π/n), which is 1.154701 for three, and a flood fill of the free space agrees with the formula at every radius sampled.

part 5 · holding

A joint with no way out in the plane it is drawn in. A dovetail. The tail is wider at its far end than at the mouth it went in through, so every direction out of the mouth is blocked by a slanted face and every direction further in is blocked by the floor. In the plane of this drawing the joint cannot be taken apart at all, and the direction it does come apart in is the one the drawing does not show. The moving part touches the rest at 3 faces, each contributing one inequality on the direction it may be translated in — the direction must not have a negative component along that face's inward normal — and the set of directions that satisfy all of them is a cone in two dimensions rather than three, because a translation has no moment term. That is why a removal cone can be drawn as an angle where a mobility cone cannot. Here it is empty: every direction is refused by one face or another, so the part cannot be taken out by any translation and cannot have been put in by one either. That is a statement about the assembly and not about the part, and the direction the joint does come apart in is perpendicular to this drawing. positioned by solving, not by drawing.

Which way it comes out

Drop the rotation from the inequalities and the cone lives in two dimensions rather than three, so it can be drawn as an angle: a block in a vee has ninety degrees of directions out, a key in a slot has exactly one and no arc around it, and a dovetail has none at all. Three answers, and each of them is a different kind of joint.

part 5 · holding

Neither one comes out, and the two of them do. Two congruent Z-shaped parts in a tray that is open at the top. Each has a step that lies over the other's, so part A's four contacts with part B have normals at all four points of the compass and leave it no free direction at all — and the same is true of B, for the same reason and by symmetry. The blocking is mutual and there is no order in which the two can be taken out one at a time. Together they have 6 contacts, all of them with the tray, and exactly one direction out: straight up. So the removal cone of a set of parts is not built from the removal cones of its members, and which part comes out first is a question with no answer here. positioned by solving, not by drawing.

Neither part comes out first

Two congruent Z-shaped pieces in a tray open at the top. Each has four contacts with the other, with normals at all four points of the compass, so each alone is blocked in every direction there is — and the pair lifts straight out. The removal cone of a set of parts is not built from the removal cones of its members, and *which part comes out first* is a question with no answer.

part 6 · holding

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.

part 6 · holding

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.

part 7 · holding

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.

part 8 · holding

A cone has no size

What a set of contacts permits is a cone of twists, and a cone is closed under positive scaling by definition — so nothing about it changes when the part it holds is made bigger. Except that a twist is a screw, a screw has a pitch, and a pitch is a length.

part 13 · holding

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.

part 14 · holding

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.

part 15 · holding

A pin with friction lets the cam's force miss the pivot, by the friction circle. A rope pulled to the right, gripped by a cam whose pivot pin has a radius of 0.16 (drawn larger) and friction 0.5, so the pin's force on the cam may lie anywhere tangent to or inside its friction circle, radius 0.072. The dashed lines from the contact touch that circle; the cam's force on the rope can lie anywhere between them, where a frictionless pin would allow only the line through the centre. The shaded wedge is the contact's friction cone, 21.80° either side of the normal for μ = 0.4. At a lean of 23° the cleat holds a forward pull (it holds up to 24.14°, against 21.80° with a frictionless pin) and lets go when pulled back (it locks backwards below 2.56°).

A stiff pin buys grip with release

A cam cleat on a frictionless pin holds a rope while the line from the contact to the pivot leans no more than the contact's friction angle. Give the pin friction and the cam's force may miss the pivot by the pin's friction circle, so the cleat holds to a lean steeper by exactly the angle that circle subtends: tan γ ≤ μ + (ρ/h)√(1 + μ²). But the same friction makes a cleat leaning less than arctan(ρ/h) lock against a backward pull as well, and refuse to let go. Both edges of the working window rise, the lower at a slope of one and the upper at √(1 + μ²): the grip gained is almost all paid for in release.

part 16 · holding

A cleat's lever, and the squeeze the load has left it. A cam cleat at a lean of 10°, its pivot pin with friction 0.5 and a radius a tenth of the pivot's height (drawn larger, with its friction circle), and a release lever on the cam. Along a load path the rope's pull rises to one, falls to nought and rises again, and the squeeze follows it between the band's edges, 4.50 and 7.62 times the pull. Here the pull is 1, the squeeze 4.50, and the lever must turn the cam with a moment of 0.412 times the pivot's height to free the rope.

A cleat set hard lets go easiest

A cam cleat on a frictionless pin is freed by a lever with no effort at all. With pin friction the lever must beat h(T − N tan γ) + ρ√(T² + N²), and since a stiff pin lets the squeeze N sit anywhere in a band, the effort is set by where in the band it sits: nought at the top, twice the pin's friction moment at the bottom. A rising load drags the cam in to the bottom, so a cleat holding the load it was set by is the hardest to free — 0.412 times the pull times the pivot's height at a lean of 10° — while one jammed hard by hand, or one whose load has eased, frees itself. Below the backward-lock lean nothing jumps in the effort. What changes is the state: the slack cleat keeps the squeeze its largest pull left.

part 17 · holding

A cam cleat whose pin swings on a sprung link. A cam cleat at a lean of 2°, below its pin's lock lean of 2.56°, its pin carried on a link that swings about a frictionless pivot 2.2 pin heights above the rope, held by a spring. The link's lever is chosen so that, with the spring a quarter as stiff as the rope at that lever, cam and link turning together squeeze the rope 12.53 times the pull, the pin's own bottom edge. Along a path the pull rises to one, falls to nought and rises again. At this step the pull is 0: a fixed pin squeezes 12.53, this link 0.00, and a link at half that ratio 6.27.

A cleat on a swinging pin has two leans

Carry a cam cleat's pin on a sprung link and, while the pin holds, cam and link turn together about the link's pivot: a second cam, frictionless, whose squeeze per unit pull is the cot of its own lean softened by the spring against the rope. A rising load is met at the pin's edge and a falling one at the link's, so a slack cleat keeps only what the link falls short of the edge — 12.5 times the pull on a fixed pin at 2°, 6.3 on a link at half the edge, nothing on a link at the edge itself. Nothing in it depends on the load's size: every force doubles with the pull.

part 18 · holding