Field

Contacts that only push

Every constraint here so far has been an equation. A part resting against another part is not one: the contact says do not come closer and nothing at all about going away, so what the part may do is a cone rather than a subspace and a freedom stops being two-sided. Six constraints fix a body in space and six contacts fix nothing — the number is seven — and whether a part is held stops being a rank and becomes a question about where the origin sits inside a hull.

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

What changes when a constraint is an inequality: the cone of permitted twists, seven arrangements measured by one routine, and the two of them that hold.

What one contact forbids

One unilateral contact, and the half-plane of rotation centres it leaves: the three contact states, the boundary line, and why the picture is a region rather than a direction.

Four in the plane and seven in space

Why the minimum number of frictionless contacts is d + 1 rather than d, proved from the arithmetic and then measured: the 3-2-1 scheme that locates and does not hold, and a census of what a count is worth.

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

How a hold is decided: extreme-ray enumeration, the hull radius, and a linear program, with the defect that came of taking the program's answer alone.

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

Reading an escape cone as a region of rotation centres: its corners, its unbounded directions, and how the picture decides where a contact must be added.

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

Where the first-order test stops: a rank deficiency that means a genuine spin and one that means nothing at all, separated only by a second derivative.

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

Redundancy in a unilateral arrangement: which contacts can be removed, why a spare one is the one that may fail to touch, and how it differs from a redundant bilateral constraint.

Free at every instant and going nowhere

The boundary of everything a cone measures: a part locally free at every configuration and globally trapped, with the threshold computed two ways.

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

The removal cone of a finished part: the arc, the single direction and the empty set, computed from the faces in contact and nothing else.

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

Disassembly as a search over subsets rather than over parts: two pieces each blocked by the other, and a pair whose contacts are not the union of its members'.

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

Where three closing jaws leave a bar's axis, in closed form and again by linear program: the sections that centre exactly, and the gcd that decides which.

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

What a clearance leaves once a hold has done its work: a pose polyhedron, bounded exactly on the arrangements that hold, whose size is linear in the clearance and whose shape is not.

A cone has no size

Why the restraint field's central object is scale-free, where the one length in it hides, and what a measurement of a fixture can and cannot recover.

Six hold nothing

The site's own exact-constraint arrangements read as unilateral contacts: rank six on every one of them, and an escape on every one of them.

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

What a hold does to a tolerance budget: Farkas's condition as a single inequality on the contact errors, its weights, and the extra one a spare contact adds.

The hold is in the corners

The largest hold margin four frictionless contacts can give a regular polygon, from 3 to 256 sides: exhaustive search over edge ends with continuous refinement, the quarter-turn arrangement it finds, a margin that approaches the half-edge sin(π/n) with a shortfall times n² levelling near 24.7, odd polygons below both even neighbours, and the triangle and pentagon as the two whose best contacts leave the corners.

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

Two frictional contacts on a regular polygon with their friction cones, held or not as the sides change; the least friction for two jaws against the number of sides, with tan(π/2n); the load two opposite jaws on a disc can balance per unit of squeeze, with its closed form μ/√(2 + μ²); the field's length-normalised margin reading friction backwards past μ = 1; the friction at which two jaws overtake the best four frictionless contacts, falling as √2·π/n; and a cube held by three frictional contacts where seven frictionless were needed.

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

Self-locking holds with no preload: a cam cleat whose pivot line leans inside or outside the friction cone; a wedge dragged into a narrowing gap between a bar and an incline; the measured thresholds of both against arctan μ and the sum of two friction angles; the range of load directions a cam cleat holds, unchanged below its threshold and gone past it; and the squeeze each device applies per unit of pull across its window.

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 whose pivot pin has friction: the pin's friction circle, the forward limit arctan(μ + (ρ/h)√(1 + μ²)) as the sum of the contact's and the pin's friction angles, the backward lock below arctan(ρ/h), the pin solved as a single contact with its own friction cone for eight designs, the working window between the two limits, and the squeeze per unit pull, indeterminate forward and unbounded backward.

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 with a release lever and a pivot pin with friction: the squeeze on the rope against the pull as a load rises, falls and rises again, between the band's two edges; the lever moment per unit pull against the lean, by the friction circle and by a pin touching its bore at one point; the moment a slack cleat needs below the backward-lock lean; and a ledger of the band, the release at its edges and the slack cleat, lean by lean.

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

A cam cleat whose pivot pin is carried on a sprung, swinging link, with a rope that compresses: the cleat drawn along a load path; the squeeze against the pull for a fixed pin and two links; what a slack cleat keeps against the link's own squeeze ratio at three leans, by a closed form and by relaxing the cleat's two angles; how far the cam must turn on its pin; and a ledger of squeeze, memory, turn and lever, lean by lean.

All essays