Two jaws and a coefficient

Every hold in the holding field so far has been frictionless, and the field has been explicit about it. Four in the plane and seven in space proved that a part needs four frictionless contacts to be held in the plane and seven in space, and then said that most holds people meet are not frictionless — a vice grips a part with two jaws and holds it — and that every number in the field is therefore a lower bound on what a real fixture achieves. The hold is in the corners found that four frictionless contacts hold a nearly round polygon only through its corners, with a margin that falls as the side does, and ended by noting that a contact with friction regains an arm of order the friction coefficient rather than of order the side.

Both essays left friction out deliberately, because a frictionless hold is a statement about geometry and a frictional one is not. This essay puts it back and measures what changes. The count changes, the kinds of part that can be held change, and one of the field’s own instruments turns out to read friction backwards. But the most important change is in what “held” means.

A contact that pushes in a cone #

A frictionless contact can push only along its normal. A constraint that only pushes wrote that as a row: the contact point’s moment about a reference, over a characteristic length, followed by the normal, and a set of contacts holds a part exactly when its rows positively span the space of loads — when every load has a combination of pushes that cancels it.

A contact with Coulomb friction $\mu$ can push in any direction within a cone about its normal of half-angle $\arctan\mu$. In the plane that cone is exactly the positive span of its two edges, $\mathbf{n} + \mu\mathbf{t}$ and $\mathbf{n} - \mu\mathbf{t}$, with $\mathbf{t}$ the tangent. So a frictional contact contributes two rows instead of one, one for each edge, and every test the field has built applies to them without change. The hull test that decides whether rows positively span, the linear program that decides it in six dimensions, the rank check that catches a set lying in a subspace: all of them take friction edges as they take normals.

One thing friction does not change is worth saying at once. What one contact forbids is a half-space of motions — those that would push the part into the surface — and that half-space is the same with friction as without. Friction adds nothing to the geometry of motion. It adds forces that resist motions the contact already permits, sliding along the surface, and it supplies them only while the contact is pressed. Everything below follows from that sentence.

That is the whole of the modelling. Coulomb friction is itself a model, with a single coefficient that real surfaces only approximate, and nothing here refines it.

Two jaws in the plane, and the line between them #

With two rows per contact, two contacts give four rows in a three-dimensional space of planar loads, which is enough to span it positively. Four frictionless contacts were the minimum because a frictionless count needs one more row than the dimension; two frictional contacts meet that with room over.

Whether two particular contacts hold is a condition anyone who has used a vice already half-knows. Two contacts hold a part exactly when the line between them lies inside both friction cones. Then a squeeze along that line is available, the friction at each end can supply any tangential force up to $\mu$ times its share, and between them the contacts can balance a force in any direction and a turn either way.

For a regular polygon the condition depends on whether it has parallel faces. An even polygon has: two contacts directly opposite each other on parallel faces have the line between them along both normals, inside any cone at all, and the polygon is held at any friction above nought. An odd polygon has none. Its most nearly opposed faces are $\pi/n$ short of parallel, and the best the line between two contacts on them can do is split that angle, sitting $\pi/2n$ from each normal. So the least friction is

$$\mu_{\min} = \tan\frac{\pi}{2n} \quad (n \text{ odd}), \qquad \mu_{\min} = 0 \quad (n \text{ even}).$$

That is a derivation, and it is checked by a search that does not use it: for each polygon, every pair of edges, with each contact at eleven positions along its edge, tested by the hull at a friction found by bisection. For every odd polygon from three to fifteen sides the search lands on $\tan(\pi/2n)$ to $2 \times 10^{-10}$. A friction one per cent below the law holds no odd polygon at any placement the search tries, and every even polygon is held at a friction of a millionth.

The triangle needs 0.577, which is more than a dry steel-on-steel contact reliably gives, and that is why a triangular bar in a two-jawed vice slips: its faces are sixty degrees from parallel, and no amount of squeezing makes a cone wide enough. A pentagon needs 0.325. By fifteen sides the requirement is 0.105 and falling as $\pi/2n$.

A disc, held #

The part the frictionless field could not hold at all was the disc. Every contact normal passes through its centre, no combination of them resists a turn about it, and the escape that is a place is a rotation. The ledger records it as the one part nothing holds.

With friction, two jaws directly opposite each other on a disc have the line between them along both normals, so they hold it at any friction above nought. The turn that nothing frictionless could stop is resisted by the tangential forces the friction supplies, each with an arm equal to the radius.

The lathe’s three-jaw chuck is the same arithmetic with one more jaw. Where the jaws put it treated it as a centring device and found where its jaws place a bar’s axis. On round stock its three jaw normals all pass through the axis, so without friction the chuck could not resist the cutting torque at all; with friction each jaw supplies up to $\mu$ times its normal force at an arm of the bar’s radius, and the chuck resists a pure turn of $\mu R$ times its total grip. A pure turn is not the worst direction, which is why the disc’s $\rho$ below is smaller than $\mu$, but it is the load a chuck is mostly asked to carry.

That is where the difference between the two kinds of hold stops being a count and starts being a quantity.

A hold per unit of squeeze #

A frictionless hold is a fact about geometry. If a set of frictionless contacts positively spans, the part cannot move at first order whatever pushes on it; a load produces its own reaction forces, and nothing has to be pressed beforehand. Remove every load and the contacts can be just touching, and the part is still held in the sense that it cannot move.

A frictional hold is not like that. Friction supplies a tangential force only in proportion to the normal force at the same contact, so the two jaws on a disc resist a turn only while they are pressed. With no squeeze the disc turns between them as freely as between two frictionless contacts, and in the field’s own terms it is not held at all: its escape cone contains the rotation. So the honest measure of a frictional hold is not a margin but a ratio — how much load it can balance for each unit of normal force its contacts press with.

That ratio can be computed with the same hull the field uses. Write each friction edge with a unit normal component, $\mathbf{n} \pm \mu\mathbf{t}$ as it stands, and take the convex hull of the edge loads. Positive combinations of edges with weights summing to one are exactly the contact forces whose normal parts sum to one, so the hull is the set of loads the contacts can supply at unit total squeeze, and the distance from the origin to its boundary — call it $\rho$ — is the largest load, in its worst direction, that they can balance. With $\mu = 0$ the edges are the normals and $\rho$ is the same measure for a frictionless set, so the two kinds can be compared in one unit.

For two opposite jaws on a disc of unit radius the hull is a tetrahedron with vertices $(\mp\mu, -1, \mp\mu)$ and $(\mp\mu, 1, \pm\mu)$ in coordinates of turn, force along the line of the jaws and force across it, and its nearest face is at

$$\rho = \frac{\mu}{\sqrt{2 + \mu^2}}.$$

The measured value matches it to $3 \times 10^{-16}$ at seven frictions from 0.01 to 5. For small friction it is $\mu/\sqrt 2$, rising towards one as the cone widens; the $\sqrt 2$ is the worst direction being a combination of a turn and a sideways push rather than either alone. The square gripped at the middles of opposite faces balances less at every friction, because its jaws sit at its inscribed radius, $\cos(\pi/4)$ of its circumradius, and their friction has that much less arm.

And on the other side of the comparison, eight frictionless contacts round the same disc give $\rho = 0$ exactly. The instrument agrees with the field’s ledger about the disc, and gives the frictional grip a number the ledger had no way to give it.

The field’s margin, applied as it stands, reads friction backwards #

Before settling on $\rho$ it is worth recording what happens with the instrument the field already had, because the obvious move is to use it unchanged, and the result is wrong in an instructive way.

The field’s margin is the distance from the origin to the hull of the rows with each row first scaled to unit length. For frictionless rows that is sensible: every row is a unit push, and the scaling only balances the moment term against the force terms. For a friction edge it is not, because the edge $\mathbf{n} + \mu\mathbf{t}$ has length $\sqrt{1 + \mu^2}$ before the moment term is counted, and scaling it to unit length shrinks its normal component by that factor. A wider cone is scored as a weaker push.

The result is a curve that rises to exactly one third at $\mu = 1$ and then falls — to 0.208 at a friction of 3 and 0.135 at 5. Nothing about the grip gets worse as friction rises: every set of forces available at $\mu = 1$ is available at $\mu = 5$, and more besides. The instrument says it is less than half as good. The load per squeeze only rises.

It is the failure a cone has no size warned about in general terms: a quantity computed from rows that carry different dimensions has a scale in it whether or not anybody chose one. Here the scale is how each row is normalised, and a normalisation chosen for one kind of generator is not neutral for another. The margin was never wrong for frictionless rows. It is the wrong instrument for friction edges, and without the comparison against $\rho$ nothing would have said so — both numbers are positive exactly when the part is held, so every held-or-not verdict agrees.

When two jaws beat four ideal contacts #

The hold is in the corners found the best four frictionless contacts for every regular polygon and a margin that falls in proportion to the side. With $\rho$ as a common unit, the question its last paragraph raised can be answered: at what friction do two jaws balance as much load per unit of contact force as the best four ideal contacts?

For a square it takes a friction of 0.775, and two jaws on a square are a worse fixture than four ideal contacts at any ordinary friction. For an octagon it takes 0.470, for sixteen sides 0.264, for thirty-two 0.137 and for sixty-four 0.069. The product of the side count and that friction rises through 3.76, 4.08, 4.22, 4.34, 4.38 and 4.43, towards $\sqrt 2\,\pi = 4.443$.

The limit has a reason. Four frictionless contacts on a nearly round polygon balance load through their corners, with an arm of about the half-edge, $\sin(\pi/n) \approx \pi/n$. Two jaws balance it through friction, at about $\mu/\sqrt 2$ as on the disc. They are equal when $\mu \approx \sqrt 2\,\pi/n$. So the corners essay’s closing claim — that friction regains an arm of order $\mu$ where a frictionless contact has one of order $1/n$ — is right, and the crossover is where the two orders meet.

The practical reading is a threshold on roundness. A part with more than about fifteen facets, gripped with a friction of 0.3, is better held by two jaws than by any four ideal locating contacts; a part with fewer is better located than gripped. That is not the same as saying either is sufficient — a frictional grip still needs its squeeze — but it says which kind of fixture is doing the work.

Three in space #

In space the frictionless count was seven: one more than the six dimensions of a spatial load. A point contact with friction contributes a round cone, which a finite test cannot hold exactly, so it is replaced here by an inscribed pyramid of eight edges — a conservative stand-in, since anything the pyramid can push the cone can too — and the test is the six-dimensional linear program the field already uses.

Two contacts on opposite faces of a cube never hold it, at any friction. The program reports the reason exactly: the rows are rank-deficient, because every force either contact can supply passes through the line joining them, and nothing resists a turn about that line. A point contact with friction cannot resist a twist about its own normal; a soft finger or a flat pad can, and that is a different contact model. Add a third contact on a side face and the cube is held at every friction tried, from 0.1 to 1. The same three contacts without friction hold nothing.

So friction takes the planar count from four to two and the spatial one from seven to three. In both cases the frictional count is the smallest number of cones whose edges can positively span, and in both the geometry decides which placements work while the squeeze decides how much they can take.

What the jaws do not settle #

Friction is a coefficient here, and a real one is not. Coulomb’s law with one coefficient is a model. Real contacts have a static coefficient larger than the sliding one, depend on surface finish and lubrication, and deform under load; a hold computed at $\mu = 0.3$ is a hold at a friction nobody measured.

The squeeze has to come from somewhere. A frictional hold is conditional on its normal forces, and those are supplied by a spring, a screw or a clamp that is not in the analysis. A frictionless hold with the same contacts would need none. Which fixture is better depends on whether the loads are known and bounded.

A hold is still not a location. Held is not located separated whether a part is held from where it ends up, and friction does nothing to close that gap: two jaws on a disc hold it at whatever angle it happened to be turned to when they closed.

The margins are not all comparable. The load per unit of squeeze, $\rho$, is used for every comparison in this essay because it is the only one of the three measures here that rises with friction; the program’s share in the spatial table is reported only as a verdict with a number attached, and is not compared with anything.

Still open: a squeeze that the load itself supplies #

A two-jawed vice supplies its squeeze with a screw, fixed before the load arrives. Some holds are designed so that the load supplies it instead: a wedge collet that closes as it is pulled, a cam cleat that bites harder as the rope pulls, a self-locking clamp. For those the load and the squeeze are not independent, and the question is not how much load per unit of squeeze but whether the hold is self-sustaining at all.

The distinct argument would be the condition for self-locking: a geometry in which every load in some set produces, through the contacts, a normal force large enough to resist it, so that no preload is needed. It is the frictional counterpart of form closure — a hold that is a fact about geometry again, but only for loads in a cone rather than for all loads. The measurement would take the wedge and the cam cleat, compute for each the set of loads it self-locks against as a function of the wedge angle and the friction, and find the angle at which the cone of self-locked loads closes — which should be where the wedge angle equals the friction angle, the classical self-locking condition, measured rather than quoted.