A hold the load squeezes shut

Two jaws and a coefficient held a body between jaws squeezed by a screw. Friction let each contact push anywhere within a cone about its normal instead of along the normal alone, and that took the number of contacts needed from four to two in the plane. The measurement there was how much load the hold resists per unit of the screw’s squeeze, and it assumed the squeeze came first: fixed before the load arrived, and independent of it.

Some holds have no screw. A cam cleat on a sailing boat grips a rope between two toothed cams on springs so light they only keep the cams touching; the rope’s own pull turns the cams inwards and they bite harder the harder it pulls. A wedge collet closes on a bar as the bar is drawn out. For these the squeeze and the load are not independent, and the question the essay ended on is whether such a hold sustains itself at all: whether, with nothing pressing beforehand, the contacts can carry the load.

The question as statics #

With no preload, the contacts’ forces are unknowns and the load is given. Each contact can push anywhere in its friction cone, so its force is a non-negative combination of the cone’s two edges. Each body — the rope, the cam, the wedge — has to balance. The hold sustains itself against a load if some choice of contact forces balances it.

That is a linear feasibility question, solved here by linear programming. Everything in it scales: if a set of contact forces balances a load, twice those forces balance twice the load. So the answer is not a magnitude but a set of directions of load, the ones the hold resists with no preload whatever their size. A cam cleat is useful because the direction a rope pulls is in that set.

A cam can push only along its pivot line #

A cam on a frictionless pin, touching the rope at one point, has two forces on it: the pin’s and the contact’s. A body with two forces balances only if they are equal, opposite and along one line, so the contact force lies along the line from the contact to the pivot. The cam cannot choose its force’s direction. The geometry chooses it.

The contact, for its part, can carry only forces inside its friction cone. So the cam holds exactly when the line from contact to pivot leans from the contact’s normal by no more than the friction angle, arctan μ. Lean it less and the cam’s force has a component along the rope, resisting the pull, inside what the contact can carry. Lean it more and no force the contact can carry points along the only line the cam can push on: the cam turns, and the rope runs.

The linear program measures this without being told it. With the rope a free body, a frictionless backstop below it and the cam’s contact force constrained to lie along its pivot line, the largest lean at which the rope’s pull can be balanced is, at friction coefficients from 0.1 to 1, arctan μ to $10^{-6}$ radians: 5.711° at 0.1, 21.801° at 0.4, 45.000° at 1. At no lean at all — the pivot straight over the contact — the cam holds nothing either, since its force then has no component along the rope. The window is between the two.

This is the frictional form of what a contact turns and slides at once found for a stop on a four-bar. There a contact holds a motion in proportion to how far its normal misses the motion’s pole, and a normal through the pole holds nothing. Here the cam’s pivot is its pole, and the cam holds only while the line to the pole lies inside the cone of forces the contact can carry. What one contact forbids drew the same line from the other side: the centres a single contact cannot resist lie on its normal line, and friction widens that line into a wedge.

A wedge passes its squeeze through two contacts #

A wedge collet, or a door wedge dragged under a door, works differently. The bar pulls the wedge along by friction, the wedge is carried into a gap that narrows, and the gap’s reaction presses it back onto the bar. The wedge has two contacts: one on the bar, whose force can lean from the bar’s normal by the bar’s friction angle $\phi_1$, and one on the incline, whose force can lean from the incline’s normal by $\phi_2$. The wedge balances only if these two forces are equal and opposite. The incline’s normal is tilted from the bar’s by the incline’s angle α, so the two cones can share a line exactly when

$$\alpha \le \phi_1 + \phi_2.$$

The measurement agrees at every pair of coefficients tried, to $10^{-6}$ radians: 11.42° when both coefficients are 0.1, 33.40° when both are 0.3, 22.41° when one is 0.3 and the other 0.1, whichever way round.

The figure usually quoted for a self-locking wedge is α ≤ arctan μ, a single friction angle. That is the condition for a block resting on one inclined face, which has one contact that can lean. A wedge between a bar and an incline has two, and each lets the force lean by its own angle. The friction angles add along the chain of contacts the squeeze passes through, and a frictionless pivot — the cam’s — adds nothing. The quoted figure is the right answer to a different mechanism and it is half the right answer here.

What is not in the chain does not matter #

The rule says which contacts count: the ones the squeeze passes through on its way from the load to the ground. It also says which do not, and the measurement checks both.

A real cam cleat has two cams, one on either side of the rope, and no backstop. Each cam is a two-force member on its own pivot line, and the rope between them is squeezed by both. Built that way and solved the same way, the pair holds up to a lean of 21.801° at a coefficient of 0.4 — arctan μ, exactly what one cam against a smooth backstop holds. The second cam doubles the grip and changes nothing about when the grip exists, because its own force has to lie inside its own cone just as the first one’s does. A cam cleat with one cam leaned inside the cone and the other outside holds with the good one and is merely carried by the bad one.

The wedge’s bar rests on a flat. In the measurements above the flat is frictionless, so the whole pull goes through the wedge. Give the flat a coefficient of 0.5 and the bar is held partly by the flat’s friction, pressed by the wedge’s squeeze — and the wedge still jams up to exactly 38.501° with coefficients of 0.4 and 0.3 on its own two contacts, the same as with a smooth flat. The flat’s friction lowers how hard the wedge must squeeze for a given pull. It does not change whether the wedge can squeeze at all, because the flat is not between the wedge and the ground.

That distinction is the design rule in one line. To make a self-locking device lock over a wider range of angles, raise the friction at the contacts the squeeze passes through; friction anywhere else only shares the load.

The set of held loads does not close #

The lead that asked for this measurement expected the set of loads a self-locking device holds to be a cone that closes as its angle approaches the threshold, and the threshold to be where it has shrunk to nothing. The measurement says otherwise.

Below its threshold the cam cleat holds every load from straight into the backstop round to 90° less its lean, measured from the direction of the pull. At a fifth of the friction angle that is 85°; at 99% of it, 68°. The range narrows only by the lean itself, never to nothing. Just past the threshold it holds none of the 179 directions tried, not even the ones pressing the rope towards the backstop, because every one of them needs the cam to carry some of it and the cam can carry nothing. The wedge behaves the same way: just past $\phi_1 + \phi_2$ it holds none either.

So self-locking is all or nothing. Below the threshold the device holds a wide cone of loads, including the one it was designed for, and above it holds none. There is no warning margin in which it holds a few loads less securely. That argues for a lean chosen well inside the friction angle of the worst rope the cleat will see — a wet one, a worn one — because a lean chosen for a dry rope gives no hint that it will fail when the coefficient falls, until it does.

The squeeze is least at the edge of slipping #

A self-locking hold’s squeeze is not set by a screw but by the geometry, and it is worth knowing how large it is.

The cam’s force lies along its pivot line, so its component along the rope is the pull and its component across the rope is the squeeze. The squeeze is $\cot\gamma$ times the pull, with γ the lean. As the pivot comes over the contact the squeeze grows without bound: a cam cleat whose pivot is nearly over its teeth crushes the rope. At the threshold it is least, $1/\mu$ — 2.5 times the pull at a coefficient of 0.4 — and the hold is at the edge of slipping. Halfway into the window, at a lean of about 11°, it is 5.2 times the pull; at 5°, 11.4 times. A cleat leaned for security pays for it in how hard it bites, and the bite is a load the rope has to carry across its fibres as well as along them. The wedge’s squeeze is $1/\tan(\alpha - \phi_2)$ times the pull, without bound as the incline’s angle falls to the incline’s own friction angle and least, $1/\mu_1$, at its threshold.

Every self-locking design therefore sits in a window between crushing what it holds and letting it go, and the two ends are the same fact seen twice. A lean well inside the cone is secure and squeezes hard; a lean near the cone’s edge squeezes gently and fails without warning when the friction drops. A designer picks the lean for the worst friction the device will see and accepts the squeeze that comes with it at the best.

A lean is a least coefficient of friction #

Turned round, each threshold says what a given geometry demands of its surfaces. A cam cleat whose pivot line leans γ holds any rope whose coefficient against the cam is at least $\tan\gamma$, and no rope whose coefficient is less. A lean of 10° demands 0.18; 15° demands 0.27; 20° demands 0.36; 30° demands 0.58. The geometry fixes a single number, and the rope either meets it or does not.

The wedge demands something subtler, because two coefficients share the demand. An incline at 25° is held if the bar’s and the incline’s friction angles add to at least 25°: coefficients of 0.3 and 0.15 will do (16.7° and 8.5°), and so will 0.47 and nought. A wedge whose incline has worn smooth can still lock if its face on the bar is rough enough, and the other way round. That is a freedom a designer can use — putting the grip on whichever surface is easier to keep rough — and one the single-angle rule hides.

And the squeeze at the demand is fixed as well. A cam at the edge of its demand squeezes with $1/\mu$ times the pull, a wedge with $1/\mu_1$. Neither can be made to squeeze more gently than that and still hold, which is the sense in which these holds are cruel to what they grip: the least squeeze they can hold with is set by friction, not by the designer, and for a slippery rope it is large.

The roller clutch and the two jaws, seen again #

A ratchet with no teeth found a roller clutch holding one way and running free the other, and left open whether it grips, because that is a question about friction. This is that question answered for the simplest cases. A roller wedged between a race and a sloped face is a wedge with two contacts, and by the rule measured here it should grip while the slope is no more than the sum of the two contacts’ friction angles; the roller’s own rotation, which this statics leaves out, is what that essay’s geometry adds. The angle that holds the lock met a locking face whose draw decided whether an escapement tooth held itself; the draw is a lean, and friction widens the range of leans that hold.

The cam cleat also completes two jaws and a coefficient in a way that essay could not. There a squeeze was bought with a screw and the question was what it bought. Here the squeeze is bought with the load, and the question is only whether it can be: a matter of whether one line lies inside one cone. A V holds the wheel to first order found a Geneva lock’s hold set by how far its contact normal misses the wheel’s pivot. A cam cleat’s is set by how far its pivot line misses the contact normal, and friction decides how far is too far.

What the statics does not settle #

Rigid bodies. Every contact is rigid and every force is found from statics alone, the same idealisation the hold is in the corners made for frictionless holds. A real rope is soft, a real cam tooth digs in, and the effective friction of a toothed cam on a rope is not a single coefficient. The thresholds are for the idealisation.

How the hold is entered. Statics says whether an equilibrium exists, not whether the device reaches it. A cam cleat needs a light spring to keep the cam touching the rope before the pull arrives, and a wedge needs to be moving with the bar to be dragged in. Both are assumed.

Release. A self-locking hold is released by pulling the rope out and up, or by driving the wedge back, and how much force that takes is not computed. A clearance is a link would count the play a real cam has before it bites; the statics here has none. For the cam it is the lean’s other side: pulling the rope the other way turns the cam out of the cone.

The wedge’s own rotation. The wedge is balanced in force only; it is assumed to seat flat on both faces, which a real wedge does by rotating slightly until it does.

Still open: a pivot with friction #

The cam’s pivot is frictionless here, and a real pin is not: a pin with clearance and friction carries its force not through its centre but tangent to a small circle about it, the friction circle, whose radius is the pin’s radius times the pin’s coefficient. The two-force member’s line then touches that circle instead of passing through the centre.

The distinct argument would be that pin friction counted along the chain. By the rule measured here the threshold should grow by the angle the friction circle subtends from the contact, and a cam cleat with a stiff pivot should hold at a lean a plain one would not. The measurement would be the threshold against the pin’s radius and coefficient, and whether pin friction also makes the cleat harder to release by the same amount — which would make it a trade, not a gain.