A stiff pin buys grip with release

A cam cleat holds a rope with nothing but its own geometry. The rope runs between a fixed backstop and a toothed cam on a pivot, touching each at a single contact of the kind what one contact forbids measured, and pulling the rope drags the cam’s face with it, which turns the cam into the rope and squeezes it harder. A hold the load squeezes shut measured when that works. A cam on a frictionless pin can push on the rope only along the line from the contact to its pivot, because a body with a pin and one other force is a two-force member. It holds exactly while that line leans from the contact’s normal by no more than the contact’s friction angle, $\arctan \mu$ — the one-sided constraint of a constraint that only pushes, given a cone of friction. It holds at any load, since everything scales with the pull, and it lets go at once when the rope is pulled back.

That essay’s pin was frictionless, and it ended by asking what a real pin does. A pin with a little clearance in its bore touches at one point — a pin is not a point and a clearance is a link both measured what that does to a mechanism’s geometry — and the force there leans from the bore’s normal by up to the pin’s own friction angle. So the pin’s force on the cam no longer passes through the pivot’s centre. It lies tangent to, or inside, a small circle about it — the friction circle, of radius $\rho = r\mu_p/\sqrt{1 + \mu_p^2}$ for a pin of radius r and friction coefficient $\mu_p$. The cam is no longer a two-force member of the strict kind. Its contact force may pass anywhere within ρ of the pivot.

The question was whether that friction angle adds to the chain the way the contact’s did, whether a stiff pivot therefore holds at a lean a free one would not, and whether it also makes the cleat harder to release.

It does all three, and the third nearly cancels the second. What follows measures each, by a closed form and by a model of the pin that shares none of its algebra.

The forward limit is a sum of two angles #

Put the contact at the origin and the pivot a height h above it, leaning γ from the contact’s normal, so that it sits at $(h\tan\gamma, h)$. A forward pull of one on the rope has to be carried by the cam’s friction, since the backstop only pushes straight up. So the cam presses on the rope with a squeeze N and drags it back with a force of one, and the contact’s friction cone requires $N \ge 1/\mu$. The cam, for its part, feels the opposite of that force, and its moment about the pivot must be carried by the pin. The contact force has moment $h(1 - N\tan\gamma)$ about the pivot. The pin can supply at most ρ times the size of the force, $\rho\sqrt{1 + N^2}$.

So the cleat holds while some squeeze $N \ge 1/\mu$ satisfies $|1 - N\tan\gamma| \le (\rho/h)\sqrt{1 + N^2}$. With no pin friction the right side is nought, the squeeze is forced to $\cot\gamma$, and the contact’s condition becomes $\tan\gamma \le \mu$, the earlier essay’s rule. With pin friction the condition is easiest to meet at the smallest squeeze the contact allows, $N = 1/\mu$, and the largest lean that holds is

$$\tan\gamma^{*} = \mu + \frac{\rho}{h}\sqrt{1 + \mu^2}.$$

That has a geometric reading, and it is the one the earlier essay’s rule predicted. Let δ be the angle the friction circle subtends at the contact at that lean: the angle between the line to the pivot and a line from the contact just touching the circle, $\sin\delta = (\rho/h)\cos\gamma^{*}$. Then $\gamma^{*} = \arctan\mu + \delta$ exactly. The two expressions agree to $10^{-12}$ on every design below. The friction angles add along the chain of contacts the squeeze passes through: the contact’s, and now the pin’s, as seen from the contact.

For a contact friction of 0.4, the contact’s friction angle is 21.80°. A pin whose friction circle is 0.045 of the pivot’s height — a pin a tenth of the pivot’s height in radius with a friction coefficient of 0.5 — adds 2.34° and takes the limit to 24.14°.

The earlier essay found the same addition in the sliding wedge, whose squeeze passes through two frictional contacts and which jams to the sum of their two friction angles. The cam on a frictionless pin had only one frictional contact in its chain, the rope’s, and its limit was that one angle. The pin with friction is a second frictional contact in the same chain, seen through the lever the cam makes between them, and its angle adds in the same way. What is different is the size: a pin’s friction angle is large, but the angle its friction circle subtends from the contact is small, because the circle is small and far away. That is the lever’s work, and it is why pin friction moves the limit by two or three degrees rather than by twenty.

Two routes to the limit #

The closed form rests on one piece of modelling, the friction circle, and so a second route was built that does not use it. The pin was modelled as what it is: a round pin touching its bore at a single point, the force there confined to that point’s own friction cone. Every one of 1,440 places round the pin was tried in turn, each as a small linear feasibility problem. The rope’s balance was imposed with the contact’s cone and the backstop, and the cam’s with its force and its moment about the pivot. The largest lean that holds was then found by bisection.

The first version of that route sampled the pin at many places at once and let them all push, and it held at every lean. Opposite points on a bore pushing together make a couple of any size, which is a pin clamped in its bore, not a pin with clearance. Allowing one contact point at a time is what makes it a pin.

On eight designs, with contact frictions from 0.2 to 0.8 and pins from none to a fifth of the pivot’s height, the contact route reaches the closed form’s forward limit to within 0.013°, always from below. The remainder is the sampling. With places 0.25° apart the best available contact point can miss the exact tangent to the friction circle by half a step, which costs a small part of the pin’s lever, and the gap shrinks as the places are made finer.

The pin’s angle, seen from the contact #

The pin’s own friction angle is large, and the angle it adds to the cleat is small, and the table shows by how much. A pin of friction coefficient 0.5 has a friction angle of 26.57°. A tenth of the pivot’s height in radius, it adds 2.34° to the cleat’s limit; a fifth of the pivot’s height in radius, 4.60°. A pin of coefficient 0.2, a friction angle of 11.31°, adds 1.04° at a tenth and 0.52° at a twentieth. In every case the sine of what reaches the cleat is exactly the pin’s radius over the pivot’s distance from the contact, times the sine of the pin’s own friction angle. A small, well-lubricated pin far from the contact adds almost nothing, and a large, dry one close to it can add more than the contact’s own angle would suggest from its size.

The contact’s friction changes δ too, though only a little, because it changes the lean at which the limit falls and so the distance from contact to pivot. For the same pin, with a friction circle 0.029 of the pivot’s height, δ is 1.61° when the contact’s coefficient is 0.2 and 1.26° when it is 0.8: a steeper limit puts the pivot further from the contact along its line, and the circle subtends less. The two friction angles add, but they are not independent. The contact’s sets the geometry from which the pin’s is seen.

The backward limit shows the opposite dependence. It is $\arctan(\rho/h)$, with no μ in it at all: 1.65° for that pin whatever the contact’s friction, because a backward lock is held by the pin alone. The contact only has to carry the squeeze without slipping, and a squeeze large enough to hold backwards is far above the contact’s floor of $1/\mu$.

Pulled backwards, it holds too #

A frictionless cleat lets go when the rope is pulled back. The pull then drags the cam’s face away from the rope, the cam turns out of the way, and there is no force the pin can supply to stop it. That is what makes a cleat useful: it grips one way and releases the other.

With pin friction the backward pull meets the same condition with the sign changed. The cleat holds backwards if some squeeze satisfies $1 + N\tan\gamma \le (\rho/h)\sqrt{1 + N^2}$. As the squeeze grows, the left side over the root on the right tends to $\tan\gamma$, so a large enough squeeze always works when

$$\tan\gamma < \frac{\rho}{h},$$

and none does otherwise. A cam leaning less than the friction circle’s own angle, seen from the contact, can be squeezed until the pin’s friction alone holds it, and the rope cannot be pulled back out. For the design above that is every lean below 2.56°. The contact route finds the same boundary to 0.019°, from below again, and here the reason is not sampling. The squeeze a backward lock needs grows as $1/(\rho/h - \tan\gamma)$, without bound at the limit, and a solver asked for feasibility stops short of an unbounded answer.

The window moves more than it widens #

A cleat that works grips a forward pull and lets go of a backward one, so it has to lean between the two limits:

$$\frac{\rho}{h} < \tan\gamma \le \mu + \frac{\rho}{h}\sqrt{1 + \mu^2}.$$

Pin friction raises both edges. The lower edge rises at a slope of one in $\rho/h$ and the upper at $\sqrt{1 + \mu^2}$, which is 1.077 for a contact friction of 0.4. So the window’s width grows only by $(\rho/h)(\sqrt{1 + \mu^2} - 1)$: from 0.400 in $\tan\gamma$ with a frictionless pin to 0.419 at a friction circle a quarter of the pivot’s height, which would be a very stiff pivot indeed. Almost all of the extra grip is taken back at the bottom of the window.

The same trade appears wherever a hold relies on friction somewhere other than at the contact that carries the load. A ratchet with no teeth left a roller clutch’s grip as a question about friction, and the answer here suggests its shape: any friction that helps the clutch lock one way is also available to stop it freeing the other way. A hold that is meant to be one-way is a hold whose friction is allowed to act only one way, and friction in a pin acts both ways.

That answers the question the earlier essay asked, whether a stiff pivot is a gain. It is mostly a shift. A designer who wants the cleat to hold at a steeper lean can get it from pin friction, and must then keep the lean above a floor that pin friction also raised. A designer who instead relies on the free-pin rule and builds a stiff pivot has a cleat that holds a little more than expected and releases a little less, which at shallow leans means not at all.

How hard it bites is no longer decided #

The frictionless cleat also said how hard it squeezes: exactly $\cot\gamma$ times the pull, since the force must lie along the pivot line. Pin friction takes that away.

With a stiff pin a forward pull can be held by any squeeze in a band about $\cot\gamma$. For the design above, at a lean of 10° any squeeze from 4.50 to 7.62 times the pull balances the cam, around the frictionless 5.67. The statics cannot say which squeeze the rope actually feels. That depends on how the cleat was set, how hard the cam was pushed in and how far the pin had turned in its bore, just as a joint with friction in a clearance is a link has a position that depends on its history. The band narrows as the lean grows and closes at the forward limit, where only one squeeze is left. At 20° it runs from 2.50 to 3.15 against a frictionless 2.75; at 24°, from 2.50 to 2.52. Its lower edge sits at 2.50, the contact’s own floor of $1/\mu$, over most of that range: the smallest squeeze the pin allows would be lower still, and it is the rope’s friction that refuses it. The band is therefore set by the pin from above and by the contact from below, and a cleat near its forward limit is held at almost exactly the contact’s floor. That is the squeeze at which the rope is on the point of slipping, which is why a cleat at the edge of its window holds but gives no warning before it stops holding.

The backward lock needs more than that, and more as the limit approaches. At 0.5° a backward pull is held only by a squeeze of at least 28 times the pull. A fiftieth of a degree short of the limit it needs more than 2,800. That is the sense in which the backward lock is real but fragile: near its limit it holds only if the cleat was jammed in hard, and in practice it shows up as a cleat that sometimes releases and sometimes does not.

What the pin model leaves out #

A rigid cam and rope. A rope compresses under the squeeze and a cam’s teeth dig in, which changes the contact’s effective friction and moves the contact point. Both are outside a rigid-body model, and both matter to real cleats more than the pin does.

Static friction only. The pin’s friction here is a limit on a stationary pin. Once the cam turns, the pin slides, and a kinetic coefficient lower than the static one gives a smaller friction circle. A cleat at the edge of its window may start to turn under a steady pull and then not stop.

One pull direction at a time. The pull is along the rope, forwards or backwards. A hold’s verdict depends on the direction of the load, which is one of the lessons six things a hold is not collected. Two jaws and a coefficient measured holds against every direction of load. The set of directions a cleat with pin friction holds was not mapped, and it would show how much of the backward lock survives a pull a little off the rope’s line.

Still open: releasing a cleat on purpose #

A cleat is released in practice not by pulling the rope back but by lifting the rope out, or by turning the cam away with a lever. With a frictionless pin the lever needs no force at all to start. With pin friction it must overcome the pin’s friction moment, and if the cleat is locked backwards, it must also undo the squeeze.

The distinct argument would be the release force as a function of how the cleat was set: the squeeze the rope was left at, the lean, and the pin’s friction circle. The measurement would add a release lever to the cam and ask, for each squeeze in the band, how much lever force frees the rope. If that force grows with the squeeze, a cleat set hard is harder to release in proportion. If it jumps at the backward-lock limit, the window’s lower edge is a cliff for the person using it and not only a line in the statics.