A cleat set hard lets go easiest

A stiff pin buys grip with release gave a cam cleat’s pivot pin friction and found two things. The cleat holds a forward pull to a steeper lean than a free pin allows, and it may lock against a backward pull too, refusing to let the rope go the way a cleat is supposed to let it go. In between, it found the squeeze no longer decided. A frictionless pin forces the squeeze to exactly $\cot\gamma$ times the pull, because the cam’s force must pass through its pivot. A pin with friction lets the force miss the pivot by the friction circle, and any squeeze in a band about $\cot\gamma$ balances the cam.

That essay ended with how cleats are actually released. Nobody pulls a jammed rope backwards; they lift the cam away with a lever, or flick it open with a thumb. On a frictionless pin that takes nothing to start. With pin friction it must take something, and the essay asked whether that something grows with how hard the cleat was set, and whether it jumps at the backward-lock lean, the line below which the cleat will not let go of a backward pull.

It does neither. It depends on how the cleat was set, but in the opposite direction, and the lock lean is a change of state rather than a change of effort.

The moment a lever has to beat #

Put the rope’s contact at the origin and the cam’s pivot a height $h$ above it, its line leaning $\gamma$ from the contact’s normal. The rope touches the cam at a single contact of the kind what one contact forbids measured, pushing only, as a constraint that only pushes requires. Under a pull $T$, squeezed with $N$, it pushes on the cam with $T$ along the rope and $N$ up into the cam. About the pivot that force has a moment $h(T - N\tan\gamma)$, and its sense is the one that closes the cam: the pull drags the cam’s face forwards, which turns it into the rope. That is the cleat’s whole principle, and a frictionless cleat is balanced when the squeeze’s moment cancels the pull’s, $N = T\cot\gamma$.

A lever on the cam is a couple. It adds a moment and no force, so the pin still carries exactly the rope’s force, $\sqrt{T^2 + N^2}$. A pin with a little clearance touches its bore at one point, as a pin is not a point and a clearance is a link measured, and with friction it can resist a moment up to $\rho$ times that force, $\rho$ being the radius of its friction circle. To turn the cam open the lever must beat the rope’s closing moment and the pin’s friction together:

$$M = h\,(T - N\tan\gamma) + \rho\sqrt{T^2 + N^2}.$$

With a frictionless pin, $\rho = 0$ and the squeeze forced to $T\cot\gamma$, the first term is nought and so is $M$: the earlier essay’s free release. With a stiff pin the squeeze can sit anywhere in the band, and $M$ depends on where.

At the band’s bottom edge the pull’s closing moment is as large as the pin can hold, $h(T - N\tan\gamma) = \rho\sqrt{T^2+N^2}$, so the lever must beat twice the pin’s friction moment, $2\rho\sqrt{T^2 + N^2}$. At the top edge the squeeze’s opening moment is as large as the pin can hold, the first term is $-\rho\sqrt{T^2+N^2}$, and $M$ is nought: the cam is on the point of being pushed open by the rope it is squeezing, and the slightest touch frees it.

So a cleat set harder is easier to release, not harder. The squeeze is what pushes a cam open. Its moment about the pivot is opposite to the pull’s, and a cleat jammed in hard has stored more of it.

The formula was checked by a route that does not use the friction circle. The pin was modelled as the earlier essay modelled it, a round pin touching its bore at a single point, the force there confined to that point’s own friction cone. At each of 1,440 places round the pin, the rope’s force was tested for whether it lies in that place’s cone. Where it does, the moment the place can resist was computed, and the largest of those is what the lever must beat. At every lean from 1° to 24° the two routes agree to the sampling, 0.412 against 0.412 at 10° and 1.436 against 1.439 at 1°, the contact route always from below by at most a step’s worth of lever.

Which edge the load leaves it at #

A squeeze anywhere in the band balances the cam, so which one the rope actually feels is a question about history. The earlier essay said so and left it there. For release it has to be settled, and it can be, by following a load as it changes.

As a pull comes onto the rope, it drags the cam’s face forwards and the cam turns in, raising the squeeze. It keeps turning while the pull’s closing moment beats what the pin can hold, and it stops at the first squeeze where the pin can just hold it. That is the band’s bottom edge. So a rising load leaves the squeeze on the lower line, $N = 4.50\,T$ at a lean of 10° for the earlier essay’s design, a contact friction of 0.4 and a pin whose friction circle is 0.0447 of the pivot’s height. At a pull of one the lever needs 0.412 times the pivot’s height, which is twice the pin’s friction moment at that squeeze.

As the pull falls, the squeeze stays where it was, since a rigid cam cannot squeeze less without turning out, and turning out needs the squeeze’s opening moment to beat the pin. That happens only when the band’s top edge comes down to meet the squeeze. On this design that is at a pull of 0.591, and from there the cam is pushed out along the top line as the pull falls further. There the lever needs nothing. Raise the pull again and the squeeze waits until the bottom line comes up to it, then rides that line.

The path is a loop between two lines, the shape of every frictional hold that has a range rather than a value. Where the input stops deciding found an escapement whose state cannot be read off its input because it depends on which face a tooth last met; the cleat’s squeeze cannot be read off its pull for the same kind of reason. A seat with friction stays where it is pushed found the same behaviour in a stop on a four-bar, which keeps whatever turn its last load gave it. The cleat’s squeeze is its memory of the load, and the band’s edges are the limits of what it can remember.

A ratchet with no teeth left a roller clutch’s grip as a question about friction, and the loop answers part of it for any such hold: a frictional one-way device has no single state at a given load, only a band, and where it sits in the band is what decides how it lets go.

So the cleat that is hardest to release is the ordinary one: holding the load it was set by, with the load still on. A cleat whose load has eased a little, or one jammed in by hand before the load came on, sits at the top edge and opens at a touch. A sailor who has jammed a sheet into its cleat and then eased it slightly has, without knowing why, made it easy to throw off.

The steeper the cleat, the easier the lever #

The lever moment for the squeeze a rising load leaves can be followed across every lean the cleat works at.

It falls steadily as the lean steepens: 1.439 times the pull times the pivot’s height at 1°, 0.923 at 3°, 0.602 at 6°, 0.412 at 10°, 0.256 at 18°. Near the forward limit, 24.14° on this design, it falls faster, to 0.110 at 22° and 0.007 at 24°. There the band has closed to a single squeeze, the cam is held with almost nothing to spare, and almost nothing frees it.

Two reasons combine. A steeper cam needs less squeeze to hold the same pull, since $\cot\gamma$ falls as $\gamma$ grows, and the pin’s friction moment is proportional to the force it carries, so a smaller squeeze leaves less friction to beat. And the band narrows as the lean steepens, so the gap between the bottom edge and the frictionless balance, which is what the pin has to hold, shrinks. Above about 20° the bottom edge stops being the pin’s at all. It becomes the rope’s own floor, $N = T/\mu$, the least squeeze at which the contact can carry the pull without slipping, and there the lever needs less than twice the pin’s moment.

The figure answers the earlier essay’s question about a cliff directly. The backward-lock lean on this design is 2.56°, and the lever moment goes through it without a step: about 1.00 times the pull times the pivot’s height from either side. Under load, the lock lean is not visible in the effort at all.

The effort is also not small. At 10° the lever must supply 41% of the pull’s own moment about the pivot, and at 3° it must supply 92%. A lever three pivot heights long, a sensible size for a thumb, needs 0.14 of the pull at 10° and 0.31 at 3°. Those numbers are the price of the pin’s friction. A frictionless pin of the same cleat needs nothing.

Below the lock lean, a slack cleat remembers #

The cliff the earlier essay suspected is real, but it is in a different place. It appears when the load comes off.

Above the lock lean the band has a top edge, and a falling load reaches it. The cam is pushed out as the pull falls, and at no pull the squeeze is nought: the slack rope lies in an open cleat. Below the lock lean the band has no top, because a squeeze of any size can be held by the pin’s friction alone. That is exactly what locking against a backward pull means. So a falling load never pushes the cam out. The squeeze the largest pull left stays when the rope is slack, and the lever must then beat the pin’s friction against the squeeze’s own opening moment, $N(\rho - h\tan\gamma)$.

At 1° a pull of one leaves a squeeze of 16.1, and the slack cleat needs 0.438 to free it; at 2°, a squeeze of 12.5 and a moment of 0.123. Both grow in exact proportion to the largest pull the cleat has carried: double the pull and the slack cleat holds twice as hard and needs twice the lever. Below the lock lean a cleat is a clamp that remembers the hardest pull it ever held.

The moment, though, falls to nought at the lock lean while the squeeze does not. Just below the line a slack rope is still clamped at about eleven times the largest pull it carried, and a touch on the lever frees it. Just above the line the rope is not clamped at all. That is the change of state the earlier essay’s question was reaching for. For the person using the cleat it means a rope that, after the load is off, sometimes lies loose and sometimes is gripped hard. Which one depends on a fraction of a degree of lean that wear, a bent bracket or a swollen rope can move.

Two routes and one loop #

The friction circle carries every result above, so it matters that the pin as a single contact gives the same numbers, and the ledger sets them side by side.

The two routes agree at every lean to within the single-contact pin’s sampling, and the contact route is always the lower. That ordering has a reason: places 0.25° apart can miss the best one by up to a step, which costs a little of the lever the pin can offer, and a coarser sampling would make the gap larger.

The ledger also corrects one reading of the earlier essay’s band. Below the lock lean a forward pull is held not by any squeeze from the rope’s floor upwards but only by a squeeze large enough for the pin to hold the cam: 16.1 times the pull at 1° and 12.5 at 2°, against a floor of 2.5. Below that the pull’s closing moment is too large for the pin, and the cam turns in until it reaches that value. The band there is bounded below by the pin and unbounded above, which is why a slack cleat below the lean keeps everything.

What a designer can set #

Three things follow for anyone choosing a cleat’s geometry.

The lean above the lock line is not only about the backward pull. It decides whether a slack rope is free. Return and stiffness come from the same aim found the same rule for a stop on a four-bar: whether a contact lets go when its load is taken off is decided by which side of a friction cone a pivot lies, and margin on the right side costs nothing. A cleat meant to let a rope run as soon as the load comes off must lean more than $\arctan(\rho/h)$ with margin for the wear and lubrication that change $\rho$.

The lever’s size is set by the load’s own setting, not by a hard jam. A thumb lever sized for $2\rho\sqrt{T^2 + N^2}$ at the bottom edge of the band will free the cleat however it was set, and an operator who jams the cleat in hard has made it easier, not harder, to release.

And a steep cleat is easy to free under load and also close to slipping. The lever moment falls to nought at the forward limit because the band collapses there, which means the cleat is held with no margin. The ease and the risk are the same fact.

A hold the load squeezes shut found that a cleat’s grip scales with the load and needs no setting. With a frictionless pin its release does not scale with the load either, since it is nought. With a stiff pin the release scales with the load too, as $2\rho\sqrt{1 + n^2}$ times the pull, and the question of how hard to make the lever is a question about the largest load it must free.

What the lever model leaves out #

A rigid cam and rope. The squeeze stays exactly where it was while the load falls only because a rigid rope cannot relax. A real rope compresses under the squeeze and creeps, so a cleat left loaded for a long time drifts down the band and becomes easier to free, and a cleat below the lock lean slowly loses some of what it remembers. Neither is modelled.

One instant of release. The lever moment is what starts the cam turning. Once it turns, the contact lifts and the rope is free in this model. A real cam’s teeth drag the rope back slightly as they lift out, and the rope’s tension then works against the lever for a little way. How much depends on the teeth and the rope’s surface, which a rigid contact cannot see.

Holds against other directions. The pull here is along the rope. Two jaws and a coefficient measured holds against every direction of load, and a pull a little off the rope’s line changes both the closing moment and the squeeze’s opening moment, so it moves the band’s edges and the lever with them.

Static and kinetic friction. The pin’s friction circle is a limit on a stationary pin. Once the cam turns, the pin slides at a lower coefficient, so a lever that just starts the cam will finish the job with room to spare. The band’s edges, in contrast, are set by static friction, and so is the slack cleat’s memory.

The lift-out release. A rope can also be freed by lifting it up and out of the jaws, which loads the cam along the rope’s normal rather than through a lever. That release, and the force it takes, is not computed. Six things a hold is not collected how much a hold’s verdict depends on the load’s direction; the direction of a release matters in the same way.

Still open: a cleat whose lean moves with its load #

Everything here holds the lean fixed, and the lean decides almost everything: the band, the lever, and whether a slack rope is free. A cam pivoted on a lever arm or a spring rather than a fixed pin would change its lean as the load changes it.

The distinct argument would be a cleat whose pivot is carried on a short swinging link, so that a rising pull draws the pivot forwards and steepens the lean while a falling one lets it return. The measurement would set the link’s length and spring rate, and follow the same rising-falling path as here. It would ask whether a cleat can be made steep under load, where it is easy to release and needs little squeeze, and shallow when slack, where it is above the lock lean or below it by design. If the swinging pivot can put the cleat above the lock line whenever the load is off and keep it gripping whenever the load is on, it removes the slack cleat’s memory without losing the hold. If it cannot, the fixed lean is a trade a designer makes once and lives with.