A cleat on a swinging pin has two leans

A cleat set hard lets go easiest found that a cam cleat’s lean decides almost everything about it. The lean sets the band of squeezes its pivot pin’s friction will hold. It sets the lever moment that frees the rope. And below one particular lean, the backward-lock lean of $2.56°$ on the design measured there, it decides that a slack rope stays clamped. The cam keeps the squeeze of the hardest pull it ever held, 12.5 times that pull at a lean of $2°$, because the pin’s friction is enough to hold the cam against the squeeze pushing it open.

That essay ended by asking for a cleat whose lean moves with its load: a pin carried on a short swinging link, so that a rising pull draws the pivot one way and a falling pull lets it return. If the swinging pin could keep the cleat steep under load and shallow when slack, or the reverse, it might remove the slack cleat’s memory without losing the grip.

A swinging pin does remove the memory, and it keeps the grip. But it does not do it by making the lean move with the load. Every force in the cleat is in proportion to the pull, so nothing in it can depend on how hard the rope is pulled. What the link adds is a second lean, and the cleat uses one lean or the other according to whether the load is rising or falling.

Two motions press the cam into the rope #

The cleat is set out as before, with the rope’s contact at the origin and the cam’s pin a height $h$ above it, the line between them leaning $\gamma$ from the contact’s normal. The pin’s friction lets the rope’s force on the cam pass anywhere within the pin’s friction circle, radius $\rho$, $0.0447$ of the pin’s height on this design. As a stiff pin buys grip with release found, that frees the squeeze from the single value $T\cot\gamma$ a frictionless pin would force. Any squeeze in a band $n_{lo}T \le N \le n_{hi}T$ balances the cam, and below the lock lean the band has no top.

The new part is that the pin no longer stands on the frame. It sits at the end of a link that swings about a pivot $G$, frictionless, $2.2$ pin heights above the rope and a little forward of the contact, held by a torsion spring.

That raises a question the fixed cleat never had to answer: what lets anything move? Once the rope is gripped, a rigid cam on a rigid rope cannot press further into it. The pin’s swing is only a change of the cam’s pose, and a pose change that pushed the contact into the rope would be refused. So the cleat needs the rope to give. Here the rope is a spring across its thickness, squeezed in proportion to how far the cam presses into it. That is what a real rope does over the small range a cleat uses: a braided line flattens visibly under a cam. Along the rope, the rope moves with the contact and nothing resists it but the pull.

With that, two motions press the cam into the rope. The cam can turn on its pin, which moves the contact down by $\tan\gamma$ per unit of turn, or cam and link can turn together about $G$, which moves it down by $a$, the link pivot’s distance forward of the contact. Call the squeeze each motion contributes $c$ and $L$, so that $N = c + L$. The link balances the rope’s force against its spring:

$$k_s\theta = gT - aN \quad\Longrightarrow\quad K L = bT - N,$$

with $g$ the link pivot’s height, $b = g/a$, and $K = k_s/(k_r a^2)$, the spring’s stiffness over the rope’s, both referred to the link’s lever. While the pin holds, $c$ is fixed and the squeeze follows the pull along

$$N = \frac{Kc + bT}{1 + K}.$$

While the pin holds, the cleat is a cam about the link’s pivot #

That formula has a plain reading. With the pin holding, cam and link are one body that turns about $G$, so they form a cam pivoted at $G$, frictionless, leaning $\arctan(a/g)$ from the contact’s normal. A frictionless cam squeezes the rope with the cot of its lean times the pull: a hold the load squeezes shut found that for the cleat’s own pin, and it is $b$ here. The spring softens it. Pressing the rope harder means turning the link against the spring as well as compressing the rope, so the squeeze per unit pull the link gives on its own is

$$n_{link} = \frac{b}{1 + K}.$$

A cleat on a swinging pin is therefore two cams in series. One is the cam on its own pin, which holds any squeeze in a band because the pin has friction. The other is the cam about the link’s pivot, which has no friction and therefore one ratio. The pin slips only when the second cam’s squeeze would leave the first one’s band. If it would fall below the band, the cam turns in on its pin and $c$ rises; if it would rise above, the cam is pushed out and $c$ falls. Inside the band the pin holds, and the link answers alone.

The design drawn in the opening figure is a cam leaning $2°$. On a fixed pin its band runs from $12.53$ times the pull upwards with no top, and the slack rope keeps $12.53$ after a pull of one. Its link puts $G$ at a lean of $3.65°$, with a spring a quarter as stiff as the rope at the link’s lever, so that $n_{link} = 12.53$, the pin’s bottom edge exactly. In the frictionless language of a hold the load squeezes shut that is a cam leaning $4.56°$, softened from $3.65°$ by the spring. That is above the pin’s lock lean, but the lock lean is about the pin’s friction, and this second cam’s pivot has none.

A rising load meets the pin’s edge, a falling one the link’s #

Follow a load up and down, as the earlier essay did, and the two cams take turns.

On the fixed pin the result is the one already known. The rising load drags the cam in along the band’s bottom edge, the falling load leaves the squeeze where it was, and since the band has no top below the lock lean, the cam is never pushed out. The slack rope is clamped at $12.53$, and freeing it takes a lever moment of $0.123$ times the pin’s height per unit of the largest pull.

On the link at half the edge, $n_{link} = 6.27$, the rise looks the same. The link alone would squeeze too little, so the cam turns in on its pin until the squeeze reaches the band’s bottom, $12.53$ at a pull of one. On the fall the pin holds, because a falling squeeze moves into the band, not out of it, and the link answers alone, with its squeeze falling at $6.27$ for every unit the pull falls. At no pull the rope keeps $6.27$, half of what the fixed pin kept.

On the link at the edge, rise and fall are one line. Rising, the link alone squeezes at $12.53$ per unit pull, which is already the band’s bottom, so the cam never needs to turn. Falling, the link gives the squeeze back at the same rate, and at no pull the rope keeps nothing.

That is the answer to the question the earlier essay asked, and it is not the answer it expected. The lean does change, but with the load’s direction, not its size. A rising load is met at the pin’s lean, through the band’s bottom edge, whenever the link would give less. A falling load is always met at the link’s lean, because a falling squeeze never leaves the band from below. A cleat that was hoped to be steep under load and shallow when slack is instead one thing when the load is coming on and another when it is going off.

Every force doubles with the pull #

The reason the size of the load cannot matter is in the equations, not in this design. The rope’s squeeze is in proportion to its compression, the spring’s moment is in proportion to its turn, and the pin’s friction is in proportion to the force it carries. Doubling every pull along any path therefore doubles every squeeze along it, and the check confirms this exactly, at leans of $1°$, $2°$, $10°$ and $18°$, on links at four tenths, one and one and a half times the edge.

So the squeeze per unit pull after a rise does not depend on how large the rise was. It is the larger of the pin’s bottom edge and the link’s own ratio, capped by the band’s top where there is one. No choice of link length, spring or pivot in this model makes a heavy load see a different lean from a light one.

The general lesson is older than cleats. A ratchet with no teeth and a seat with friction stays where it is pushed both found frictional holds whose state is a band and whose position in the band is history. Neither found a hold whose band moved with the load’s size, and for the same reason: Coulomb friction, linear springs and rigid geometry are all homogeneous. Making a load’s size matter needs something that is not in proportion to the load. A spring with a preload is the obvious candidate: it does nothing until the moment on it passes a set value, and then it behaves as here. That is where this argument leaves off.

What a slack cleat keeps #

The memory follows directly. After a rise to $T_1$ on the band’s bottom edge the cam has turned in to $c = (1+K)(n_{lo} - n_{link})T_1/K$. On the fall the pin holds and the squeeze falls along the link’s ratio, so at no pull the rope keeps

$$N_{slack} = (n_{lo} - n_{link})\,T_1,$$

or nothing, if the link’s ratio reaches the edge. Above the lock lean the band has a top, and a falling load reaches it and pushes the cam out, so nothing is kept on any link, as on a fixed pin.

Below the lock lean the memory falls in a straight line from the fixed pin’s value to nothing: from $16.06$ at a lean of $1°$, from $12.53$ at $2°$. Each line reaches nought exactly where the link’s ratio reaches the pin’s bottom edge, and beyond that it stays at nought.

The dots come from a second route that does not use the closed form. The cleat is given a real rope stiffness, a real link lever and a real spring, and the pull is walked along the path in forty small steps between each pair of stops. At each step the pin is tried as holding, as slipping in and as slipping out. For the two slipping cases, Newton’s method on the cam’s own moment finds the squeeze at which the rope’s moment about the pin equals the pin’s friction couple, without the band’s closed form. The admissible case is then taken. Across 84 designs, 28 link ratios at each of the three leans, the two routes agree on every slack squeeze to about $10^{-14}$ of the pull. They agree so closely because both are exact for this model: the walk solves each step’s balance rather than approximating it, so a coarser walk would arrive at the same squeezes.

The rule holds whatever the spring is. The spring’s stiffness enters $n_{link}$ and nowhere else in the memory, so a stiffer spring needs a lever placed to give a larger $b$, but the slack squeeze depends only on how $n_{link}$ compares with the edge. That makes the rule a design rule in one line: a cleat leaning below its pin’s lock lean will free a slack rope if its link, acting as a frictionless cam with the spring counted, squeezes at least as hard per unit pull as the pin’s bottom edge.

What the cam’s own pin is left doing #

The obvious worry is that something has been given up, and the next figure shows what.

On a link that gives less than the edge, the cam has to turn further on its pin than it would on a fixed pin, by a factor $(1+K)(1-r)/K$, with $r$ the link’s ratio over the edge. As the cam turns in, the extra squeeze pushes the link back against its spring and undoes part of the turn. On a link at half the edge with a spring a quarter of the rope’s stiffness, the cam turns $2.5$ times as far as a fixed pin’s. With a spring four times as stiff, it turns $0.63$ times as far, because the link barely moves. A real cam has a limited arc of profile, so that factor is a cost a designer must pay in profile.

On a link at the edge or beyond, the cam does not turn on its pin at all. The link supplies the whole squeeze, and the cam’s pin, whose friction made the band, the lever and the memory, never slips. In that design the pin is a passenger. The cleat has become a frictionless cam about the link’s pivot, and its own pin only fixes the cam to the link.

That is the honest reading of what the swinging pin achieves. It does not make the fixed-lean trade disappear. It moves the trade to a pivot that has no friction to remember with. A cleat with a frictionless pin of its own would have no memory either, as the earlier essays showed; the link is a way to get there with an ordinary pin on the cam. The friction is moved to where it is never asked to act.

Two things count against the edge design. The first is the squeeze: a link at a quarter past the edge squeezes the rope $1.25$ times as hard as it needs, $15.67$ against $12.53$ at $2°$, and a rope crushed harder than its grip requires wears faster. The second is the link’s own pivot. It is taken as frictionless here, and a real one is not. By the rule a stiff pin buys grip with release measured, a pin with the same friction circle at $G$, $2.2$ pin heights from the rope, would subtend an angle of only $\arctan(0.0447/2.2) = 1.16°$, below the $3.65°$ at which this design leans. That suggests a real link pivot would keep nothing either. The two-pin model was not built, so the suggestion is unproved.

The lever moves less than the memory #

Freeing the rope under load is the other half of a cleat’s use, and the ledger sets every design’s lever beside the fixed pin’s.

A lever on the cam is a couple, and on a fixed pin it must beat the rope’s closing moment and the pin’s friction together. On a swinging pin the same couple first turns cam and link together about $G$, against the spring, which eases the squeeze before the cam turns open. That lowers the pin’s friction moment but raises the rope’s closing moment, because less squeeze means less of the squeeze’s opening moment. The two nearly cancel.

At $2°$ the lever under load moves from $1.125$ times the pin’s height per unit pull on a fixed pin to between $1.067$ and $1.095$ on the links. At $10°$ it moves the other way, from $0.412$ to between $0.398$ and $0.567$, the link at the edge needing the most. At none of the leans and designs tried does the squeeze fall below the rope’s floor, a pull over $\mu$, before the cam opens, so the rope never runs out from under a lever that has only started to work.

The slack lever, by contrast, follows the memory down. At $2°$ it is $0.123$ on the fixed pin, $0.060$ on the half link, $0.029$ on the three-quarter link, and nothing on the edge link, because there is nothing to release. Return and stiffness come from the same aim found that whether a stop lets go when its load is taken off is decided by a pivot’s place relative to a friction cone, and margin on the right side costs nothing. The edge link is the cleat’s version: the pivot’s place decides the slack state, and the effort under load hardly notices.

What a designer can set #

A cam cleat that must grip below its pin’s lock lean, which is where a cleat holds a backward pull, can be made to free a slack rope by carrying its pin on a sprung link. The rule is one inequality: the squeeze per unit pull the link gives as a frictionless cam, with the spring counted, must reach the pin’s bottom edge. At that point the slack rope keeps nothing, the cam never turns on its pin, and the grip under a rising load is the fixed pin’s.

Short of it, the memory falls in proportion. A link at half the edge halves what the slack rope keeps and costs the cam two and a half times its turn on a soft spring.

The size of the load buys nothing. A cleat designed this way frees a rope after a light pull and after a heavy one alike, and keeps the same share of each. A designer who wanted a cleat that behaves one way for small loads and another for large ones needs a preload, and the linear cleat drawn here cannot supply one.

The fixed pin’s backward lock is not computed here for the swinging pin, and neither is a backward pull. Two jaws and a coefficient and six things a hold is not both found that a hold’s verdict changes with the direction of the load, and a link that eases the squeeze when the load is taken off will very probably ease it under a backward pull too. Whether it lets the rope run backwards before the pin locks is a separate question.

What the two-cam model leaves out #

Small motions. The lean at the contact is taken as fixed while the rope compresses. A cam cut as a logarithmic spiral meets the rope at one lean however far it has turned, so for the cam’s own turn this is exact. The link’s swing moves the contact along the rope by a distance of the order of the rope’s compression, and that is taken as small next to the cam.

A linear rope. A real rope stiffens as it is squeezed, so its compression is not in proportion to the squeeze, and a stiffening rope is exactly the kind of departure from homogeneity that would let the load’s size in. How much it moves the memory line has not been computed.

A frictionless link pivot, and no stop. The spring holds the link with no preload and no stop, so the link swings both ways about its rest pose. A real cleat’s link would rest against a stop, and a stop is a unilateral constraint like a constraint that only pushes. It would change the falling branch wherever the link tries to swing past it.

The cam’s own spring. A real cam carries a light spring that holds it against a slack rope. It is omitted, as in the earlier essays, and it would add a small squeeze at no pull on every design.

Still open: a spring with a preload #

Everything here is in proportion to the pull, so a light load and a heavy one see the same cleat, scaled. A link held against a stop by a preloaded spring would not be. Below the moment the preload resists, the link stays on its stop and the cleat is the fixed-pin cleat. Above it, the link swings and the cleat is the two-cam cleat drawn here.

The distinct argument would be that threshold made a design variable. The measurement would choose a preload and a stop, follow loads of several sizes up and down, and ask whether a cleat can keep the fixed pin’s backward lock for light loads while freeing itself after heavy ones, or the reverse. It would also ask where in the load range the switch should sit for a cleat that holds a sheet: a light breeze leaving the rope locked, a gust leaving it free.