A ratchet with no teeth

A joint that works one way is where this field begins: a mechanism whose reachable set from a given state is a half-line rather than a neighbourhood, and which does that with a pawl and a tooth. The resolution is the pitch then measured what the teeth cost. A ratchet is held at a tooth and pulled to anywhere between two, so its lost motion is one pitch — $360°/z$ — and the only way to improve it is more teeth or more pawls, each of which divides it by a count.

There is a one-way mechanism with no teeth on it at all. A cylindrical bore, a star inside it whose faces are flats rather than arcs, and a roller in each wedge: turn the star one way and each roller rolls into the narrowing part of its wedge and jams the pair, turn it the other and the rollers run back and the star is free. Every bicycle freewheel built since about 1970 is one.

The mechanism is worth describing once more slowly, because its one-wayness works by a route the field has not met. A ratchet is one-way because a tooth has two faces and they are different shapes: one of them is square to the pawl and one of them is a ramp the pawl rides over. A roller clutch’s two directions are the same surfaces. Nothing about the bore or the flat is asymmetric; what is asymmetric is which way along the flat the gap narrows, and the roller rolls one way into that and the other way out of it.

So the asymmetry is in the gradient of a gap rather than in the shape of a tooth, which is why the mechanism has no pitch — a gradient has no period.

Half of it is not this site’s question #

The obvious question about a roller clutch is whether it grips, and it is a question of a kind this collection leaves alone.

One test, three mechanisms sorted three one-way verdicts — a pawl’s hold, an escapement’s draw, a four-bar’s transmission — into one cross product, and then explicitly excluded a fourth candidate on the ground that decides this essay’s scope:

A friction clutch holds one way and slips the other, exactly as a ratchet does. Its verdict looks like the same kind of thing: a wedge angle, a contact, a decision about whether the mechanism grips. And it is not the same kind of thing at all, because the answer is $\tan\theta < \mu$ — a comparison between an angle the drawing sets and a coefficient the drawing does not.

That is exactly right and it applies here. A pawl holds on steel, on brass, on ice and oiled; a roller clutch’s wedge angle has to be under a friction angle, and change the materials and the same drawing gives a different verdict.

So the gripping is out. What is left is the lost motion, and it has no material property anywhere in it.

The split is worth naming because it is the same one the field keeps making rather than a convenience adopted for this mechanism. The angle that holds the lock computes an escapement’s draw as a moment arm and stops before the force; where the tooth lets go budgets a pallet’s angles and declines to integrate its motion through the drop. Each time the line falls between a ratio and a magnitude. Here it falls between how far the mechanism moves — a length divided by a slope, a pure number once the bore is the unit — and whether it then holds, which is a comparison against a coefficient.

A gap that narrows as a secant #

A flat at perpendicular distance $a$ from the axis is at radius $a/\cos\psi$ when measured $\psi$ round from its own perpendicular, and the bore is at $R$ everywhere. So the gap between them,

$$\text{gap}(\psi) = R - \frac{a}{\cos\psi},$$

is widest at the middle of the flat and narrows towards its ends. A roller of radius $r$ is nipped where the gap equals its diameter, which fixes its seat at $\cos\psi^\star = a/(R - 2r)$ — and $\psi^\star$ is the wedge angle, because the bore’s normal at the contact is radial and the flat’s is the flat’s own perpendicular, and the angle between them is $\psi$.

At rest a light spring holds each roller where the gap is its diameter plus a clearance $c$ — the fit the parts were made to. To hold, it must roll from there to its seat. On the drawn clutch that is 0.658° of roller travel, and the star turns 2.204 times as far, because the roller is rolling between a still bore and a turning flat and its centre moves at half the flat’s surface speed. 1.449° of lost motion, against a 24-tooth ratchet’s 15°.

A length divided by a slope #

Linearising the gap at the seat gives the whole of it in closed form. The gap’s slope there is $(a/\cos\psi)\tan\psi$, so the roller’s travel is $c$ divided by that, and the star’s is that times the rolling gain $2(R-r)/a$. Substituting $a = (R-2r)\cos\psi^\star$ the cosines cancel and

$$\text{lost} \;=\; \frac{2c\,(R-r)}{(R-2r)^2\,\sin\psi^\star}.$$

The two converge: 0.13% apart at a clearance of 0.0005 and 17% apart at 0.05, the departure growing in proportion to the clearance because that is what a leading term does. The exact travel always exceeds the linear one, since the gap’s slope is steepest at the seat and slackens behind it.

What matters is what the axis is. A ratchet’s lost motion is a pitch divided by a count and takes values from a list. A roller clutch’s is a length divided by a slope and takes any value a fit can be held to. One is improved by making the parts more numerous and the other by making them better, and those are different kinds of engineering with different costs.

The convergence is a two-route check and not decoration. The exact travel is found by asking where the gap equals two stated numbers and subtracting the two angles, which involves an arccosine and nothing else; the closed form is found by differentiating the gap once and dividing. Neither uses the other, and a mistake in the rolling gain — the factor of 2.204 that turns the roller’s travel into the star’s — would appear as a constant ratio away from one rather than as a ratio approaching it. It approaches it, so the gain is right.

The one number that is not the designer’s #

Everything in the formula but the sine is fixed by the bore, the roller and the fit. The star’s shape enters once, through the wedge angle.

The product converges on $2.608 \times 10^{-3}$, which is $2c(R-r)/(R-2r)^2$ exactly, approached from above — 12.8% high at a 4.4° wedge and 0.46% at 22.7°. The departure is largest at the shallow end because a shallow wedge makes the roller’s travel a large fraction of its own seat angle, so the linear form is worst precisely where a clutch is built.

And the wedge angle is the one quantity the designer does not choose. It has to be under the friction angle of the two materials, which is a few degrees for hardened steel, and the whole of that bound comes from outside this site. So the division of labour is exact: geometry says the lost motion is inversely proportional to the sine of the wedge angle, and something else says how small the wedge angle has to be. Halving the wedge angle to be safe doubles the lost motion, and neither half of that sentence can be checked with the other’s instruments.

It is also worth seeing what the formula says about size, because it says something a tooth count never can. Every length in it — $c$, $R$, $r$ — is a length, and the combination $2c(R-r)/(R-2r)^2$ has the dimensions of one over a length times a length, which is to say it is dimensionless only because $c$ is divided by something of its own order. Double the whole clutch and hold the fit the same and the lost motion halves; double the clutch and scale the fit with it and the lost motion is unchanged. So a roller clutch’s resolution improves with size at a fixed manufacturing standard, which is the opposite of a ratchet, whose resolution is a shape and is the same at every size.

More rollers buy nothing #

A ratchet has an obvious way to improve its resolution without changing its teeth: several pawls, staggered against the tooth pitch by a fraction of it, so that one of them is always nearer the next tooth than the others. Three pawls on a 24-tooth wheel give 5° of lost motion where one gives 15°.

The clutch has no such move. Its rollers sit on identical flats at identical angles, so they all reach their seats at the same instant — measured across three, four, five, eight and twelve faces the lost motions differ by nought, exactly. The number of rollers is a question about how much load each carries and not about how far the star turns before they carry it.

That is the two mechanisms answering one design move in opposite directions, and the reason is worth stating. A ratchet’s pawls do not have to take turns. They are all riding the same teeth at once and only one of them happens to be the one that catches, so staggering costs nothing. A clutch’s rollers all nip, and nipping is the mechanism working, so there is nothing to stagger — the improvement a ratchet gets from redundancy is unavailable to a machine whose elements are not redundant.

A clutch could of course be built with its flats cut at slightly different depths, so that the rollers nipped in sequence. That would divide the lost motion as a ratchet’s pawls do, and it would mean that only one roller carried the load until the star had turned far enough to bring in the next — which is the load argument again, arriving as the price of the resolution.

Where each mechanism belongs #

Setting the two side by side gives a short answer to which to reach for, and it is not the one the drawings suggest.

A ratchet’s index is exact: it is held at a tooth, and where a tooth is, is a shape. Its lost motion is coarse — 15° on a common 24-tooth wheel — and is improved only by counts. So a ratchet is the mechanism for a machine that must know exactly where it is and does not mind waiting a pitch to get there, which is a hand tool, a hoist pawl, a cable tie.

A roller clutch’s holding position is not exact at all: it is wherever the roller happened to nip, which depends on the fit and moves as the parts wear. Its lost motion is fine — a degree and a half on the one drawn — and is improved by making the parts better. So it is the mechanism for a machine that must take up drive quickly and does not care where it is, which is a freewheel, a starter drive, a one-way bearing in a gearbox.

Neither can be made into the other. The ratchet cannot get its lost motion down without teeth so fine they stop being teeth, and the clutch cannot be made to hold at a stated place at all, because there is no place in it that is stated.

What this does not settle #

The gripping, entirely. The wedge angle’s bound, the coefficient it is compared against, whether the clutch grips under load or slips, and what happens to the contact as the surfaces wear are all outside this. The essay is about a mechanism’s lost motion and not about whether it works.

The rollers do not slide. The rolling gain $2(R-r)/a$ assumes pure rolling at both contacts. A roller that slips against one surface travels a different distance for the same star rotation, and the lost motion changes by whatever that slip is. Nothing here computes it, and slip near the seat is exactly where friction comes back.

The star’s faces are flat. A flat is what makes the gap a secant and gives the wedge angle a closed form, and it is what most freewheels have. A cammed face — a logarithmic spiral, which is the shape that holds the wedge angle constant as the roller travels — would make the wedge angle independent of where the roller sits and would change the lost motion’s dependence on the clearance entirely, since the gap’s slope would then be proportional to the gap.

One clutch geometry. Flats on the inner and a cylinder outside is the commonest arrangement and is the one measured. The inverse — a cylinder inside and a cammed outer — and the sprag clutch, whose elements are shaped blocks that rotate rather than roll, both have the same character and different formulae.

The spring is not modelled. Each roller is held at rest against a light spring, and where it sits at rest is where that spring puts it against the fit. The clearance $c$ is treated as a given; what actually sets it is the tolerance stack of a bore, a star and a roller, which is the practice field’s kind of question and would turn the one number this essay’s answer depends on into a distribution.

What the mechanism has instead of a state #

One more difference, and it is about the state space rather than about a number.

A joint that works one way described a ratchet’s configuration space as a set in which being joinable has stopped being symmetric: the strap can be pulled to anywhere and only held at a tooth, so the held states are a discrete set inside a continuum. That discreteness is the whole of what a tooth is for.

A roller clutch has no held states at all in that sense. Its star can be stopped anywhere, and wherever it is stopped it holds, because the rollers will have nipped wherever they were. So its reachable set is still a half-line and its holding set is the whole of the line rather than a lattice in it.

That is a genuinely different object, and it is why the lost motion is the only number the two mechanisms can be compared on. A ratchet has a resolution and a lost motion, and they are the same quantity. A clutch has a lost motion and no resolution, because there is nothing to resolve.

Still open: the lost motion as a tolerance stack #

The whole answer here is proportional to one clearance, and that clearance is not a number a drawing carries. It is what is left over after a bore, a star and a roller have each been made to their own fits, so it is a sum of three tolerances and has a spread rather than a value.

Its distinct argument would be that stack carried through: the three fits a clutch is actually made to, the distribution of clearance they leave, and therefore the distribution of lost motion — with the worst case and the root-sum-square both quoted, on the convention the practice field already uses. Two things would come out of it that a single number cannot give. The lost motion of a batch has a spread as well as a mean, and a mechanism whose resolution varies from one unit to the next is a different design proposition from one whose resolution is a tooth count and is the same in every unit ever made. And the stack would say which of the three parts to tighten, which is the question a manufacturer asks and which the formula above answers only in the aggregate.