One belt for every step

Before every machine in a workshop had its own motor, a lathe or a drill press took its power from a line shaft overhead through a flat belt, and changed speed by moving that belt from one step to another of two stepped pulleys — cone pulleys, one on the countershaft and one on the spindle, turned in opposite directions so that a large step on one faces a small step on the other. Each step pair gives a different speed ratio. But there is only one belt, and it has to be taut on every step.

That makes the stepped pulley a pure strand problem. A member with no length of its own found that the only thing a strand constrains is its own length, and the tensioner is the unknown found that a run closes when a number comes back, not a vector. So the design question is: which pairs of radii give the same belt length? The usual rule is to keep the two radii’s sum the same on every step. This essay finds that the rule is exact for one of the two ways a belt can be run and wrong for the other, and by how much.

Why a crossed belt only knows the sum #

The wraps add up to a turn found the fact everything here rests on. A crossed belt wraps its two pulleys in opposite senses, its signed wraps sum to nought, and so its two wraps are equal: both are $\pi + 2\gamma$, where $\sin\gamma = (r_1 + r_2)/C$. The two straight runs are each $C\cos\gamma$ long. So the whole length is

$$L_{\times} = 2C\cos\gamma + (\pi + 2\gamma)(r_1 + r_2), \qquad \sin\gamma = \frac{r_1 + r_2}{C},$$

and the radii appear in it only through their sum. Keep the sum and the length is kept, exactly, whatever the two radii are individually. A crossed belt fits every step of a pair whose steps’ radii sum to the same value.

That is a derivation, and it is checked here by building the belt instead. Every length in this essay is the taut run the strand field constructs — two tangent lines between the pulleys, each leaving its pulley at the right angle a shortest path has to, found from the signed radius difference, and the two arcs between the tangency points — never the formula, which is kept as the second route and agrees with the run on every step to $2 \times 10^{-13}$ mm. For 266 random pairs of steps, with radius sums from 60 to 260 mm, centre distances from 1.2 to 4.2 times the sum and speed ratios from 0.3 to 3.3, each pair built with a common radius sum, the two crossed belts’ lengths differ by at most $4 \times 10^{-16}$ of their length. That is the last bit of the arithmetic: the two lengths are the same number.

An open belt knows the difference as well #

An open belt wraps both pulleys the same way round. Its signed wraps sum to one turn, and they are not equal: the larger pulley gets $\pi + 2\gamma$ and the smaller $\pi - 2\gamma$, with $\sin\gamma = (r_1 - r_2)/C$ now depending on the difference. Its length is

$$L_{\parallel} = 2C\cos\gamma + \pi(r_1 + r_2) + 2\gamma\,(r_1 - r_2),$$

and a pair of steps with the same sum but different differences needs different belts.

The pair in the figures has five speeds a factor of 1.6 apart — ratios of 1/2.56, 1/1.6, 1, 1.6 and 2.56 — with radii summing to 160 mm and centres 400 mm apart. Crossed, every step needs 1,367.552327 mm of belt. Open, the middle step, where the two radii are equal, needs 1,302.655 mm, the next steps out need 3.411 mm more, and the outer steps 12.321 mm more. An open belt made for the middle step is 12.3 mm too short for the outer steps — about 0.9% of its length, to be found by stretching it or by re-tensioning at every change of speed. Made for the outer steps, it is 12.3 mm slack on the middle one.

Where the difference goes #

The two formulas are easier to trust once it is clear where the length goes when a belt moves from equal radii to unequal ones at the same sum.

Two things happen. The straight runs tilt: with the radii unequal, the tangent lines are no longer parallel to the line of centres, each is $C\cos\gamma$ long instead of $C$, and the belt saves $2C(1 - \cos\gamma) \approx C\gamma^{2}$ of straight run. And the wraps change: an open belt gives the larger pulley $2\gamma$ more wrap and the smaller $2\gamma$ less, so the arcs grow by $2\gamma r_1 - 2\gamma r_2 = 2\gamma d$. With $\gamma \approx d/C$ the arcs gain $2d^2/C$ and the runs give back $d^2/C$, and the belt is longer by the difference, $d^2/C$. The wrap redistribution wins, by exactly a factor of two at leading order.

A crossed belt’s straight runs tilt too, and by an angle set by the sum, the same on every step. Its wraps do not redistribute at all — both pulleys get $\pi + 2\gamma$ whatever the split — so nothing about the split can change the length. The difference between the two belts is not that one tilts and the other does not; it is that one moves wrap from one pulley to the other and the other cannot.

How much, to leading order and beyond #

Expanding the open length about equal radii says what the excess is. With $d = r_1 - r_2$ the difference, $2C\cos\gamma = 2\sqrt{C^2 - d^2} \approx 2C - d^2/C - d^4/4C^3$, and $2\gamma d \approx 2d^2/C + d^4/3C^3$. At a fixed sum the $\pi(r_1+r_2)$ term does not move, so

$$\Delta L_{\parallel} = \frac{d^{2}}{C} + \frac{d^{4}}{12\,C^{3}} + \dots$$

The leading term is the whole story for most pulleys: the excess grows as the square of the radius difference and falls with the centre distance.

Measured on the run itself, at radius differences from 2 to 140 mm on a 400 mm centre distance: at $d = 2$ mm the excess is $d^2/C$ to $2 \times 10^{-6}$ of itself. What is left over grows in proportion to $(d/C)^2$ with slope $1/12$, the next term of the expansion, and the measured points sit on that slope to $3 \times 10^{-3}$. A plausible slip — writing the law with $2C$, as if both straight runs contributed the whole of it — is refused by a factor of two from the first point. On the pair in the figures the outer steps have $d = 70.1$ mm, and $d^2/C = 12.3$ mm is the excess to the precision the figure shows.

Long belts forgive, but not entirely #

The excess falls as one over the centre distance, which is the source of the workshop’s confidence in the constant-sum rule: a line shaft is a long way from the machine it drives. The same five steps on a 1 m centre distance give an outer-step excess of 4.918 mm; on 2 m, 2.458 mm; on 3 m, 1.639 mm; on 5 m, 0.983 mm. Each is $d^2/C$ to three parts in a thousand, and the product of excess and centre distance is the same number every time.

So even a belt five metres long, between a ceiling countershaft and a lathe, misses by a millimetre on its outer steps if its pulleys were sized by the crossed-belt rule, and a belt from a countershaft on the machine’s own frame — half a metre or so — misses by ten. A millimetre is inside what a leather belt can take up by stretching when it is laced a little tight. Ten is not.

There is a second way out, which the strand field has already met. A belt made for the outer steps and slack on the others can be taken up by a tensioner — an idler whose position is solved for the one length the belt has — and on the pair in the figures the idler would have to absorb 12.3 mm between the middle and outer steps. That is the modern answer on machines that still use belts. The stepped pulleys of the line-shaft era had no idler, and the steps themselves had to be right.

The steps an open belt needs #

So an open belt’s steps cannot keep their sum. To take the middle step’s belt, a step with a radius difference $d$ has to shed about $d^2/C$ of length, and at a fixed ratio the only way to do that is to shrink both radii together. The $\pi(r_1 + r_2)$ term gives the exchange rate: a millimetre off the sum is $\pi$ millimetres off the belt. So the sum must fall by about $d^2/\pi C$.

Found exactly — for each speed ratio, the driven radius that makes the open run’s length equal the middle step’s, by bisection — the outer steps need radii summing to 156.260 mm, 3.740 mm less than the crossed belt’s 160, and the estimate $d^2/\pi C$ gives 3.731. The inner steps need 1.071 mm less against an estimate of 1.070. Each open step then takes the middle step’s belt to $10^{-9}$ mm.

That is the answer the workshop rule misses. “Keep the sum constant” is the crossed-belt rule, and it is exact there. For the far commoner open belt it is a first approximation whose error is second order in the step’s radius difference, and the correction — shrink the outer steps’ sum by $d^2/\pi C$ — is small enough to be missed on a drawing and large enough to matter on a belt, which is a few millimetres of flat leather that either stretches or does not fit.

The speed ratio is not affected by the correction: each open step keeps exactly the ratio it was designed for, because both radii shrink in proportion. What changes is the size of the pulleys, not their ratio, and a drum is a size, a wrap is a shape is the reason that is harmless to the speed: a belt drive’s ratio is a ratio of radii and transfers across sizes.

Why the error is all at the ends #

The speeds on a machine tool’s stepped pulley are chosen a constant factor apart — here 1.6 — because a cutting speed matters in proportion, and a geometric series gives every step the same proportional change. Symmetric about one to one, that series puts the steps’ radius differences at 0, 36.9 and 70.1 mm on the pair in the figures: the difference grows faster than the step number, because a ratio of $q$ at fixed sum gives $d = S(q - 1)/(q + 1)$.

The excess then grows as the square of that. The inner steps are out by 3.4 mm and the outer by 12.3, nearly four times as much for less than twice the difference. So on a geometric series the constant-sum rule is nearly right in the middle and wrong where the steps are most unequal, and a pair designed with more speeds, a factor of 1.4 apart say, is no better at its extremes: the outermost step’s ratio, and so its difference, is what the error depends on, not how many steps lie between.

What the crossed belt also gives #

The crossed belt’s other property is less often mentioned, and it comes from the same equal-wraps fact.

A flat belt slips first on the pulley it wraps least, because friction grip grows with wrap. Crossed, both wraps are $\pi + 2\gamma$ with $\gamma$ fixed by the sum, so on a constant-sum pair every step wraps both pulleys by the same angle: 227.16° on every step of the pair in the figures. Open, the wrap on the smaller pulley is $\pi - 2\gamma$ with $\gamma$ growing with the difference: 180° on the middle step, where the radii are equal, falling to 159.81° on the outer ones, while the larger pulley’s rises to 200.19°. The open belt gives the smaller pulley least grip on exactly the steps where that pulley is smallest.

The price of crossing a belt is the rubbing where the two runs cross and the twist in each run, which is why open belts are commoner. The crossed arrangement is not exotic geometry, though: a tooth flank is an unwound strand found that the line of action of an involute gear pair is exactly a crossed strand on the two base circles. But a stepped pulley is the one place where crossing buys two things at once: an exact fit on every step and an even grip on every step.

Across many pulleys #

The pair in the figures is one design. The claim that a crossed belt fits every constant-sum step, and an open belt does not, is about every design.

For all 266 random pairs, the crossed belts’ two lengths differ by at most $4 \times 10^{-16}$ of the length, which is rounding. The open belts’ differ by at least $7 \times 10^{-5}$ and by as much as $4 \times 10^{-2}$ — four per cent of the belt at the worst. There is no pair among them for which the constant-sum rule gives an open belt a fit, and none for which it fails a crossed one. And the open mismatch is the leading-order law at work on every one: the difference of the two steps’ $d^2/C$ matches the measured difference to a median of 0.27% and a worst of 2.9%, the worst being pairs on short centres where the $d^4/12C^3$ term is no longer small.

What the stepped pulleys do not settle #

The belt is an ideal strand. It has no thickness, no bending stiffness and no stretch. A real flat belt runs with its neutral line outside the pulley face, which adds the belt’s half-thickness to every radius — on a crossed belt it adds the same to both radii and so to the sum, which is still constant; on an open belt it changes the difference not at all. Neither result moves.

Crowned faces and belt shifters are not modelled. Real stepped pulleys have crowned faces to keep the belt centred, and moving the belt between steps is done with a shifter while it runs. The lengths here are for the belt at rest on its step.

A chain is a different problem. A bicycle’s derailleur moves a chain between sprockets and takes up the length difference with a sprung arm, because a chain is not a strand: its length comes in whole links and its sprockets are polygons, so no choice of tooth counts makes every gear take the same chain.

Tension is not computed. A belt’s grip depends on its tension and on the wrap through the capstan relation; the wraps here say which pulley slips first, not at what torque.

Still open: a cone rather than steps #

A stepped pulley offers a few speeds. The same idea taken to its limit is a pair of tapered drums — true cones — with a belt shifted continuously along them, which gives a continuous range of ratios; it is how some early variable-speed drives worked. A crossed belt on two opposed cones whose radii sum to the same value at every station along them fits everywhere, for the reason above.

The distinct argument would be the open-belt version of the continuous drive — a cousin of the drum that is not round, where a demanded rate fixed a drum’s profile — where the correction $d^2/\pi C$ has to be applied at every station along the drums, so that the drums are no longer cones but curved profiles — and whether a belt running on such a surface stays at one station at all. A flat belt on a taper climbs towards the larger end, and a belt whose two runs meet the two drums at different stations is twisted; the measurement would be the profile the drums need for one open belt to fit at every station, and the length error at the ends of a travel when the profile is approximated by a straight cone, as the drawings of those machines show.