A cam moves the law off the drums

A drum for every law cut cone-drive drums for a family of demanded speed laws and found that what each law really costs is its grading, the proportional change of speed per millimetre of shift, at its steepest point. A belt of width $w$ spans a stretch of drum $w$ long, and its two edges want ratios a factor $e^{wg}$ apart where the grading is $g$. That spread has to be taken up by uneven stretch across the belt, so the drums must be long enough for the law’s steepest stretch to be gentle. The even proportion, which grades evenly, needs 300 mm on the drive those essays used. The speed changing evenly with the shifter needs 884 mm, and a fan’s power changing evenly needs 13.4 m.

A tolerance shortens the drum then let each law depart from its demand by a stated proportion and found that softening the steep end buys a great deal of travel and never turns the law into the even proportion. It ended on a different idea. The law could be taken off the drums altogether. Cut the drums for the even proportion, the gentlest any range allows, and put a cam between the operator’s handwheel and the fork that guides the belt, so that the belt’s station is a chosen function of the handwheel. One pair of drums would then serve any law, and changing the law would mean changing a cam.

The question it left was whether the trouble had simply moved. The steep end might reappear on the handwheel side, as a steep end of the cam. The answer is that it reappears exactly, factor for factor, and that this is still a very good trade, because of what the factor buys in each place.

The groove is the law redrawn #

On drums cut for the even proportion the speed ratio grows by the same factor for every millimetre of shift. Its logarithm is therefore a straight line in the station: $\ln q$ runs from $-\ln 2.56$ at one end of the 300 mm to $+\ln 2.56$ at the other. Whatever law is wanted at the handwheel, written $f(u) = \ln q$ against the handwheel’s share $u$ of its travel, the station that delivers it is found by reading that straight line backwards:

$$t_s(u) = \frac{f(u) - f(0)}{2 \ln 2.56},$$

with $t_s$ the station as a share of the drums’ length. Nothing has to be solved. The cam’s groove is the demanded law drawn on new axes, shifted and scaled so that its ends meet the ends of the drums.

The figure draws the groove for each law in turn. For the even proportion it is the straight diagonal, since the handwheel and the belt then move together. For the speed law it climbs steeply at first and then flattens. The speed law changes speed by equal amounts per unit of handwheel, so at the slow end, where a given amount of speed is a large proportion, it must cover a large share of the drums for a small turn. The first 8.3% of the handwheel carries the belt across the bottom fifth of the drums. The roll law, the speed’s reciprocal changing evenly, is the same groove turned end for end and steep at the fast end. The fan law is the extreme. Its power changes evenly, so the bottom fifth of the drums is crossed in 0.7% of the handwheel and the rest of the travel is spent on the top speeds, where a small change of speed is a large change of power.

The slope of that groove is where the question lives. Differentiating,

$$\frac{dt_s}{du} = \frac{f'(u)}{2\ln 2.56},$$

which is the law’s grading over the even proportion’s, the very ratio whose peak set the law-cut drum’s length. The steepest point of the speed law’s groove is 2.95 times as steep as the even groove, and 884 mm is 2.95 times 300 mm. That is not a coincidence that holds for one law; it is the same number computed twice.

Steepness is conserved #

The figure tests that statement across the whole family the earlier essays measured. It takes three laws, each exact and softened within five tolerances by the taut-string construction of the tolerance essay. For each it plots the cam’s steepest slope against the length the same law would need cut into its own drums. Every point lies on the diagonal. The largest departure is a part in $10^{12}$, which is rounding in the arithmetic, and the agreement holds as well for the softened laws as for the exact ones. A softened law makes a gentler groove by exactly the factor it made a shorter drum.

So the suspicion was right in one sense. A law’s steepness is a property of the law as the operator meets it, the way speed changes against the control. It cannot be designed away by moving the mechanism that delivers it, because the even drums make the station a fixed function of the ratio, and the cam must then turn the handwheel’s motion into that station at whatever rate the law demands. Somewhere in the drive a member has to move fast where the law is steep. On law-cut drums that member was the belt’s position against the ratio. On even drums with a cam it is the belt’s position against the handwheel.

What changes is where that member’s steepness is paid for, and what it costs there. On law-cut drums the steep end costs three things at once. It costs length of drum, which is metal turned to an accurate profile. It costs the belt, which must survive the drum’s steepest stretch across its width. And it costs a drum pair for every law, since the law is cut into them. In the cam, the same factor costs a groove that winds more turns round its barrel.

A barrel pays in turns #

A cone drive’s shifter is commonly a fork on a carriage, moved by a screw or by a barrel cam turned by a handwheel. The barrel’s groove is a helix of varying lead: a pin on the carriage rides in it, and each turn of the handwheel moves the carriage by the lead of the groove where the pin happens to be. The groove can be given any law of lead along its length, which is exactly the freedom a cam on the shifter needs.

The limit on a barrel groove is the one every cam has, the pressure angle. It is the angle between the groove’s wall and the barrel’s circumference at the pin, and it decides how much of the handwheel’s effort pushes the carriage along and how much pushes it sideways into its guides. A lift is a size and a law is a shape found that a cam’s law decides its shape and its size is a choice. A barrel is the same: for a groove of pitch radius $R$ spread over $N$ turns,

$$\tan\alpha = \frac{1}{R}\frac{ds}{d\theta} = \frac{300\,\text{mm}}{2\pi N R}\,\frac{dt_s}{du},$$

so the pressure angle is largest where the groove is steepest, and keeping it under a limit takes

$$N \geq \frac{300\,\text{mm} \cdot P}{2\pi R \tan\alpha_{\max}},$$

where $P$ is the groove’s steepest slope over the even groove’s. On a barrel of 40 mm pitch radius held to 30°, the even proportion needs 2.07 turns. The speed law needs 6.1, the roll law the same, the law with $\lambda = 2$ needs 22.7, and the fan law 92. Each was checked twice: the formula, and the largest pressure angle measured on the groove so built, which sits on the limit to the last digit, while one per cent fewer turns break it.

Turns are cheap. A handwheel that goes round six times instead of twice costs the operator four turns’ time and the shop nothing. A barrel twice the radius halves every count, and a motor on the handwheel makes ninety-two turns as easy as two. The 884 mm of drum the speed law needed was 584 mm more of profiled metal, a belt that crosses it, a frame to carry it and a carriage to traverse it. The cam takes the same factor and pays it in something that has no weight.

What the belt still feels #

There is one cost the cam does not move, and the figure finds it. A belt is shifted along its drums by being led on at a slight angle. Each turn of the drum walks it sideways by an amount set by that angle, and a belt forced to walk faster climbs the fork rather than following it. So a belt has a top axial speed, and a change of speed takes at least as long as the belt needs to walk across the drums.

On drums cut for the speed law and shifted directly, the belt moves with the handwheel. Turned at a uniform rate, it walks at a uniform speed, 2.95 times what it would on the even drums turned the same way, because the drums are 2.95 times longer. On the even drums with the cam, its speed follows the groove’s slope: 2.95 at the slow end and falling to 0.45 at the fast. Both peak at the same speed, which is the conservation found above appearing again. If the operator insists on a uniform rate, the belt’s top speed limits both drives equally, and a full change takes the same time on each.

But the areas under the two curves differ. They are the belt’s whole walk: 300 mm on the cam’s drums against 884 mm on the law’s. An operator who turns quickly where the groove is flat and slowly where it is steep, or a motor governed by the belt’s own rate, makes the full change on the cam’s drums in 34% of the time the law’s drums allow at any rate of turning. For the fan law the figure would be 2%. The steepness came back on the handwheel side, as the earlier essay suspected. There, unlike on the drum, it can be met by turning faster where the groove is gentle, which the law-cut drum never permits, because its belt has to walk the whole length.

The belt-width limit is gone entirely, and that is a larger change than it looks. On even drums the grading under the belt is the even proportion’s at every station, whatever the law. A belt chosen for the even proportion, at the width and the edge spread that made 300 mm enough, serves every law the cam can be cut for. Where a strand leaves a body found the belt meeting each drum at a right angle whatever the shape. With the cam, the shape it meets never changes.

The drums every law can share #

The figure puts the two pairs side by side to one scale. The even pair is the logistic drum a cone drive is not two cones found, with its two drums one shape reversed, so the same profile is turned twice. The speed law’s pair is nearly three times as long, and its two drums are different shapes. A drum and its mirror serve a law and its mirror, which is why the speed and roll laws share a pair of drums. The speed law and the even proportion need different pairs.

That difference answers the essay’s starting problem. The tolerance essay pictured a lathe that also wound material, wanting the even proportion for cutting and a constant surface speed for winding, the roll law. On law-cut drums it needs two drive pairs, 300 mm and 884 mm long, or one pair and a compromise. On even drums with a cam it needs one pair of drums and two grooves: a 2.1-turn straight helix and a 6.1-turn groove of varying lead, cut side by side on one barrel with a selector for the pin, or on two barrels swapped. A machine that serves three duties needs a third groove and still one pair of drums.

One belt for every step found that a stepped pulley’s steps had to be chosen so that one belt fits them all, a constraint on the steps imposed by the belt. The cam lifts the matching problem off the drums. Every station of the even drums takes the same crossed belt, since their radius sum is constant. The wraps add up to a turn makes the radii a function of the ratio alone, so the same belt fits whichever law the groove asks for.

The ledger #

The ledger sets the two prices side by side. Read along a row, the drum and the cam pay the same factor, since the turns column is the drum column divided by 300 and multiplied by the even proportion’s 2.07. Read down, the rows say which laws a cone drive can deliver at all. On law-cut drums the speed and roll laws are buildable and the steeper ones are not: λ = 2 needs 3.3 m exact and 1.7 m softened, and the fan law needs 13.4 m exact and 6.1 m softened. On even drums with a cam every row is buildable. The fan law asks for 92 turns of a 40 mm barrel exact, or 42 softened within 11%. That is a motorised handwheel’s work rather than an operator’s, but it is a cam a shop can cut, on drums the shop already has.

The softened columns show that the taut-string softening still has a use, in a different place. On law-cut drums it shortened the metal; on the cam it shortens the groove. The tolerance essay found that a few per cent of tolerance takes the sharpest part off a steep end, and the same few per cent take the same share off the cam’s turns, from 6.1 to 3.8 for the speed law at 11%. A designer has the same lever as before and now applies it to the cheaper part.

The paced column is the belt’s walk again. On even drums the belt walks 300 mm whatever the law, so a change paced to the belt’s top speed takes 34% of the law-cut drum’s time for the speed law, 9% for λ = 2 and 2% for the fan. The steeper the law, the more the cam saves. That is the reverse of what might be expected from a mechanism whose difficulty grows with the law’s steepness.

Why the factor cannot be dodged #

It is worth stating the conservation as a rule, because it constrains any other arrangement a designer might try. Between the operator’s control and the delivered ratio there is a chain of members: a handwheel, perhaps a cam, a carriage, a belt, two drums. The law the operator meets is the composition of every link’s own law, and its logarithmic slope at each point is the product of their slopes. The drum fixes how the ratio grows with the belt’s station; the cam fixes how the station grows with the handwheel. Their product is the demanded law, so making either one gentler forces the other to be steeper in proportion. The even proportion is the gentlest drum, so with it the cam carries the whole factor, and with a straight-sided cam the drum carries it.

Any split between them is possible, and the ledger’s two columns are the extreme cases. A drum cut for half the law’s steepness and a cam for the other half would have intermediate length and intermediate turns. On a drum the factor costs metal, belt and a pair per law; on a cam it costs turns. Since every cost on the drum side is worse than every cost on the cam side, the extreme that puts it all in the cam is the right one. This is the same pattern the drum that is not round found for a winding drum: a law can be put in the shape of whichever member is cheapest to shape, and the shape is forced once the member is chosen.

What the cam does not settle #

The cam’s own wear and play. A pin in a groove has a running clearance, and a roller in a groove changes walls with the speed found that a groove’s follower crosses from one wall to the other as the load reverses. Where the groove is steep, the clearance becomes a band of uncertainty in the belt’s station that is wider in ratio by the groove’s own slope. Whether that matters depends on the drive’s purpose, and it is not measured here.

The belt’s top walking speed. The paced comparison is in units of a top speed whose value depends on the belt, the drum and the fork’s angle. Only ratios are claimed. A real belt’s top speed may itself depend on the grading under it, and on even drums that would favour the cam further.

Holding the station. A belt on a cone drive pulls sideways as well as round, towards the larger radius, and something must hold the carriage against that pull. A barrel cam whose groove is flat over most of its length holds well there. Where the groove is steep it is more easily back-driven, and the pressure-angle limit is also the angle at which the carriage could start to drive the handwheel. That is not computed.

A strand of finite stiffness. As throughout this field, the belt is a member with no length of its own, inextensible along its run, and the edge spread it can absorb is a stated allowance rather than a property computed from a material.

Still open: a groove that is softened only where it is turned slowly #

The cam’s turns are set by its steepest point, and a barrel groove of constant pitch radius spends them evenly along the handwheel. Most of the speed law’s groove is far flatter than its peak, with a pressure angle under 9° over its fast half and 5° at its fast end, so most of its turns are spent where the limit is nowhere near binding.

The distinct argument would be a groove whose pitch radius changes along the barrel, larger where the law is steep and smaller where it is flat, so that every point of the groove sits at the same pressure angle. The handwheel’s turns would then be spent in proportion to the law’s own grading rather than its peak, and their number would be set by the law’s average steepness rather than its steepest. The measurement would find the least-turns barrel for each law under a cap on its largest radius and compare it with the constant barrel here. It would also ask whether the fan law, whose peak is 45 times its average, then comes down to the turns of a practical handwheel. If it does, a conical barrel is to the cam what the logistic drum was to the belt. In both, the even distribution of a cost along the member is what makes it short.