A cam behind a bent joint

Every motion law in the cam field is written in cam angle, which is the point of a cam: it gives the motion it was asked for, specified against its own rotation. The law that costs least compared them that way: a rise of given lift over a given span of cam rotation, drawn under constant acceleration, simple harmonic motion or a cycloid, with the peak acceleration of each read off its second derivative with respect to the cam angle. A lift is a size and a law is a shape went further and showed that the whole comparison is about dimensionless shapes, so that it transfers between cams of any size.

A follower does not feel a derivative with respect to cam angle. It feels one with respect to time, and the two are the same object only when the camshaft turns at a constant speed. Multiply a steady speed through and the ranking of laws is untouched — every law’s acceleration is scaled by the same speed squared. The question here is what happens when the speed is not steady, and the simplest honest source of an unsteady camshaft is one that everybody has met: the joint that is not constant velocity. A camshaft driven through a Hooke’s joint at a bend turns fast and slow twice in every revolution, with nothing wrong with any part.

Two terms where there was one #

Write the rise as $s(\theta)$, the follower’s displacement as a function of cam angle, and let the cam angle be a function of time. Differentiating twice,

$$\ddot{x} = s''(\theta)\,\dot\theta^{2} + s'(\theta)\,\ddot\theta.$$

The first term is the law’s own acceleration scaled by the square of the shaft’s speed. On a steady shaft it is the whole answer, and it is what every cam essay here has computed. The second term is the law’s velocity multiplied by the shaft’s angular acceleration. A steady shaft has no angular acceleration, so the term is invisible to a designer who works in cam angle — and it is largest where the law’s velocity is largest, which on every smooth rise is the middle, exactly where the law’s own acceleration passes through nought.

For a Hooke’s joint both of the shaft’s derivatives are closed forms. With the driving yoke at angle $t$ and turning at unit speed, the camshaft’s angle satisfies $\tan\theta = \tan t/\cos\beta$ for a bend $\beta$, and

$$\dot\theta = \frac{\cos\beta}{1 - \sin^{2}\!\beta\,\cos^{2}t}, \qquad \ddot\theta = -\frac{2\cos\beta\,\sin^{2}\!\beta\,\cos t\,\sin t}{\big(1 - \sin^{2}\!\beta\,\cos^{2}t\big)^{2}}.$$

Those formulas are checked against something they share nothing with except the law’s displacement. The rise is composed with the joint’s output angle — the follower’s position as a function of drive time, and nothing else — and differenced twice. For all three laws at a 30° bend, over the whole of a rise, the two-term formula matches the differenced position to $2 \times 10^{-7}$ of the steady peak, the size of the differencing step’s own error. Drop the second term, as a steady-shaft habit would, and the match fails by 0.118 of the peak.

The joint’s half-turn #

The joint does not turn the camshaft unevenly at random. It does it on a fixed programme with a period of half a turn.

At a 30° bend the camshaft’s speed runs between $\cos\beta = 0.866$ and $1/\cos\beta = 1.155$ of the drive’s: fastest where the driving yoke is at nought or half a turn in the formula above, slowest a quarter-turn later. The angular acceleration is nought at the fastest and slowest points and largest between them, ±0.295 of the drive speed squared. Everything repeats after 180°.

So a rise has a clocking: where in that half-turn it begins. It is measured here as the cam angle of the start of the rise, from the cam angle at which the shaft turns fastest. A cam keyed to its shaft has one clocking, fixed when it was assembled, and nothing about the cam’s own profile says what it is.

The half-turn period has a practical consequence worth stating first. A rise and a return half a turn apart see the same clocking. Measured over all three laws and every clocking in steps of 10°, a rise moved by 180° has the same peak to $10^{-7}$; moved by a quarter-turn, a cycloidal rise’s peak changes by 0.46 of its steady value. A cam with its events half a turn apart can therefore be clocked once for both.

The same rise, 35% worse or 17% better #

The first figure shows what the clocking does to a cycloidal rise of 120° at a 30° bend. At the worst clocking, beginning 100° past the fastest point, the follower’s peak acceleration is 1.352 times its steady-shaft peak. At the best, beginning at 30° — so that the rise is centred on the shaft’s slowest point — it is 0.828. The same cam on the same joint gives its follower a peak acceleration 35% above or 17% below what its designer computed, depending on a keyway.

The better case is not a trick of cancellation. The cycloidal law’s acceleration peaks a quarter and three quarters of the way through the rise, and at the best clocking both of those points fall where the shaft is turning slower than average, so the first term, scaled by the square of the speed, is smaller there than on a steady shaft. The second term is largest in the middle of the rise, where the first term is nought, so it adds a hump where there was none rather than raising a peak that was already there.

At the worst clocking the picture is the reverse. The rise’s second half runs across the shaft’s fastest point, so the first term is scaled up there, and the second term, reaching 0.168 of the steady peak, is spread over the part of the rise where it has the same sign as the first. The two terms reinforce rather than interleave.

Both effects vanish on a straight shaft and grow as the square of the bend. At 5° the cycloidal rise’s peak ranges from 0.994 to 1.008 of its steady value over all clockings, and at 10° from 0.976 to 1.033 — four times the spread for twice the bend. That is small enough to explain why cam designers can ignore a slightly misaligned drive, and large enough by 30° that they cannot.

A small bend, worked by hand #

The numbers so far were computed. For a small bend they can also be predicted, and the prediction says where the two clockings come from.

To the order of the bend squared, the joint’s speed and angular acceleration are $\dot\theta \approx 1 + \tfrac{1}{2}\beta^{2}\cos 2\theta$ and $\ddot\theta \approx -\beta^{2}\sin 2\theta$: a ripple at twice the shaft’s frequency, with the angular acceleration a quarter of a ripple out of step with the speed. A cycloidal rise has two acceleration peaks, a quarter and three quarters of the way through its span, where its velocity is exactly $1/(2\pi)$ of its peak acceleration times the span. Put the ripple into the two-term formula at those two points and each peak changes by

$$\beta^{2}\big(\cos 2\theta_{1} - k\sin 2\theta_{1}\big) \quad\text{and}\quad \beta^{2}\big(\cos 2\theta_{3} + k\sin 2\theta_{3}\big), \qquad k = \frac{\text{span}}{2\pi},$$

with $\theta_1$ and $\theta_3$ the cam angles at which the two peaks fall. The first part of each is the speed ripple acting on the law’s own acceleration; the second, weighted by $k$, is the new term. For a 120° rise $k$ is a third.

The best clocking makes both changes as negative as possible at once, and by symmetry it centres the rise on the shaft’s slowest point, which puts the peaks 30° either side of it. Both changes are then $-\big(\tfrac12 + k\sin 60^\circ\big)\beta^{2} = -0.789\,\beta^{2}$. The worst clocking pushes one of the two peaks to the largest value its combination can reach, $\sqrt{1 + k^{2}}\,\beta^{2} = 1.054\,\beta^{2}$, and it does so at a clocking of 99.2° — or at 140.8°, which does the same to the other peak.

Measured at a bend of 1°, the best and worst clockings are 30° and 99° and the changes are $-0.790\beta^{2}$ and $1.054\beta^{2}$: the hand calculation to within half a per cent. Leave out the new term — set $k$ to nought — and it predicts $-0.5\beta^{2}$ and $\beta^{2}$, which the measurement refuses. More than a third of the benefit of clocking the rise well, and a twentieth of the harm of clocking it badly, comes from the term a steady shaft does not have. By 30° the bend is too large for a second-order formula, and the computed 0.828 and 1.352 are where the higher orders have taken it.

Every law has its own clocking, and one cannot be helped #

The three laws respond to the clocking differently, because their accelerations have different shapes.

At a 30° bend the cycloidal rise runs from 0.828 to 1.352 of its steady peak; the simple harmonic rise from 1.027 to 1.333, best clocked at 120° rather than 30°, because its acceleration peaks at the two ends of the rise rather than inside it; and the constant-acceleration rise from 1.171 to 1.475, with no clocking at which it does as well as on a steady shaft.

The constant-acceleration law is the one no clocking helps, and the reason is the very property that makes it the least on a steady shaft. Its acceleration is a constant magnitude across each half of the rise, so its peak on a steady shaft is not a peak at all but a plateau. An uneven shaft scales that plateau by $\dot\theta^{2}$, and somewhere in any 120° window the shaft is turning faster than average. Its best clocking puts the window’s edges, where the law’s velocity is small and the second term nearly nothing, at the fastest speed the window has to contain — and the peak is that speed squared, 1.082² = 1.171, to the precision of the table. A law designed to spread its acceleration flat has nowhere to hide from a speed that varies.

The ranking changes #

The laws have been compared so far against their own steady peaks. The question a designer asks is the other one: which law gives the follower the least peak acceleration, now that the shaft is bent and each can be clocked at its best? That needs one unit for all three — lift per radian of span squared — in which the steady peaks are 4, 4.935 and 6.283, the constant-acceleration law least by a clear margin.

As the bend grows, the constant-acceleration law’s best-clocked peak rises and the cycloidal law’s falls. They cross at a bend of 34.50°, where both are 4.980. Beyond it the ranking is reversed: at 40° a well-clocked cycloidal cam gives its follower a peak of 4.78 against the constant-acceleration cam’s 5.47. The law that exists to minimise peak acceleration is not the one that does, on a shaft bent that far, and the law that is usually chosen for its smoothness at the price of a higher peak turns out to pay no price at all.

The crossing is measured, not interpolated: the peak at each clocking is computed over the whole rise, the clocking searched in whole degrees, the bends scanned in whole degrees to bracket the change of sign, and the bend inside the bracket found by bisection.

A 34.5° bend is large for a Hooke’s joint driving a camshaft, and the practical reading is not that cam drives should be bent. It is that a cam’s law and its drive are a single design, and a comparison of laws made in cam angle assumes a drive that does not exist once the shaft is not straight.

Where each law overtakes, by the span of the rise #

The crossing depends on how much of the joint’s half-turn the rise occupies.

A short rise sees the shaft’s speed as nearly steady across it. For a 60° rise the cycloidal law needs a bend of 48.3° before clocking pays for its higher steady peak, and for 90° it needs 39.0°. The crossing falls to 34.5° at 120° and 33.6° at 150°, then rises again to 37.9° at 180°, where the rise spans the whole of the joint’s period and cannot avoid its fast half.

The simple harmonic law does not overtake at all on the three shorter rises — within 60° of bend — and on the two longer ones overtakes first: at 21.6° for a rise of 150° and 18.4° for 180°. Its acceleration peaks at the two ends of the rise, and a rise nearly half a turn long can place both ends near slow points of the shaft, which come round every half-turn. So the answer to “which law, behind this joint” depends on the span as well as the bend, and the steady-shaft answer is the limit of both going to nought.

What a bent joint carries into the dwells #

Peak acceleration is one of the two things the laws are ranked by. The other, in the law that costs least, is what happens where the rise meets a dwell: a law whose acceleration is not nought at its ends delivers a step in acceleration to the follower, which is an impulse in jerk.

The cycloidal law arrives at its dwells at nought behind any joint at any clocking, measured to $10^{-11}$. Both terms of its acceleration are products — the law’s acceleration times something, and the law’s velocity times something — and both of the law’s derivatives vanish at the ends. The joint has nothing to multiply.

The other two laws arrive with a step on a steady shaft, and the joint changes the size of the step without changing its existence. It scales each by the square of the shaft’s speed at that instant, since the law’s velocity is nought there and the second term drops out: the harmonic law’s step at the start of its rise grows from 1.125 to 1.500 at its worst clocking, the full factor $1/\cos^{2}\beta$ of a third. A step whose size depends on where the cam is keyed is a step whose effect on a valve train — where the cam is not the valve even on a steady shaft — or on a feed mechanism will vary from one assembly to the next.

What the joint does not change #

The lift and the timing. A Hooke’s joint does not change where the camshaft is at the start and end of each revolution, and it does not change where the cam’s rise begins and ends in cam angle. It changes only when, in time, the follower passes through each part of the rise. A drawing of the displacement against cam angle is identical on a straight shaft and a bent one.

The profile. Pressure angle and the smallest base circle that can be cut are geometric properties of the cam and its follower, computed in cam angle, and so is the profile as an envelope of the roller; the joint does not touch any of them.

The fix. A second Hooke’s joint, bent equally and phased correctly, cancels the first, which is why real drives that must be bent use two; two errors, one ellipse measured how well a real pair cancels when its phasing or its angles are slightly wrong. Everything here is about the single joint, or equally about the residual unevenness of an imperfect pair, whose period is also half a turn.

What this does not settle #

Dynamics. The follower’s acceleration here is kinematic, computed as if the camshaft’s motion were imposed. A real camshaft behind a joint is also loaded by the follower’s inertia and spring, and its speed is decided by those loads as well as by the joint.

One lift, one programme shape. The rises here are single events of unit lift with dwells either side. A programme whose rise and return are not half a turn apart gets two different clockings, and its worst event decides.

Three laws. The modified-trapezoid and polynomial laws used in practice have their own acceleration shapes and would each have their own best clocking and crossing.

Still open: a cam on a chain #

The joint’s unevenness has a period of half a turn, which is long compared with a rise, and that is what made clocking possible: a rise can be placed inside the joint’s slow stretch. The chain is a polygon found a drive whose unevenness has a period of one tooth — a fluctuation of $1 - \cos(\pi/n)$ in the effective radius, once for every link that engages.

Its distinct argument would be what happens when the shaft’s unevenness is fast rather than slow. The second term scales with the shaft’s angular acceleration, and a fluctuation of fixed size at a period of one tooth has an angular acceleration that grows as the square of the tooth count. So a sprocket with more teeth has a smaller fluctuation that changes faster, and whether the product rises or falls with the tooth count is a question with an answer. And since many periods fall inside one rise, no clocking can avoid them: the measurement would be the peak follower acceleration of each law against the sprocket’s tooth count and the rise’s span, with the chain’s own geometry supplying the shaft’s motion, to find which law is least hurt when its drive ripples inside every event.