A roller in a groove changes walls with the speed

A cam that holds its follower both ways closed a yoke round a cam and found that two parallel faces demand a programme which is its own reflection half a turn later. It ended on the form-closed cam most often built instead: a roller running in a groove. The groove’s two walls are the pitch curve moved out and in by the roller’s radius, so they are the same distance apart along every normal whatever the programme, and a groove needs no reflection rule and no second disc.

What a groove has instead is a choice the yoke never faces. A roller touches one wall or the other, never both, and nothing about the cam’s shape decides which. The essay named the question and left it: where the roller changes walls, how often, and what the motion law’s steps do to it.

The answer turns on a fact that makes the question different in kind from every other on the cam’s profile. The count of crossovers is not a property of the cam. It is a property of the cam at a speed, carrying a particular follower against a particular load, and the same groove crosses over no times at one speed and four at another.

The wall is chosen by a sign #

A follower in a groove is carried along its line of travel with whatever acceleration the programme and the speed dictate. If its displacement is $s(\theta)$ at cam angle $\theta$ and the cam turns steadily at $\omega$, its acceleration is $s''\omega^2$, with $s''$ the programme’s second derivative per radian squared. Something must supply the force for that acceleration, and something else is usually pressing on the follower as well: a load from the work it does, gravity on a vertical follower, or the reaction of whatever it pushes. Call that load $F$ and take it as steady and directed toward the cam.

The only thing that can supply force is the groove. So along the follower’s travel it must provide

$$N = m\,s''\omega^2 + F,$$

outward positive. When $N$ is positive the groove pushes the follower away from the cam, which only the inner wall can do. When $N$ is negative the groove pulls the follower toward the cam, which only the outer wall can do. A crossover is a change in the sign of $N$: at that instant the roller leaves one wall, crosses the groove’s working clearance, and lands on the other.

Every consequence below comes from reading that one expression. On a dwell $s''$ is nought and $N$ is the load, so the roller rests on the inner wall. Where the follower accelerates outward, $s''$ is positive and $N$ is larger still. $N$ can go negative only where the follower decelerates, and there only once $m\,|s''|\,\omega^2$ outweighs $F$.

The follower used throughout has a mass of 0.5 kg and is pressed in by 20 N, on the cam field’s prime circle of 40 with a roller of 10.

The same crossovers off the drawn pitch curve #

The angles at which $N$ changes sign are computed from the programme’s second derivative, differenced from the displacement the programme generates. The second route never consults the programme’s acceleration: it samples the pitch curve, the path the roller’s centre is made to follow, at 14,400 points, reads the roller’s distance from the cam’s centre off each one, subtracts the prime circle, and second-differences the result. That is the follower’s displacement as the drawn cam delivers it, and $N$ is built from it with the same mass, load and speed.

On the quick-rise programme at 600 rpm both routes find four crossovers, at 47.869°, 87.131°, 151.354° and 193.646°, and they differ by at most $2.8 \times 10^{-6}$ degrees. On the standing programme they differ by $6.3 \times 10^{-7}$. Given half the load and everything else unchanged, the drawn route’s crossovers move by 11.54°, so the agreement is about the force the groove must supply, not about the curve alone.

Nothing below a speed, and two above it #

The expression predicts a speed below which nothing happens, and computes it. Over a stretch where the follower decelerates, $N$ is most negative where $s''$ is. It first touches nought when $m\,|s''_{\min}|\,\omega^2 = F$, so the threshold speed is

$$\omega^* = \sqrt{\frac{F}{m\,|s''_{\min}|}}.$$

Below it, the stretch passes without a crossover. Just above it, $N$ dips below nought over a short interval around the deepest deceleration, which is two crossovers: out to the outer wall, and back. As the speed rises the interval widens but it stays one interval, so the stretch never contributes more than two.

The quick-rise programme rises 20 over 90° and returns over 170°, with dwells of 40° and 60° between. Its rise decelerates from 45° to 90°, most sharply at 67.5° with $s''$ of −50.93 per radian squared; its return decelerates from 130° to 215°, most sharply at 172.5° with −14.27. Those give thresholds of 267.6 rpm for the rise and 505.5 rpm for the return. At 200 rpm the force never reaches nought. At 400 rpm the rise has crossed it and the return has not, and there are two crossovers. At 800 rpm there are four.

The staircase is exact. At one per cent below each threshold and one per cent above it the count rises by two, and nowhere else. The standing programme, whose rise and return are mirror images over 120° each, has both thresholds at 356.8 rpm and goes from none to four in a single step. The quick-rise cam goes in two steps, because its fast rise decelerates three and a half times as hard as its slow return and reaches the load at a lower speed.

Which is to say that a groove designed at one speed and run at another is a different machine. A quick-rise groove that never crosses over at 250 rpm crosses over twice at 300 and four times at 550, and nothing about its drawing changes.

The threshold is a spring’s lift-off speed #

The threshold has a familiar shape, and it is worth seeing why. The first cam essay set a groove against the cheaper way of bringing a follower back, a spring, and warned that a spring-closed follower leaves its cam at speed, when the spring can no longer supply the deceleration the programme demands.

That is the same inequality. A follower held against its cam by a spring and a load leaves the cam wherever $m\,|s''|\,\omega^2$ exceeds the spring’s force and the load together. The groove’s threshold speed is exactly the speed at which a follower held by the load alone would leave a cam that had no outer wall. Above it, that follower would lift off. A grooved cam’s roller meets the outer wall instead, and the meeting is a crossover.

Read the other way, the expression prices the spring that would prevent every crossover. To keep the quick rise’s roller on its inner wall at 600 rpm, the force pressing the follower in must be at least $m\,|s''_{\min}|\,\omega^2$: 100.5 N against the rise’s deepest deceleration, and 28.2 N against the return’s. A preload of a little over a hundred newtons, from a spring or from the work the follower does, turns this groove at that speed into a cam that never uses its outer wall, and the outer wall becomes a guard against the follower leaving rather than a surface it runs on. At 1,200 rpm the same guarantee needs 402 N, four times as much, because the price rises with the square of the speed.

So the groove and the spring are not two answers to one question. The spring prevents crossovers up to a speed and fails above it; the groove lets them happen and survives them. A designer choosing between the two is choosing between a load that grows with the square of the running speed and an outer wall that the roller will hit, and the expression gives both prices in the same units.

Where the crossings fall #

A threshold says when crossovers begin; the law of motion decides where in the stretch they sit and how they move.

Under the cycloidal law the rise’s deceleration is a smooth half-wave, nought at mid-rise, deepest at three-quarters, and nought again at the rise’s end. Just above the threshold the two crossovers appear close together near the deepest point: at 300 rpm they are at 58.18° and 76.82°. As the speed rises the interval where $N$ is negative spreads toward the two zeros, and the crossovers follow: 51.65° and 83.35° at 400 rpm, 46.03° and 88.97° at 1,000, 45.11° and 89.89° at 3,000. They approach mid-rise and the rise’s end and never reach them, because at those two places $s''$ is nought and the load wins at any speed.

Under the constant-acceleration law the rise’s deceleration is not a wave but a plateau: $s''$ is a constant $-A$ over the whole second half and jumps to it from $+A$ at mid-rise. Its deepest value is also its only value, so once the speed passes the threshold, 335.4 rpm on this cam, the whole half-rise is on the outer wall at once. The crossovers sit exactly at 45.00° and at 90.0° and do not move with speed at all.

The simple harmonic law is half of each. Its deceleration grows smoothly from nought at mid-rise to its deepest at the very end, where it stops abruptly as the dwell begins. So its first crossover moves, from 52.34° at 600 rpm to 45.29° at 3,000, and its second is pinned at the rise’s end.

A law with steps throws the roller across #

Where a crossover sits is less important than how it happens, and the figure’s dashed lines mark the difference.

Under the cycloidal law $N$ passes through nought at a finite rate at every crossover. The force on the inner wall falls away, reaches nought, and builds up on the outer wall, so the roller is unloaded before it leaves and loaded gradually after it lands. Whatever clearance the groove has, the roller crosses it while very little force is acting.

Under the constant-acceleration law $N$ does not pass through nought at all. At mid-rise the acceleration jumps from $+A$ to $-A$, so the force jumps by $2mA\omega^2$ in no time; at the rise’s end it jumps back by $mA\omega^2$. On the quick rise $A$ is 32.42 per radian squared, and at 1,000 rpm the steps are 356 N and 178 N. At 3,000 rpm they are 3,200 N and 1,600 N, and the crossovers’ step sizes, measured across each one, match the formula to better than two per cent. The roller is not eased across the groove; it is thrown across it by a load that was pressing it into one wall an instant before.

The harmonic law steps once per stroke: 219 N at the rise’s end at 1,000 rpm and 1,974 N at 3,000.

This is the comparison of motion laws seen from inside a groove. The constant-acceleration law has the smallest peak acceleration of any law for a given rise and time, which is why its thresholds are the highest of the three — it is the last to start crossing over — and it pays for that with steps that make every crossover an impact. The cycloidal law begins crossing over at the lowest speed and never steps. A groove is the one place where a law’s discontinuity in acceleration is felt directly as a force changing sign, rather than as vibration somewhere in the train.

Two strokes with nothing between them #

The quick-rise and standing programmes both have a dwell between the rise and the return, so their two stretches of deceleration are separated and each contributes its own pair. A programme with no dwell between the strokes asks a sharper question: the rise decelerates into its end and the return decelerates out of its start, and whether that is one stretch or two is decided at the join.

The back-to-back programme here rises 20 over 150°, returns over 150° immediately, and dwells for 60°. Under the constant-acceleration law the rise’s last half has $s''$ of $-A$ and the return’s first half has the same $-A$, so the deceleration runs straight through the join from 75° to 225°. It is one stretch, and at any speed above its threshold of 559.0 rpm the roller crosses over twice, at 75° and 225°, both as steps. The harmonic law also decelerates through the join at its deepest, one stretch, two crossovers, both smooth.

The cycloidal law does something the others cannot. It brings the acceleration to nought at the end of every stroke, including a stroke that is immediately followed by another, so at the join the load wins at every speed. The programme has two stretches, touching at 150°, and above 446.0 rpm it has four crossovers. At 1,000 rpm the middle two are at 145.22° and 154.78°, and the roller spends 9.56° of the turn back on the inner wall. At 3,000 rpm they are 1.06° apart, and at 5,000 rpm 149.81° and 150.19°, 0.38° apart.

A crossover the rigid model counts and a real groove may not #

That last number is where the model this essay uses reaches its edge, and it should be said plainly.

At 5,000 rpm, 0.38° of the turn lasts thirteen millionths of a second. The rigid model says the force on the roller changes sign twice in that time, so the roller should leave the outer wall, cross the clearance, touch the inner wall and come back. A real roller has mass of its own and a groove has clearance, and in thirteen microseconds a roller under a few newtons of net force moves far less than any working clearance. The kinematic crossover is real — the force does change sign — but whether the roller travels anywhere in response is a dynamic question the model does not ask.

The same caution applies with less force everywhere near a threshold. Just above 267.6 rpm the quick rise’s two crossovers are close together, and the force between them is barely negative. The staircase counts them exactly, and a physical roller might ride through that interval on the wall it was already on.

The roller must reverse its spin #

A crossover costs something even when it is perfectly smooth, and the cost is in the roller rather than the follower.

A roller rolls on whichever wall it is pressed against. The groove slides past the roller’s centre at the pitch curve’s speed, and the two walls slide past in the same direction, so a roller rolling without slipping on the inner wall spins one way and on the outer wall the other. At every crossover its spin must reverse, by twice the groove’s sliding speed divided by the roller’s radius.

On the quick-rise cam at 3,000 rpm, a crossover at the rise’s end, where the roller’s centre is 60 from the cam’s centre, asks a roller of radius 10 to change its spin by 36,000 rpm — twelve times the cam’s own speed. The mid-rise crossover asks 33,694 rpm. None of that spin can reverse instantly, so for some part of every crossover the roller slides on the new wall while its spin catches up. That sliding is where a grooved cam wears, and it happens at exactly the places this essay locates.

The table gathers the three questions at one speed. The count is set by the programme and the join; the thresholds by the law’s deepest deceleration; the steps by whether the law’s acceleration jumps; and the spin reversal by where on the cam the crossovers fall, which is at the pitch curve’s largest radius whenever a crossover is pinned to the end of a rise.

A rigid follower under a steady load #

The follower is rigid and the roller massless. Neither the roller’s own inertia, nor the elasticity of the follower and its drive, nor the groove’s clearance enters the force $N$. They decide how long a crossover takes and how hard the roller lands, and they are what would make the 0.38° interval physical or not.

The load is steady. A process load that varies through the cycle moves every threshold and every crossover, and a load that changes direction adds crossovers of its own.

The pressure angle and friction are left out of the force’s size. The force along the follower’s travel is the part of the groove’s normal force that lies along that travel, equal to it only where the pressure angle is nought; elsewhere the normal force is larger by the angle’s secant, and friction at the wall adds to it again. Neither changes the force’s sign, so neither moves a crossover, but both enlarge the steps and the load the roller lands with.

One follower and one load. Every number scales, in the way a lift is a size and a law is a shape sorts a cam’s numbers: the thresholds with the square root of the load over the mass, the steps with the mass and the square of the speed. The shapes of the staircase and of the crossover curves do not change.

What comes next: the time a crossover takes #

The time a crossover takes. With a clearance $c$ and the net force $N$ known at every angle, the roller’s flight across the groove can be integrated, and the landing speed on the new wall follows. That would turn this essay’s steps from a force into an impact velocity, and it would say at what speed the cycloidal law’s narrow return at the join stops being a crossover at all.

A roller in a groove on a swinging arm. An arm is an offset that grows with the lift put the roller on an arm. In a groove the arm’s own moment of inertia joins the follower’s mass, the load acts about the pivot, and the sign that chooses the wall becomes a torque. Whether a pivoted follower’s crossovers fall where a sliding follower’s do is the next question the two essays together can answer.

Two rollers instead of one groove. A follower with two rollers, one on each side of a rib, is a groove turned inside out, and it has no crossover at all because each roller keeps its own wall. What it has instead is the yoke’s breadth condition, measured along a line through two rollers rather than between two faces.