The time a crossover takes

A roller in a groove changes walls with the speed found where a grooved cam’s roller leaves one wall for the other. A follower of mass $m$, pressed toward the cam by a load $F$ and carried at an acceleration $s''\omega^2$, needs a force $N = m s''\omega^2 + F$ from the groove; the inner wall can only push and the outer wall can only pull; so the wall in use is decided by the sign of $N$, and the places where the sign changes are the crossovers. Their count steps up by two at each stretch of deceleration’s threshold speed, and the motion law decides whether $N$ passes through nought or jumps across it.

That essay treated the change of wall as instantaneous, which is what a groove with no clearance would do, and closed on the one thing the treatment could not see. A groove that can be cut has a gap between the roller and the wall it is not bearing on. Across that gap nothing holds the roller: it is in flight. How long the flight lasts, how fast the roller arrives, and whether it arrives at all are questions about a small, well-defined motion, and each has an answer.

The flight, written down #

At the instant $N$ changes sign, the wall the roller is bearing on can no longer supply the force the programme demands. Before that instant the roller moved with the wall; after it, the only force on the follower is its load, so the follower decelerates at $F/m$ while the wall carries on along the programme.

Measure the gap $g$ from the wall the roller left, in millimetres, and take derivatives with respect to the cam angle in radians, so that the cam’s speed appears only where time does. The roller’s own acceleration is $-F/m$, which per radian squared is $-(F/m)/\omega^2$; the wall’s is $s''$. The gap therefore obeys

$$g'' = -\frac{F/m}{\omega^2} - s''(\theta), \qquad g = g' = 0 \text{ at the crossover}.$$

At the crossover itself the right side is nought, because that is the definition of the crossover: $N = 0$ is exactly $s'' = -(F/m)/\omega^2$. After it the wall’s deceleration outweighs the load’s, $g''$ turns positive, and the gap opens. The roller lands when $g$ reaches the clearance $c$, and its speed relative to the far wall is $g'\omega$.

That equation is integrated here from each sign change of $N$ that the earlier essay found, by a Runge–Kutta step of two thousandths of a degree and less at small clearances, with the landing interpolated inside the step where the gap crosses $c$. The follower is the same one throughout: half a kilogram, a load of 20 N, on the quick-rise programme that rises 20 mm over 90° and returns over 170° — a lift is a size and a law is a shape, and the flight needs both.

At 600 rpm with 0.05 mm of clearance the four flights of the turn take 6.6°, 7.1°, 14.6° and 13.4° of cam rotation — between 1.8 and 4.1 milliseconds — and land at 80, 63, 35 and 42 mm/s. The first figure draws them, and its dial runs the speed from 280 to 1,400 rpm.

A flight measured in degrees hardly depends on the speed #

Written per radian of cam, the flight equation has the cam’s speed in one place only: the load’s term, $(F/m)/\omega^2$. Well above a threshold that term is small beside the programme’s deceleration, and the equation becomes a statement about the programme alone — the gap opens as the wall’s own motion dictates, measured in degrees of cam, and the load merely decides where the flight starts.

So the angle a flight occupies should settle to a constant as the speed rises, and it does. The quick rise’s first flight takes 6.64° of cam at 600 rpm and 6.56° at 1,200 — a change of one per cent for a doubling of speed, most of it because the crossover itself moves from 47.9° to 45.7° as the load’s share shrinks. In time, the flight halves: 1.8 ms at 600 rpm, 0.9 ms at 1,200.

The same reasoning fixes how the landing speed scales. If the flight in degrees is fixed, the rate at which the gap closes on the far wall is fixed per radian, and converting it into millimetres per second multiplies by $\omega$. Outward landings grow in proportion to the cam’s speed, which is a stronger statement than “faster is harder”: it says a groove that lands at 80 mm/s at 600 rpm will land at about 160 at 1,200 and 320 at 2,400. Flown, it lands at 163 mm/s at 1,200 rpm and 328 at 2,400, and the figure on landing speeds further on shows the same proportion for each outward flight of a stepped law. The flights that break the rule are the ones whose equation keeps the load term, and they are the subject of a later section.

How hard it lands depends on the clearance, by an exponent #

A smaller clearance means a shorter flight and a gentler landing; that much is obvious. What is not obvious is how much gentler, and the answer is a power law whose exponent belongs to the motion law.

Flown across clearances from a thousandth of a millimetre to one millimetre at 1,200 rpm, the first crossover of the quick rise lands at speeds that fall on two straight lines. Fitted over the smallest four clearances, the cycloidal law’s line has slope 0.664 and the constant-acceleration law’s has slope 0.5000.

The two exponents mean different things for a designer tightening a groove. Halving the clearance under a cycloidal law cuts the landing speed by a factor of $2^{2/3} = 1.59$; under constant acceleration by $2^{1/2} = 1.41$. Over two decades of clearance the difference compounds: at one millimetre the two laws land within fifteen per cent of each other — and there the cycloidal law is the harder of the two — at a hundredth of a millimetre the stepped law lands at 97 mm/s and the smooth one at 56, and at a thousandth the stepped law’s landing is two and a half times the smooth law’s.

Where the two exponents come from #

The exponents are not fitted constants. They follow from what $g''$ does in the first instants of a flight, and the next figure shows exactly that.

Under the cycloidal law, $s''$ passes smoothly through the crossing value, so $g''$ starts at nought and grows in proportion to the angle since the crossover. Call the rate $j$ — it is the programme’s jerk at the crossover, in the same units. Then $g'' = j\psi$, $g' = j\psi^2/2$ and $g = j\psi^3/6$. The flight ends when $j\psi^3/6 = c$, at $\psi = (6c/j)^{1/3}$, and the landing rate is

$$g' = \frac{j}{2}\left(\frac{6c}{j}\right)^{2/3} = \tfrac12\,(6c)^{2/3}\,j^{1/3},$$

two thirds of a power of the clearance. The figure shows the linear start directly: 56 m/s² a degree after the sign change.

Under constant acceleration the crossing sits at the jump in the programme’s acceleration from $+A$ to $-A$ at mid-rise. The wall’s deceleration is at its full value from the first instant, so $g''$ is a constant $\Delta$ — here 472 m/s² — and $g = \Delta\psi^2/2$. The flight ends at $\psi = \sqrt{2c/\Delta}$, and the landing rate is

$$g' = \sqrt{2c\Delta},$$

one half of a power. The fitted 0.5000 is that exponent to four figures. The fitted 0.664 sits a little short of two thirds because the cycloidal law’s $g''$ is linear only near the crossover and bends over as the flight lengthens, which is also why the slope is fitted on the smallest clearances and why the two lines converge at a millimetre.

So a landing exponent is a measurement of the programme’s smoothness at the point where the roller leaves. It is the same distinction the law that costs least draws between a law with finite jerk and one with an impulsive jerk, arriving as a number a groove’s clearance multiplies. It also says where the “good” law’s advantage lives: entirely in the small-clearance limit. A sloppy groove makes the laws equivalent.

A flight into a dwell lands at the same speed at every cam speed #

The dial on the first figure shows something the exponent does not: flights at different crossovers respond to the cam’s speed differently. One kind of flight does not respond at all.

Under the constant-acceleration law — the lowest peak acceleration of any law, bought with an impulsive jerk — the quick rise’s second crossover is pinned at 90°, where the rise ends and the dwell begins. The roller has been riding the outer wall through the rise’s decelerating half, and at 90° the programme’s acceleration jumps to nought. From that moment the wall is still, and the only thing moving the follower is its load. A body starting at rest relative to a stationary wall and falling across a gap $c$ under an acceleration $F/m$ arrives at

$$v = \sqrt{2cF/m},$$

which with 0.05 mm, 20 N and half a kilogram is 63.2 mm/s. The integrated flights land at exactly that — to seven parts in a thousand million — at 800, 1,200 and 2,000 rpm.

The flights that leave the inner wall do the opposite: 137, 217 and 372 mm/s at the same three speeds, roughly in proportion to the cam’s speed, because what pushes the gap open is the programme’s acceleration at that speed, which grows as $\omega^2$, and the landing goes as its square root.

The practical reading is a prescription for where a cam’s worst impact is. The flights that land hardest at speed are the ones that leave the load’s wall while the programme is still decelerating hard. The flight into a dwell is gentle, bounded and speed-independent, and a designer can price it in advance from three numbers without integrating anything. The one that grows without bound is the outward flight, and its size is set by the programme’s step at that crossover and the clearance.

A sign change that sends the roller nowhere #

The last result is the one the earlier essay suspected and could not measure. Just above a stretch’s threshold speed, $N$ dips below nought over a short interval around the deepest deceleration and comes back. The count of sign changes goes up by two. But a flight needs time to cross the clearance, and if $N$ returns positive before the gap has reached $c$, the load is again winning, the gap closes, and the roller comes back to the wall it left.

Integrating each flight through to either landing or return finds that band exactly. With 0.05 mm of clearance, the quick rise’s force first changes sign at 267.6 rpm; at 270 rpm the roller rises 4.4 µm, less than a tenth of the clearance, and falls back at 6.3 mm/s; the first speed at which it reaches the outer wall is 275.8 rpm. The return stroke’s force changes sign at 505.5 rpm and its flight first lands at 521.0 rpm. Between each pair of speeds a count of sign changes records two crossovers per stretch and the roller makes neither.

The band grows with the clearance — for the rise from 271.2 rpm at a hundredth of a millimetre to 284.8 rpm at 0.2 — and it is the reason flights must be taken in order rather than one at a time. A flight that falls back leaves the roller on its original wall, so the next sign change, which would have sent it back, finds it already there. Flown in sequence, the turn at 270 rpm has one real event, a lift and a fall-back of a few microns, where the count of sign changes has two.

That changes what “a crossover” should mean to anyone listening to a grooved cam. The earlier essay’s staircase is exact about the force and misleading about the roller: at speeds just above a threshold the roller touches down on the wall it never left, gently, and an impact on the far wall begins at a slightly higher speed that depends on how much clearance there is. The gentler of the two landings is still an impact, and 6.3 mm/s of fall-back is far from nought — but it is a different noise at a different speed from the one a count of sign changes predicts.

What this model does not include #

The roller does not spin up. The earlier essay showed that at every crossover the roller has to reverse its spin to roll on the new wall. The flight here treats the roller as a point on the follower’s line; the sliding at landing while the spin reverses is a separate loss, and at speed it is the larger one.

The walls and the follower are rigid. A landing at 372 mm/s is a speed, not a force; what it does to the groove depends on the contact stiffness, the follower’s own compliance and the damping in both. The speed is the input to that question.

The load is steady. A follower whose load varies within the turn, or a load from a spring whose force changes with lift, changes $F/m$ along the flight, and the dwell result in particular — $\sqrt{2cF/m}$ — uses the load at landing.

The clearance is one number. A real groove’s clearance varies around the cam with the machining, and a cam that cannot be cut is a reminder that the profile is only as good as the tool that makes it. A landing speed that grows as a power of the clearance makes the widest point of the groove the one that matters.

What a designer does with it #

To make a grooved cam quiet at speed, the law and the clearance work together. Reducing the clearance helps under every law and helps most under laws that are smooth at the crossover — by a factor of $2^{2/3}$ for each halving against $2^{1/2}$. A stepped law cut to a tight groove still lands harder than a smooth law cut to the same groove.

The flight into a dwell is priced without integrating anything. It is $\sqrt{2cF/m}$ and independent of speed, so it sets a floor on the noise of any cam whose rise decelerates into its dwell.

The speeds just above a threshold are quieter than the staircase suggests. A groove designed to run just above its threshold speed, where the force barely changes sign, may never make the roller cross at all — and the margin is set by the clearance, so a worn groove moves the onset of real crossovers to a higher speed rather than a lower one.

The same sign-and-flight reasoning applies to the other form-closed cams. A cam that holds its follower both ways closed its yoke with no clearance by construction; any real yoke has some, and the flight equation above, with the yoke’s breadth error for $c$, is what it would obey.

Still open: a roller in a groove on a swinging arm #

Everything here assumes the follower slides along a straight line. A grooved cam with its roller on a pivoted arm has the same sign rule with a torque in place of a force: the arm’s moment of inertia about its pivot joins the follower’s mass, the load acts about the pivot, and the clearance is measured along a normal that turns as the arm swings.

Its distinct argument would be the flight on the arm: the same free-flight equation written for the arm angle, with the gap measured along the moving normal. Two things would come out of it that the sliding follower cannot show. Whether the crossovers fall at the same cam angles as on a sliding follower with the same programme, since an arm is an offset that grows with the lift and the programme the roller actually follows is not the one the arm is cut to; and whether the landing exponent is still the motion law’s alone, or whether the arm’s geometry adds a term — a swinging normal changes the gap’s direction during the flight, and at small clearances that effect is of the same order as the one measured here.