An offset trades the rise for the return

The essay on the cam that cannot be cut ends with four ways to rescue a cam without making it bigger, and the fourth is to offset the follower: move its line of travel off the cam’s centre, which “changes the pressure angle asymmetrically, improving it on the rise at the expense of the return”. That sentence was written as a list item and never computed. It is the kind of claim that is easy to believe in the wrong form, because the word improving travels further than the words at the expense of.

This essay computes it. The short version is that an offset is a trade, not an improvement, and the trade can be priced exactly: how much one stroke gains, how much the other pays, where the two break even, and what a follower that is not moving at all is charged for the privilege.

Where the offset enters #

A roller follower is driven along a straight guide, and the cam pushes it through the roller’s contact. The push acts along the contact normal, the line through the contact point perpendicular to the cam’s surface, which also passes through the roller’s centre. The pressure angle is the angle between that normal and the guide. The part of the push along the guide lifts the follower; the part across it presses the follower sideways into its guide, and past about 30° the sideways part is large enough that friction in the guide can stop the follower altogether. That is the familiar limit, and it is the cam’s version of a linkage’s transmission angle.

Put the cam’s centre at the origin, let the cam turn anticlockwise, and let the follower travel straight up along the line $x = e$. The prime circle is the circle the roller’s centre sits on during the lowest dwell; call its radius $R_0$. With the offset, the roller’s centre on that dwell is not directly above the cam’s centre but at height $d = \sqrt{R_0^2 - e^2}$, and when the follower has risen by $s$ it is at $(e,\; d + s)$.

Everything the cam does to the follower is in how that point moves relative to the cam. In the cam’s own frame the roller’s centre traces the pitch curve, and differentiating that curve with respect to the cam angle and turning the result back into the fixed frame gives a tangent of $(d + s,\; s' - e)$, where $s'$ is the follower’s velocity per radian of cam rotation. The normal is perpendicular to the tangent, so

$$\tan\varphi = \frac{s' - e}{\sqrt{R_0^2 - e^2} + s}.$$

With no offset this is the textbook $\tan\varphi = s'/(R_0 + s)$, and the site’s cam library has computed exactly that since the first cam essay. The offset does two things to it, and they are of very different sizes.

It is subtracted from the velocity. On the rise $s'$ is positive and a positive $e$ makes the numerator smaller. On the return $s'$ is negative, and the same $e$ makes the numerator’s magnitude larger. This is the first-order effect and it is the whole trade: the offset is a constant taken off a quantity that changes sign between the two strokes.

It shortens the lever. The denominator loses a little, because $\sqrt{R_0^2 - e^2}$ is less than $R_0$. At an offset of 5.88 on a prime circle of 40 it is 39.565, a shrink of 1.1%, and it makes both strokes very slightly worse. It is second order in $e$ and it is the reason the trade turns out not to be quite even.

The sign convention matters enough to state in words. With the cam turning anticlockwise and the follower above it, the cam’s surface under the follower is moving from right to left, so a rising flank arrives from the right. A positive offset moves the follower toward that side, which is toward the flank that is about to lift it.

The same angle, read off the drawn curve #

A formula derived in three lines is still a claim about three lines of algebra, and a sign slip in the second of them would produce a formula that looks equally plausible and favours the other stroke. So the angle is also measured without the formula, which is the site’s standing habit for any number a figure prints.

The second route builds the pitch curve the figures draw: at each cam angle it places the roller’s centre at its offset position, turns that point into the cam’s frame, and records it. Then, at any cam angle of interest, it differences two neighbouring points of that recorded curve, turns the difference back into the fixed frame, and reads off the direction of the normal. It uses no pressure-angle formula and no follower velocity. The only thing it knows is where the roller’s centre was put.

The two agree to $1.1 \times 10^{-6}$ degrees at every angle of the turn, for an offset of 6 on the quick-rise programme below. The discipline that makes the agreement worth anything is the refusal: rebuild the curve with the offset on the other side, keep the formula as it is, and the two differ by 17.25°. A formula with its sign backwards would have been caught by exactly that disagreement, and it would not have been caught by looking at a picture of a cam.

A pitch curve measurement stands in for the cam’s surface here, and that is not a shortcut. The cut surface is the pitch curve moved inward by the roller’s radius along its own normal, so the two curves share every normal. A pressure angle read off one is the pressure angle of the other.

Mirror-image strokes, and an offset worth nothing #

Start with the programme the cam field uses everywhere: a cycloidal rise of 20 over 120°, a dwell of 30°, a cycloidal return of 20 over 120°, and a dwell of 90°. On a prime circle of 40 with no offset, the rise’s pressure angle peaks at 21.22° and the return’s at 21.22°. The two strokes are the same motion run in opposite directions, so their peaks are the same number.

Offset the follower by 8 and the rise’s peak falls to 12.84° while the return’s climbs to 29.44°. Offset it by −8 and the numbers swap exactly: 29.44° on the rise, 12.84° on the return. The offset has not improved the cam. It has moved 8.38° of pressure angle off one stroke and put 8.22° of it on the other, and the worse of the two strokes is now 8.22° closer to the 30° limit than it was.

That is the claim in its sharpest form. The offset that minimises the worse stroke’s peak is found by search, and on this programme it comes back at $6 \times 10^{-12}$ — zero, to the precision of the search. There is no offset that helps this cam, because every unit of help on one stroke is paid for on the other, and the payment falls on whichever stroke was not helped.

Why the trade is not quite one for one #

The sweep shows two curves crossing at zero, one falling and one rising, and at first sight they look like straight lines with opposite slopes. They are not quite, and the reason is instructive rather than incidental.

Add the two peaks together. Centred, the sum is 42.45°. At an offset of 4 it is 42.41°; at 8 it is 42.28°. On this programme the stroke that gains gains slightly more than the stroke that loses, so the sum drifts down. Two things are at work.

The first is the arctangent. The offset subtracts the same amount from the numerator on both strokes, but an angle is not proportional to its tangent. The helped stroke’s tangent falls from 0.39 toward 0.23, where the arctangent is steep, and the hurt stroke’s rises from 0.39 toward 0.56, where it has begun to flatten. The same change in tangent buys more degrees on the way down than it costs on the way up, so the stroke being helped moves a little further than the stroke being hurt.

The second is the lever. The denominator $\sqrt{R_0^2 - e^2} + s$ shrinks as the offset grows, which pushes both angles up, and it does so by an amount that depends on the follower’s height at each stroke’s peak. On mirror-image strokes the two peaks happen at the same height, so this term costs both strokes the same and does not tilt the balance.

Neither effect is large, and neither turns an offset into a free improvement. What they establish is that the trade is not an accounting identity. It is a property of one formula applied to two particular stroke shapes, which is why it has to be computed on the cam in hand rather than assumed. On the next programme the sum goes the other way.

A rise that is faster than its return #

Now change the programme into the shape an offset exists for: a cycloidal rise of 20 over only 90°, a dwell of 40°, a cycloidal return of 20 over a leisurely 170°, and a dwell of 60°. This is a cam whose working stroke is fast and whose return is idle — the opposite arrangement to a linkage’s quick return, which works slowly and comes back fast — and it is very common.

Centred, on the same prime circle of 40, the rise’s pressure angle peaks at 27.37° and the return’s at 15.33°. The rise is uncomfortably close to the 30° limit, and the return has twelve degrees of room it will never use. That is the situation the offset trade was designed for, because the stroke being charged has margin to spare.

The balancing offset is where the two peaks become equal, and a search finds it at 5.88, where both are 21.81°. The worse stroke has gone from 27.37° to 21.81°, which is 5.56° of relief. The return has gone from 15.33° to 21.81°, which is 6.48° of cost. The sum of the two peaks has risen from 42.70° to 43.63°, and the cam as a whole is slightly worse by that measure while being substantially better by the only measure that decides whether it jams.

That is the correct reading of an offset. It does not lower pressure angles. It moves them from a stroke that has too much to a stroke that has too little, at a small premium, and it is worth doing exactly when the two strokes are unequal and the one being relieved is the one that matters.

The premium is also where this programme parts company with the mirror-image one. There the sum of the two peaks drifted down as the offset grew; here it rises, by 0.93° at the balance point and by 1.36° at an offset of 9. The difference is the lever term. The two strokes now peak at different follower heights and at very different velocities, so the arctangent’s flattening and the shortened denominator no longer fall evenly, and on this cam they add up against the offset rather than for it.

The balance point assumes both strokes matter equally #

The crossing in the sweep is a minimax: it minimises the larger of the two peaks, which is the right target when both strokes carry comparable force and either could jam. It is not the only defensible choice, and it is worth being clear about what it assumes.

A spring-returned follower is pushed back onto the cam by its spring during the return, and the force at the contact is the spring’s rather than the load’s. If the load is large and the spring gentle, the rise is the only stroke on which a steep pressure angle is dangerous, and a designer would reasonably accept a higher return angle to take more off the rise.

The sweep answers that question too, because it is the same two curves read at a different place. At an offset of 9 the rise peaks at 18.77° and the return at 25.29°. The offset can be pushed until the return itself reaches the 30° limit, and on this programme that happens at 13.13, where the rise has fallen to 14.60°. Beyond that the trade has spent the whole of the return’s margin, and the cam is worse on the return than it was on the rise before any offset was applied.

So the useful range of offsets for this cam runs from 0, which favours nobody, to 13.13, which spends everything the return had. The balance point sits inside it at 5.88. Where in that range to stand depends on which stroke carries the load. That is a question about the machine and not about the cam, which is the same boundary the motion-law comparison runs into when it declines to name one law as best.

A follower leaning while it stands still #

The table has a column that the pressure-angle literature rarely prints, and it is the cost of an offset that no sweep of peaks shows.

On a dwell the follower does not move, so $s' = 0$, and for a centred follower the pressure angle there is exactly zero: the cam pushes straight up the guide. For an offset follower the formula gives $\tan\varphi = -e / (\sqrt{R_0^2 - e^2} + s)$, which is not zero. A follower at rest on an offset cam is pushed sideways into its guide the whole time it waits.

On the quick-rise programme at the balancing offset, that lean is −8.45° on the low dwell and −5.64° on the high one, where the extra 20 of lift lengthens the lever. At an offset of 9 the low dwell’s lean is −13.00°. Nothing about the motion law or the rise produces it; it is the geometry of a push that does not pass through the cam’s centre.

Whether it matters depends on how long the follower dwells and what it is holding. On a cam whose dwells are half the cycle and whose follower is clamping a workpiece, a permanent side load of a sixth of the contact force is a steady wear source on one side of the guide. On a cam whose dwells are brief and unloaded it is nothing. Either way it is an entry in the offset’s bill that the stroke-by-stroke comparison does not contain, and it is the reason a centred follower is the default rather than a missed opportunity.

What the offset does to the metal #

Undercutting is the other reason a cam has to be big: where the pitch curve bends more tightly than the roller is wide, the offset surface crosses itself and the cam cannot be made. An offset changes the pitch curve, so it changes the curvature, and the question is by how much.

The pitch curve’s radius of curvature for an offset follower has a closed form that extends the polar formula the cam field already uses. With the tangent $v = (d + s,\; s' - e)$ it is

$$\rho = \frac{|v|^3}{(d+s)^2 + (s'-e)(2s'-e) - (d+s)\,s''}$$

and at $e = 0$ it reduces to the familiar expression. It is checked the same way the centred one is, by fitting circles through three neighbouring points of the drawn pitch curve, and the two agree to four decimal places at every offset in the table.

The answer is that on this cam the offset barely touches it. The tightest convex radius is 31.611 centred, 30.506 at the balancing offset and 29.850 at an offset of 9. With a roller of 10 that leaves between 21.61 and 19.85 to spare, and the cam can be cut at any of them.

That is a result about this cam rather than a law, and it is worth saying so. The curvature is set mostly by the follower’s deceleration at the top of the rise, which the offset does not change, and only slightly by the tilt the offset puts into the curve. A cam already close to undercutting, with a large roller on a small prime circle, would have less room, and there the one-unit shift in the tightest radius could decide. The claim this table supports is narrower: an offset is a tool for the pressure angle, and it should be re-checked against curvature rather than expected to help or hurt it.

Reverse the cam and the offset changes sides #

One consequence of the formula is so practical that it deserves its own statement, and it is the one most easily forgotten on a machine that runs in both directions.

The offset helps the stroke whose velocity has the same sign as $e$. Run the cam the other way round and every velocity changes sign: the flank that used to lift the follower now lowers it, and the flank that lowered it now lifts it. The same physical offset, on the same cam, now helps what used to be the return and charges what used to be the rise.

For the quick-rise cam that is not a small effect. Run backwards, it rises over the 170° flank and returns over the 90° one, so the fast stroke becomes the return. The offset of 5.88 that balanced the strokes when the cam ran forwards now puts its charge on that fast return, and the peak there climbs to 32.85° — past the 30° limit, where the same cam with no offset would have had 27.37° and the same offset running forwards had 21.81°. The slow stroke, meanwhile, falls to 8.80°, a margin nobody asked for. A cam that is jogged backwards for setting up, or that reverses as part of its cycle, should not be offset. Nor should it be designed with the direction of rotation left for somebody else to decide.

The figure above shows the geometry of it without any reversal: at the return’s peak the normal leans the other way, and it is the sign of that lean, not its size, that an offset exploits.

What a pressure angle stands in for #

The whole of this essay treats the pressure angle as the quantity to be managed, and it is a geometric stand-in for a force question it does not compute.

Whether a follower jams depends on the friction in its guide, the length of the guide’s bearing relative to the follower’s overhang, the load, and the spring. A 30° limit is a rule of thumb that bundles all of them into one angle, and a guide with long bearings and good lubrication tolerates more. The comparison between strokes is robust to that, because the same guide serves both. The absolute numbers are not, and a designer who has measured a guide’s friction should use that measurement rather than 30°.

Nor does anything here model dynamics. The side load on a dwell is computed as a direction, not as a force, and a follower moving fast has inertial loads that change which stroke is loaded at which instant. What a cam competes with at speed is a dynamics problem, and an offset chosen for a static load can be the wrong one at speed.

What the geometry does settle is the structure of the trade, and that part does not depend on any of the missing physics. The offset is subtracted from the velocity, the velocity changes sign between strokes, and so whatever one stroke gains the other pays for. Every refinement of the force model changes the price and none of them changes the fact that it is a price.

What comes next: a follower that swings #

The offset roller is the first departure here from the follower that travels through the cam’s centre, and two more follow from it directly.

The first is to take the pressure angle away altogether. A flat-faced follower is always pushed at right angles to its face, so its pressure angle is zero at every angle and no offset is needed. What binds instead is whether the cam is convex, which moves the constraint from the follower’s velocity to its acceleration.

The second is the next step, and it is the natural generalisation of this one: the oscillating roller follower, carried on an arm that swings about a pivot instead of sliding in a guide. Its line of travel is no longer a line but the arc of the roller’s centre about the pivot, so the offset of this essay becomes two numbers — the arm’s length and where the pivot sits — and the pressure angle depends on both. A swinging arm has an offset built into it that changes through the stroke. Whether the two strokes can be balanced by choosing the pivot, as they were here by choosing $e$, is a question with a definite answer, the same two-route check will price it, and it has not yet been computed.

Behind that is a question the scaling essay makes easy to ask: the balancing offset is a length, and a length scales with the cam. Whether the balancing offset divided by the prime circle is a property of the programme alone is the kind of shape-and-size claim that essay sorts, and it is the check that would turn this essay’s 5.88 into a number anybody could reuse.