The flank that trims the root

A cycloidal drive has a ring of pins and a lobed disc inside it, carried round an eccentric, and the disc is the envelope its pins leave as the eccentric turns. A disc whose roots are not sockets moved that eccentric below the classical proportion, the pin circle’s radius over the pin count, and found that the classical proportion was the worst place to be. At it the pins cut into the disc’s roots, leave round sockets, and two or three of the twelve pins stand clear of the disc they generated. Below 0.974 of it, every root is the smooth offset of a smooth curve and every pin touches.

That number belonged to one drive, twelve pins of radius 5 on a ring of 60. The essay ended by asking whether it stays near one across the drives people build. If it does, the classical proportion is a hair from the best and its sockets are harmless. If it falls well below one for fat pins or many of them, a drive built the classical way is giving away contacts.

To answer that for every drive, the trimming has to be understood rather than bisected, and it turns out to be simpler than that essay said. It is a bend, and the bend has a closed form.

The tip bends away, the flank bends back #

Seen from the disc, each pin’s centre traces a curtate trochoid. Near a root the trochoid runs in towards the disc’s centre, turns round a tip that points straight at it, and runs out again. The disc’s edge there is that curve offset towards the disc by the pin’s radius. Where the offset stays a simple curve, the pin touches the disc along it. Where the offset folds over itself, the pin coming back takes away material its own offset left, and the root is trimmed.

The tip cannot be what folds. It bends away from the disc, like the outside of a bend in a road seen from the verge, and a curve offset towards its convex side spreads out rather than folding. A test of the tip’s own curvature, asked when it becomes tighter than the pin, puts the twelve-pin threshold at 0.42 of the classical eccentric. The offset test contradicts that at every eccentric between.

Either side of the tip the flank changes direction of bend. As the path runs back out from the root it curves towards the disc, concave on the disc’s side, and the offset towards a concave side does fold once the bend is tighter than the offset distance. The hero figure shows it on a twelve-pin drive. At 0.97 of the classical eccentric the flank’s tightest bend has a radius of 5.36, the pin’s radius measured along the same normal stops short of the centre of that bend, and the offset stays simple. Move the slider past 0.974 and the pin’s radius reaches past the centre of curvature, the offset crosses itself in a small swallowtail, and the root is trimmed.

The offset test finds exactly this. On the twelve-pin drive at 0.98 of the classical eccentric, every point of the offset that lies within a pin’s radius of the path belongs to the same flank as the point it is too close to: 25,106 such pairs, and none across the root. So the trimming is not the two flanks of a narrow root approaching each other, as the earlier essay put it. It is one flank bending too sharply for the pin.

The bend in closed form #

A tight bend is a curvature, and a curvature can be written down. With the pin circle’s radius as the unit, the eccentric $e = k/N$ and $L = N - 1$, the pin’s centre in the disc’s frame is the complex curve

$$z(\psi) = e^{i\psi/L} - e\,e^{i\psi N/L},$$

where $\psi$ is the eccentric’s angle from the root. Differentiating twice and taking the usual ratio gives its curvature as

$$\kappa = \frac{1 + k^2 N - k(N+1)\cos\psi}{(1 - 2k\cos\psi + k^2)^{3/2}}.$$

The numerator changes sign where the curve changes its direction of bend. Near the root it is negative, the tip bending away. Further out it is positive, the flank bending towards the disc. Substituting $u = 1 - 2k\cos\psi + k^2$, $a = (N-1)(1-k^2)$ and $b = N + 1$, the flank’s radius of curvature is $2u^{3/2}/(bu - a)$ wherever $u > a/b$. That is least at $u = 3a/b$, where

$$\rho_{\min} = \frac{3\sqrt{3}\,\sqrt{(N-1)(1-k^2)}}{(N+1)^{3/2}}.$$

The formula says three things at a glance. The bend’s radius falls to nought as $k$ approaches one, as the square root of $1 - k^2$. That is why every pin radius trims the classical disc, and why the classical proportion is the boundary of the family rather than a member of it. The radius shrinks as the pin count grows, roughly as $1/N$. And the pin’s own radius does not appear, because the path does not depend on it.

The formula was checked against the curve itself rather than against its own algebra. For each pin count and eccentric, the path was sampled, its curvature taken by second differences, and the least concave radius found by a search along the flank. The two agree to $9 \times 10^{-5}$ of the radius, at thirty-six combinations of pin count and eccentric. The dashed line in the figure is a pin of radius 5 on a ring of 60; each curve crosses it at that drive’s threshold.

The threshold for every drive #

The root is untrimmed while the pin’s radius is below the flank’s tightest bend. Setting $r/R = \rho$ equal to $\rho_{\min}$ and solving for the eccentric gives the largest eccentric, as a share of the classical one, at which the roots stay whole:

$$k^* = \sqrt{1 - \frac{\rho^2\,(N+1)^3}{27\,(N-1)}}.$$

For twelve pins of radius 5 on a ring of 60 this is 0.9740, the number the earlier essay found by bisecting its offset test. That was a check on one drive. The claim here is for every drive, so it has to be checked the same way on others.

The offset test is the definition. It walks a root’s offset and asks whether any of it comes within a pin’s radius of the path, and it knows nothing of curvature. Bisected on the eccentric for thirty-three drives — five pin counts, pin radii from 0.02 to 0.15 of the pin circle, wherever the pins fit on the ring — it agrees with the formula to $2 \times 10^{-4}$ everywhere. Its answers sit a hair above the line, because a path sampled at nine hundred points misses close approaches by a little, never below it.

So the question the earlier essay left can be answered as a curve rather than a guess. For the thin pins of a small drive the threshold hugs one. Six pins of 0.02 of the pin circle stay whole up to 0.9995 of the classical eccentric. For fat pins or many of them it does not. Twelve pins at 0.15 of the pin circle trim above 0.913. Twenty at 0.12 trim above 0.860, and thirty at 0.1 above 0.787, more than a fifth below the classical proportion. Because $(N+1)^3/(N-1)$ grows as the square of the pin count, doubling the pins with the same pin radius roughly quadruples how far below one the threshold falls. The same fraction of pin circle that is harmless on a coarse drive trims deeply on a fine one.

What sets the threshold is how crowded the ring is #

The formula has a simpler reading than its fractions suggest. A pin’s radius cannot be anything: pins of radius $\rho$ on a ring of $N$ touch their neighbours once $\rho$ reaches $\sin(\pi/N)$, roughly $\pi/N$. So write the pin’s radius as a share $f$ of that limit, $\rho = f\sin(\pi/N)$, which says how crowded the ring is. Then $\rho^2(N+1)^3/(27(N-1))$ tends to $\pi^2 f^2/27$ as the pin count grows, and the threshold tends to

$$k^* \to \sqrt{1 - \frac{\pi^2 f^2}{27}}.$$

The pin count has dropped out. A ring whose pins are half-way to touching trims above about 0.95 of the classical eccentric whatever its pin count: 0.917 with six pins, 0.936 with twelve, 0.947 with thirty, closing on 0.953. Pins nine-tenths of the way to touching, the crowded rings of compact reducers, trim above 0.70 to 0.81 and close on 0.84 as the count grows. The twelve-pin drive of the earlier essays has pins a third of the way to touching, which is why its threshold sat so near one.

So the earlier question — does the threshold stay near one across the drives people build? — has an answer that needs no table. It stays near one for sparse rings and falls for crowded ones. Adding pins with the same pin radius makes the ring more crowded, which is why the table’s fine drives trimmed so far below one. A designer who keeps the ring equally crowded while adding pins keeps the same threshold.

Near one the formula also has a rule a designer can carry. For small pins, $1 - k^* \approx \rho^2(N+1)^3/(54(N-1))$, so the margin below the classical eccentric grows as the square of the pin’s radius. Halving the pins’ radius quarters it. That is the curvature argument again: the flank’s tightest bend shrinks as the square root of $1 - k$, and the pin’s radius has to fit inside it.

What the classical proportion gives away #

A trimmed root is a shape, and the essays on this drive have been about what shape costs: contacts that carry load and levers that turn force into torque. The earlier essay counted both at one drive. The formula now places any drive on the curve, so the counting can be repeated where the threshold is far from one.

Twenty pins of radius 6 on a ring of 60 have a threshold of 0.905. At the classical eccentric their disc touches ten to twelve of its twenty pins over a pin pitch. Nearly half stand clear, where the twelve-pin drive lost only two or three. At 0.878 of the classical eccentric, just below the threshold, every one of the twenty touches at every angle sampled. What a second contact is for found that many contacts are what let a pin drive work where two conjugate rotors cannot, because some contact always has a lever. A proportion that throws away half the contacts is throwing away that insurance.

It throws away lever too, which is the surprise. A smaller eccentric shortens the pitch offset, the distance from the disc’s centre to the pitch point, $(N-1)e$, and every contact’s lever is a share of it. The earlier essay found the share nearly fixed below the threshold, so it expected a smaller eccentric to cost lever in proportion. Across the threshold, though, the share jumps. The contacts the trimming removes are the ones nearest square to the pitch point’s line, because they sit nearest the roots. So the worst lever the classical disc offers over a pin pitch is 35.3. The disc at 0.878 of the eccentric, with a pitch offset an eighth shorter, offers 49.5.

The table repeats that on four drives spread across the curve. Every one keeps all its pins at 0.97 of its threshold and loses some at the classical proportion: six pins of radius 7.2 drop to four at worst, twelve of radius 5 to nine, thirty of radius 3.6 to seventeen. And every one has a longer worst lever just below its threshold than at the classical eccentric. The gain runs from 42.2 against 35.4 on six pins to 52.0 against 42.6 on thirty. The levers are measured directly, by turning the exact disc a hundred-thousandth either way and reading how fast each touching pin’s gap closes, so they include whatever the trimming did to the contacts that remain.

A drive built at the classical proportion is therefore not trading contacts for lever, which is what the earlier essay’s question supposed. It gives up both, and the more pins it has the more it gives up. The lever argument for the largest eccentric the geometry allows is still sound, but the largest the geometry allows is the threshold, not the classical proportion.

The same rule as a roller on a cam #

None of this is peculiar to cycloidal drives. A roller follower is a pin that rides a cam, and the cam’s profile is the roller’s path offset by its radius. A curvature is a size with a minus sign stated the cam designer’s rule: a roller cannot follow a stretch of profile that bends towards it more tightly than its own radius. Where it would have to, the profile undercuts, and the cam that gets cut is not the cam that was drawn. The law that costs least priced that margin for motion laws.

A cycloidal disc is a cam with many rollers, all at once, and its profile is fixed by the drive’s motion rather than chosen. The flank’s tightest bend is the tightest concave stretch of that cam’s pitch curve, and the threshold is the roller rule solved for the eccentric. The earlier essay’s picture of two flanks approaching each other across a narrow root is not what happens on this curve. Here the trim is always the roller rule, a single flank against the pin that rolls along it. It is the same event the cutter takes back the tooth described in a gear, a later part of a generating motion passing back through material an earlier part left, and the curvature says exactly when it begins.

Where the tightest bend is #

The formula also says where on the flank the trouble is. At $u = 3a/b$ the eccentric’s angle from the root is

$$\cos\psi^* = \frac{1 + k^2 - 3(N-1)(1-k^2)/(N+1)}{2k},$$

22.6° of the eccentric from the root on the twelve-pin drive at 0.97. In the disc’s own frame that is about two degrees, since the disc turns once for eleven turns of the eccentric. The trim happens close beside each root, on the part of the flank where the pin is arriving at or leaving the root. That is also where the pins that stand clear at the classical proportion are: the two or three nearest the roots, which were the best-levered pins in the drive.

As the eccentric falls from one, $\psi^*$ moves out along the flank as the square root of $1 - k$. So a drive a little above its threshold has a shallow trim close to each root, and one far above it is trimmed along a longer stretch. The earlier essay’s four close-ups of a root showed the first kind. The twenty-pin drive at the classical proportion is the second.

What the curve does not settle #

Strength and wear. The threshold is the largest eccentric that keeps every contact and every lever, and nothing here weighs what a smaller eccentric does to the lobes’ bending stiffness, the pins’ contact stress, or how fast a pin slides along a flank. The earlier essay’s case for about 0.95 of the classical eccentric on its drive now has a general form, just below the threshold. Whether just below is where a designer stops is a question about materials.

Pins that can turn. Real drives often put a sleeve on each pin, or make the pins rollers, so that a pin rolls rather than slides. That changes the friction and none of the geometry; the threshold is the same.

Clearance. Every disc here is exact. A disc cut undersize for a running clearance is a slightly different offset of the same path. The flank’s bend then competes with the pin’s radius plus the clearance, which lowers the threshold by a hair, and the first contact under clearance was measured only on the twelve-pin drive.

Still open: a disc cut to the threshold on purpose #

The curve says where the trimming starts and that a drive just below it keeps everything. A drive built exactly at its threshold has one flank point per root where the bend equals the pin’s radius. There the offset is on the edge of folding, and the contact there has a curvature match of the kind a flat holds the wheel to second order found in a lock.

The distinct argument would be the contact at that point. The pin and the flank there have equal curvatures, so the pin’s gap to the disc near that contact grows as the fourth power of the distance along the flank rather than the square. That would make it the most conformal contact the drive can have, and the least stressed. The measurement would take a drive at its threshold, compute the gap profile near each contact and the contact’s width under a stated load by Hertzian estimate, and ask whether cutting a disc exactly at the threshold buys a contact stress no other eccentric can. If it does, the threshold is not only the limit of safe proportions but the proportion to choose.