A disc whose roots are not sockets

A cycloidal drive has a ring of pins and a lobed disc inside it, carried round an eccentric, and the disc is cut as the envelope its pins leave as the eccentric turns. Every drive measured in these essays puts the eccentric at exactly the pin circle’s radius divided by the pin count — five on a ring of sixty with twelve pins. At that proportion the curve each pin’s centre traces in the disc’s frame is an epicycloid with a cusp at every root, and the disc’s roots are round sockets of the pin’s own radius.

A clearance leaves one contact cut that disc undersize and turned it until a pin touched. One pin touches each way, always the one with the best lever about the disc’s centre, and at the roots the socket made that pin the first both ways. It closed its gap whichever way the disc turned, at up to the whole pitch offset. That essay ended with the obvious doubt. Drives are also made with a smaller eccentric, which gives smooth roots and no sockets. Does a designer who does that to avoid sharp roots give something up that no count of contacts shows?

The measurement below sets the eccentric at $k$ times the pin circle’s radius over the pin count, for seven values of $k$ from a half to one, and measures each with the same instruments. The answer is the reverse of the worry. The sockets were never doing work. They are what is left of a disc its own pins have cut into.

What changes when the eccentric shrinks #

The curve a pin’s centre traces relative to the disc is a trochoid. Seen from the disc, the pin moves because the eccentric carries the disc past it and because the disc turns. At a root the two motions oppose each other, and when the eccentric times the pin count equals the pin circle’s radius they cancel exactly. The pin’s centre stops for an instant, relative to the disc, and goes straight back out: a cusp. With any smaller eccentric the disc’s turning wins, the pin’s centre never stops, and the curve turns back round a smooth bend. That is a curtate trochoid.

The disc is the material no pin reaches. Its boundary is the curve’s inner offset by the pin radius, with every point removed that some other stretch of the curve comes within a pin radius of. Every gap in this essay is measured against that boundary, computed from the definition to about a millionth of a unit, and the disc is turned about its own centre until the smallest gap reaches nought. Nothing about sockets or smooth roots is assumed. They come out of the definition.

The root, close up #

At $k = 1$ the root is the socket that essay described: a circle of the pin’s radius about the cusp, the disc cut back to it on both sides. Just below one, the cusp has become a narrow tip, and either side of it the flank bends back towards the disc more tightly than a pin’s radius. The flank’s offset then folds on itself, a pin coming back along that flank removes material its own offset had left, and the root is trimmed to a point. Below a threshold the flank’s tightest bend is wider than the pin and nothing is trimmed: the root is simply the smooth offset of a smooth curve.

The threshold is found two ways. A local test walks one root’s inner offset and asks how close it comes to the rest of the root’s curve; bisected on the eccentric, it puts the threshold at k = 0.974. The whole curve, searched for any point within a pin’s radius of those offsets, finds none at 0.964 and some at 0.984. Between 0.974 and one the roots are pointed; below 0.974 they are smooth.

The first test written for this was a different question: whether the curve bent more tightly than a pin’s radius at the root’s tip. It reported the threshold below a half, which the boundary contradicted at every eccentric. The tip bends away from the disc, so its own offset never folds. What trims the root is the flank either side of the tip, which bends towards the disc; the flank that trims the root finds its tightest bend in closed form, gets the same 0.974 from it, and extends the threshold to every drive. An earlier version of this paragraph put the trimming down to the two flanks meeting across a narrow root. Every point the offset test flags is too close to its own flank, not to the other one.

Every pin touches below the exact proportion #

The first measurement is the plainest and the most surprising. Over a pin pitch, sampled every two and a half degrees of the eccentric, the socketed disc touches nine or ten of its twelve pins. At every smaller eccentric measured — 0.95, 0.9, 0.8, 0.7, 0.6 and 0.5 — it touches all twelve, at every sample.

The difference is the undercut. A conjugate profile touches every generating element it is conjugate to, unless some other part of the generating motion has removed the place where it would touch. At $k = 1$ the cusps do that: near each root the pin’s own later motion takes away the flank an earlier position would have touched, and the two or three pins nearest the roots stand clear by between three and thirteen thousandths. The cutter takes back the tooth described the same event in a gear: a later part of the cut passing back through material an earlier part had left. Below 0.974 nothing is taken back and every pin keeps its contact. None of the seven eccentrics measured falls between 0.974 and one, so how the contact count behaves as the trimmed point turns into a socket is not measured.

What a second contact is for found that many contacts are why a cycloidal drive can drive where two conjugate rotors cannot: some contact always has a lever. The proportion every drive measured here has used turns out to be the one proportion that loses contacts.

The pitch point decides the arms #

Each touching pin’s arm — the rate at which its gap closes as the disc turns, which is its contact normal’s lever about the disc’s centre — is measured directly, by turning the disc a hundred-thousandth either way. It can also be predicted from nothing but where the pins are. Every contact normal of a conjugate pair passes through the pitch point, which sits on the line of centres at the pitch offset $(N - 1)e$ from the disc’s centre. So a pin’s arm is the pitch offset times the sine of the angle its direction from the pitch point makes with the line to the disc’s centre, and the best arm each way is the largest such sine of each sign — provided every pin touches.

Below the exact proportion, where every pin does touch, the prediction matches the measured best arms to $7 \times 10^{-4}$ of the pitch offset at every sample and every eccentric. At $k = 1$ it misses by 17%, and it misses in one direction: it predicts arms the disc does not have. The pins it counts on are the ones the undercut took away. The best-armed pins at a cusped drive are the ones nearest the roots, and those are the ones it loses.

Why every contact’s normal knows the pitch point #

The prediction rests on one fact, and it is worth saying why it holds, because the whole argument about sockets turns on it. The disc and the ring of pins are a conjugate pair: the second shape is not a choice once the first shape and the motion are fixed, and the motion here is fixed by the eccentric. At any instant the disc’s motion relative to the ring is a rotation about one point, its pole, and for this pair that pole is the pitch point, where the disc’s pitch circle rolls on the ring’s. A contact point can only slide along the contact, never into it, so its relative velocity is square to the common normal. A velocity produced by a rotation about the pole is square to the line from the pole. So the normal lies along that line: every contact normal passes through the pitch point.

That makes the arms a matter of angles. The disc turns about its own centre, and each contact resists or drives that turning with the distance of its normal from the centre — the same lever a stop on a four-bar holds with, measured from the pole. The normals all radiate from the pitch point, so the lever of each is the pitch offset times the sine of its angle there. Nothing about the disc’s shape enters except the question of which pins it actually touches.

Two identical rotors are conjugate in the same way and have one contact at a time, so they have one normal through their pitch point, and twice a lobe it points straight at a shaft and its lever is nought. A ring of pins has many normals through the same point, spread round it, and some are always near square. Where two shapes stop touching traced how each profile’s working arc hands one contact to the next; with twelve pins all touching, the hand-over is continuous and one pin is always near the best angle.

So the shares tell the same story. The exact disc’s worst best-arm is 0.819 of the pitch offset with sockets and at least 0.966 at every smaller eccentric. Cut 0.02 undersize and turned until a pin touches, one pin touches each way at every eccentric, as it did with sockets, and its arm never falls below 0.968 of the pitch offset below the exact proportion. With sockets it falls to 0.886. The proportion with the sockets is the worst of the seven, on the measure the sockets were supposed to help.

The 0.97 has a reason. With twelve pins all touching, their directions from the pitch point are spread round it, and one of them is always within a few degrees of square to the line of centres. Its sine is then close to one. Take away the pins near the roots and the nearest-to-square direction can be well off square.

What a smaller eccentric does cost #

The worry that prompted the measurement was about the wrong number. The share of the pitch offset hardly moves. The pitch offset itself moves with the eccentric: it is $(N - 1)e$, 55 at the exact proportion and 27.5 at half of it. The first contact’s arm, in length, falls with it: 48.7 at $k = 1$, 50.6 at 0.95, 42.7 at 0.8 and 26.6 at 0.5. A contact force produces a torque through that length, so a drive with half the eccentric needs about twice the contact force for the same output torque.

One detail runs the other way. At 0.95 the first contact’s least arm in length is larger than at $k = 1$, because the share has risen by more than the offset has fallen. A drive a twentieth smaller in eccentric has both a better worst-case lever and all twelve contacts.

The backlash shows the cost most directly. Backlash is an allowance treated free turn as something a designer chooses to spend, and a clearance is how it is spent here. The disc’s free turn, both ways together, is the clearance divided by the forward pin’s arm plus the clearance divided by the backward one’s. With the share fixed near 0.97, the arms scale with $k$ and the free turn with $1/k$. Measured, the least free turn per unit of clearance at every eccentric times $k$ runs from 0.0364 to 0.0367, a spread under 1%. At half the eccentric, a clearance buys twice the backlash.

The first-order law — free turn equals the clearance times the sum of the two best arms’ reciprocals — is exact once the sockets are gone. Below the exact proportion it matches the measured free turn to 0.2% at every sample. With sockets it was out by up to 9%, because a pin sliding across a socket’s round wall closes its gap faster than any flank can. The sockets did not make the drive tighter on average, only harder to predict: its free turn breathes by 12% over a pin pitch, against between 1% and 3% at every smaller eccentric.

Load sharing does not come back #

The other thing the earlier essay measured was how many pins are within reach of carrying if the contacts give under load. A disc touching all twelve pins at nominal size might be expected to share better once it is cut undersize.

It does not, much. With the contacts allowed to give by the whole clearance, six or seven pins are within reach with sockets and six at every smaller eccentric; with half the clearance, three to five with sockets and four or five without. The clearance spreads the other gaps from nought to more than itself whatever the proportion, and that spread, not the count of contacts at nominal size, sets how many pins a given give brings into play. The twelve contacts of the smooth-rooted disc are a property of the disc as drawn, as the nine or ten of the socketed one were.

What a designer should take from it #

The exact proportion is not a natural choice that happens to have sockets. It is the boundary of the family, where the pin’s centre comes to rest against the disc, and it is the only proportion at which the pins cut into their own roots enough to lose contacts. A designer choosing an eccentric has two things to weigh, and neither is the sockets.

The eccentric sets the lever. Every contact’s arm is a share of the pitch offset near 0.97, and the pitch offset is proportional to the eccentric, so the drive’s torque per unit of contact force, and its stiffness against backlash, grow in proportion to it. That argues for the largest eccentric the geometry allows.

The eccentric also sets the root. Above 0.974 the roots are trimmed to a point and at one they are cut into sockets, which costs contacts and makes the free turn breathe. Just below 0.974 the drive keeps nearly all of the pitch offset, all twelve contacts and a free turn that obeys the first-order law. On this twelve-pin, ring-of-sixty drive, the measurements point to about 0.95 of the classical eccentric — a judgement they support rather than a result they prove, since nothing here weighs the lobes’ strength or the cost of cutting them.

What the eccentric does not settle #

Pin count and pin radius. The threshold of 0.974 is for twelve pins of radius 5 on a ring of 60. A root’s width scales with the eccentric and the pin count, and the threshold moves with the pin radius. Other drives have their own, and none is measured here.

Load. How the load divides among the pins within reach is a statics problem with the contacts’ stiffness in it, and the count of pins in reach says nothing about the division.

The disc’s lobes. A smaller eccentric makes shallower lobes, which changes the disc’s own bending stiffness and the pin’s sliding speed along the flank. Neither is measured.

Manufacturing. A pointed root between 0.974 and one is a sharp internal corner in a hardened disc, which is a stress raiser. A socket is a round one. That may be the real reason drives are made at the exact proportion, and it is not a kinematic reason.

Still open: the eccentric that balances lever against root #

The measurement brackets a choice: the largest eccentric whose roots are not trimmed, 0.974 of the classical one on this drive, keeps nearly the whole pitch offset and all its contacts. That number came from one pin radius.

The distinct argument would be the threshold as a design curve: for each pin count and ratio of pin radius to pin circle, the largest eccentric at which the roots stay untrimmed, found the same two ways — the local flank test and the whole-curve search — and the first contact’s arm in length at that eccentric against its value at the exact proportion. If the threshold stays near one across the range drives are built at, the classical proportion is a hair from the best one and the sockets are a harmless habit. If it falls well below one for fat pins or few of them, a drive built at the classical proportion is giving away contacts to keep a lever it does not need, and the curve would say by how much.