A Rzeppa cage adds its clearances

Constant velocity is a mirror explained the Rzeppa joint without solving it. A shaft coupling keeps its output turning at exactly its input’s speed when its configuration is its own mirror image in the plane that bisects the two shafts, and the Rzeppa joint is that mirror built in steel. Each ball runs in a groove of the inner race, which turns with the input, and a groove of the outer race, which turns with the output. A cage holds all six balls in the bisecting plane. A ball in that plane is its own mirror image, so the joint is.

Two errors, one ellipse then asked what a small departure from the mirror costs a double Cardan joint, and found that only the part of an error the mirror reverses costs anything, and that two such parts add as the sides of a right triangle. It ended by asking the Rzeppa’s version of the question, which is different in kind. A real cage has clearance: its windows are a little wider than the balls, so each ball can sit a little off the bisecting plane. That is not two errors but six, each able to take up its clearance or not. Is the lag the joint then allows set by the worst ball, by the average, or by a root sum of squares?

It is set by none of those. The balls’ clearances add.

Where each ball has to be #

The model is the joint’s own geometry and nothing else. The input shaft runs along one axis, the output is turned from it by the bend, 20° here. Ball $k$’s inner groove lies in a plane through the input’s axis, at the input’s angle plus $k$ sixths of a turn; its outer groove lies in a plane through the output’s axis, at the output’s angle plus the same. The ball’s centre is on both, so it is on the line where the two planes meet, at the ball circle’s radius from the joint’s centre. Everything below is measured in that radius.

When the output’s angle equals the input’s, each pair of groove planes is a plane and its mirror image, and the line they meet on lies in the mirror plane. All six balls are then in the bisecting plane, to $10^{-12}$ at every input angle tried. The cage is a plane through the joint’s centre whose normal can tilt two ways, and with no clearance it can hold the balls only where the mirror puts them: the output cannot lag.

With a clearance $c$ each ball may sit up to $c$ off the cage’s plane — a short free link of the kind a clearance is a link described, one per window. The question is how far the output angle can then move from the input’s — how much lag — while some tilt of the cage keeps every ball within its clearance.

Six balls are three constraints #

For small clearances the question is linear. A lag of ε moves ball $k$ off the bisecting plane by $a_k\,\varepsilon$, and a tilt $t$ of the cage moves its plane past ball $k$ by $b_k \cdot t$. The constraints are

$$|a_k\,\varepsilon + b_k \cdot t| \le c \qquad (k = 1, \dots, 6).$$

The first thing the measurement shows is that opposite balls give the same constraint. Ball $k$ and ball $k + 3$ are half a turn apart, so a lag moves them off the plane equally and oppositely, and so does a tilt. Their two inequalities are one inequality with its sign changed. The six balls are three constraints, measured to agree pair by pair to $2 \times 10^{-8}$. A Rzeppa joint is overconstrained by design, the way three parallel bars between two frames are: its constraints repeat each other, and that repetition is what a clearance can be spent against.

Three constraints and two tilts leave one combination the cage cannot touch. There are three weights λ, fixed by the geometry, with $\lambda \cdot b = 0$: moving the cage cannot change $\sum_k \lambda_k (a_k\varepsilon + b_k\cdot t) = (\lambda\cdot a)\,\varepsilon$. Each term of that sum is at most $c|\lambda_k|$ in size, so the lag is bounded by

$$E = c\,\frac{\sum_k |\lambda_k|}{|\lambda \cdot a|},$$

and the bound is reached, since a linear program’s best is always at such a vertex. It is the same one-obstruction arithmetic that the slide five poses almost have found in five prescribed poses, read the other way round. There the obstruction was fixed and the question was how small a miss could share it. Here the misses are fixed at $c$ and the question is how large an obstruction they can absorb.

The allowances add #

Read term by term, the bound is a sum. Each pair of balls contributes $c|\lambda_k|/|\lambda\cdot a|$ of lag, and the joint’s free lag is the three contributions added. At a 20° bend, with the balls symmetric about the plane of the bend, they are 0.116, 0.117 and 0.117 of the clearance, and they add to 0.350. The linear program over every lag and tilt, solved directly with no knowledge of the pairs or the weights, gives 0.350 as well, to $10^{-9}$.

The alternatives are both wrong, and by a lot. The most sensitive ball, which a lag drives off the plane fastest, would allow $0.176\,c$ if it were the limit — half the answer. A root sum of squares of the balls’ sensitivities gives $0.203\,c$. What neither sees is the cage. It is not a fixed plane that the balls must stay near. It tilts, and it spends its two tilts on two of the three constraints, so that every pair’s clearance ends up contributing to the lag. The clearances add as clearances in series do: each one lets the next one move.

Two errors, one ellipse found errors adding as a root sum of squares in a double Cardan joint, because its two costs left their marks a quarter-period apart and the lag’s peak was the length of a vector. Here there is no vector sum. The cage makes the three clearances act together, in one direction, and the result is their plain sum.

At the limit every ball is at its window #

The sum has a physical reading, and the linear program’s solution shows it directly. At the largest lag the joint allows, every one of the six balls is pressed against an edge of its window: its distance from the cage’s plane is exactly the clearance, at the symmetric position and at every other angle tried. The sides alternate round the circle — one ball against the window’s leading edge, the next against its trailing edge, and so on — and the cage has tilted by about two clearances’ worth of angle, $2c$ radians per unit of the ball circle’s radius, about the axis across the bend, to put them all there.

That is what distinguishes the sum from the worst ball. A worst-ball reading imagines one ball reaching its edge while the others still have room. The cage does not allow that to be the end of the story: while any ball has room, the cage can tilt to spend it, and the lag grows until there is no room anywhere. The alternating pattern — each ball at the limit, the signs alternating — is the same one the slide five poses almost have found in the residuals of its best compromise, and it arises for the same reason: one obstruction, shared as widely as the constraints allow.

It also says what a loaded joint does. A torque that tries to wind the output relative to the input takes up the free lag, and at the end of it every ball is bearing on its window. The clearance is taken up everywhere at once, not ball by ball.

How much lag that is #

The numbers are small, which is why the question is rarely asked. A cage whose windows are a thousandth of the ball circle’s radius wider than the balls — three hundredths of a millimetre on a radius of thirty — lets a joint bent 20° lag or lead by up to 0.020°. A single Hooke joint at the same bend, which does not keep constant velocity at all, swings its output up to 1.78° ahead of and behind its input twice every turn: ninety times as much, and not free play but a fixed law of motion.

So a Rzeppa joint’s cage clearance buys a tiny, bounded free play in exchange for a joint that can be assembled and can turn under temperature changes. What the measurement adds is the shape of that play: largest when the balls straddle the plane of the bend, nought as each crosses it, and set by all six balls together.

The lag breathes six times a turn #

The free lag depends on where the balls are. It is largest, 0.350 c, when they sit symmetrically either side of the plane that contains both shafts — the plane of the bend. It falls to nought every sixty degrees, when a ball crosses that plane.

The reason is in the grooves. A ball in the plane of the bend has its inner groove and its outer groove in that same plane, since both shafts lie in it. The two planes coincide, and the line where they meet — the ball’s position — is undefined: the grooves alone would let the ball sit anywhere round its circle in that plane. Lag the output by the smallest amount, and the two planes are no longer the same; they now meet along a line that has swung far from where the ball was. That ball’s sensitivity to lag is enormous, and no cage tilt can accommodate it. At the crossing, that one ball holds the joint.

So a Rzeppa joint’s backlash from cage clearance is not a number but a curve. Backlash is an allowance treated free play as something a designer spends; here the same spend buys different amounts of play at different angles of the shaft. Averaged over a turn it is about two thirds of its peak — 0.228 c over a ball pitch — and six times a turn it vanishes.

Three balls and six hold alike #

The pairing explains what different ball counts do, and the measurement confirms it. A joint with an even number of balls pairs them, and each pair is one constraint. A joint with an odd number does not pair them, and each ball is its own. The cage’s two tilts always absorb two constraints, and the number of obstructions left over is what holds the lag.

Three balls are three constraints and leave one obstruction. Six balls are three pairs — and the three pairs are exactly the three balls of a three-ball joint turned half a turn — so they hold identically: 0.350 c at most in both. Four balls are two pairs, and the cage’s two tilts absorb both. Nothing is left, and a four-ball joint’s cage cannot hold its lag at first order at all: the linear program finds no bound. Five balls leave three obstructions and hold tighter, at 0.157 c; seven leave five, at 0.101 c; eight are four pairs and two obstructions, at 0.229 c.

That is a statement about the cage, not about torque capacity, and six balls are chosen for reasons this model does not see: load sharing, the space round the ball circle, the grooves’ depth. But it says that the sixth, fifth and fourth balls of a six-ball joint add nothing to how tightly its cage holds the lag. Their pairs duplicate the first three.

The bend makes the lag #

The free lag depends on the bend, and in a simple way. At the symmetric position it is 0.087 c at a bend of 5°, 0.175 c at 10°, 0.350 c at 20° and 0.706 c at 40° — the bend in radians to within half a per cent up to 20°, rising to 1.2% above it at 40°.

A straight joint allows no lag at all. With the shafts in line, each ball’s two groove planes are the same plane at equal angles, and a lag makes them meet along the shared axis, where no ball can go. The bend is what separates the planes, and the lag a clearance buys grows with it. A Rzeppa joint running nearly straight is held almost rigidly by its grooves; one at a steep bend relies on its cage.

Checked at finite clearance #

The rule is linear, and a cage’s clearance is small but not nought. The exact question — the largest lag for which some tilt of the cage keeps every ball within $c$, with the balls’ positions computed from the grooves at that lag rather than from their rates — can be answered by searching the tilt for the least worst ball and bisecting the lag until that least worst equals the clearance.

At a clearance of a hundredth of the ball circle’s radius the exact answer is 0.350096 c, against the rule’s 0.349980. At a thousandth it is 0.349981, and at a ten-thousandth 0.349980. The difference shrinks in proportion to the clearance, as a first-order rule’s should, and at any clearance a real cage has — a few hundredths of a millimetre on a radius of thirty — it is invisible.

What this says about the mirror #

Constant velocity is a mirror found that the mirror explains why a Rzeppa joint turns at constant velocity. The clearance measurement adds how firmly the mirror is held, and the answer is not the one the picture suggests. The picture is six balls, each pinned to one plane. The reality is three constraints and a plane that moves, and the three constraints are enough to hold the mirror only because the cage cannot tilt three ways. A coupling that only translates found the Oldham coupling’s constant velocity in a group; the Rzeppa’s is in a count, and the count comes out one.

The same count also says why the joint that is not constant velocity, the single Hooke joint, has no such tolerance to spend: it has no redundant constraints at all, and its lag is fixed by its angle, not left free within a clearance.

What the cage does not settle #

The cage’s own float. Everything here keeps the cage centred on the joint and lets it only tilt. A real cage also floats along the shafts by the clearance between its spherical faces and the races, which is a third freedom. That would absorb a third constraint and leave six balls, like four, with no obstruction at first order, unless the grooves’ offset — which real Rzeppa joints have — holds the cage axially. Neither is modelled.

Load. Which balls actually press on which side of their windows depends on the torque. The free lag is the range the geometry allows, not where a loaded joint sits in it.

Grooves that are not in planes. The model puts each groove in a plane through its shaft’s axis, which is the classical Rzeppa design. Joints with offset or skewed grooves steer their balls differently, and their sensitivities would differ.

Still open: a cage that floats #

The count above rests on the cage having two freedoms. A real cage has at least three, since it can slide along the shafts as well as tilt, and the designs that stop it doing so — grooves whose centres are offset either side of the joint’s centre, so that the balls are steered to the bisecting plane by the grooves as well as by the cage — are exactly the designs that make the Rzeppa joint work without a separate steering linkage.

The distinct argument would be that third freedom measured. Let the cage slide along the bisector as well as tilt, and ask what the free lag becomes: by the count, six balls would leave no obstruction and the lag would be unbounded at first order, which a working joint is not. Then add the groove offset as a parameter and measure the free lag against it. The prediction is that the offset turns each ball’s constraint into one that depends on the cage’s axial position, restoring an obstruction, and that the free lag falls as the offset grows. How large an offset is needed for a given clearance to hold the lag to a given angle would be a design curve no picture of a mirror supplies.