A floating cage frees three balls and not six

A Rzeppa joint keeps its output turning with its input because a cage holds six balls in the plane that bisects the two shafts, and constant velocity is a mirror showed that a configuration lying in that plane is its own mirror image. A Rzeppa cage adds its clearances then gave the cage’s windows a clearance $c$ and asked how far the output could lag or lead. At a 20° bend the answer was 0.350 $c$, in radians per unit of clearance measured in ball-circle radii. It came from a count. Opposite balls give one constraint between them, so six balls are three constraints; the cage tilts two ways and spends those tilts on two of the three; the one left over holds the lag, and the three pairs’ clearances add along it.

That essay held the cage on the joint’s centre. A real cage can also slide along its own normal, by however much its spherical faces allow, and the count it used has an obvious next line. Three constraints against three freedoms leave nothing over. A four-ball joint, with two pairs against two tilts, could not hold its lag at all, and a six-ball joint with a floating cage ought to be in the same position.

It is not, and the reason is the same symmetry that made opposite balls one constraint in the first place. A slide does not count as a freedom against a pair. It costs one.

A slide moves both balls of a pair one way #

The model is the earlier essay’s with one more number. The input shaft runs along one axis and the output is turned 20° from it. Ball $k$ lies where the plane of its inner groove, through the input’s axis, meets the plane of its outer groove, through the output’s axis, at the ball circle’s radius from the joint’s centre. The cage is a plane whose normal may tilt two ways, and now its distance from the centre is a third unknown $s$. Each ball’s distance from the cage must stay within $c$, the short free link of length $c$ that a clearance is a link put in every joint with play.

Balls $k$ and $k + 3$ are half a turn apart. A plane through the input’s axis at an angle and the plane at that angle plus half a turn are the same plane, and the same is true on the output side. So the two balls of a pair sit on one line through the joint’s centre, at opposite ends of it. A lag that moves one of them a distance $d$ off the bisecting plane moves the other $-d$, and so does any tilt of the cage. A slide moves the plane by $s$ past both, the same way. The pair’s two conditions are therefore

$$|d - s| \le c \quad\text{and}\quad |d + s| \le c,$$

which together say exactly $|d| \le c - |s|$. Whatever room a slide makes for one ball of the pair it takes from the other.

The figure above shows it at the largest lag the joint then allows. With the cage centred, all six balls are pressed against an edge of their windows, alternating sides, which is the state the earlier essay found at its limit. Slide the cage by 0.4 $c$ and one ball of each pair is still against its edge while the other has 0.8 $c$ of its window spare, room the lag cannot use because using it would push the partner further. The lag falls from 0.350 $c$ to 0.210 $c$. Measured at ten forced slides from nought to 0.9 $c$, it is 0.350 $(1 - s/c)$ every time, to $10^{-7}$.

So the best slide is nought, and a linear program allowed to choose tilt and slide together finds exactly that: a six-ball joint with a floating cage holds its lag at 0.350 $c$, the same number to nine figures as the centred cage. An eight-ball joint, four pairs, keeps its 0.229 $c$ the same way. The count’s next line was wrong because the slide was never a freedom against the pairs. It is the same kind of miscount, made in the other direction, that the mechanism Grübler says cannot move found in three parallel bars: a count of constraints against freedoms is right only when every freedom can act on every constraint independently, and here one of them cannot. In the language of two errors, one ellipse, the half-turn that pairs the balls sends a lag and a tilt to their own negatives and leaves a slide where it is. Only the part of a motion that the symmetry reverses can be spent against a pair of constraints, and a slide has none.

Three ball centres always lie in a plane #

An odd count has no pairs. No ball of a five-ball joint has a partner half a turn away, so the half-turn argument says nothing about it, and a slide is a genuine third freedom against its five constraints.

With the cage centred, a five-ball joint at a 20° bend holds its lag to 0.157 $c$ at the symmetric position. Slide its cage by 0.6 $c$, the right way, and the lag the windows allow rises to 0.207 $c$. Three balls stay pressed against the edges of their windows and the other two move into the room the slide makes. The slide is buying lag here, the opposite of what it did to six balls.

The three-ball joint is the extreme case and the clearest. Three ball centres always lie in some plane. A cage that may tilt two ways and slide is a plane that can be put anywhere through the joint, so it can be put through the three balls exactly, whatever the lag. With nothing to stop the slide, a three-ball joint with a floating cage has no bound on its lag at all, and the linear program duly reports that its lag is unbounded.

A cage cannot slide without limit. Its spherical faces fit the races with some clearance of their own, and the slide is bounded by that. Allowing the cage to slide up to σ window clearances either way gives the figure above. Three balls hold to 0.350 $(1 + σ)$, a straight line with no ceiling: the lag a three-ball joint allows is set by how far its cage can float as much as by its windows. Five balls loosen from 0.157 $c$ to 0.228 $c$ and seven from 0.101 $c$ to 0.153 $c$, both reaching their final value once the cage may slide about 0.8 $c$ to 0.85 $c$; beyond that, the best slide has been found and more float is not used. Six and eight balls are flat lines from the left edge of the figure to the right.

The earlier essay found that “three balls and six hold alike”, both at 0.350 $c$, because a six-ball joint’s three pairs behave exactly like a three-ball joint’s three balls. That is true of a cage held on the joint’s centre and of nothing else. Let the cage float and the two joints part completely: the six-ball joint does not move, and the three-ball joint has no bound. The difference between them is the half-turn. Three balls have no opposites, and so nothing to cancel a slide.

Grooves that steer #

The earlier essay’s grooves were classical: each groove’s centre-line a circle about the joint’s centre, so that every ball sat on one sphere and the grooves alone said nothing about the lag. The cage did all the steering. Many Rzeppa joints do not work that way. Their inner race’s grooves are circles about a point a little way along the input’s axis, and their outer race’s about a point the same distance the other way along the output’s axis, so that the two grooves a ball runs in diverge and steer the ball into the bisecting plane without help.

The model is direct. Put the inner track’s centre at $f$ along the input axis and the outer track’s at $f$ the other way along the output axis, both of radius $\rho = \sqrt{1 + f^2}$, lengths in ball-circle radii, and allow a radial play δ between each ball and each track. A ball lies on its line through the joint’s centre at some radius $r$, and its distances from the two track centres differ, to first order, by

$$\frac{2\,r f \cos(\beta/2)}{\rho}\,(\ell \cdot n),$$

where $\ell \cdot n$ is its distance off the bisecting plane and β the bend. With no play the two distances must both be ρ, so the ball must sit in the plane: the grooves steer. With play δ the distances may differ by $2\delta$, so each ball is held to $|\ell\cdot n| \le \delta\rho/(r f\cos(\beta/2))$. A lag ε puts ball $k$ a distance $a_k\varepsilon$ off the plane, with the same $a_k$ the earlier essay used, so the grooves allow each ball a lag of $\delta\rho / (r_k f \cos(\beta/2)\,|a_k|)$.

The grooves hold through their worst ball #

Nothing in the grooves’ condition involves the cage, and nothing in one ball’s condition involves another’s. There is no shared freedom for a groove limit to spend, so there is nothing to add: the joint’s groove limit is simply its worst ball’s. At a 20° bend and an offset of 0.1, with the play equal to the window clearance, balls 2 and 3 allow 1.766 $c$, balls 5 and 6 allow 1.820 $c$, and balls 1 and 4, which a lag moves off the plane only half as fast, allow 3.544 $c$. The joint’s grooves hold the lag at 1.766 $c$.

This is the contrast the earlier essay drew between a worst-ball reading and a sum, with the two halves of it now belonging to two different parts of the joint. The cage’s limit is a sum, because the cage tilts and spends its freedom so that every ball’s clearance contributes. The grooves’ limit is a worst ball, because the grooves have no freedom of their own: each one holds its ball and nothing else. A designer reading a Rzeppa joint’s free play has to know which of the two is binding before either reading can be used.

The offset changes the cage’s limit as well, a little. A ball’s radius is set by its inner track: $r = f p + \sqrt{1 + f^2p^2}$, with $p$ its direction against the input axis. Opposite balls have opposite $p$, so they sit at radii $1 \pm f p$ to first order, 1.015 and 0.985 for the pair carrying balls 2 and 5 at an offset of 0.1. A pair whose two balls are at different radii is no longer exactly symmetric under the half-turn, and a slide can now even out one pair’s two radii — but only one pair’s, since the three pairs want different slides. The cage’s limit falls from 0.3500 $c$ to 0.3482 $c$ at an offset of 0.05, 0.3465 $c$ at 0.1, 0.3430 $c$ at 0.2 and 0.3330 $c$ at 0.5: a slope of about 3.5% per unit of offset, small enough that the cage is still, for every practical offset, the joint’s tighter limit.

The joint takes the lesser of the two. A linear program over every freedom at once — the lag, two tilts, the slide and all six ball radii, with every constraint differenced from the exact geometry and no knowledge of either rule — gives the lesser of the cage’s limit and the grooves’ at every offset tried, to better than one part in a million. The grooves and the cage do not interact at first order, because a ball’s freedom to move radially enters the cage’s condition only multiplied by its distance off the cage’s plane, which is itself first order. On the logarithmic axes above the grooves’ limit is a straight line of slope minus one, and the joint’s limit follows the cage’s nearly flat line until the two cross, at an offset of 0.548.

Tight tracks take over near the crossings #

The symmetric position is where the cage is loosest. Over a ball pitch both limits fall to nought where a ball crosses the plane of the bend, since there that ball’s two groove planes coincide and the smallest lag swings the line they meet on. But they do not fall in step.

At the symmetric position the grooves’ limit is 5.10 times the cage’s; near the crossings it is 3.40 times. With tracks as loose as the windows the grooves therefore never bind at an offset of 0.1, anywhere in the pitch: the cage holds the joint throughout. Tighten the tracks to a quarter of the windows’ clearance and the grooves’ limit, now 0.441 $c$ at the symmetric position, still sits above the cage’s 0.346 $c$ there, but it crosses below the cage’s limit on the way to each crossing. From 11° to 49° of input the grooves hold the joint and the cage holds it elsewhere. Which part of a Rzeppa joint sets its free play depends on where in its turn it is, as well as on how its parts are toleranced. Backlash is an allowance treated free play as something a designer spends; here two different allowances are spent at once, and the joint spends whichever is smaller at that angle.

How much offset lets the grooves hold #

A design curve falls straight out of the two limits. For each ratio of track play to window clearance, the offset at which the grooves first become the tighter limit somewhere in a ball pitch, and the offset at which they are tighter everywhere, can be found by bisection on the exact first-order limits.

Both are almost exactly proportional to the play: the lower offset is 0.034 at a play of a tenth, 0.102 at three tenths and 0.351 at equal play, and the upper is 0.049, 0.149 and 0.548 at the same plays. The proportion is 0.351 to 0.340 for the lower curve and 0.548 to 0.495 for the upper, falling slightly as the play shrinks, because the cage’s own limit falls slowly with the offset. Below the lower curve the cage holds the lag at every input angle and the grooves’ steering adds nothing to how tightly it is held. Above the upper curve the cage could be removed, as far as the lag is concerned, and the grooves would hold it.

The curves hardly depend on the bend. With tracks as loose as the windows the lower offset is 0.351 at a 5° bend, 0.350 at 10° and 15°, 0.351 at 20°, 0.356 at 30° and 0.367 at 40°. Both limits grow in proportion to the bend at small bends, as the cage’s limit was found to, and the bend cancels from their ratio.

So for a six-ball joint the reading is plain. The offset’s job is steering: it is what lets the balls find the bisecting plane without a separate linkage to steer the cage. It is not, unless the tracks fit much more tightly than the windows, what holds the lag. A joint with an offset of a tenth of its ball circle’s radius has its free play set by its cage unless its tracks are under about 0.3 times as loose as its windows.

The three-ball joint relies on its grooves #

For a three-ball joint the reading reverses. Its floating cage, as found above, holds nothing, so its grooves hold everything, through their worst ball. At an offset of 0.1 and tracks as loose as the windows that is 1.766 $c$ at the symmetric position, the same number as the six-ball joint’s groove limit, because three balls are exactly the first three of six. But for six balls that number never binds, since the cage holds them at 0.346 $c$. For three it is the answer, and it is five times as loose.

This is the cleanest statement of what the half-turn buys. A constraint that has been said already found that a repeated constraint is a dependency a clearance can be spent against; the half-turn is what makes a six-ball joint’s constraints repeat, and it is also what makes the slide’s freedom useless against them. A joint with opposite balls has a cage that holds its lag whether or not the cage floats, so its free play is set by its windows. A joint without opposite balls has a cage that holds its lag only if it is held centred, and if it floats, the free play is set by the grooves’ offset and play, through its worst ball.

Checked against the exact geometry #

The rules are first order, and the clearances are small but not nought. The exact problem can be solved directly: at each trial lag, place every ball on its line at the radius that best centres it between its two tracks, refuse the lag if any ball’s misfit exceeds δ, search the cage’s tilt for the least worst window distance with the slide set to the midpoint of the extremes, and bisect the lag.

For six balls the exact answer is 0.34658 $c$ at a clearance of a hundredth of the ball circle’s radius and 0.34647 $c$ at a thousandth and a ten-thousandth, against the rule’s 0.34647. For three balls, held by their grooves, it is 1.8012, 1.7685 and 1.7660 against the rule’s 1.7658. Both differences shrink in proportion to the clearance, as a first-order rule’s should. The three-ball joint’s difference is the larger, about 2% at a hundredth, and it too falls tenfold with each tenfold step in the clearance; at the clearances a real joint has, both rules hold to the figures quoted.

What a working joint adds #

The cage’s own fit. The float allowance σ stands for the clearance between the cage’s spherical faces and the races. It decides a three-ball joint’s free play and does not touch a six-ball joint’s. Its own value, and whether it is larger or smaller than the windows’ clearance, is a manufacturing fact this model does not supply.

Track shape. Each track is modelled as a circle, with a radial play of one number. Production tracks are often gothic arches, touching each ball at two points at a contact angle, and their play is not one number but depends on direction. The worst-ball reading should survive, since it comes from the grooves having no freedom to share, but the size of each ball’s allowance will change.

Load. The free lag is the range of output angles the geometry permits. Which balls bear, and in what order, when a torque takes that range up is a different question, and it is where the two limits differ most.

Still open: the balls a torque loads first #

When the cage binds, the earlier essay found every ball pressed against its window at the limit, so a torque taking up the free play meets all six balls at once. When the grooves bind, only the worst ball’s track is at its edge, and a torque would meet that one ball, or one pair, first. Between the two lies the shaded band of the pitch figure, where the binding limit changes with the input’s angle.

The distinct argument would be the load sharing that follows. With some compliance in each ball’s contacts, a torque applied after the free play is taken up would load the balls unequally, and the inequality should be large where the grooves bind and small where the cage does. The measurement would apply a torque at several input angles to joints on either side of the lower design curve, and record each ball’s share. If the shares follow the limits, a designer choosing tighter tracks to reduce free play would be buying uneven loading with it, and the offset curve here would be a curve of load sharing as much as of play.