Constant velocity is a mirror

A coupling that only translates settled the question for parallel shafts. Put a frame on each hub, take the output hub’s frame as seen from the input hub’s, and the rotation in that relative displacement is exactly the difference of the two shaft angles. So the output keeps step with the input at every instant precisely when the relative displacements contain no rotation — when they lie in a translation group. Oldham’s disc, two equal parallel cranks and any other arrangement whose hubs translate against each other couple at a ratio of exactly one, and a four-bar whose hubs’ relative motion closes into the whole planar group does not.

Shafts that meet at an angle are the case that matters on a vehicle, and the joint that is not constant velocity measured what a single universal joint does there: its output runs fast and slow twice a turn, by an amount fixed by the shaft angle alone. Two such joints on an intermediate shaft can cancel each other, if their angles are equal and the intermediate shaft’s two yokes lie in one plane. That essay stated the conditions. This one asks what kind of fact they are, using the pairs field’s instrument from twelve kinds of freedom, and the instrument’s answer is that the translation-group argument does not carry over and something else takes its place.

Every joint between meeting shafts lands in one group #

Each joint here is solved from its pins rather than from a formula. A universal joint is a cross whose two arms are square to each other; the input yoke holds one arm square to the input shaft and the output yoke holds the other square to the output shaft. So at each input angle the input arm is known, and the output arm is the line square to both the output shaft and the input arm: a cross product, normalised, with its sign kept continuous from the previous step. A double Cardan joint repeats that twice across an intermediate shaft. The output angle is then read off the output arm. For the single cross it agrees with Cardan’s formula to 9 × 10⁻¹⁶ over a turn, which is the cross-check and not the method.

The hub frames need an origin each, and any point on a hub’s own axis will do. When the input and output lines meet, put both origins at the meeting point. Then every relative displacement leaves that point where it is, so every one is a rotation about it, and the smallest group they can generate is inside the spherical group S of rotations about a point — the group whose orbit of any point is a sphere.

The first three rows are the case in point. A single cross at 30° spans all three dimensions of rotation and closes to S. A double Cardan joint in the W arrangement — input and output each 15° off the intermediate shaft and bent the same way, so that their lines meet — with its middle yokes in phase keeps its output in step to 1.1 × 10⁻¹⁵, and it also closes to S. With the middle yokes a quarter-turn apart it wanders by 3.97°, and it closes to S.

That is not an accident of which joints were picked. When the double joint keeps step, its relative displacement at input angle θ is $R_z(-\theta)\,R_y(\beta)\,R_z(\theta)$: a turn through the shaft angle β about an axis square to the input shaft, and the axis goes round with θ. Their logarithms span exactly the two rotation directions square to the input shaft, and the measurement gives two. Almost nothing is a group found that a random two-dimensional space of twists closes at its first bracket into something larger, and this one does: the bracket of two rotations about perpendicular axes is a rotation about the third, so the closure is all of S. Any group containing the constant-velocity joint’s motion therefore contains the universal joint’s. No subgroup of displacements can tell the two apart.

What stays the same is how far the relative displacement turns #

The group is the wrong object, but the parallel-shaft result was not wrong, so something must generalise it. The trace of a rotation is one plus twice the cosine of its angle, and the trace of $R_z(-\theta)\,R_y(\beta)\,R_z(\varphi)$ equals the trace of $R_y(\beta)\,R_z(\varphi - \theta)$, because a trace does not change when a product is cycled. That depends on the two shaft angles only through their difference. So the relative displacement turns through the same angle at every input exactly when φ − θ is constant — when the joint keeps step — and the angle it then turns through is β.

A constant-velocity joint is one whose hubs’ relative displacement is always a turn through the shaft angle. That is a condition on the size of a rotation, not on its axis. Every rotation through β about any axis through the point belongs to it, and the set of such rotations is not closed under composition: two turns through 30° compose to a turn of anything from 0° to 60°.

The parallel-shaft essay was the special case β = 0. A displacement that turns through nothing is a translation, and the translations happen to form a group, so there the condition on the size of the turn and the condition of lying in a group were the same condition. At any other shaft angle they come apart, and only the first survives.

The figure also shows why this is a poor instrument despite being the right invariant. The relative turn depends on the lag through a cosine, so it moves only to second order: the single cross lags and leads by 4.1°, and its relative turn changes by only 0.27°. A joint could hold that angle to a hundredth of a degree and still wander by about a degree. The invariant is exact, but it is too insensitive to measure a joint with.

The lag, measured from the pins #

The direct measurement is the lag itself, and it separates the joints cleanly.

Four facts come out of the curves, all of them known to anyone who has fitted a propshaft. The single cross at 30° wanders by 8.23° from peak to peak, twice a turn. Two crosses in phase at equal angles cancel exactly, in either arrangement. Turning the intermediate shaft’s second yoke a quarter out of line makes the two crosses add, and 3.97° is almost exactly twice what one cross at 15° does. With unequal angles, 15° and 10°, the joint is left with the difference of two crosses, 1.11°.

What the curves do not show is why the in-phase joint cancels, as opposed to that it does. Cardan’s formula for one cross followed by its own inverse for the other gives the answer, but only as an identity between trigonometric expressions. It says nothing about what the geometry has that the other arrangements lack. And averaging hides the failure entirely: every one of these joints makes exactly one output turn per input turn, which is the sense in which a ratio that is right on average is not a ratio at all.

A mirror that swaps the hubs #

The double joint in the W arrangement has a plane of symmetry. Reflect it in the plane that bisects the intermediate shaft, and the input shaft’s approach lands on the output shaft’s departure, the first cross lands where the second is, and the intermediate shaft lands on itself, end for end. The reflection is a symmetry of the mechanism: it carries the fixed geometry onto itself with the two hubs exchanged.

The single cross has one too.

Reflect a single universal joint in the plane bisecting its two shafts, and the input yoke goes to where the output yoke is, the cross’s two arms exchange, and the fixed geometry lands on itself. So both joints have a mirror, and the mirror alone cannot be what makes the double joint keep step. What differs is what the mirror does to a configuration.

A symmetry of a mechanism carries each of its configurations to a configuration — the same test composing two positions applied to displacements, applied here to whole configurations. Measured, the image of every configuration of every joint here lands on that joint’s own motion to 1.4 × 10⁻¹⁵. The question is which configuration it lands on.

For the in-phase double joints the answer is the same configuration, at every angle. Each configuration is its own mirror image. The output yoke’s arm is then the mirror image of the input yoke’s arm at every instant, the mirror carries the input hub’s mark onto the output hub’s, and the output is at the input’s angle. That is the whole mechanism of cancellation, stated without a formula: a joint keeps step exactly when every one of its configurations is its own image under the swap.

For the single cross, the mirror carries each configuration about a quarter of the input’s turn along the motion, give or take the lag, and never nearer than 86°. No configuration of a universal joint is its own mirror image.

All or none #

Those two outcomes are the only two, and the reason is short enough to state. Rotating the input carries a one-freedom joint round a closed loop of configurations. The swap maps that loop onto itself, and it keeps the direction round the loop, since the image of a joint turning forward is a joint turning forward. Done twice, the swap is nothing. A map of a circle with those three properties either moves no point or moves every point, and in the second case it is essentially a half-turn of the circle. There is no third option. If it fixed some points and not others, somewhere between a fixed point and a moved one it would have to run backwards.

So a symmetric joint is either constant-velocity or has no symmetric configuration at all. The single cross is the second case, and the rows in the table marked “none” are the half-turn type. The double joint in phase is the first. The test for which case a joint is in needs one configuration: find one position in which the joint is its own mirror image, and the whole turn follows.

This is what a phasing mark on a propshaft’s slip joint does, though it is never described that way. It lets a fitter check one configuration, the one where the two yokes of the intermediate shaft lie in a plane, and the argument above says that one configuration is enough. The equal-angle requirement is the other half. It is what makes the mirror a symmetry of the installed joint in the first place.

Twelve bars and a symmetry used a symmetry to make a framework move when the count said it could not. Here a symmetry makes a motion keep a ratio that the count, the group and the formula leave unexplained. A symmetric curve from a lopsided machine is the planar cousin: a four-bar with no symmetry in its lengths draws a curve that is its own mirror image, because a symmetry of the motion is not the same thing as a symmetry of the parts.

The Z arrangement is a point reflection #

The other common double joint has its input and output parallel and offset, bent opposite ways off the intermediate shaft, like the letter Z. It keeps step as exactly as the W, to 1.1 × 10⁻¹⁵, and it has no mirror plane: reflecting in the plane that bisects the intermediate shaft sends the input’s approach to the direction of the W’s output, not the Z’s.

The obvious candidate is the half-turn about the line through the middle of the intermediate shaft, square to the plane of the shafts. It does carry the fixed geometry onto itself with the hubs exchanged, and it moves every configuration by up to 90°. The reason is that a rotation reversing the intermediate shaft end for end also reverses the direction in which that shaft turns. The image of the shaft turning one way is the shaft turning the other way, and that is a different configuration.

What reverses a shaft end for end and keeps its sense of turning is a reflection of some kind: conjugating a turn about the shaft’s axis by an improper isometry that flips the axis gives back the same turn. For the Z arrangement that isometry is the point reflection through the middle of the intermediate shaft, which sends every point to the opposite side of that centre at the same distance. It fixes every configuration of the in-phase joint, and the table’s fourth row records it.

The essay this one continues had guessed a half-turn for the Z arrangement. That was the right element of the wrong kind. Both working arrangements are held by improper symmetries, a plane in one and a point in the other, and for the same reason: the intermediate shaft must be exchanged end for end without being made to turn backwards.

The two arrangements differ elsewhere as well, and the difference returns the argument to groups. The Z arrangement’s input and output are parallel, so its hub frames can be put at the feet of the common perpendicular. Its relative displacements then turn through nothing and translate round a circle. They span two dimensions and close to T2, the same group as Oldham’s coupling, so an in-phase double Cardan joint in the Z arrangement is, as a pair of hubs, an Oldham coupling. A quarter out of phase it closes to the planar group G, like the failing four-bars of the parallel-shaft essay. The same double joint, with the same ratio of exactly one, closes to T2 in one arrangement and to S in the other. The group records how the shafts are arranged. The ratio is decided by the symmetry.

Phasing: only three settings have a mirror #

The argument says more about the intermediate shaft’s yokes than the rule “in one plane” does.

Turn the second yoke through δ relative to the first, and the reflection in the bisecting plane is a symmetry of the joint only when the reflected second yoke lands on the first. A reflection that flips the shaft’s axis carries a yoke turned through δ to one turned through δ the other way round, so that needs a turn of 2δ to bring an arm back onto its own line. For an arm, which is a line and not a direction, that happens at δ = 0°, 90° and 180° and nowhere else. The measurement agrees. At 0° and 180° the mirror fixes every configuration and the spread is nought. At 90° the mirror is a symmetry and fixes none, and the spread is at its largest, 3.972°, which is 1.9997 times one cross at 15°. It falls short of exactly twice because the second cross adds its wobble at an angle the first has already moved. At every other setting the joint has no mirror, and its spread lies between the extremes.

So there are exactly two ways to phase a double Cardan joint that give it a symmetry. One gives the best possible behaviour and the other the worst, and every intermediate setting has no symmetry at all. The all-or-none argument says there could not be a symmetric setting that was only slightly wrong. A symmetric joint is exact or maximally wrong, and a joint that is only slightly wrong has lost its symmetry.

Unequal angles have no mirror at all #

The other condition is equal angles, and its failure is simpler.

With the first cross at 15° and the second at anything else, no isometry carries the input’s approach onto the output’s departure while keeping the intermediate shaft on itself. Such an isometry would have to preserve the angle each outer shaft makes with the intermediate one, and those angles differ. With no symmetry there is nothing to fix configurations, and the joint wanders. The pins and Cardan’s formula agree to 10⁻¹⁵ at every second angle. The spread is nought at 15° and nowhere else, and at 0°, where the second cross is straight, it is exactly one cross at 15°.

The curve is steeper on the far side of 15° than the near side. A second cross larger than the first overshoots by more than a smaller one undershoots, because a cross’s wobble grows faster than its angle. This is the reason a propshaft is specified by its angles at both ends and not by their average.

What the symmetry decides, and what it does not #

It decides the ratio exactly, and from one configuration. A joint whose fixed geometry has a hub-exchanging isometry, and one configuration that isometry fixes, keeps step at every angle and every speed. No formula is needed, and the argument does not care whether the joint is two crosses, a set of balls in grooves or anything else.

It explains the Rzeppa joint’s cage without solving it. A Rzeppa joint’s balls run in grooves in an inner and an outer race, and a cage between them holds every ball in the plane bisecting the two shafts. That plane is the mirror, and a ball sitting in it is a configuration that is its own mirror image. The cage enforces the symmetric configuration, and the all-or-none argument does the rest. The groove geometry that makes this possible is a spatial contact problem that is not solved here.

It says nothing about a joint that is nearly symmetric. Every real joint has clearance, and a double Cardan joint’s centring device holds its two angles equal only to within its own play. The argument is exact or silent: a joint that is almost symmetric has no symmetry, which is the position a parallelogram a micron wrong is in at its change point. The two sweeps above show what replaces it: near δ = 0 and near equal angles the spread grows in proportion to the error, so the loss is first order. Nothing in the symmetry argument predicts the slope.

It does not make the group useless. The group still says what the hubs do to each other, and that is a different fact from the ratio. In the Z arrangement the output hub’s frame never turns relative to the input’s and only its offset goes round, which is the relative motion of an Oldham coupling between parallel shafts. In the W arrangement it turns through the shaft angle about an axis that goes round. Both keep step. The ratio is decided by the symmetry, and the kind of relative motion by the group.

Still open: the slope of a nearly symmetric joint #

A joint with a small asymmetry — a phasing error of δ, or angles that differ by ε — loses its mirror entirely, and the sweeps above show its lag spread growing linearly from nought on either side. The distinct argument there would be that slope, derived from the symmetric joint rather than measured off the broken one. The perturbation that breaks the mirror splits into a part the mirror maps to itself and a part it reverses. Only the reversed part can move a configuration off its own image, so it alone should set the first-order lag. The measurement would be the slope of the spread against δ and against ε, predicted from that split and compared with the solved pins. The practical question is a Rzeppa cage with clearance, which lets each ball sit a little off the bisecting plane. How much lag does a given cage clearance allow, and is it set by the worst ball or by the average?