A coupling that only translates

Two shafts that should turn together are never quite in line. A motor and a pump are bolted to one bed and their axes are a fraction of a millimetre apart; a roller and its drive sit on frames that move with temperature. A coupling between them has one job — turn the output at the input’s speed — and has to do it across an offset nobody chose.

Twelve kinds of freedom established the pairs field’s instrument: any set of displacements a mechanism produces can be logged, spanned, closed under the bracket and named as one of twelve connected groups. A chain multiplies applied it to open chains, whose joints compose. A coupling is a closed loop — ground, input hub, whatever is between, output hub, ground — and the instrument has something sharp to say about it, because the question a coupling answers is a question about one component of a displacement.

The ratio is one component of a relative displacement #

Put a frame on each hub. As the input turns through θ and the output through φ, the output hub’s frame seen from the input hub’s frame is a displacement

$$D(\theta) = R(-\theta)\; T(e)\; R(\varphi),$$

a rotation back through the input angle, the fixed offset $e$ between the axes, and the output’s rotation. Its rotation part is $\varphi - \theta$ and nothing else. So the output turns at exactly the input’s speed at every instant precisely when that rotation part is constant — and for a coupling that starts in step, when it is nought throughout.

That turns a question about a ratio into a question about a set. Over a turn the displacements $D(\theta)$ trace out a curve in the group of planar motions, and the ratio is one exactly when that curve lies among the displacements with no rotation. The connected groups of planar displacements with no rotation in them are two: translations along a line, $T$, and all translations of the plane, $T_2$. A coupling has a constant velocity ratio exactly when its hubs’ relative motion lies in a translation group.

Why Oldham’s coupling is one #

Oldham’s coupling puts a disc between the hubs. The disc slides in a slot cut across the input hub’s face, and a tongue across the disc’s other face slides in a slot across the output hub. Both joints are prismatic pairs, and a prismatic pair permits the group $T$ along its own direction and nothing else.

The hubs’ relative displacement is therefore a product: a translation along the first slide times a translation along the second. A product of translations is a translation, and two translations in different directions generate the whole of $T_2$. The disc can take up any offset in the plane, and neither joint can change an orientation, so the output hub has the input hub’s orientation at every angle.

Written as a computation, that is the pairs instrument run on the coupling itself. At each half degree of input the disc’s two slide positions are solved from the offset, the output hub’s displacement relative to the input hub is formed, and its logarithm taken. The 720 logarithms span a two-dimensional space of twists; bracketing its basis adds nothing; the classifier names it $T_2$; and the composite of any two of the displacements lies inside the span to $2 \times 10^{-16}$. The output’s angle stays on the input’s with a spread of exactly nought.

The slides are not the reason #

The argument never used the slides except to say their product was a translation group. Anything else whose hubs translate relative to one another should couple as well.

Two equal cranks, one on each hub, joined by a link exactly as long as the offset, put the four pins on a parallelogram. A parallelogram’s link never turns — a parallelogram’s coupler translates — and the two cranks stay parallel, so the hubs’ relative motion is again a translation, this time round a circle. Solved as a four-bar with the output angle found from two circles at each input angle, the output stays on the input to $10^{-13}$, and the relative displacements classify as $T_2$.

Change one length and it stops. With the output crank 0.75 instead of 0.6 and a link of 0.55, or with equal cranks and a link 15% longer than the offset, the chain is still a four-bar whose cranks both turn fully, and its output lags and leads the input by up to 1.27 radians over a turn, with an instantaneous ratio running from about 0.4 to 2.7. The hubs’ relative displacements, logged and closed, span three dimensions with a rotation in them: the whole planar group $G$.

The table is the equivalence, measured. Every coupling whose relative motion is a translation group has an angle spread of nought to rounding; every coupling whose relative motion reaches $G$ has a spread above a radian. There is no intermediate case, because a connected group either contains a rotation or does not.

One column in the table deserves a sentence of honesty, because it is not the column doing the work. The composite of two displacements falls inside the span to rounding for every coupling, the failing ones included. That is not a malfunction: the failing couplings’ displacements span all three dimensions of planar motion, and a three-dimensional span in a three-dimensional group contains everything, so no composite can fall outside it. The composition test earns its place on open chains whose displacements span less than the group they generate. Here the discriminating number is the rotation count the classifier reads off the span — nought for the translation groups, one for $G$ — and it is the same number the angle spread reports in radians.

The parallel-crank coupling has a flaw the group does not show, and it is worth stating because it is why real ones are built differently. A parallelogram lies flat twice a turn, and at those positions it can fold into the crossed linkage instead, which does not translate — the change point a parallelogram is known for. A practical parallel-crank coupling therefore carries a second set of cranks and a second link a quarter-turn out of phase, so that one is always far from its flat position. That is exactly the mechanism Grübler says cannot move, and the coupling is its oldest use: the redundant link does nothing to the group and everything to getting through the change points.

A translation does not have to be straight #

The parallel-crank coupling is worth a second look for what it says about the word translation. Its link does not move in a straight line: every point of it runs round a circle of the crank’s radius. The group it belongs to is still $T_2$, because a translation is a displacement that changes no orientation, and a body can be carried round a circle without ever turning — a Ferris wheel’s cars do it, and so does the coupling rod of a locomotive.

That distinction is exactly the one the classifier draws and a drawing tends to blur. A curved path suggests rotation, and a rotation about a distant centre and a translation along a circular path look alike over a short stretch. They are different elements of the group, and only one of them leaves the output hub’s mark parallel to the input’s. The four-bars that are not parallelograms also move their links along curved paths, and their links turn as they go; the classifier separates the two cases by asking whether any rotation direction is present in the span, not by looking at paths.

Constant on average is not constant #

The two four-bars that fail have one thing in common with the couplings that succeed, and it is the thing that makes them tempting. Both of their cranks turn fully, so over a whole revolution the output makes exactly one turn for each turn of the input: the mean ratio is one. A tachometer on the output shaft averaging over a second would read the input’s speed.

Within the turn, the instantaneous ratio of the coupling with unequal cranks runs from 0.53 to 2.70, and the one with the long link from 0.38 to 2.64. Its output hub is driven ahead and then held back, twice the input’s speed at one part of the turn and half at another, and whatever the output drives sees that as a torsional oscillation at the shaft frequency. A group containing a rotation cannot be constant-velocity; it can be right on average, and a ratio that is not constant is the general version of that failure.

The right angle is not the reason either #

The textbook drawing of Oldham’s coupling has the tongue at right angles to the slot, and the natural reading is that the right angle is what makes it work. The group argument says otherwise: any two slides that are not parallel generate $T_2$.

At 60° the classification is the same — two dimensions, $T_2$, composite inside the span to $2 \times 10^{-16}$ — and the output’s angle is on the input’s with a spread of nought. Only at 0°, with the slides parallel, does the argument fail: two translations along one line generate only $T$, the disc can absorb offset in one direction and not the other, and the coupling locks except at the two input angles where the offset happens to lie along the slides.

What the angle does decide is how the disc moves.

With the slides at $\beta$, the disc’s centre is on the input slot at a distance $s_1$ from the input axis and on the output slot through the output axis. Those two lines cross at an angle $\beta$ and pass through two fixed points one offset apart; as the shafts turn, both lines turn together and their crossing runs round a circle through the two shaft centres — the inscribed-angle theorem, since the chord between the two centres is seen from the crossing at a fixed angle. The circle’s diameter is $e/\sin\beta$, and because two lines turning through half a revolution sweep every direction once, the crossing goes round it twice per turn of the shafts.

Measured over a turn at 90°, 60° and 40°, each locus is a circle to $7 \times 10^{-16}$, of diameter 1.0000, 1.1547 and 1.5557 for an offset of one unit — exactly $1/\sin\beta$ — and each is traversed twice.

What the right angle buys #

A disc running twice round a circle means its slides reciprocate twice per turn, and how fast they run is a matter of the circle’s size.

Differenced from the coupling’s solved positions at 1,440 angles a turn, the peak sliding speed is $e\omega/\sin\beta$ to three parts in a million, at every slide angle from 30° to 150°. It is least at a right angle, where each slide peaks at the offset times the shaft speed; at 60° it is 15% faster and at 30° twice as fast.

The disc’s own motion has a second consequence at speed. Its centre runs round a circle of radius $e/2\sin\beta$ at twice the shaft’s angular speed, so it accelerates towards the circle’s centre at $(e/2\sin\beta)(2\omega)^2 = 2e\omega^2/\sin\beta$. For a coupling with a tenth of a millimetre of offset turning at 3,000 revolutions a minute, that is 20 m/s² at a right angle — a shaking force at twice the shaft frequency, set by the offset and the disc’s mass, and larger again at any other slide angle. The group decides that the output turns evenly; it does not make the part in the middle move evenly, and at speed the part in the middle is what a designer has to balance.

So the right angle is a wear specification, not a kinematic one. An Oldham coupling runs with its slides sliding under load, the wear on a sliding pair grows with its sliding speed, and the geometry that minimises sliding for a given offset is the right angle. The ratio of one is bought by the slides being slides.

What the group decides, and what it does not #

It decides the ratio exactly. The equivalence between a constant ratio and a translation group is not an approximation or a small-offset result. It holds at any offset the coupling can assemble at, because it is a statement about which displacements the hubs can reach relative to each other.

It says nothing about spatial misalignment. Every coupling here joins parallel shafts. Shafts that meet at an angle cannot be joined by a planar mechanism, and the joint that is not constant velocity is the universal joint’s version of the same question: its hubs’ relative motion is a rotation about a moving axis, which is no translation group, and its ratio varies twice a turn.

It says nothing about load sharing or backlash. A real Oldham disc has clearance in both slots, and a parallel-crank coupling with a redundant second link needs its links matched in length; both are tolerance questions the group sees nothing of.

It tells a designer which couplings tolerate an offset that changes while they run. Oldham’s disc absorbs any offset in the plane, so the offset can drift as the machine warms and the coupling stays in $T_2$ throughout. The parallel-crank coupling’s link has the offset’s length built into it: change the offset and the four pins stop being a parallelogram, the relative motion leaves $T_2$ for $G$, and the coupling becomes one of the failing four-bars in the table. The group is the same; the set of offsets over which each coupling stays in it is not.

The obvious way out of that restriction is to put a second parallel-crank stage in series with the first: an intermediate disc joined to the input hub by one set of parallel cranks and to the output hub by another. Each stage’s relative motion is a translation round a circle, the product of two such translations is a translation, and the pair together can reach any offset within the sum of the two crank radii. Commercial offset couplings built on that principle carry three discs and two sets of links, and the group argument above says why they transmit a ratio of one at whatever offset they settle at — the discs never turn relative to the hubs. That arrangement is not solved here, and its measurement would be a two-loop mechanism rather than a four-bar.

It does not tell a designer which translation group to use. $T$ couples only an offset along one fixed direction; $T_2$ couples any offset in the plane. Every practical coupling for unknown offsets is therefore a $T_2$ coupling, and the difference between Oldham’s and a crank coupling is where the translation comes from — slides that wear, or pins that have to be carried through change points.

Still open: couplings for shafts that are not parallel #

A constant-velocity coupling between intersecting shafts must make the output’s rotation about its own axis equal the input’s while the two axes are at an angle, and no subgroup of displacements does that for every input angle — the relative motion necessarily includes a rotation that changes with the input. That is why constant-velocity joints for driveshafts are built from symmetry rather than from a group: a Rzeppa joint’s balls stay in the plane that bisects the two shafts.

Its distinct argument would be that symmetry, stated in the pairs field’s terms: the set of relative displacements of a double Cardan joint or a Rzeppa joint, logged and classified, compared with a single universal joint’s. Two things would come out of it. Whether a constant-velocity joint’s hub-to-hub motion is characterised by an invariance under the reflection in the bisecting plane rather than by membership of any of the twelve groups — which would make it a different kind of statement from the one this essay makes; and whether the two Cardan joints of a double Cardan need exactly that reflection to cancel, which legs intersect and the composition instrument can check without solving the joint’s angle law at all.