A bearing is a planetary with no teeth

The rolling field opened with a constraint that takes nothing away: a wheel that rolls without slipping forbids a direction of motion without removing any coordinate from the machine’s description. Everything after that — parking, towing, the ball whose orientation remembers its path — has been about bodies rolling on the ground, where one side of every contact stands still.

A bearing is rolling with both sides moving. A ball sits between an inner race on the shaft and an outer race in the housing; a cage keeps the balls spaced and carries no load worth mentioning. When the shaft turns, the balls go round. How fast is not specified by anybody: there is no gear, no pin, nothing that fixes the cage’s speed except the two rolling contacts each ball makes. So the cage’s speed is a pure consequence of rolling constraints, and it is a good place to see what they decide.

The answer is familiar from a different field, and the second half of this essay is where the familiar answer stops being enough.

Two contacts, two unknowns #

Take a bearing with pitch diameter $D$ — the diameter of the circle the ball centres run on — and balls of diameter $d$. The inner race touches each ball at radius $r_i = (D - d)/2$ from the axis and the outer race at $r_o = (D + d)/2$. The inner race turns at $\omega_i$; the outer, for now, is held.

A ball has two unknowns in the bearing’s plane: the speed $\omega_c$ at which its centre goes round the axis, which is the cage’s speed, and its spin $\omega_b$ about its own centre. Rolling without slipping says the ball’s surface matches each race’s surface at the contact:

$$\omega_c \tfrac{D}{2} - \omega_b \tfrac{d}{2} = \omega_i r_i, \qquad \omega_c \tfrac{D}{2} + \omega_b \tfrac{d}{2} = \omega_o r_o.$$

Two equations, two unknowns, and they are solved as they stand. Adding them eliminates the spin: the centre’s speed is the mean of the two contact speeds, and the cage turns at

$$\omega_c = \frac{\omega_i r_i + \omega_o r_o}{r_i + r_o}.$$

With the outer race held that is $\omega_i r_i/D$, or

$$\frac{\omega_c}{\omega_i} = \frac{1}{2}\Big(1 - \frac{d}{D}\Big).$$

Less than a half, by exactly half the ball’s share of the pitch diameter. In the figure, a pitch diameter of 40 and balls of 8 put the cage at 0.4000 of the race: turned 60°, the inner race carries the cage 24° and spins each ball 120° backwards about its own centre.

The same fraction from the gear field #

Now write the planetary relation from two inputs and one output, which is Willis’s equation computed both ways: a sun turning at $\omega_s$, a ring at $\omega_r$, and a carrier at

$$\omega_s + K\omega_r = (1 + K)\,\omega_c, \qquad K = \frac{\text{ring teeth}}{\text{sun teeth}}.$$

Put $K = r_o/r_i$. That is the bearing’s formula, term for term: inner race as sun, outer race as ring, cage as carrier, balls as planets.

The correspondence is not an analogy to be admired. A gear mesh is a rolling constraint on the two pitch circles — the pitch circles roll even though the tooth surfaces slide — and teeth are what make the rolling positive rather than frictional; tooth counts are proportional to pitch radii. So a bearing whose contact radii stand in a rational ratio is a planetary, and the gear field’s train solver — which builds a matrix of mesh constraints and finds its null space in exact rational arithmetic, a ratio is a null space — should return the cage’s speed with no idea that anything rolls.

Handed a planetary with a sun of $D - d$ teeth and a ring of $D + d$ — whose planet then has $d$ teeth by the planetary’s own coaxial rule, exactly the ball — the solver returns, for the seven bearings, 19/40, 9/20, 17/40, 2/5, 3/8, 7/20 and 13/40. The rolling solve gives 0.4750, 0.4500, 0.4250, 0.4000, 0.3750, 0.3500 and 0.3250. They agree exactly, and the “half the shaft speed” of informal description is wrong by at least 0.025 for the smallest ball in the table.

The ball’s spin comes out of the rolling solve as well, $-\omega_i r_i/d$ with the outer race held: 2.0 turns backwards per turn of the shaft for the ball of 8, 9.5 for the ball of 2. A small ball in a large bearing spins fast, which is why rolling-element speed limits are quoted against the pitch diameter as well as the shaft speed.

Why the answer is a little under a half #

The half has a picture behind it, and the correction has a picture too. With the outer race held, the point of the ball touching the outer race is momentarily still, so it is the ball’s instant centre. The ball’s centre is half as far from that point as the inner contact is, so it moves at half the inner contact’s speed. On a flat bearing — balls between two plates, one sliding — that is the whole story: the balls travel at half the plate’s speed, and a linear bearing’s cage runs at exactly a half.

A round bearing converts speeds to angular speeds, and there the radii differ. The inner contact’s speed is $\omega_i r_i$, the ball centre moves at half of that, and it moves on a circle of radius $D/2$, which is larger than $r_i$. So its angular speed is $\omega_i r_i/D$: half the race’s angular speed, reduced by the ratio of the inner contact’s radius to the pitch radius. That ratio is $1 - d/D$, and it is where the whole correction comes from. A bearing of very small balls is nearly flat as far as each ball can tell, and approaches a half; a bearing of large balls is strongly curved, and falls well below it.

The frequencies a bearing announces #

The cage’s speed is not only a design number. Every ball that passes a point on a race strikes it slightly, and a defect on the race turns those passes into a vibration at a frequency the kinematics fixes. Bearing diagnostics reads those frequencies off an accelerometer, and all of them follow from the two rolling constraints.

With $Z$ balls and the outer race held, a point on the outer race is passed by balls at $Z$ times the cage’s speed: $Z(1 - d/D)/2$ times the shaft’s. A point on the inner race moves with the shaft, so the balls pass it at $Z$ times the shaft’s speed relative to the cage, $Z(1 + d/D)/2$. For the bearing in the first figure, eight balls of 8 on a pitch diameter of 40, those are 3.2 and 4.8 times the shaft frequency — and they add to exactly $Z$, eight, because the cage’s speed and the inner race’s speed relative to it add to the shaft’s.

The ball’s own spin frequency relative to the cage is the other diagnostic number. The rolling solve’s spin is the ball’s absolute angular speed, −2.0 per turn of the shaft; relative to a cage turning at 0.4 it is −2.4, which is the handbook’s $(D/2d)(1 - (d/D)^2)$ to the last digit. A defect on a ball strikes both races once per relative turn and shows at twice that.

None of these numbers is measured on a bearing here; each is a consequence of the cage speed that is. What the derivation adds to the handbook formulas is where they come from: each is a count of passes at a speed set by rolling constraints, and each is wrong in the same direction when a lightly loaded cage skids.

When both races turn #

The same relation covers a bearing whose outer race also turns — a wheel bearing on a rotating hub, or the planet bearing inside an epicyclic, whose own outer race is carried round.

The cage turns at the mean of the two race speeds weighted by their contact radii: 0.400 with the outer race held, 1.000 when both races turn together and the whole bearing turns as a rigid body. It stands still when the outer race turns backwards at 0.667 of the inner’s — the ratio of the contact radii — which is a planetary with its carrier held: an ordinary train in which the balls are idlers carrying motion from one race to the other.

The case a car’s wheel presents is the mirror image of a shaft’s. A wheel hub turns its outer race on a stationary spindle, so the inner race is the one held, and the same formula gives a cage share of $r_o/D = (1 + d/D)/2$ of the hub’s speed: more than a half, by the same amount the shaft bearing’s is less. The balls of the bearing in the figure would go round at 0.6 of the wheel’s speed instead of 0.4. Which race turns decides which side of a half the cage is on, and a diagnostic frequency quoted for one arrangement is wrong for the other by exactly the ball’s share of the pitch diameter.

This is the differential again, and it is worth saying what it means for a bearing designer. A cage’s speed is a quantity of the bearing’s kinematics and cannot be chosen, so a cage that must, for instance, stay still to carry a sensor, or run in step with something else, has to be designed through the race speeds and the ratio $r_i/r_o$ and nothing else.

The extreme case of the rule is the needle bearing, whose rolling elements are thin rods on a pitch diameter many times their own. With $d/D$ at a thirtieth, the cage runs at 0.483 of the shaft and each needle spins fourteen and a half times per turn of the shaft, backwards; the planetary picture still holds, with planets thirty times smaller than the sun, and it predicts the needle’s high spin before any failure analysis does. A needle bearing on a shaft turning at 3,000 revolutions a minute spins its needles at over 43,000, which is why needle bearings are lubricated and caged with more care than their size suggests.

A roller is not a ball #

A ball touches each race at a point, and a point contact can roll in any direction. A roller touches along a line, and rolling along a line is a stronger demand: every point of the line must have the same velocity on both bodies.

For a cylindrical roller between two cylindrical races that is automatic, because every point of the line is the same distance from both axes. A tapered roller — a truncated cone running between two conical races, the kind that carries the combined radial and thrust load of a wheel hub — is different. Along its contact line the distance from the bearing’s axis changes, and so does the distance from the roller’s own axis. Both surface speeds are linear along the line, and they can agree everywhere only if they have the same zero.

The surface speed of the race is zero where the contact line meets the bearing’s axis — the race cone’s apex. The roller’s surface speed is zero where the contact line meets the roller’s axis. So the roller rolls without slipping along the whole line only if the roller’s axis, the bearing’s axis and the contact line all meet at one point.

That is measured rather than argued. In the plane containing both axes every surface velocity points out of the plane, so the slip at each point of the line is one number. The roller is given the rate that makes the slip nought at the middle of its contact line, which is what a loaded roller settles to, and the slip is read at every point of the line. With the roller’s axis through the apex it is nought along the whole line, to $2 \times 10^{-15}$. With the crossing point 2 along the axis from the apex, the ends slip by 5.0% of the surface speed, one forwards and one backwards.

Every axis through one point, again #

The slip is nought only when the crossing point is exactly at the apex, and it grows either side: 1.0% at half a unit before the apex, 3.6% at two, 6.3% at four; 1.1%, 2.3%, 5.0% and 12.6% at the same distances beyond. A roller axis that misses the apex slides at its ends on every revolution, which is heat and wear at exactly the place a tapered bearing carries its highest load.

The condition is not new to the rolling field. Every axis through one point found that several wheels bolted to one body can roll together only if all their axle lines meet at a single point — the instant centre of the body’s motion — and derived the steering geometry of a car from it.

A tapered roller bearing is that condition in three dimensions. The roller and the race are two bodies turning about axes in space; their relative motion at an instant is a rotation about an axis through the point where their own axes meet; and a line of contact can be free of slip only if it lies along that relative axis, which means it passes through the same point. A car’s wheels meet at an instant centre in the ground plane; a tapered bearing’s cones meet at an apex on the shaft. Both are rolling constraints that cannot be satisfied at more than one point unless the geometry is concurrent.

What the kinematics decides #

It decides speeds exactly. The cage ratio, the ball spin and the slip along a roller’s line are consequences of rolling at the contacts and nothing else, and they are what a bearing does whenever its elements roll.

It decides them for any size. Every speed above is a ratio of lengths, so a bearing scaled up or down keeps its cage share, its ball spin per shaft turn and its tapered rollers’ slip fraction. What does not scale is the load a contact carries, and with it whether the contact rolls at all.

It does not decide whether they roll. Rolling is an assumption about friction. A lightly loaded bearing at high speed lets its cage lag the kinematic speed — skidding — because the traction at the contacts is too small to accelerate the cage and balls against drag, and the cage then turns slower than $(1 - d/D)/2$. The kinematic speed is the upper limit a cage reaches when it rolls, and measuring how far below it a real cage runs is a way of measuring traction.

It treats contact as geometric. A loaded ball flattens into an elliptical contact patch, and across a patch of finite size the two surfaces’ speeds cannot agree everywhere unless the geometry is concurrent in the way the tapered roller’s is. A deep-groove bearing’s ball contacts a curved groove across a patch, and that patch has a small slip built in — the ball-bearing version of the tapered roller’s miss, and the reason the two kinds of bearing run at different temperatures.

Still open: the angular-contact ball and its spin #

A deep-groove ball touches its races where their radii are measured in the bearing’s plane. An angular-contact ball touches them along a line tilted by a contact angle, so that it can take thrust, and then its two contacts are at different distances from the ball’s own axis of rotation in a way no choice of that axis can match at both. Rolling at both contacts can then be satisfied only with some spin of the ball about the contact normal at one or both of them.

Its distinct argument would be that spin, computed from the same rolling constraints written in three dimensions: the ball’s angular velocity as a three-vector, rolling imposed at two contact points, and the component of spin about each contact normal that the constraints leave free or force. Two things would come out of it. The ratio of spin to rolling at each contact as a function of the contact angle, which is the quantity bearing makers minimise when they choose where the ball’s axis should point; and whether there is a contact angle at which the spin vanishes at both contacts, which by the tapered roller’s argument would be exactly the angle at which the two contact tangents and the bearing’s axis are concurrent.