A rotor's inertia slows the jump and cannot make it ring

A motor turns the runaway into a jump drove a self-heating oil film through a real motor — one whose torque falls in a straight line as its speed rises — and found that the thermal runaway of a film held at fixed stress never happens under such a drive. A soft drive jumps instead, from a cool steady state to a hot one, and a stiff drive does not even do that. But it made one idealisation that no machine obeys: the shaft’s speed followed the motor’s torque instantly, as if the rotor weighed nothing.

It ended by asking what a real rotor does. A rotor with inertia lags its torque by a mechanical time, and a film lags its heating by a diffusion time. Two lags that feed each other are how most oscillators are built — a mass and a spring, an inductor and a capacitor, a compressor’s duct and its plenum — and the natural guess was that a soft drive near its cusp, with the two clocks comparable, might ring, or cycle between its branches without settling on either.

It cannot, and the reason is not in the numbers. It is in the signs.

The rotor as a second state #

The film is the same one the earlier essays used: oil between two walls held at a fixed temperature, its viscosity falling by a factor of $e$ for every fixed rise in temperature, heated by its own shearing. In scaled units its temperature excess obeys $\theta_t = \theta_{xx} + \tau^2 e^\theta$ across the gap, where $\tau$ is the stress, and the speed of one wall past the other is $V = \tau S$, with $S$ the mean of $e^\theta$ — the factor by which the oil has thinned, on average.

The motor gives the stress $\sqrt{\delta_s} - kV$, a straight line from its stall value to zero at its no-load speed, with $k$ the droop that the motor’s essay found decisive. Until now the film’s drag and the motor’s torque were simply set equal. With a rotor they are not: their difference accelerates it,

$$M\,\frac{dV}{dt} = \sqrt{\delta_s} - kV - \frac{V}{S},$$

and $M$ is the one new number. It is the time the rotor would take to spin up against cold oil alone, measured in the film’s thermal diffusion times across its half-gap. $M = 0$ is the weightless rotor of the essay before; $M \to \infty$ holds the speed fixed, which, as the film that heats itself showed, has no fold at all.

It is worth knowing where real machines sit before asking what happens. A journal bearing with a 25-millimetre shaft and a clearance of fifty microns has a thermal diffusion time across its half-gap of a few milliseconds. Its cold oil drags the shaft with a torque of about a thousandth of a newton-metre per radian per second, so a modest rotor, a hundredth of a kilogram-square-metre, spins up against that drag in several seconds. That makes $M$ several hundred, and heavier rotors push it into the thousands. In practice the rotor is the slow clock, by two or three orders of magnitude. The interesting range, where the two are comparable, is reached only by rotors built to be light.

What inertia cannot change #

Two things follow before any calculation.

The steady states are the same. The operating-point construction is the one a pump on its system uses, and at a steady state the rotor is not accelerating, so the motor’s torque equals the film’s drag exactly as it did without inertia, and the operating points — cool, middle and hot on a soft drive’s S-curve — are where they were.

Their stability is the same too, and this takes one more step. Linearise the film and the rotor about a steady state and ask for the growth rates of small disturbances. Those rates are the eigenvalues of a matrix whose last row, the rotor’s, is divided by $M$. Its determinant is therefore the weightless drive’s determinant divided by $M$: it can shrink or grow but never change sign. A growth rate that changed sign as $M$ varied would have to pass through zero, which would make the determinant zero. So if the rates are real, the number of unstable ones cannot change with inertia. The middle state stays unstable and the other two stable, for any rotor at all.

That argument leaves one door open: a pair of rates could become complex, and a complex pair could cross into instability without any rate passing through zero. That is how an oscillation is born, and it is what the guess was about.

Why the door is shut #

The linearised system couples the film and the rotor through two links. A faster rotor heats the film: raising $V$ raises the stress at a given temperature, and the heating goes as its square. A hotter film lets the rotor speed up: raising $\theta$ thins the oil, and the drag $V/S$ falls. Both links are positive. Each state pushes the other the same way.

That is not only a qualitative remark. In the matrix, the rotor’s column — how each point of the film responds to the rotor — is proportional to $e^\theta$ across the gap, and the rotor’s row — how the rotor responds to each point of the film — is also proportional to $e^\theta$, with a positive ratio between the two that is the same at every point, $4M/h$ on a grid of spacing $h$. The film’s own block is symmetric: diffusion is, and the one non-local part, the way a hotter film everywhere lowers the stress everywhere, is an outer product of $e^\theta$ with itself. Rescale the rotor’s speed by $\sqrt{h/4M}$ and the whole matrix becomes symmetric. A symmetric matrix has real eigenvalues. At every steady state and every inertia, no disturbance of the driven film can oscillate.

The contrast that makes this clear is the compressor. A compressor on its plenum surges because its two states push each other in opposite senses: flow arriving in the plenum raises its pressure, and a higher plenum pressure decelerates the flow. That is a mass on a spring, where the spring pushes back, and the energy sloshes between the two. The bearing has no spring. The rotor and the film only ever encourage each other, and a system that only encourages can run away or settle, but it cannot swing.

The spectrum, computed #

The first figure is the calculation that could have refuted this. It takes the soft drive of the previous essay, droop $k = 0.04$, at a stall group of one, where there are three steady states with centre temperatures of 0.97, 2.07 and 4.69 e-folds, and computes the full spectrum of the linearised film and rotor, forty-one eigenvalues, for inertias from a hundredth to a hundred.

The cool and hot states decay and the middle one grows, at every inertia. With a nearly weightless rotor the cool state’s slowest disturbance dies at 0.72 per diffusion time and the middle state’s grows at 0.75. As the rotor gets heavier every rate slows, and past an inertia of about one each falls as $1/M$: the slowest motion in the system has become the rotor’s, and the film simply follows. Across all of it, the largest imaginary part of any eigenvalue on any branch is zero, and the rescaled matrix is symmetric to three parts in $10^{16}$, which is the arithmetic’s own rounding.

Two clocks that add #

What inertia does do shows up the moment the drive is switched on. The second figure takes the soft drive five per cent past its fold — beyond the load at which the cool state exists, so the film must jump — and starts it from rest with three rotors.

The weightless rotor is at its cold operating speed at once, and the film heats at nearly that speed until it thins, whereupon the rotor runs away from it along the motor’s line. A rotor with $M = 10$ does something quite different: it spins up so slowly that at every speed it passes through, the film has had time to reach the steady temperature for that speed. Its path lies on the curve of fixed-speed films, and the film is simply carried along it by the rotor. All three paths end at the same hot state, as the argument above requires.

The third figure plots the same events against time. Every film climbs, lingers and jumps; none overshoots its hot state or rings about it, which is what real eigenvalues promise. What differs is when. The time to pass three e-folds is 7.0 diffusion times for the weightless rotor, 16.5 for $M = 0.1$, 101 for $M = 1$ and 948 for $M = 10$.

The fourth figure shows those numbers follow a simple rule. The delay is the film’s delay plus the rotor’s: $7.0 + 94.6M$ diffusion times. The first term is the weightless drive’s own delay, the film lingering near the temperature its vanished cool state would have had — the thermal ghost that a lost steady state still holds the film found at fixed stress. The second is what a rotor heavy enough to keep the film always steady takes to pass the fold: $M$ times $\int dV/(\text{motor} - \text{film drag})$ along the film’s own torque–speed characteristic. That integral is computed directly, from the characteristic alone, and it predicts the marched time at $M = 10$ to half a per cent. The two clocks are equal at an inertia of 0.074 — so for anything heavier than a small fraction of the film’s own time, it is the rotor that sets the pace.

The ghost moves to the rotor #

Most of that integral is not spin-up. It comes from the stretch of speeds near the fold, where the motor’s line passes just above the film’s characteristic and the torque left over to accelerate the rotor is tiny. That is a saddle-node’s ghost again, the same shape of delay the film showed at fixed stress, but now in the rotor’s speed rather than the film’s temperature, and paced by the rotor’s clock.

The fifth figure puts both ghosts side by side against the overshoot past the fold. Both fall as its inverse square root, the signature of a saddle-node: at one per cent past the fold the film’s own delay is 17.5 diffusion times and the heavy rotor’s is 226 times its inertia; at twenty per cent, 2.8 and 39. The constants differ by a factor of about thirteen, so the rotor’s ghost is the longer of the two once $M$ exceeds about 0.08, and in a real machine, with $M$ in the hundreds, it is the whole of the delay.

That has a consequence an operator would notice. The rotor’s delay is $226M$ diffusion times at one per cent overshoot, and $M$ diffusion times is by definition the rotor’s own spin-up time. A heavy machine loaded one per cent past its fold takes more than two hundred spin-up times to jump — for a rotor that spins up in a few seconds, well over ten minutes, during which the bearing’s temperature and the shaft’s speed drift upwards so slowly that they look like a steady state settling rather than the start of a jump. The fold’s warning in such a machine is a drift that does not stop.

Ringing needs a link that pushes back #

The argument makes a prediction of its own: reverse one link and the film should be able to ring. The sixth figure does it with something real machines have. Many drives protect themselves by derating — lowering their torque as a temperature rises. Here the motor’s stall torque is multiplied by $1 - g(S - 1)$, so it backs off as the film warms and thins, with $g$ the gain of the protection.

That adds a negative contribution to the rotor’s row, the link from film temperature to rotor speed. At a gain of about 0.48 on the cool state of the soft drive the link changes sign: a hotter film now slows the rotor, and the rotor, as before, heats the film. The loop has a spring in it, and the symmetry is gone. The spectrum shows the consequence. For a rotor with $M = 0.3$ an oscillatory pair appears at a gain of about 0.6 and its frequency rises to three per diffusion time at a gain of three; for $M = 1$ it appears at about 1.3; for $M = 3$ not at all in this range, because a heavy rotor cannot follow a fast exchange. The sign change is necessary, not sufficient.

The pair is damped: a derating is negative feedback and it stabilises the state it makes ring. But the structure has changed from one that cannot oscillate to one that can, and that is the distinction the essay turns on. A bearing and a motor cannot ring by themselves; a bearing, a motor and a protection circuit can.

What was checked #

The seventh figure is the ledger. The rescaled matrix is symmetric to $3\times10^{-16}$ and its spectrum real on the cool, middle and hot states at inertias of 0.01, 1 and 100, with zero, one and zero unstable directions at each. A rotor with $M = 0.001$, marched from rest five per cent past the fold, passes three e-folds at 7.12 diffusion times, where the weightless drive of this model does at 7.04 and the instantaneous drive of the previous essay’s own code at 7.08. At $M = 10$ the marched delay is within half a per cent of the film’s delay plus $M$ times the quasi-static integral, which uses no march at all. And a drive derated with $g = 2$, at $M = 1$ and a stall group of 1.2, has a pair at $-1.85 \pm 1.13i$ on its cool state.

What the model leaves out #

The oil’s own momentum. The film’s velocity profile is taken to be the steady Couette profile at every instant. Momentum crosses the gap faster than heat by the Prandtl number, which for oil is in the hundreds or thousands, so this is the safest approximation in the problem.

A straight motor line. As in the previous essay. A curved characteristic changes where the folds are but not the signs of the two links, so it does not reopen the door. Nor does anything that only changes which states exist: what a flow does depending on what holds it is a question about steady states, and inertia leaves those alone.

Walls at a fixed temperature. A bush with a thermal mass of its own would be a third clock. The film heats the bush and a warmer bush heats the film — another pair of links with the same sign — so on the argument here it adds delay rather than oscillation; showing that the symmetry survives it is a calculation this essay has not done.

A planar film. A journal bearing’s film varies round the shaft and its thickness depends on the load. The shape of the gap sets its pressure; a thinner hot film lets the shaft move, which is a further state with links of its own.

Who worked it out #

The fold at fixed stress is Nahme’s, from 1940, and Frank-Kamenetskii’s for thermal explosion. That coupling a thermal fold to a mechanical degree of freedom can make a relaxation oscillator is the mechanism behind several real ones, and in the theory of dynamical systems the rule that a loop of mutually reinforcing variables cannot sustain oscillation is the subject of the theory of cooperative systems, developed by Hirsch in the 1980s. What is used here is narrower and exact for this problem: the linearised film-and-rotor matrix is similar to a symmetric one, whatever the inertia. The compressor with the opposite sign is Greitzer’s, from 1976.

Still open: a protection that acts late #

The derated drive above responds to the film’s temperature instantly, and it rings but settles. Real thermal protection does not act instantly: it reads a sensor in the housing, which warms from the film over the housing’s own thermal time, often minutes. That makes three clocks — the film’s, the rotor’s and the housing’s — with a slow negative link among fast positive ones.

That is the structure of a genuine relaxation oscillator, the same as a neuron’s: a fast variable with an S-shaped set of steady states — the film under a soft drive, which has one — and a slow variable that pushes it back across. The next calculation adds the housing’s temperature as a slow state, derates the motor on it, and asks whether a soft drive near its cusp then cycles: jumping hot, letting the housing warm until the torque falls below the lower fold, dropping cool, letting the housing cool, and jumping again, with a period set by the housing and an amplitude set by the width of the S. A drop at its fold meeting a second degree of freedom was the analogy the motor’s essay offered. What this essay found is that the second degree of freedom has to push back.