A slow bush quickens the hunt it cannot start

A warm bush stops the bearing hunting took a journal bearing whose motor is derated on its housing’s temperature — a protection that can make the film jump between a cool state and a hot one on the housing’s clock, as a slow sensor makes the bearing cycle found — and let the bush warm with the housing. A warm wall turned out to be a stiffer drive by the factor $e^{\beta H}$, where $\beta$ is the share of the housing’s excess that reaches the bush, and the stiffening closed the S-shaped characteristic the hunt runs round. Past a coupling of about a third the bearing settles, warm and derated, under any protection.

That essay assumed the bush follows the housing instantly, and named it as its weakest assumption. A real bush is a separate piece of metal with its own heat capacity and its own contact with the housing, and it lags. The lag is on the second feedback — the one that usually makes things run away — and the obvious worry is that a lagging positive feedback brings the hunt back: the housing warms, the bush is late to stiffen the drive, and in the gap the film jumps. This gives the bush its own time and finds that the worry is wrong for a reason that is general, and that the lag does something else instead.

Three clocks #

The film is fast: it settles to a steady profile in a fraction of a second, and between events it sits on its S, a Frank–Kamenetskii profile of centre excess $\theta_0$ under the effective drive, with stall group and droop both multiplied by the bush’s factor. The two slow variables are the housing’s excess $H$ and the bush’s $B$:

$$\frac{dH}{dt} = \frac{B + \bar\theta_{\text{rise}}(\theta_0) - H}{T_H}, \qquad \frac{dB}{dt} = \frac{\beta H - B}{T_B},$$

the film’s mean being the bush’s temperature plus the film’s own rise above it. With the bush’s time $T_B$ zero, $B = \beta H$ and the machine is the earlier essay’s; with $\beta = 0$ it is the fixed-wall bearing. The film stays on its branch until that branch ends at a fold and then jumps to the other. The march is second-order in the two slow variables, with the film’s branch tracked between steps, and time is counted in housing times. With the bush instant and no coupling it returns the fixed-wall period, 0.756 housing times against the exact 0.7535; with the bush instant at a coupling of 0.3, 0.086 against the earlier essay’s 0.084.

The hunt, with a slow bush #

The first result is the one the worry predicts least. A bearing that hunts with an instant bush still hunts with a slow one, and a bearing that settles with an instant bush still settles with a slow one. With a fifth of the housing’s excess reaching the bush and a derating of 0.2 both bushes hunt, with film jumps of the same size, and the slow bush’s cycle is shorter: 0.291 housing times against 0.347.

A slow bush does not move with the cycle at all. Its time is thirty times the cycle’s, so it sits at the cycle’s mean — the march holds it between 0.3779 and 0.3781 of excess while the housing swings by a tenth — and the walls the film sees are fixed. The instant bush swings with the housing, from 0.366 to 0.387, and that swing is what the slow bush removes.

Why the loop narrows #

Drawn as a loop in the plane of housing and film the cycle shows where the time went. On the cool branch the film is cooler than the housing, so the housing falls; the derating eases as it falls, the drive strengthens, and the cool branch ends at its fold, where the film jumps hot. On the hot branch the housing rises, the derating bites, and the hot branch ends at its other fold. The width of the loop in housing excess is the hysteresis the housing has to cross each way, and the period is that width divided by how fast the housing moves.

An instant bush widens the loop at both ends. On the cool branch the falling housing takes the bush with it, the cooler walls weaken the drive, and the cool branch holds on to a lower housing temperature than it would with the walls held — 1.832 instead of 1.844 of excess. On the hot branch the rising housing warms the bush, the warmer walls strengthen the drive, and the hot branch lasts to 1.937 instead of 1.932. A positive feedback inside a hysteresis loop always widens it, because it pushes each state further into its own branch; the loop spans 0.097 of housing excess with the instant bush and 0.084 with the slow one. Every fold-bounded loop behaves this way — the drop falls at a fold for the same geometric reason the film jumps at one — and every positive feedback that rides with the control widens such a loop. The housing moves at much the same rate either way, so the period falls with the width, by a sixth.

Where the time goes #

The share of the cycle the film spends hot hardly changes at a fifth coupling: 0.563 of the period with the instant bush and 0.564 with the slow one, because the narrowing is nearly symmetric — the cool branch’s end moves up by 0.013 of housing excess and the hot branch’s down by 0.005, but the housing also crosses the hot branch’s stretch more slowly. At three-tenths coupling the hot share rises from 0.305 to 0.344, because there the instant bush’s swing is a larger part of what held the film cool. In both cases the film’s two states and their temperatures are those of the fixed-wall bearing with the walls at the cycle’s mean warmth; the slow bush turns the warm-bush bearing back into a fixed-wall one, at a wall temperature it chooses itself.

That is also why the slow bush’s period is easy to predict without marching. A fixed-wall bearing’s period is the housing’s time integrated across the window on each branch, $T_H\int dH/|\bar\theta - H|$, and with the walls held at the mean the integral is the earlier essays’ with the drive multiplied by $e^{B}$ once and for all. The march and that integral give the same period because they are the same calculation; the bush’s time drops out once it is long, exactly as it dropped out of the band.

The period, against the bush’s time #

Swept over the bush’s time the period falls smoothly from the instant value and levels once the bush is slower than the cycle. At a fifth coupling it is 0.347 instant, 0.320 with a bush a twentieth of a housing time slow, 0.302 at a tenth, and 0.291 from about a housing time on. At three-tenths coupling, where the cycle is shorter, the fall is from 0.138 to 0.111, a fifth. Every march at every bush time settles onto a single period: the cycle’s successive periods agree to the step of the march.

That answers the second half of the question the earlier essay left — whether a slow bush could make a bearing hunt on two periods at once, the film jumping on the housing’s clock while the window breathed on the bush’s. It cannot, for the same reason the loop narrows: a bush slower than the cycle averages the cycle rather than following it, so there is nothing for the window to breathe with. A bush comparable to the cycle is caught partway, and the cycle simply takes an intermediate period.

The band the lag cannot move #

Whether a bearing hunts at all has a test that does not involve time. Its equilibrium — the state where nothing changes — has the bush at $\beta H$ and the film’s mean equal to the housing’s, whatever the bush’s time, because the bush’s time sets only how fast $B$ approaches $\beta H$. The film can sit at that equilibrium only if it lies on the cool branch or the hot one; if it lies on the S’s middle branch, the branch the fast film cannot hold because any disturbance carries it to one of the others, then no steady state is available and the bearing must cycle.

Parametrised by the film’s centre temperature rather than searched for, the equilibria are explicit. Each centre temperature fixes the film’s rise and so the housing, $H = \bar\theta_{\text{rise}}/(1-\beta)$; the operating-point condition then fixes the derating that holds it there; and it is on the middle branch when its centre temperature lies between the folds of the drive at that housing. At a fifth coupling the middle-branch equilibria cover deratings from 0.182 to 0.278. At zero coupling the same construction gives 0.162 to 0.300, and at three-tenths 0.205 to 0.254, the bands the earlier essays found by following the machine. No bush time enters the construction, and the marches agree: at bush times of nothing, a tenth, one and ten housing times, deratings of 0.20 and 0.26 hunt and 0.17 and 0.29 settle.

A settled bearing stays settled #

The equilibrium test says where the bearing settles if it settles; whether the settled state survives a lagging bush is a linear question with a short answer. Near the equilibrium the housing’s equation is $\dot H = aH' + bB'$, with $a$ the film’s response to the housing at fixed walls — negative, because a warmer housing derates the motor — and $b$ its response to the walls, positive. The bush’s is $\dot B = (\beta H' - B')/T_B$. The two growth rates satisfy

$$\lambda^2 - \Big(a - \frac{1}{T_B}\Big)\lambda - \frac{a + \beta b}{T_B} = 0.$$

The settled state with an instant bush is stable, so $a + \beta b < 0$ and the product of the rates is positive at every bush time. The sum is $a - 1/T_B$, negative at every bush time because $a$ is. Both rates therefore have negative real parts however slow the bush is, and the discriminant stays positive in every case computed, so they are real: a disturbance decays without ringing. Past the cusp at a coupling of a half, $a = -4.23$ and $b = 2.59$, and the slower rate goes from $-2.9$ per housing time for a bush a hundred times quicker than the housing to $-0.007$ for one a hundred times slower. The lag slows the return to the settled state; it cannot reverse it.

Why a lag needs a negative link #

The argument is not special to bearings. In a system of two variables, one of them lagged, the sum of the growth rates is the unlagged variable’s self-feedback minus the inverse lag. An oscillation that grows needs that sum positive, so it needs the unlagged variable to feed itself; and the lag is then what lets a negative loop through the lagged variable push back too late. The figure swaps the bearing’s signs — an unlagged variable that feeds itself, a lagged loop that opposes it — and that system does oscillate, once the lag exceeds the time on which the fast variable grows. The bearing has the opposite arrangement: its unlagged variable, the housing, is held back by the derating, and its lagged loop, through the bush, reinforces. A lag on a reinforcing loop only delays the reinforcement.

The same arithmetic appears wherever a lag is blamed for an oscillation. The lag that makes flutter possible is the aeroelastic case, where the lag in the wing’s own lift is what turns a restoring moment into one that feeds the motion; and a rotor’s inertia slows the jump and cannot make it ring found the bearing’s third clock equally harmless, because the rotor and the film push each other the same way. The bush joins the rotor: two clocks that can slow the bearing and neither of which can make it ring.

What an operator would see #

A bearing on the hunting side of the band hunts the same way whatever its bush; a slow bush makes the saw-tooth on the housing thermometer a little faster and a little shallower, and makes the motor current switch a little more often. On the settled side the bush’s time is invisible in steady running and shows only in recovery: a disturbance — a load change, a cold start — takes longer to die away with a slow bush, by the bush’s own time, as the slow growth rate says. Neither is a failure. What changes the bearing’s character is the coupling β, set by how the bush is seated, and the derating g, set by the protection; how fast the bush follows is a matter of detail. The hazard a motor turns the runaway into a jump identified, the jump itself, is untouched.

The fixed wall and the instant bush, as limits the march must return #

The march with no coupling returns the fixed-wall period to the step of the march, and with an instant bush at three-tenths coupling the earlier essay’s period. The equilibrium construction reproduces the earlier essays’ bands at zero and three-tenths coupling, and the marched bands at a fifth coupling to the resolution of the march. Past the cusp the marched bearing settles at a housing excess of 2.196139, and the equilibrium found by bisection on the same branch is 2.196139. And the march, run just inside and just outside the band’s lower edge with a bush ten housing times slow, cycles inside and settles outside.

The convention: three times, one ordering #

The film’s own time is taken as instant against both slow times, so the film is always on its S except at a jump. The bush’s time is measured in housing times, and it runs from zero, the earlier essay’s case, to ten. The coupling β is the bush’s share of the housing’s excess at equilibrium; the derating g is the fractional fall of the motor’s stall group per unit of housing excess, applied as $(1 - gH)^2$. All temperatures are excesses over the reference, in units of the oil’s viscosity-temperature scale, as in the film that heats itself.

What a quasi-static film leaves out #

The jump’s own duration. A film leaving its branch at a fold does not jump instantly: it lingers near the fold’s ghost, as a lost steady state still holds the film computed, for a time that grows as the housing approaches the fold slowly — and a slow bush, by narrowing the loop, changes how slowly the fold is approached. That lingering is outside the quasi-static film. A bush whose time is comparable to the film’s. The film was held quasi-static; a bush thin enough to follow the film’s own transients would couple the fast and slow scales, and the jump itself would then be shaped by the bush. A non-uniform bush. The bush here has one temperature; a real one is hotter at the loaded arc, and its wall condition varies round the journal. The rotor’s clock together with the bush’s, which were each shown harmless alone and are not computed together, though the sign argument above covers any number of lagged reinforcing loops on a negatively self-fed variable. And a protection that is not a derating — a trip, a hysteretic switch — which brings its own hysteresis into the loop and its own fold.

Who reasoned this way first #

The quasi-static treatment of a fast variable on a folded manifold, with jumps at the folds, is van der Pol’s relaxation oscillation of the 1920s, made rigorous by the geometric singular perturbation theory of the 1970s. The statement that a two-variable system’s oscillations need a positive trace — a self-reinforcing unlagged variable — is the Bendixson–Dulac and Hopf reasoning applied to the simplest case, and is how chemical and biological oscillators are classified into activator–inhibitor pairs; a lagged activator, which the bush is, is the arrangement that does not oscillate. The bearing problem itself is the thermohydrodynamic instability of plain journal bearings, studied from the 1960s on with the bush’s temperature usually held fixed or tied to the housing’s.

Still open: a protection that remembers #

Every protection here is a smooth derating, which acts on the housing’s temperature as it stands. Real protections often trip and reset: they cut the motor at one temperature and restore it at a lower one, a hysteresis of their own laid on top of the film’s. The next calculation replaces the smooth derating with a two-level protection, and asks whether two hysteresis loops — the film’s and the protection’s — can lock together, so that the bearing hunts on the protection’s band rather than the film’s, and whether a trip set inside the film’s own loop can make a bearing hunt that a derating of the same average strength would have settled.