A torn film still pulls

A ball that bounces in water and not in oil builds the rebound threshold of a sphere under liquid out of three steps. In through the film, losing a fixed fraction of the arrival speed for every decade of gap closed; a dry bounce at a contact gap $x_c$ that the film equation does not contain; and back out through the same film, at the same cost per decade. The threshold that results,

$$\mathrm{St}_c = \frac{L\,(1 + e_d)}{e_d}, \qquad L = \ln\frac{x_0}{x_c},$$

is 10.18 for a sphere of radius three millimetres with a contact gap of twenty microns, and it matches the measured value of about ten because the unknown gap sits inside a logarithm.

That essay closes on the weak point of its own model. On the way out, the film is being pulled apart, and its pressure falls below ambient — by a lot, near the wall. A liquid cannot follow a pressure much below its vapour pressure; it cavitates, and a torn film would seem to charge the sphere nothing for the part of the gap where it has torn. The essay estimates that a film tearing at a tenth of a millimetre would lower the threshold by about a third, and suggests an experiment in a partially evacuated tank to see it.

This essay does the calculation. The estimate turns out to be wrong by a factor of seven, and the reason is worth more than the number.

The suction a liquid is asked for #

Near a sphere of radius $R$ at a gap $x$ from a wall, the film is $h = x + r^2/2R$ thick at a distance $r$ from the axis, and Reynolds’ equation for the gap gives its pressure when the sphere moves at speed $u$:

$$p - p_\infty = \mp\frac{3\mu R u}{h^2},$$

positive on the way in and negative on the way out. Integrated over the film this is the $1/x$ force of the last of the oil, $6\pi\mu R^2 u/x$, and on the axis it is $3\mu R u/x^2$. For a sphere leaving a twenty-micron contact gap at ten centimetres a second through a light oil of 0.02 pascal-seconds, that is 45 kilopascals below ambient — nearly half an atmosphere. Through an oil ten times as viscous it is four atmospheres of tension, which no liquid with a speck of dust or a trace of gas in it will supply.

The first figure draws the film’s pressure against the distance from the axis, in units of the tension the liquid can bear, $\Delta p = p_\infty - p_{\text{cav}}$ — an atmosphere, less the vapour pressure, for a liquid that cavitates at its vapour pressure. In the case drawn, the lubrication solution asks for four times that tension on the axis. The simplest thing a liquid can do is refuse: wherever the solution asks for more than $\Delta p$, a disc of vapour opens at the cavitation pressure, and outside it the liquid carries on exactly as before. The disc’s edge is where the demand equals the floor, $h = h_{\text{cav}} = \sqrt{3\mu R u/\Delta p}$, and at that radius the outer solution already has the floor’s pressure, so it needs no adjusting.

That rule — a floor on the pressure and no other change — is the Gümbel condition, the same one that turns Sommerfeld’s journal bearing, whose exact solution demands a suction as large as its load, into the half-film that real bearings run on. It is not the only rule: the Reynolds and Jakobsson–Floberg–Olsson conditions fix up the flow at the edge of the cavity more carefully. It is the one that gives the largest cavity for a given demand, which makes it the generous test: if cavitation does not matter under this rule, it matters less under the others.

What a torn film still charges #

Integrated over the film, the floored pressure gives the suction on the way out:

$$F = 2\pi R\,\Delta p\,(h_{\text{cav}} - x) + \frac{6\pi\mu R^2 u}{h_{\text{cav}}} \qquad (x < h_{\text{cav}}).$$

The two terms are the two halves of the first figure. The first is the vapour disc: its area is $2\pi R(h_{\text{cav}} - x)$, and over all of it the pressure is $\Delta p$ below ambient, so the disc pulls with the full tension the liquid can bear. The second is the liquid round the disc, and it is exactly the force an intact film would exert at a gap of $h_{\text{cav}}$ — the torn film pulls as though the sphere were already at the gap where it tore.

The closed form was checked by integrating the floored pressure over the film directly, in four cases, torn and intact; the worst disagreement is three parts in $10^9$. Both routes give a force that is always smaller than the intact $1/x$, and the ratio shows where the saving goes. Very close to the wall, $x \ll h_{\text{cav}}$, the torn force is $2/h_{\text{cav}}$ in the units where the intact one is $1/x$: the intact film’s force grows without limit as the gap closes, and the torn film’s stops growing at twice the value an intact film would have at the tear gap. The saving is real, and it is the whole of the logarithm between $x$ and $h_{\text{cav}}$ — less a constant.

The disc term is the one intuition leaves out, and a suction cup is the everyday version of it. A cup pressed onto glass holds because the air under it is at a lower pressure than the air outside, over the whole of its area; it does not matter that there is nothing under the cup to pull with. A vapour disc under a sphere is a suction cup a few hundred microns across, holding at the liquid’s full tension, and the naive picture confuses a film that has torn with one that has let go. A film lets go only when the gap under the disc refills, which happens from the edge inwards at the rate the liquid can flow into a gap that is still only tens of microns thick.

The saving, in closed form #

The second figure is that constant. The outward leg costs, in the earlier essay’s accounting, $1/\mathrm{St}$ of the arrival speed for every factor of $e$ in gap: that is what $F \propto 1/x$ integrates to. With the sphere’s speed held fixed across the torn region, the torn force integrates from the contact gap $x_c$ to the tear gap $h_{\text{cav}} = r\,x_c$ to

$$\frac{1}{\mathrm{St}}\left[2 - \frac{2}{r} - \frac{1}{2}\left(1 - \frac{1}{r^2}\right)\right]$$

instead of $\ln r/\mathrm{St}$. The saving is the difference,

$$S(r) = \ln r - \frac32 + \frac{2}{r} - \frac{1}{2r^2},$$

which is zero at $r = 1$, grows like $(r-1)^3/3$ for a small tear, and approaches $\ln r - \tfrac32$ for a large one. The naive count, $\ln r$, is the figure’s dashed line, and the gap between the two approaches a full one and a half factors of $e$: however large the tear, the torn region charges the sphere as much as an intact film would across a factor of about four and a half in gap.

The earlier essay’s case is marked. A film that tears at a tenth of a millimetre, five times a twenty-micron contact gap, saves 0.49 rather than 1.61 — less than a third of what the naive count gives. The difference is the disc. The naive count imagines the sphere pulling away from a vapour bubble; the calculation has it pulling against a disc a few hundred microns across at nearly an atmosphere of tension, which is a large force on a sphere three millimetres across.

The outward leg, marched #

The saving formula holds the speed fixed through the torn region; the third figure lifts that. It marches the outward leg from the contact gap to the starting gap of three millimetres, with the tear gap recomputed at every step from the current speed. The group that decides everything is

$$C = \frac{\mu v_0}{R\,\Delta p},$$

the viscous suction scale of the impact against the tension the liquid bears; the tear gap is $h_{\text{cav}}/R = \sqrt{3C\hat u}$, with $\hat u$ the speed as a fraction of the arrival speed. The Stokes number does not appear in it, and neither does the solid’s density — $C$ is a property of the liquid, the speed and the size.

At a Stokes number of 12, just above the threshold, the sphere leaves the film with 0.147 of its arrival speed if the film holds. With $C = 10^{-4}$ it leaves with 0.152; with $10^{-3}$, 0.198; with $10^{-2}$, 0.275. In every case the curves are parallel once the sphere is clear of the torn region: the tear is small and closes as the sphere slows, and after that the film charges its usual rate per decade. All the saving is made in the first fraction of a decade.

The threshold moves slowly, and late #

The fourth figure is the threshold, found by bisection on the Stokes number at each value of $C$. It is exactly the earlier essay’s 10.176 until the tear can open above the contact gap at all — the tear gap at the start of the outward leg is $R\sqrt{3C\hat u}$, and with $\hat u$ near one half at the threshold that reaches twenty microns at $C \approx 3\cdot10^{-5}$. Beyond that it falls, by 0.4 per cent at $C = 10^{-4}$, five per cent at $10^{-3}$ and fourteen at $10^{-2}$, and it keeps falling by about one for every decade of $C$ towards 5.01. That floor is $L$ itself, the threshold if the outward leg cost nothing: the sphere then needs only to reach the contact gap with any speed at all.

The liquids of the earlier essay’s table are marked, for a three-millimetre sphere arriving at 0.2 metres a second at an atmosphere. Water, at $C = 6.6\cdot10^{-7}$, is fifty times too weak to tear its film above the contact gap. Light oil is below the onset. Heavy oil moves the threshold by 0.7 per cent and glycerol by four — and in both of those the spheres the earlier table put in them sit far below the threshold, where the answer to “does it rebound” is no whatever the outward leg costs.

The slowness is the same logarithm the earlier essay found in the contact gap, arriving a second time. There, a decade in the unknown contact gap moved the threshold by 4.68; here the tear gap grows only as the square root of $C$, so a decade in $C$ is half a decade in the tear and moves the threshold by about half as much again, divided by $e_d$ — near 1.2 in the limit, and a little less at the values of $C$ that can be reached. The threshold is insensitive to cavitation for the same reason it is insensitive to the roughness of the surfaces: every length in the problem enters it through a logarithm. A damper that turns into a spring found the same stubbornness from the other side, where a gas film’s compressibility, not its tension, is what the squeeze runs into.

There is a cleaner way to see why the measured threshold did not notice. At the threshold, the Stokes number and $C$ are tied: their product is $\tfrac29\rho_s v_0^2/\Delta p$, so at $\mathrm{St} \approx 10$, $C \approx \rho_s v_0^2/45\Delta p$. A five per cent shift needs $\rho_s v_0^2$ of four and a half kilopascals — a steel sphere arriving at three-quarters of a metre a second, in a liquid chosen to put it at the threshold. The threshold is quoted as “about ten” rather than to a decimal place because the measurements near it scatter by more than five per cent. Cavitation on the way out is a real effect in the faster of those impacts, and it is of a size that the scatter would hide.

An experiment in a partial vacuum #

The earlier essay’s proposed test was to lower the ambient pressure and look for a change in the rebound, since nothing else in the problem depends on it. The fifth figure prices that test for a three-millimetre sphere in a light oil at 0.2 metres a second, taking the oil to cavitate at zero absolute pressure. Because $C$ does not contain the solid’s density, the threshold for this oil at this speed is the one a family of spheres of different densities would show; a sphere of about 1,500 kilograms per cubic metre — a dense plastic — sits at it.

The threshold does not move by one per cent until the tank is below 8.3 kilopascals, and does not move by five until it is below 1.5 — a vacuum of about ninety-nine per cent. At that pressure the oil’s own dissolved gas and its lightest fractions start to matter, and the cavitation pressure is no longer zero. The test is possible, but it is not the easy one the earlier essay hoped for: the effect is small because the physics is logarithmic twice over, once in the gap and once in the tear.

A gas-saturated liquid does not change this as much as it might seem. Gas comes out of solution by diffusion into a nucleus, and the torn region lives for a few microseconds to milliseconds; a vapour cavity forms in that time, a gas cavity grown from dissolved air largely does not. The tension to use is the one at which vapour forms, and it depends on temperature — the vapour pressure of the film is that of the liquid at whatever temperature the cavity has left it, though for oil at room temperature it is negligible against an atmosphere. It is also the premise of a bubble that hammers on its way down, and of the jet pump whose curve stops at the suction the liquid can bear.

What was checked #

The sixth figure is the ledger. The closed-form force agrees with the floored pressure integrated over the film to three parts in $10^9$. The threshold with no cavitation reproduces the earlier essay’s closed form, 10.1760 against 10.1762, the difference being the step size of the march; and with an enormous $C$ it approaches $L = 5.011$ — 5.036 at $C = 10^6$, the residue of the part of the leg outside the tear. The threshold is unchanged at $C = 10^{-5}$, where the tear cannot open above the contact gap, and falls monotonically from there.

What the picture cannot show #

The simplest cavitation rule. The Gümbel floor lets a vapour disc appear and vanish instantly with the pressure. Real cavities take time to grow and to collapse, and a cavity that outlives the demand for it goes on shielding the sphere a little longer; the more careful Jakobsson–Floberg–Olsson rule conserves the liquid and gives a smaller cavity. Both move the answer towards the uncavitated one.

The approach is unchanged. On the way in the film is in compression and nothing tears, but a film squeezed hard enough changes its viscosity with pressure and deforms the solids elastically — the elastohydrodynamic regime, which moves the contact gap itself.

The contact is a length. As in the earlier essay, the dry bounce happens at a single gap and returns a fixed fraction of the speed. A cavitating film could change the bounce too, if the vapour disc is still open when the solids touch.

Lubrication all the way out. The film theory holds while the gap is small against the radius, and the leg is followed out to a gap equal to the radius, as the earlier essay did. The cost of the last decade is the least certain, and it is the same in every case compared here.

The convention the numbers depend on #

Lengths are in sphere radii, speeds in the arrival speed. The sphere is three millimetres in radius, starts its approach at a gap of three millimetres, touches at twenty microns and has a dry restitution of 0.97, all as in the earlier essay. The tension limit $\Delta p$ is the ambient pressure less the pressure at which the liquid cavitates, taken as zero absolute for the oil in the fifth figure. The Stokes number is $m v_0/6\pi\mu R^2$, the cavitation group $\mu v_0/R\,\Delta p$.

Who found it, and when #

The floor on a bearing’s pressure is Gümbel’s, from 1914, and the mass-conserving cavitation conditions are Jakobsson and Floberg’s of 1957 and Olsson’s of 1965. Cavitation during the rebound has been raised repeatedly in the literature on lubricated collisions, usually as a reason the measured restitution might differ from the model, and the measured threshold near ten is from the experiments of Joseph, Zenit, Hunt and Rosenwinkel in 2001 and Gondret, Lance and Petit in 2002. The saving formula here follows directly from the floored film and is given for the flat-to-sphere geometry of the earlier essay.

Still open: whether the vapour disc outlives the bounce #

The pressure floor treats the cavity as appearing and vanishing with the demand, and at the contact gap the demand is reversing in microseconds: in on the approach, out on the rebound. The next calculation lets the vapour disc have a life of its own — growing at the rate the film’s inflow allows, collapsing as a Rayleigh cavity does once the demand is gone — and asks whether a disc opened on one bounce is still there on the next, so that a sphere bouncing repeatedly in a viscous liquid leaves a train of cavities that makes each later rebound cheaper than the first.