Nothing but the shape of the gap

A journal in a turbine carries several tonnes on a film of oil that is thinner than a sheet of paper and never touches the shaft. Nothing pumps that oil. There is no valve, no accumulator, no supply pressure worth the name — the only moving part is the shaft itself, and the pressure under it is measured in tens of megapascals.

The mechanism is a shape. A film dragged into a narrowing gap has to leave faster than it arrived, and the only thing that can accelerate it is a pressure gradient, so the film builds one. The consequence is exact and slightly startling: the same oil in a gap of the same size that does not narrow builds nothing at all.

What is left of the equations when one length is tiny #

A lubricating film is about a thousand times longer than it is thick. That ratio is not a detail; it is the whole reason the problem has a closed-form answer, and the argument is the same one that makes the boundary layer tractable — write the Navier–Stokes equations, compare the size of each term, and delete what cannot matter.

Across the film, viscous shear is of order $\mu U/h^2$ and inertia is of order $\rho U^2/B$. Their ratio is a Reynolds number multiplied by the aspect ratio,

$$\frac{\rho U^2/B}{\mu U/h^2} = \mathrm{Re}\,\frac{h^2}{B^2},$$

and with $h/B \approx 5 \times 10^{-4}$ the second factor is a quarter of a millionth. Inertia is gone even when the Reynolds number is respectable, which is why a bearing turning at two thousand revolutions a minute is still solving the equations of a creeping flow. The pressure cannot vary across so thin a film either, so $p$ is a function of $x$ alone, and what remains is one ordinary differential equation:

$$\frac{d}{dx}\left(h^3 \frac{dp}{dx}\right) = 6\mu U \frac{dh}{dx}$$

The right-hand side is the mechanism, and it is a shape. Where $dh/dx$ is zero the equation reads $(h^3p')' = 0$, whose only solution vanishing at both ends is $p \equiv 0$. Nothing about the oil appears in that argument. The viscosity sets how much pressure a given taper produces; the taper sets whether there is any.

The pad that carries nothing #

This is the essay’s claim in one picture, and it is worth being precise about what “zero” means here. It is not a small number reported as zero. The quadrature returns a value of order $10^{-13}$ times the scale of the converging pad’s load, which is the arithmetic’s own noise floor. There is no regime, no speed and no oil in which a parallel film generates lift.

That fact has a practical shadow which is easy to miss. A flat thrust face cannot be made to work by polishing it. Every real flat pad that carries a load is carrying it on a taper it acquired somewhere — thermal distortion, elastic deflection, or wear — and a manufacturer who removes those imperfections in the name of quality removes the bearing along with them.

Where the peak sits, and why it is not in the middle #

Integrating the equation once gives $h^3 p' = 6\mu U h + C$, and the constant follows from requiring zero pressure at both ends. For a linear film it comes out as

$$C = -\frac{12\mu U h_1 h_2}{h_1 + h_2},$$

so $dp/dx$ vanishes where $h$ equals the harmonic mean of the inlet and outlet clearances. For the pad drawn above, 55 µm and 25 µm give a harmonic mean of 34.4 µm and a peak 69 per cent of the way along — downstream of the middle, because the harmonic mean leans towards the smaller of its two arguments.

The centre of pressure is a different point again, at 58 per cent of the length, and it is the one a designer needs: a tilting pad pivoted anywhere else will not sit at the angle it was designed for. The two are frequently confused, and this site computes both rather than either.

The backflow is the pressure #

The profiles carry the mechanism more directly than the pressure plot does. Near the inlet the film is thick and the pressure is rising, so the pressure gradient opposes the drag and the fluid next to the stationary surface is moving backwards. Near the outlet the gradient has reversed and pushes the flow along.

What is conserved through all of it is the flux. Mass conservation in a film is the statement that

$$q = \frac{U h}{2} - \frac{h^3}{12\mu}\frac{dp}{dx}$$

is the same at every station, and the solver checks it at ten of them before drawing anything. This is the film’s version of the mass conservation asserted on every field on this site, and it is the assertion that would catch a plausible-looking pressure field describing a film that quietly creates fluid. The check is cheap and the failure it prevents is invisible: a leaking film draws exactly as smoothly as a conserving one.

There is a tidy consequence. Setting the two expressions equal at the peak, where $p' = 0$, gives $q = U\bar{h}/2$ with $\bar{h}$ the harmonic mean — so the whole film flows as though it were a plain Couette flow in a gap of the harmonic-mean thickness, and the pressure that does all the work transports nothing on balance.

The taper that carries most, found rather than quoted #

Writing the closed-form load as $W = (6\mu U B^2/h_2^2)\,G(k)$ with $k = h_1/h_2$ leaves

$$G(k) = \frac{\ln k - 2(k-1)/(k+1)}{(k-1)^2},$$

a function of the taper ratio and of nothing else. Every textbook prints its maximum as “about 2.2”.

The two searches return 2.1887047404 and 2.1887047960. That agreement is the point of doing it twice: a maximum found by sampling a curve is a property of the sampling until something independent confirms it, and the same discipline is applied to every root on this site.

The curve’s shape matters more than its maximum. A bearing built at $k = 2$ loses 0.9 per cent of the available load and one built at 2.5 loses 1.6 per cent, so the optimum is a broad one and nobody has to hit it. What is not broad is the other end: at $k = 1.1$ the pad carries just over a quarter of what it could, and at $k = 1$ it carries nothing.

What the support costs #

A bearing is a bargain and the arithmetic says how good a one.

At the film drawn here the coefficient is 0.0024. Dry steel on steel is around 0.6, so the film has bought a factor of about 250, and it has done it while carrying 12.8 MPa of mean pressure. Both the drag and the load rise as the film thins — the drag as $1/h$ and the load as $1/h^2$ — and it is that difference of one power that makes the machine worth building.

The comparison to draw is with the friction factor of a pipe, where the useful number is also a ratio of two things the flow supplies rather than an absolute. In both cases the dimensionless group hides an enormous range of absolute forces, and in both cases quoting the coefficient without the geometry it belongs to is meaningless.

Round a shaft, the exact solution asks for something impossible #

Wrapping the same equation round a journal gives a film $h(\theta) = c(1 + \varepsilon\cos\theta)$ and a closed-form pressure that Sommerfeld published in 1904. It is exactly antisymmetric.

The suction is the interesting part. Sommerfeld’s solution is an exact solution of the right equation, and the assumption hiding inside it is that the liquid will sustain whatever negative pressure the arithmetic asks for. It will not: below the vapour pressure it boils, which is cavitation, and the diverging half of a real bearing is a region of streaming ribbons at roughly ambient pressure rather than a region of tension.

Gümbel’s amendment is to keep the positive half and discard the negative. It is crude, it is the standard first answer, and it changes the load by nearly half and its direction by twenty-five degrees. This is a good example of a general point that this site keeps meeting: an exact solution of the correct equation can be far from the machine on the bench, and the distance is set by a boundary condition rather than by the equation. The same shape of error appears when the ideal theory predicts no drag at all: the mathematics is faultless and the physical hypothesis is not.

There is one more thing the antisymmetry says, and it is worth stating because it is a genuine prediction rather than an artefact. Because the full solution’s load is exactly perpendicular to the line of centres, an uncavitated bearing would carry its load at ninety degrees to the direction the shaft is displaced. The measured attitude angle of a real bearing runs from about 70 degrees when lightly loaded down to a few degrees when heavily loaded, and the whole of that range is cavitation.

The bearing that is allowed to be parallel #

The claim that a parallel film carries nothing is exact, and it is a claim about a film with nothing driving it but the moving surface. Supply the pressure from outside and the whole argument is void — which is not a loophole but a second class of machine, built for the cases where the first will not do.

A hydrostatic bearing has a pump. Oil or air is fed through a restrictor into a shallow pocket in the middle of a deliberately parallel pad, and it escapes across the surrounding land. The pressure in the pocket is whatever the restrictor and the escape resistance settle on, and it carries the load directly. There is no taper anywhere, the film is a gap rather than a wedge, and Reynolds’ equation’s right-hand side is zero throughout — because the pressure is not being generated by the flow at all.

The stiffness comes from the restrictor and is worth following, because it is the design’s whole subtlety. Push the pad down and the land’s clearance falls; the escape resistance goes as $1/h^3$, so the flow out drops steeply; so the pocket pressure rises towards the supply pressure and pushes back. The restrictor is what makes that a stiff response rather than a soft one — with no restrictor the pocket sits at supply pressure whatever the gap and the bearing has no stiffness at all — and choosing its resistance against the land’s is the entire design.

Two properties follow that a hydrodynamic film cannot offer. It works at zero speed, so a machine that starts under full load never touches, and a telescope weighing hundreds of tonnes can be floated before it is asked to move. And its stiffness is set by the supply rather than by the operating point, which is why precision machine-tool spindles and metrology stages are built this way: the film’s thickness is a designed number rather than an outcome.

The costs are equally plain. There is a pump, and its power is a permanent parasitic loss. There is a filtration requirement, since a restrictor is a small hole. And the failure mode is abrupt: a hydrodynamic bearing that loses pressure has been slowing down, while a hydrostatic one that loses its supply loses everything at once, at whatever speed it happened to be turning.

Gas changes the arithmetic again. Air is three orders of magnitude less viscous than oil, so an air film generates correspondingly less pressure and runs at gaps of a few microns — but it is also compressible, and that brings a new group. Writing the compressibility number $\Lambda = 6\mu U B/(p_a h^2)$, the film behaves like an ordinary incompressible one when $\Lambda$ is small, and as $\Lambda$ grows the density variation limits how much pressure the film can build, so the load saturates rather than rising without bound with speed. A gas bearing therefore has a ceiling that an oil bearing does not, and it is reached at practical speeds.

What it buys in exchange is everything oil costs: no contamination, no viscous drag worth measuring, no temperature rise, and operation at temperatures where any liquid would boil or freeze. Which is why the same argument that says a flat pad carries nothing has produced two entirely different industries — one that shapes the gap, and one that pays a pump to make the shape unnecessary.

The gap that will not close #

Turn the geometry through ninety degrees — surfaces approaching rather than sliding — and the same equation gives the squeeze film.

The integral is finite and unhelpful. Pressing two 40 mm discs together with a hundred kilograms takes 38 seconds to squeeze the film from a tenth of a millimetre to a micron, and 3,800 seconds to get to a tenth of a micron. Closing the gap completely takes infinite time at any finite load.

That is why a bearing survives being started and stopped, why a wet plate sticks to a table, and why two gauge blocks wrung together are so hard to separate. It is also why the film in a bearing that has been standing for a month is still there.

What the model does not contain #

Cavitation is not solved anywhere here. The journal figures draw Sommerfeld’s exact solution and then discard its negative half, which is Gümbel’s rule and is a decision rather than a computation. Where the film ruptures, and along what line it reforms, is a free-boundary problem; the standard better answer is Reynolds’ own condition — pressure and its gradient both zero at the rupture — and nothing on this site solves it.

The oil has one viscosity. A real film is hottest where the shear is greatest, the viscosity of mineral oil falls by roughly half for every twenty-five degrees, and the thermal problem is coupled to the hydrodynamic one through exactly the term that generates the load. Bearing design is more a thermal calculation than a hydrodynamic one, and none of that is here.

The surfaces are rigid, smooth and perfectly aligned. Elastic deflection matters at the pressures computed above — 20 MPa deflects steel by microns, which is a substantial fraction of the film — and that coupling has its own subject, elastohydrodynamics, in which the pressure is high enough to change the viscosity by orders of magnitude. Roughness matters too: an average machined surface has asperities of a micron or so, which is a twenty-fifth of this film.

Nothing here says whether the film exists. All of the above assumes a full film. Whether one forms at all is a question about starting, starvation and supply, and the parameter that decides it is not in any equation on this page.

Who found it, and when #

The order of discovery is unusually clean and unusually instructive. In 1883 Beauchamp Tower was asked by the Institution of Mechanical Engineers to measure the friction of railway journal bearings. He drilled a hole in the top of a bearing to admit lubricant, and the oil came out of the hole. He plugged it with a cork; the cork was pushed out. He fitted a pressure gauge and found pressures several times what the load, divided by the projected area, could account for — which is to say he measured a distribution rather than an average, by accident, while trying to fill an oil hole.

Osborne Reynolds explained it in 1886 with the equation above, three years before the paper on turbulence that his name is better known for. Nikolai Petrov had computed the friction of a concentric bearing in Russia in 1883 without the pressure; Arnold Sommerfeld solved the journal in closed form in 1904; and Anthony Michell patented the tilting pad in 1905, whose whole content is that a pad free to pivot finds its own taper and therefore cannot be built flat by mistake.

Tower’s cork is the best experiment in this subject. It measured a pressure nobody was looking for in an apparatus built to measure something else, and the theory arrived three years later to explain a number that was already on the table.

Where the ladder goes next #

The film here is dragged along a wedge. Take the wedge away and let the two surfaces be still, and another exact solution appears: the corner between two stationary walls, where a slow flow outside sets up an infinite sequence of eddies inside, each hundreds of times weaker than the one before it. That sequence is decided by a complex exponent, and no experiment has ever seen more than three of them.