The bath that was only ever a wait

The bath does not know the hemisphere computes the Coriolis acceleration at bath scale and finds it four orders of magnitude below the asymmetries the tub already has: the shape of the basin, the residual motion from filling, the plug being pulled sideways. The conclusion is right and the essay is right.

It also leaves a loose end, because the experiment has been done and it does show the hemisphere. Ascher Shapiro’s tank in Massachusetts drained anticlockwise, and Trefethen’s in Sydney drained clockwise, and both were repeatable.

This essay is about what the difference is between those two situations, and the answer is not a force.

What the earth is actually worth #

Start with the quantity that matters, which is not an acceleration.

A parcel of water sitting at radius a in a tank at latitude φ is going round with the earth. Relative to the tank it has a small circulation — the vertical component of the earth’s rotation times the area it encloses. When it drains inward, its angular momentum is conserved, so its azimuthal velocity rises as it approaches the hole.

That gives a velocity at the drain of the Coriolis parameter times the difference of the squares of the two radii, over the drain radius. It is small at the wall and it is not small at the hole, because the hole is small.

For a washbasin it is 0.06 millimetres a second. For a bathtub, 0.22. For a garden pool, 5.5. For a 1.8-metre tank draining through a 4.8-millimetre hole, 8.6 millimetres a second — a hundred and forty times the washbasin’s, from the same earth.

Two vessels at the same latitude of 42.4 degrees, where the Coriolis parameter is 4.917·10⁻⁵ s⁻¹:

The spread comes from the geometry: the velocity goes as the tank’s area over the drain’s radius. That is why the answer to “does the Coriolis force steer the drain?” depends on which vessel is being argued about, and why arguing about it in the abstract cannot terminate.

The number that settles it #

A velocity at the drain is still not the observable. What is watched is a rotation, and the question is how much of one happens before the water runs out.

Divide the draining time by the rotation period at the drain.

A washbasin empties in nine seconds and its Coriolis rotation period is thirty-four minutes: five thousandths of a turn. A bathtub empties in fifty-six seconds with a period of nine and a half minutes: 0.098 of a turn. There is nothing to see, and no amount of care with the plug changes it, because the vessel is gone before the rotation has started.

The 1.8-metre tank empties in an hour and three quarters with a rotation period of three and a half seconds: 1,873 turns.

That factor of nineteen thousand between the tub and the tank is the whole of the disagreement, and it is bought almost entirely by the drain. Shrinking the hole slows the draining as the square of the radius and speeds the rotation as the first power, so the number of turns goes as the cube of the ratio of tank to drain. The apparatus is not a bigger bath; it is a machine for making that ratio large.

The wait #

The remaining requirement is the one that makes this a memory problem, and it is the one Shapiro’s account is mostly about.

Water put into a tank is left circulating. That residual swirl is enormous compared with the earth’s contribution — filling a tank leaves centimetres a second, against the tank’s 8.6 millimetres — and it decays, because the walls and the floor exert a stress on it.

The decay is a diffusion problem with a known slowest mode: a first-kind Bessel function across the radius, whose first zero is 3.8317059702, and a quarter wave from the no-slip floor to the free surface. The rate is the viscosity times the sum of the two eigenvalues, and comes out at 1.249·10⁻⁴ s⁻¹. For the tank that is a decay time of 8,008 seconds — two hours and thirteen minutes.

The Coriolis contribution does not decay. It is set by conservation of angular momentum from the wall to the drain and it is the same on the first day as on the second.

So the experiment is a race that the experimenter wins by waiting, and the wait is a logarithm: the decay time times the log of the ratio of the starting swirl to the earth’s contribution.

For five centimetres a second of residual swirl in that tank the wait is 3.9 hours. Shapiro waited twenty-four, which is eleven decay times — a margin of five hundred thousand in the residual, and a reasonable choice for an experiment that allows one attempt a day.

The logarithm is what makes the experiment possible at all. Filling the tank a hundred times more roughly costs 2.3 more decay times, which is five hours, rather than a hundred times as long.

What the solver computed, and how it was checked #

The arithmetic is short and the checks are on the parts that are usually quoted rather than computed.

The Bessel root. The first non-zero root of the first-order Bessel function is summed from the power series and located by bisection at 3.8317059702, which matches the tabulated value to ten figures. It is quoted in every treatment of this decay and it is two dozen lines of arithmetic, so it is computed here rather than typed in.

The logarithm. The wait is computed at seven starting swirls and each interval is checked against the decay time times the log of the ratio, to a part in a million. That is the check that the wait really is a logarithm rather than something that happens to look like one over the range drawn.

The turns. A bathtub is required to complete less than a quarter of a revolution and the apparatus to complete more than a hundred times as many. If either failed, the essay’s reconciliation of the two positions would be wrong, and it is the assertion that would say so.

And a refusal. A tank whose drain is bigger than the tank is refused rather than returning a negative area.

Against the observation #

One number here is checkable against what was actually reported, which is unusual for a calculation of this kind.

The Coriolis velocity at the drain of the tank comes to 8.6 millimetres a second, and the drain radius is 4.8 millimetres, so the fluid at the edge of the hole is going round with a period of 3.5 seconds. Shapiro’s account describes watching a cross made of floating chips turn slowly, over a few seconds per revolution, in the last stage of the drain.

That is a prediction with no fitted parameter in it — the latitude, the tank’s radius and the drain’s radius are all measurements of the apparatus — landing in the range that was reported: a residual swirl of five per cent needs a wait of 14,070 seconds, which is 3.9 hours against the twenty-four Shapiro allowed. It is the best evidence available here that the mechanism identified is the one that was observed.

Where the two essays meet #

Both essays are right and they are answering different questions, which is worth stating explicitly because the disagreement is usually conducted as though one of them must be wrong.

The previous rung asks whether the Coriolis force dominates. It does not, anywhere in a domestic vessel, and the ratio it computes — a Rossby number of about ten thousand for a bath — is the correct answer to that question.

This one asks whether the Coriolis contribution is the only one left. That is a different question, and it is answered by waiting rather than by comparing forces. A Rossby number of ten thousand says the earth cannot overpower the residual motion; it says nothing about what happens when the residual motion has been allowed to decay to nothing, because a ratio to zero is not a ratio.

So the resolution is temporal rather than dimensional. The competing effect is not weaker than the Coriolis one; it is transient, and the Coriolis one is not. Any experiment that can outlast the transient sees what is left.

The same structure appears wherever a small persistent effect competes with a large decaying one: sedimentation in a stirred tank, a slow drift under a fluctuating force, a secondary flow in a pipe that only appears when the entry disturbance has died. The question is never which is bigger; it is which one is still there.

The company this argument keeps #

Three other essays in this collection have the same shape, and putting them beside each other is the fastest way to see what kind of statement is being made.

A threshold that is also a duration finds that a cavitation threshold quoted as a pressure is the long-hold limit of a curve, and that a short pulse needs forty times as much. The tabulated number is right and applies to a case the machine is not in.

The part of the closure a pipe cannot see finds that a surge peak is a functional of the valve’s whole history and that a duration does not determine it. The specification is not wrong, it is incomplete.

The outlet is the inlet, a while ago finds that a bed’s mean residence time says nothing about when it first breaks through, and the quantity that decides the design is the one nobody measures.

In every case a number that is correct is being used to answer a question it does not address, and the repair is the same: identify what is actually being asked, find the time it depends on, and compare. That is the procedure counting what matters sets out in general, and the bath is its simplest instance — the question is not “how strong is the Coriolis force” but “how many turns before the water is gone”.

The spin-down time, and what this collection already knows about it #

The decay time used above is the same object this collection computes elsewhere under a different name.

How long a fluid takes to forget it was not rotating prices spin-up in a rotating container and finds the Ekman time — the depth over the square root of viscosity times rotation rate — which is far shorter than the diffusive time and is the reason a spinning container settles in a hundred seconds rather than ten thousand.

That mechanism is not available here, and the reason is instructive. Ekman spin-down needs a background rotation to produce its boundary layer, and a tank sitting on a laboratory floor at 42 degrees north is rotating at 5·10⁻⁵ radians a second. Its Ekman time is longer than the diffusive one, so the slow mechanism wins by default and the decay is the diffusive mode used above.

That is a real difference between a laboratory tank and, say, a cup of tea being stirred and then released on a turntable: the second has a background rotation and settles in seconds, and the first does not and takes hours.

Where the number belongs #

The comparison here is a residence time against a rotation period, which is the same construction as every other group in this collection — every memory number is one time over another — with the draining time as the process and the Coriolis rotation period as the thing being compared against it.

It is worth noticing that the wait is a second ratio, of a different kind: the decay time against the patience of the experimenter, mediated by a logarithm. A vessel can fail the experiment in two independent ways — by emptying before the rotation completes, or by being used before the filling has been forgotten — and the four vessels in the ledger fail in different ones.

The washbasin fails on turns: even after a week of standing it would drain in nine seconds and rotate five thousandths of a turn. No amount of waiting fixes it, which is the honest reply to anyone proposing to do the experiment at home in a sink.

What the picture cannot show #

The decay figure draws the residual swirl as a single exponential, which is the slowest mode. Immediately after filling there are faster modes on top of it, so the real decay starts steeper and only becomes the drawn line after about one decay time. That makes the wait computed here slightly conservative, which is the safe direction.

The vessels’ rotation periods are computed at the drain radius, where the velocity is largest. Further out the fluid is going round more slowly, so a picture of the whole surface rotating uniformly is wrong — the rotation is concentrated near the hole, which is exactly where Shapiro put his floating cross.

The vessels are also drawn at one instant of a process whose whole content is a history, which is the difficulty the shutter is part of the answer is about: the observable is the accumulated rotation over the draining, and no picture of the surface at one moment carries it.

And the draining time is Torricelli’s, which ignores the vortex that forms. Once a visible funnel develops the discharge changes, and the last part of the drain is not the process modelled here.

Where the model stops #

No drain vortex. The calculation is a kinematic statement about angular momentum, and the actual flow near a draining hole is a concentrated vortex with its own core and its own dynamics. What is computed is the circulation available to it, not the vortex.

One decay mode. The higher radial and vertical modes are dropped, and the wall’s own boundary layer is treated as no-slip diffusion rather than as a turbulent layer, which it may be immediately after filling.

Nothing about temperature. A tank of water a degree warmer at the top than the bottom convects, and convection is a far more effective destroyer of residual swirl than viscosity — which is why Shapiro’s protocol includes covering the tank, and which this calculation does not represent.

And no other asymmetry. A tank that is a millimetre out of round, or whose drain is a millimetre off centre, imposes its own circulation. That is the previous essay’s list of confounders and none of them decays with time, so the wait does not help against any of them. Building the apparatus symmetric enough is the part of the experiment that is craft rather than arithmetic.

Who found it, and when #

Shapiro’s experiment is 1962, in Nature, one page. Trefethen and colleagues repeated it in Sydney in 1965 and got the mirror result. Both are careful papers whose method is mostly a protocol for making the initial conditions negligible: fill through a diffuser, cover, wait a day, pull the plug from below.

The theoretical framing — that this is a Rossby-number problem — is older and is standard geophysics. The part that is rarely stated in the popular retelling is the one this essay computes: the experiment is not delicate because the force is small, it is delicate because the memory of the filling is large, and the entire protocol is about erasing it.

Limits recorded rather than smoothed over #

The starting swirl is assumed. Five centimetres a second is a plausible residual after careful filling and is not measured. The wait scales with its logarithm, so a factor of ten in it costs 2.3 decay times, which is why the number is quotable at all despite the uncertainty.

The vessels are stylised. A cylinder with a hole in the middle. Real baths are rectangular and their drains are at one end, which changes the angular momentum accounting and generally reduces it.

The latitude is one value. Everything is computed at 42.4 degrees north, which is where the original experiment was. The Coriolis parameter goes as the sine of the latitude, so the tropics are harder and the poles easier by a factor that reaches 1.5 either way.

And the drain time is a free-surface fall. It ignores the discharge coefficient of the hole, which is about 0.6 for a sharp-edged orifice and would lengthen every draining time by about sixty per cent — making the turns count larger and the argument stronger, which is why the conservative version is the one computed.