Rossby Number

Everything on a rotating planet has to answer one question before anything else: does the rotation matter here? The Rossby number is the answer, and it is a comparison of two timescales dressed up as a ratio of forces.

The inertial timescale is $L/v$, how long the flow takes to cross its own system. The rotational timescale is $1/f$, how long the Coriolis force takes to turn it through one radian. Divide the second into the first and you have

\[\mathrm{Ro} = \frac{v}{fL}\]

Small Ro means the flow is deflected many times over while it crosses the system, so rotation runs everything. Large Ro means the flow is across and gone before rotation has bent it measurably, so rotation is irrelevant.

Small Rossby number is why weather maps look the way they do. A mid-latitude system has $v \approx 10$ m/s and $L \approx 1000$ km, and at 45° latitude $f \approx 1.03\times10^{-4}$ s⁻¹, giving $\mathrm{Ro} \approx 0.1$. Below about that value the pressure-gradient force is balanced not by acceleration but by the Coriolis force, which is geostrophic balance, and the immediate consequence is that the wind blows along the isobars rather than across them. Air does not rush into a low to fill it; it circles the low indefinitely. That single fact, which every forecast depends on, is a statement about a dimensionless number being small.

Push the flow faster or make the system smaller and Ro rises. Near a hurricane eyewall, and at tight jet-stream curvature, the number climbs to order one and the centrifugal term joins in — gradient-wind balance instead of geostrophic. At order ten and above, rotation is a footnote.

Which brings us to the bathtub, and to the pleasure of settling a famous argument with arithmetic instead of assertion.

The claim is that water drains anticlockwise in the northern hemisphere and clockwise in the southern. Compute the Rossby number. A bath is about half a metre across and the water near the plug moves at something like 0.1 m/s. At 45° latitude:

\[\mathrm{Ro} = \frac{0.1}{1.03\times10^{-4} \times 0.5} \approx 1900\]

Nearly two thousand. Set that beside the weather system: $\mathrm{Ro} = 10/(1.03\times10^{-4} \times 10^{6}) \approx 0.097$. The same formula, two scales, and a factor of twenty thousand between them. The Coriolis force on your bathwater is not zero — it is about four orders of magnitude weaker than the residual swirl already in the tub from how it was filled, how you got out of it, the shape of the basin and where the outlet sits. It loses, every time, to effects nobody controls and nobody notices.

The honest footnote is that the experiment can be done. Ascher Shapiro at MIT in 1962, and Lloyd Trefethen in Sydney in 1965, both did it: a symmetrical circular tank, filled carefully, covered, and left to settle for a full day so that every trace of the filling motion had died away, then drained very slowly through a small central hole. The predicted rotation duly appeared, anticlockwise in Boston and clockwise in Sydney. That result confirms the folk claim in one sense and demolishes it in another: it took twenty-four hours of stillness and a purpose-built tank to make Coriolis the largest remaining effect. Your bath is not that tank, and the demonstration at the equator where a man with a bucket shows you the water turning both ways ten metres apart is a conjuring trick with a wrist flick in it.

Two further uses of the group. Run it at $\mathrm{Ro} = 1$ and solve for length and you get the Rossby radius of deformation, the scale above which a flow of a given speed organizes itself into a rotating structure and below which it does not. In the atmosphere it is of order 1000 km, which is the size of the highs and lows on a chart; in the ocean, where speeds are far lower, it is 10 to 50 km, which is why ocean eddies are so much smaller than storms and why resolving them takes so much more grid. And note how strongly latitude enters through $f$: at 10° north, $f$ is a quarter of its 45° value, so the same flow has four times the Rossby number. Tropical meteorology is a different discipline rather than the same discipline in warmer air, and this is why.

The choice hiding here is $L$, and the group is linear in it, so a system quoted on its radius and the same system quoted on its diameter differ by a factor of two. Since the thresholds are order-of-magnitude judgements that rarely changes a conclusion — but state the scale you used, as you would for a Reynolds number.