# Resolving water sink problem

There is a common myth that water flowing out from a sink should rotate in direction governed by on which hemisphere we are; this is shown false in many household experiments, but how to show it theoretically?

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I would go about this by computing the magnitude of the Coriolis effect in a typical sink drain and comparing it to other effects that might change the direction of the drain, e.g. some tilt in the sink or faucet. –  j.c. Nov 2 '10 at 20:32
Exactly what @J.C. said. Other factors, namely the angle of the sink's axis relative to gravity and how the water enters the sink or bowl. –  Mark C Nov 22 '10 at 18:07
@Mark For breaking 1k? Thanks. Now its time to make some edits [-; –  mbq Nov 22 '10 at 20:35
Yes, I like to be the one to give the "deciding" vote. –  Mark C Nov 22 '10 at 23:06

The calculation of the Coriolis force is dependent on latitude:
$F = m a$ where $a = 2 \Omega sin(lat)$, with $\Omega$ being the Earth's angular velocity
$m$ is the mass of the object in question

The Earth's angular velocity is (about) $7.29 \times 10^{-5}$ rad/sec

So, for a sink with a couple gallons of water in it at 45 degrees north... the Coriolis force is about $7.57 \times 2 \times 7.29 \times 10^{-5} = 1.10 \times 10^{-3}$ N.

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The Coriolis force is dependent on velocity. You can't get it just from the latitude. –  Mark Eichenlaub Mar 29 '11 at 12:45
One way of seeing that this answer is wrong is by noting that the units don't make sense. –  Ben Crowell Jul 25 '13 at 17:39

The sink demonstration is not an optimal one to show the effect of Coriolis force, mainly because the water in the sink will never be perfectly still enough to start with in order for that force to be the dominant one to determine the direction it will swirl (you could force it to spin either direction.

Better demonstrations are easy, however. One effective demo would be to fire a high powered rifle at a distant target (neglecting windage, of course). The bullet should always curve to the right in the Northern Hemisphere, and it should always curve to the left in the Southern Hemisphere.

Anyone who flies a powered lighter than air craft (blimp or dirigible) can attest, it makes a considerable difference when flying Westward or Eastward. Flying Westward in the Northern Hemisphere will gain altitude / buoyancy. Flying Eastward will lose altitude / buoyancy. In the Southern Hemisphere, the opposite effects will prevail.

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The Coriolis acceleration goes like $-2\omega \times v$, which for the sake of an order of magnitude estimate we can take to be $a\sim \omega v$. But in order to get an observable effect, we don't just need an acceleration, we need a difference in acceleration between the two ends of the tub, which are separated by some distance $L\sim 1$ m. The accelerations differ because $v=\omega r$, and $r$ differs by $\Delta r\sim L$. The result is that the difference in acceleration is $\omega^2 L$, which is on the order of $10^{-8}$ m/s2. This is much too small to have any observable effect in an ordinary household experiment.

This explains why the Coriolis effect is important for hurricanes (large L) but not for bathtub drains (small L).

Detecting the Coriolis effect in a draining tub requires very carefully controlled experiments (Trefethen 1965; also see this web page by Baez). Lautrup 2005 gives numerical estimates showing that in order to see the Coriolis effect, the the water must be very still ($v\lesssim 0.1$ mm/s), the water must also be allowed to settle for several days, and precautions have to be taken in order to prevent convection.

Lautrup, Physics of Continuous Matter: Exotic and Everyday Phenomena in the Macroscopic World, p. 289

Trefethen, Letters to Nature 207 (1965) 1984, http://www.nature.com/nature/journal/v207/n5001/abs/2071084a0.html

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