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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    $\begingroup$ 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. $\endgroup$
    – j.c.
    Nov 2, 2010 at 20:32
  • $\begingroup$ 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. $\endgroup$
    – Mark C
    Nov 22, 2010 at 18:07
  • $\begingroup$ Good, self contained answer here: physics.stackexchange.com/a/748980/226902 $\endgroup$
    – Quillo
    Feb 7 at 8:41

4 Answers 4


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

  • $\begingroup$ It's misleading to call it an acceleration. It is indeed a change of velocity but not per unit time, rather, per unit distance. Your east-west velocity will have changed by the same amount after going a certain distance to the north, and it doesn't matter whether you whether you walked, ran or flew. So the proper units are 1/s. $\endgroup$
    – Adrian May
    Oct 27, 2017 at 9:28
  • $\begingroup$ $\omega \times v$ is an acceleration. The units are m/s$^2$. If you want to call acceleration $v dv/dx$ that's ok too, but the units are still m/s$^2$. @AdrianMay You appear to be concerned that what matters is how much the velocity changes by, not how long it takes. I dealt with that in my answer to the duplicate. $\endgroup$
    – ProfRob
    Oct 27, 2017 at 10:23
  • $\begingroup$ I don't dispute that $\omega \times v$ is an accelaration. I just don't think it's relevant that's all. Funnily enough we get the same number though. $\endgroup$
    – Adrian May
    Oct 27, 2017 at 10:35
  • $\begingroup$ Anyway Rob, Can you help with physics.stackexchange.com/questions/365171/… $\endgroup$
    – Adrian May
    Oct 27, 2017 at 10:38
  • $\begingroup$ A both-hemispheres experiment documented in Youtube videos. $\endgroup$
    – rob
    Feb 7 at 15:59

If you stand at a latitude of $\theta$ radians north of the equator, then the Earth's rotation is carrying you eastwards at


where $\omega$ is the angular velocity of the Earth's rotation (i.e. $2\pi$ in 24 hours) and $R$ is its radius.

If you throw a stone a distance $D$ due north its $\theta$ will change by $\Delta\theta = D/R$ and its $V$ by

$\Delta V = \frac{dV}{d\theta}.\Delta\theta = -R\omega.sin(\theta) . (D/R) = -D\omega.sin(\theta)$

For a typical bath at middling latitudes $D.sin(\theta)$ is of order 1 meter, so the change of velocity for an element of water going from one end to the other is $\omega$ meters. ($\omega$ is in $s^{-1}$ so this is a velocity.)

$\frac{2\pi}{24*60*60 *seconds}$.$meters$ = $73\mu m/s$

You'd have to be incredibly careful to set up an experiment where this played any role. A pondskater would be like a tsunami in comparison. Bear in mind, that stone was assumed to be unimpeded by anything, but in a bathtub the elements of water are very much impeded by the mass and viscosity of the rest of the water, so that value is a huge overestimate.


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.


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.

Interesting reading about this over this way, and (of course) here.

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    $\begingroup$ The Coriolis force is dependent on velocity. You can't get it just from the latitude. $\endgroup$ Mar 29, 2011 at 12:45
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    $\begingroup$ One way of seeing that this answer is wrong is by noting that the units don't make sense. $\endgroup$
    – user4552
    Jul 25, 2013 at 17:39