From: linden@positive.Eng.Sun.COM (Peter van der Linden)
Subject: The pure truth on the Coriolis Farce (sic).
Date: 7 Dec 1992 20:19:13 GMT
Here, specially for that thrusting young Antipodean Wayne, is
the straight dope on why the Coriolis force doesn't bend the water
flowing out of a tub.
--------- reposted from 1 year, 4 months ago ---------------
>>>>> On 30 Aug 91 04:34:40 GMT, moroney@ramblr.enet.dec.com said:
mike> The story that water in a bathtub spirals in a certain way due
mike> to the coriolis force came up again. Can anyone provide me with
mike> a formula that, given the diameter of a bucket and the diameter
mike> of a small hole in the center of its bottom, and it's filled
mike> with water to a given depth, and the latitude of the bucket (on
mike> Earth) and any other variables necessary, what the observed
mike> rotation rate of the water will be as it drains? Can this rate
mike> ever exceed one revolution per day? After all, Earth rotate
mike> only once per (sidereal) day.
It is possible to make an experimental apparatus that actually
measures the influence of the Coriolis force, but it is not easy!
The governing equations for a homogeneous, incompressible inviscid
fluid are the Euler equations. When you add the vertical component of
the Coriolis force, you get:
du du du dh
-- + u -- + v -- - f*v = -g --
dt dx dy dx
dv dv dv dh
-- + u -- + v -- + f*u = -g --
dt dx dy dy
dh dh dh du dv
-- + u -- + v -- + h -- + h -- = 0
dt dx dy dx dy
Where u and v are the two horizontal components of the velocity and h
is the thickness of the fluid. 'f' is 2*Omega*sin(Phi), with Omega
being 2 Pi/(1 day) and Phi being the latitude.
When you assume that the velocity scales like U, and the horizontal length
scale like L, then the ratio of the nonlinear terms to the Coriolis
terms is
U
----
f L
For a bathtub, we have U=O(0.1 m/s), L=O(0.1 m), and at mid-latitudes
we have f=O(.0001/s). So the ratio is O(10,000), meaning that the
nonlinear terms are 4 orders of magnitude bigger than the Coriolis
terms. So for a quasi-steady swirling flow, the dominant balance is
going to be the nonlinear (centrifugal) terms against the pressure
gradient. The Coriolis force will be utterly negligible....
An alternate scaling contrasts the size of the Coriolis term with the
size of the acceleration term. The ratio of du/dt over fv is
(1/(Tf)), where T is the time scale of the flow. In order for the
Coriolis terms to be O(1), the time scale would have to be of the
order of (1/f) or 10000 seconds (3 hours). Most of us don't put up
with bathtub drains that slow!
--
John D. McCalpin mccalpin@perelandra.cms.udel.edu
Assistant Professor mccalpin@brahms.udel.edu
College of Marine Studies, U. Del. J.MCCALPIN/OMNET
From exodus!cronkite.Central.Sun.COM!sun-barr!cs.utexas.edu!swrinde!zaphod.mps.ohio-state.edu!cis.ohio-state.edu!rutgers!zodiac!tiscione Sun Sep 1 15:01:27 PDT 1991
In article <MCCALPIN.91Aug30204215@pereland.cms.udel.edu>, mccalpin@perelandra.cms.udel.edu (John D. McCalpin) writes:
> [Impressive quantitative analysis deleted]
> For a bathtub, we have U=O(0.1 m/s), L=O(0.1 m), and at mid-latitudes
> we have f=O(.0001/s). So the ratio is O(10,000), meaning that the
> nonlinear terms are 4 orders of magnitude bigger than the Coriolis
> terms. So for a quasi-steady swirling flow, the dominant balance is
> going to be the nonlinear (centrifugal) terms against the pressure
> gradient. The Coriolis force will be utterly negligible....
This reminds me of a letter to the editor of Science News, when there was
a real furor over how the Coriolis force interacts with this "sensitive
dependence on initial conditions" water drainage thing. Some guy who was a
tourist in Africa wrote that he was at the equator. One of the native
folk had a pan full of water, and some leaves floating on the top. He would
hold the pan in the air WITH HIS HANDS and let the water drain out the
bottom. When he did this twenty feet from the equator, the leaves swirled
clockwise, and twenty feet on the other side, it swirled counterclockwise,
much to the astonishment of the tourists. When he stood directly on the
equator, the leaves and water did not swirl around but just flowed radially
into the hole. Someone who responded to this letter made the calculation
that unless the guy held it to within one millionth of an arc second to the
horizontal, the Coriolis force could not possibly be responsible for the
direction of the drainage. He speculated that he just imperceptibly
twisted the pan one way or the other, and let the water do the rest.
(Anything for the tourists.)
--
* THIS SPACE FOR RENT!!! *
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From exodus!cronkite.Central.Sun.COM!sun-barr!cs.utexas.edu!uunet!munnari.oz.au!bruce!monu0.cc.monash.edu.au!monu1.cc.monash.edu.au!map Tue Sep 3 12:34:26 PDT 1991
In article <1991Sep2.210401.23439@athena.mit.edu>, aeosawa@athena.mit.edu (Atsushi E Osawa) writes:
> The Coriolis effect can be observed in a bath-tub vortex if you wait long
> enough for the residual momentum of the filling action to die down. In
> an article in Nature (196:1080-1081, Dec. 15, 1962), Ascher Shapiro reports
> that a counter-clockwise rotation of 0.25-0.33 sec-1 was observed when
> a circular tank of water (filled clockwise) was drained after allowing the
> water to sit for about 24 hrs. If the tank was drained after only 1 or 2 hrs,
> the vortex went clockwise. His tank was 6 ft in diameter, 6 in high, with
> a 3/8 in diameter drain, for the experimentally minded of you out there. He
> was at MIT, approximately 42 N latitude.
>
> Another fun fact from the "goofy articles" file of
>
> -the Edster
I've been waiting to see if anyone actually answers the guy's original question
and at last someone comes close. John McCalpin gave us a good outline of scale
analysis (see Pedlosky's Geophysical Fluid Dynamics for more mathematical
detail) but still no formula.
So here goes. If we assume that the fluid is inviscid (dubious on long time
time scales, but the order of magnitude should be correct) then it can be shown
that the vorticity zeta=du/dy-dv/dx of the flow satisfies
D( (zeta+f)/H )/Dt = 0
where f = 2*Omega*sin(theta) in terms of the angular velocity Omega = 2*pi/day
of the earth and latitude theta, and where H is the depth of the fluid. For
a fluid initially at rest in a bath of depth H0 we have that
(zeta+f)/H = f/H0
for every fluid particle in the bath. If we drain a bath of area roughly
1 m^2 through a hole of area (0.05)^2 m^2 = 2.5E-3 m^2 then we are effectively
rearranging this volume into a tube of height H0/2.5E-3 = 400*H0. Since
(zeta+f)/H is conserved for fluid particles (see D/Dt equation above), we
conclude that the vorticity zeta of the fluid in the tube of height 400*H0
is
(zeta+f)/(400*H0) = f/H0 => zeta = 399*f (call it 400*f).
This corresponds to an angular velocity of 0.5*zeta = 200*f, which has a
period of 1/200 of a day when theta = 30 degrees. That's about a 7.5 minute
rotation period, which is indistinguishable by most of us. It is smallest
near the poles, where it is half that, but gets longer as you approach the
equator (infinite as theta -> 0).
As has been pointed out earlier, the observed swirl is due to residual rotation
from when the bath is filled, which can take a LONG time to completely die
out. It's easy to rig the results, particularly in a bath with two separate
outlets (for hot and cold). To make the earth's rotation noticeable you
would need to make the bath larger by a factor of 500 or so, by extending
the linear dimensions by 20 or so AND leaving the whole thing to settle for
a long time, as they do for the experiments.
Sure, there is no formula above, but it's easy to work it out from the
information above (for any sized bath and plug hole).
