How large is the Coriolis effect on water flowing down a sink drain, and on an artillery shell?

Look here, it's not that hard to calculate the magnitude of the Coriolis
effect in a simple case.  So let's do one.

Take the Earth as a sphere of circumference 25000 miles = 40000 km, and
consider a point, point A, at latitude 45 degrees north.  How fast is
point A moving with the rotation of the Earth?  Well, the circumference
of that parallel of latitude is obviously 1/sqrt(2) times that of the
equator, or in other words, about 18000 mi = 28000 km.  And the point
travels that distance in 24 hours; thus, 750 mph = 1200 km/h.  For
pedagogical purposes I will pretend that the 750 mph is exact.

Now consider another point, point B, 1 foot = 30 cm south of the point
A, this distance representing the radius of a typical sink.  This point
is 1/sqrt(2) foot farther from the Earth's axis than point A, and so the
circumference of ITS parallel of latitude is 2*pi/sqrt(2) feet greater
than the 45th parallel.  That's about 4.5 feet = 140 cm.

So, point B is traveling at 4.5 feet PER DAY faster than point A.  That
is 0.000035 mph, or 0.000057 km/h, difference in speed of motion.

Now let that plug out and watch a particle of water as it flows from
point A to B.  At A, it's traveling eastward at 750 mph.  At B, it's
still traveling eastward at about 750 mph, but the Earth is now moving
under it at 750.000035 mph.  For a viewer on the Earth, it looks as
though the water was deflected and acquired a westward velocity of
.000035 mph in addition to its original southward velocity.

THAT is the Coriolis effect.  Not a mystical twist; just the Earth
moving under you, once a day.

Do you still think it will have a measurable effect on a drain under
ordinary circumstances?  Dream on.

(The above calculation contains a number of approximations, of course,
but it really is good enough for this purpose.)

> By the way, that artillery story is more of the coriolis mythology.
> Unless the guns could shoot a shell over 100 miles or so, the
> coriolis effect would be negligible. ...

In 1918 the German army shelled Paris with guns over 75 miles (120 km)
away.

Say that such a shell was fired on an airless infinite plain at an
angle of 45 degrees to the horizontal.  Then its velocity has equal
vertical and horizontal components, call that amount v.  The horizontal
component doesn't change and if it reaches the target after time t,
then (working in SI metric) we have v = 120000/t.  But the shell must
be rising for half that time and falling for half that time; therefore
in time t/2 = 60000/v, the acceleration of gravity (value 10 in SI
units) must reduce its vertical component of velocity to 0.  That is,
v = 10*(60000/v), so v = sqrt(600000) = 775 m/s = 1700 mph = 2800 km/h.

So the shell is traveling over the ground at about 1700 mph.

Now let's take Paris to be at 45 degrees latitude and the trajectory to
be north-south, and plug in 75 miles = 120 km in place of 1 foot =
30 cm in the first calculation.  Then the difference between the
rotational speeds of A and B is also 400000 times larger than in the
first calculation, or about 14 mph = 23 km/h.

This is about 1/120 of the shell's horizontal speed, or equivalent to
an error of a substantial fraction of a degree in pointing the gun.

-- Mark Brader <msb@sq.com>

In article <CMHt5w.B9J@news.chalmers.se> lubin@fy.chalmers.se writes:
>In article 1180@sq.sq.com, msb@sq.sq.com (Mark Brader) writes:
>>In 1918 the German army shelled Paris with guns over 75 miles (120 km)
>>away.
>
>Your data cannot be correct. Modern artillery cannot fire shells at
>almost three times the speed of sound, and to such ranges. It was
>certainly impossible in 1918.

In fact the Funk & Wagnalls' encyclopedia article on Krupp, the German
munitions maker, states that their most famous cannon, Big Bertha, had
a range of over 15 km or 9 miles, but evidently nowhere near 120 km.

The range (from y=0 to y=0) of a projectile fired at elevation A is
approximately x = (gt^2 cot A)/2 where t is the flight time.  Set
x = 15000m, assume the elevation is 45 degrees to get maximum range, and
solve for t:

t^2 = 2x/g = 2*15000m/10ms^-2 = 3000s^2
t = sqrt(3000)s = 55s

For a north-south flight in the northern hemisphere, a target 15 km away
is approximately 15*sin(L) km farther from the axis of the Earth, where
L is the latitude.  For Paris, L is about 48 degrees, and sin(L) = .74
approximately, so 15*sin(L) = 11, approximately.  The rate of rotation
of the Earth is w = 7.3*10^-5 s^-1 approximately, so the linear
deflection in 55 seconds is

  w*11*55 km = .044 km = 44 m

Now 44 meters doesn't seem like a lot of difference at that range, but
if you're trying to hit a fortified bunker (one of the uses of Big
Bertha) I suspect it matters.  Also, the angular deflection is about
44/15000 or .003 radians, or .17 degrees, or 10 minutes of arc, which in
that sort of gunnery I suppose you might notice.

I'd think the effect might be even larger for the big naval guns of
World War II, which I believe could shoot much further than 15 km.  Does
anyone know for a fact whether a Coriolis effect was accounted for by
the mechanical analog computers that were used to help aim those guns?

-- David A. Karr <karr@cs.cornell.edu>

[ Editor: Mark Brader's figure is right, and the objection confuses two
  guns.  Big Bertha was a 42 cm howitzer of short range; Paris was
  shelled from about 120 km away in 1918 by a different Krupp weapon,
  the "Paris Gun" (Paris-Geschuetz), which was often miscalled Big
  Bertha. ]