# Hess's Pool Table Problem

Updated: 27 November 2000. Added an alternative solution, provided by Matt Parker (currently a grad student at Colorado State University). Thanks, Matt!

Problem #6, p 174 in Introduction to Theoretical Meteorology by S.L. Hess:

A billiard table is 102 inches long. Two billiard balls, 2 in. in diameter, are placed at opposite ends of the table as shown. Neglecting friction, how fast does one ball have to be propelled directly at the other at the initial moment in order that it shall just barely miss striking the second ball due to the Coriolis force? The table is at 43o N (f =10-4 s-1) The problem requires us to find the speed of the ball in the y-direction, such that the Coriolis force will deflect it (in the x-direction) by a total of 2 in. after having travelled 100 in. (since the balls are 2 in. in diameter -- see the figure). ## Detailed Solution:

The Coriolis Force (per unit mass) and y-velocity are given by The x-displacement (where the acceleration, a, is constant) is given in general by The y-displacement (where there is no acceleration so the speed is constant) is given in general by Using the initial conditions and the known information Then it can be seen that we have the two equations in the two unknowns, , the initial (and final since the table is frictionless!) y-velocity, and t, the time it takes to transit the 100 in. across the table. Solving the first for the time gives and substitution into the second gives so back-substitution yields With the values given, then Now, the Coriolis force per unit mass (force per unit mass = acceleration) associated with this speed is found: How large is this? If we compare it to another well-known acceleration, that due to the Earth's gravity (which is 32 ft s-2, or 384 in. s-2), Earth's gravitational acceleration is about one million times larger! Thus, it is safe to say that this is not a very strong acceleration.

Another comparison we might make is the Coriolis force on a parcel of air moving with the relatively modest speed of 10 knots (10 nautical mi. h-1 = 16.89 ft s-1 = 202.7 in. s-1). For air moving at this speed, the Coriolis acceleration is roughly 1000 times larger than that on our hypothetical pool ball, but is still about 1000 times smaller than that due to Earth's gravity.

Such small forces have to act for a long time to have an observable effect on things, which is the whole point of doing this problem in the first place. Many people might think that you need to make the pool ball move fast in order to achieve the correct displacement. However, given reasonable speeds, the amount of acceleration you can create is so small, that it takes time for that acceleration to move the ball far enough to achieve the goal. A fast-moving ball simply does not have enough time for the weak accelerations to act.

## An alternative approach ... a more general solution

I received the following e-mail from Matt Parker:

```I was feeling curious and looked over your online pool table (Coriolis)
problem.  Your method of solution gives the correct answer (at least to
an appropriate order of accuracy), but it can't be generalized to
similar problems in which much larger lateral deflections (let's say, 50
inches) are required of the cue ball.  In such cases, assuming that
Dv/Dt=0 is a bad idea (the ball's kinetic energy will increase in a very
unsatisfactory manner).  Holton (p. 64, 3rd ed.) shows that the radius
of the frictionless ball's trajectory is given by R=|V|/f.  This is a
nice, physical way for me to think about the pool ball problem before I
even pick up my pencil: the slower the ball moves, the smaller the
radius of its trajectory.  From simple geometry, one can figure out the
radius of curvature required for the cue ball to miss the target ball
(or whatever even larger object  is sitting out there), after which the
appropriate velocity is computed for any given value of f.  I don't know
if you're interested in revising the online solution or not, but perhaps
the `geometry technique' would be helpful to any fledgling students who

The reference to Holton is:

Holton, J.R., 1992: An Introduction to Dynamic Meteorology, 3rd Ed. Academic Press, 511 pp.

I appreciate Matt offering this rather interesting correction to the solution of the problem. For the problem as I originally cast it, the circle passes through the points (x = 0, y = 0) and (x = 2, y = 100), and the center of the circle(xc, yc) is on the x-axis (yc = 0). Since the center of the circle must be equidistant from the two points, ,

it is easy to find that .

Using the fact that yc = 0 implies that the center of the circle is at xc = 2501 inches (or about 63.5 m). This is also the value of the circle's radius. From Holton's formula given by Matt, this means that ,

which is obviously very close, but not exactly the same as the answer found by my proposed method. Matt's method is a more general result than mine. Although Matt's suggestion is indeed both more accurate and more general than my first solution, care will be needed to treat problems on scales larger than a pool table, because the variation of Coriolis with latitude would eventually need to be incorporated in the solution. Neither of the two methods applied above considers the variation of f .

Students should think about Matt's point concerning the increase of kinetic energy when ignoring dv/dt ... do you understand why? I leave it up to my readers to work that out, if they don't already know.