# At what starting angle will a billiard ball hit a pocket? [closed]

Let's say you have a square pool table with one pocket.

At what angles can you hit the ball to make it hit the pocket? What relation do these angles hare?

Constraints:

• Table is square: $\ell=b$.
• Frictionless
• Bouncing of ball of the sides is ideal
• Ball starts in the lower left corner.
• Pocket is in upper left corner.
• Pocket is 1x ball diameter.
• Pocket will accept a ball from any approaching angle.
• If the whole thing is frictionless and ideal wouldn't the ball keep on bouncing untill it pockets ? So isn't a better question: at what angles can't you pocket? Or at what angles will the ball pocket in 5 bounces or less ?
– Nick
Jul 23 '13 at 20:38
• I was just writing a similar comment to @Nick's: on a frictionless square table with ideal bouncing off the sides, I'd say any direction that isn't parallel to any of the sides will eventually lead the ball to that top left corner. Jul 23 '13 at 20:39
• If there might be more pockets the case will probably change, maybe a sketch or something ?
– Nick
Jul 23 '13 at 20:44
• "Pocket is 1x ball diameter." and "Pocket will accept a ball from any approaching angle." seem contradictory (by a factor of maybe even $\sqrt{2}$). I also think that this question, depending on how it is corrected, belongs to Math SE, given that everything is ideal and this gives all sorts of considerations regarding rational/irrational numbers, measures, etc. Methinks. (My intuition is that almost none of the angles will get you there precisely.) Jul 23 '13 at 21:13
• Agreed with Gugg. There is no physics in this question. Jul 23 '13 at 22:57

This is more like a math question. Babou's answer is correct. $$\forall i,j \in W | \theta = \tan^{-1}\left( \frac{2j-1}{2i} \right)$$

Where $\theta$ is the angle with the axis to the right. Here is an illustration for one example: Where the pool is the square "MVWA".

If I understand correctly, you want all the angles that work. You just pave the plane with pool tables so that 2 tables that share a side are mirror images with respect to that side. Then you put a ball on one of the tables, and aim for all the pockets.

You actually notice that since the pocket is in a corner, it will be shared by four tables. This amount to putting the ball at coordinate (0,0) in the plane, and the pockets at coordinates (2i, 2j+1) for all integer values (positive and negative) of i and j. So the angles are arctg(2j+1 / 2i) for all integer values of i and j, with i non null, plus the angles $\pi/2$ and $3\pi/2$, in the positive angular direction from the X axis.

To avoid special cases here, it is probably better to use arccotg(2i / 2j+1), so as to avoid zero as denominator.

The first comment by @Nick is not quite true. If you require exact angles, you will not have all of them because you have only a denumerable number of them that work, as the above set of ratios 2i / 2j+i (reversed to avoid infinity) is a denumerable set, but you have a continuous set of angles in [0, 2$\pi$]. However, it is almost true as tangent (or rather cotangent here) is a continuous function, and the set of ratios is a dense subset of the reals. That means that any angle that is not the above defined set can be approximated as precisely as you want by an angle in the set.

However, to approximate angles, you need to consider pockets that are very far away, so that whatever small error is made on the angle will be amplified by distance, so that the pocket may actually be missed by a wide margin. I have not yet worked out whether that can be be approximated too, but I very much doubt it.

Indeed, the angles 0 and $\pi$ follows the X axis and never get arbitrarily close to a pocket, even by going far enough. This is also clearly true for the diagonals (x=y). The same is true, though with increasingly smaller minimal distance to pockets, for all angles defined by arctg (2j+1 / 2i+1), for all i and j. These angles are as many as the succesful ones.

These very special pockets are very tolerant on velocity angle, but probably not on position. The idea is that the ball is point-like.

In this abstract problem, where the balls is dimensionless, I considered that it is possible to bounce from the lowest corner sides. If you do not want that, you just keep the first quadrant with arctg(2j-1 / 2i) for i and j positive integers, and you add $\pi/2$ for the direct shot.

Another interesting fact is that the number of bounces is necessarily an odd number, if you do not allow bouncing in the initial corner, i.e., if you consider only the first quadrant.

Given a fixed positive number n of bounces, and considering only the first quadrant, there are $(n+1)/2$ solutions which are defined as $arctg((2j-1) / 2i)$ for $i$ and $j$ positive such that $i+j=(n+3)/2$

Use a long cue, some pockets are far away, and as they say, infinity is very far, especially at the end.

• ... and thanks for all the votes. Not that science was ever a matter of opinion, so that voting should rather be reserved to politics. Jul 24 '13 at 16:22
• Thanks for the elaborate answer. If the successful and unsuccessful set are both denumerable, will they form a continuous set together? If not, what are the missing angles? Jul 24 '13 at 20:44
• ,@Docman No, finite union of denumerable sets only give denumerable sets. The fact that these two sets are dense means that any direction may be approximated with whatever precision you want both by a succesful direction and by an unsuccessful one. Life is full of mystery. Essentially this is because the tangents for both sets represent each time a sizeable part of the rationals. Actually there is a third (dense) set, with unsuccessful direction, defined by $arctg(2j/(2i-1))$ for $i,j\gt 0$. The three sets are disjoint, and their union is the set of all angles with a rational tangent. Jul 24 '13 at 23:32
• @Docman So what is missing. Very simply all the angles with a tangent that is not rational. And there are continuously many of them. They correspond to directions that get arbitrarily close to the pocket if you wait long enough, without ever quite making it ... else we would be in the rational case (I did not check formally the "arbitrarily close", but I think the proof is easy). Jul 24 '13 at 23:35