Billiard Geometry

To hit a bank shot:

If the cue ball and the red target ball are the same distance from the rail, then you just aim half-way between them. Referencing the diagram, the angle to aim at is y=ax/(a+b) if they are not the same distance from the rail (the top). To hit a one bank shot the answer is y=ax/(a+b). I don't need the solution for a 1 bank shot since this is the solution.

Neglecting initial velocity (force of hitting the ball), and pretending the ball would travel forever until it hit another ball on the pool table, how can you calculate what angle you need to shoot at to hit the red ball in a double-bank (two side hits) before hitting the red ball? How about 3, 4, or 5 hits first?

I have a general idea of what the solution might be, but it's too complex for me.


For multi-rail shots, just expand the solution of Alex, above. Reflect the pool table left and right, above and below the original pool table. Include reflections of reflections. It might help to color the original four sides in four different colors, to keep track of which side is which in the reflected pool tables. Be sure to also reflect the target ball. Now draw a straight line from the cue ball to any reflected target ball. That line will represent a possible direction for the cue ball to hit the target; the real and reflected sides that get crossed are the ones the real ball bounces off in its path to the target.

EDIT: Using mathematics to analyze the theoretical solution:

Take the "billiard table" as being the region one ball radius in from the actual bumpers on the real table (Thank you, Johannes). Let the origin be at the lower left corner of this table, with width $W$ in the $x$ direction and height $H$ in the $y$ direction. Place the cue ball at $X_c,\,Y_c$, and the target at $X_t,\,Y_t$.

Assume you want to do a bounce off the left cushion and then the top cushion, before hitting the target. The reflection in the left cushion places the reflected target at $-X_t,\,Y_t$, while the reflection of this reflected target in the top cushion places the reflected reflected target at $-X_t,\,(2H-Y_t)$

The line from the cue ball to the doubly reflected target is at an angle $\theta$ where:$$\theta = \tan ^{-1} \left( \frac{(2H-Y_t)-Y_c}{-X_t-X_c} \right)$$

Note that the angle may need to have $\pi$ radians added, depending on the actual quadrant.


To solve a 1 bank hit problem, imagine there is a mirror on the top bank. Then the red ball and bottom right corner pocket get reflected across the top bank. Then, you would simply aim at the red ball in the mirror as if there was no bank, and the cue ball will continue "through" the bank into the red ball. For multiple bank hits, can you see how to extend this reasoning?

enter image description here

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    $\begingroup$ It is essential that the mirror is placed inside the billiard, about a ball radius away from the edge. $\endgroup$ – Johannes Aug 12 '13 at 17:27
  • $\begingroup$ I've a seen a similar reasoning in one of Martin Gardner's books. $\endgroup$ – TZakrevskiy Aug 12 '13 at 17:40
  • $\begingroup$ To solve a 1 bank problem is y=ax/(a+b) as illustrated in the diagram in the original post. $\endgroup$ – Luke Allen Aug 12 '13 at 20:11

For a reflection, the angle coming in is the same as the angle going out. That means that the two triangles in your diagram are similar. Can you figure out the rest?

  • $\begingroup$ Yes that is what y=ax/(a+b) is derived from. No I can't figure out the rest. $\endgroup$ – Luke Allen Aug 12 '13 at 20:11
  • $\begingroup$ @LukeAllen do you know trigonometry? You now know all 3 side lengths of both triangles. You can figure out the angles using those and trigonometric functions. $\endgroup$ – Robert Mastragostino Aug 13 '13 at 3:02
  • $\begingroup$ no, I don't know everything about trigonometry, but as I said in my original post, I have a general idea what the solution would be - like you, and everyone else who can visually see the triangle diagram I posted, the solution involves triangles... trigonometry = The branch of mathematics dealing with the relations of the sides and angles of triangles and with the relevant functions of any angles. Please stop regurgitating the information I've already given in hopes for points, I'm looking for actual help. $\endgroup$ – Luke Allen Aug 13 '13 at 11:36
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    $\begingroup$ I've given all the information in the original post that you "gave" me. I found out how to solve the problem for one bank shot, and I'm asking a question for additional information I don't have in order to solve a more complex subset of the problem I'm stumped on. You've given 0 additional information. "Do you know trig? You can use trig." is like someone asking "How do you build a house?" "Do you know math? Use math." It's a vague answer that doesn't practically help at all. $\endgroup$ – Luke Allen Aug 13 '13 at 12:05
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    $\begingroup$ @LukeAllen Both approaches are common on this site, as you would know if you stayed for awhile and observed how the site is used in practice before deciding what is and isn't acceptable behaviour. Your initial two comments were valid, and if you had continued being respectful and detailed your particular confusion I would have had no problem elaborating and potentially providing a solution. The later ones are you just digging your heels in and deliberately refusing to do 5 minutes of clearly directed reading. "GTFO" certainly doesn't label you as someone deserving of assistance. $\endgroup$ – Robert Mastragostino Aug 21 '13 at 14:02

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