gtMath

Prerequisites: Pool Part 1, Similar Triangles

In part 1, we showed how to pocket an object ball with some basic physics. In part 2, we'll derive simple methods for making bank and kick shots using similar triangles.

For all of the below, we'll assume, unless explicitly stated otherwise, that we can ignore the effects of spin on the balls that would alter their trajectories. In practice, English (side-spin) as well as topspin and backspin (sliding) will affect the trajectories and need to be taken into account for more complicated shots. For example, if the target ball is to be banked off a rail very close by, it will still be sliding upon contact, which changes its trajectory coming off the rail. Ball speed also affects the trajectory off the rails, since a faster ball compresses the rail more and changes the angle a bit. We'll ignore these effects in our analysis, which should suffice for most easier bank and kick shots you'll encounter, where you just need to hit the cue ball with medium speed and at center. If any good players are reading this, commentary is welcome on how to use spin on some of these shots.

The one-rail bank

In part 1, we looked at a straight shot into the side pocket. Now, suppose we have the following situation where the purple 4 ball is blocking the line to the side pocket:
We need to find an alternative to the shot from part 1. One solution would be to take the long-distance shot to the top-right corner pocket, but we're going to look at an alternative, the bank shot, where we hit the 2 off the top rail and into the bottom side pocket.

The key to executing this shot is the reflection principle, that the 2 will bounce off the rail at the same angle with which it bounces into the rail- angle in equals angle out. The problem is to find the correct point on the rail to hit the 2 into, where angle in = angle out will get the ball on a trajectory to the side pocket.

Take a look at the following diagram, which is the same layout, but with two triangles drawn on.
The angle $\theta$ is the angle in and out. If the contact point with the rail, $Q$, is chosen correctly, then the angle $\theta$ will be the correct one to send the 2 to the target pocket. Now, because the two $\theta$ angles are equal and the yellow line is drawn at a 90 degree angle to the rail, we know that angles $OQP$ and $SQR$ are also equal. Since angles $OPQ$ and $SRQ$ are both right angles and thus equal, the triangles $OPQ$ and $SRQ$ are similar by angle-angle.

Let's use the notation $L(\overline{AB})$ for the length of a line segment $\overline{AB}$. The similarity of triangles $OPQ$ and $SRQ$ implies that the ratios $\dfrac{L(\overline{PQ})}{L(\overline{QR})}$ and $\dfrac{L(\overline{OP})}{L(\overline{SR})}$ are equal.

Remember that $Q$ is still at an unknown position, but given the equality of those ratios, we can find it. $L(\overline{OP})$ and $L(\overline{SR})$ are known lengths, which can be measured in units of segments. One segment is the distance between two of the white diamonds (or sometimes they are circles) running along the sides of the table. In the diagram above, $L(\overline{OP})$ is just shy of 2 segments, and $L(\overline{SR})$ is just shy of 4 segments. Thus $\dfrac{L(\overline{PQ})}{L(\overline{QR})} \approx \dfrac{1}{2}$. Since $L(\overline{PR})$ is about 2 segments, we see that the correct position for $Q$ puts $L(\overline{PQ}) \approx \dfrac{2}{3}$ and $L(\overline{QR}) \approx \dfrac{4}{3}$.

Alternative method: parallel shifts

The ratio calculation above is relatively simple and works, but can be a bit annoying to work out while playing. There is an alternative method that achieves the same result without having to think about any ratio calculations.

Below is the same diagram as the previous one, but with the blue lines added in:
The slope of line $OQ$ is the same as the slope of line $MR$, i.e. the lines are parallel. You can prove this by assigning coordinates to each of the points and calculating the two slopes (it's probably easiest to assign $O=(0,0)$ and go from there). This being the case, we have an easier method for finding point $Q$:

1. Find the midpoint $M$ between the 2 ball and the target pocket.

2. Make a line (with your cue stick) from $M$ to the opposite pocket.

3. Move your cue, keeping it parallel, over to the 2 ball, and where it intersects the top rail is the target contact point $Q$.

A similar method will work for kick shots as well.

The one-rail kick

The kick shot is similar to the bank, except we hit the cue ball off a rail and into an object ball, as opposed to the bank, where we hit the cue ball straight into the object ball, which then bounces off a rail.

Take a look at the following diagram:
Here, we have the cue ball and 2 ball in the same places as before, but the cue ball's line to the 2 is blocked by the purple 4. In order to make contact with the 2, we need to hit the cue ball off a rail first to avoid hitting the 4 before the 2 (for example, if we're playing 9-ball, where you need to hit the lowest-numbered ball first or else give your opponent ball-in-hand).
The red triangles above show a set-up just like the bank above, with the two red triangles being similar. Now, instead of going to a pocket, the blue line goes to the point where the target ending spot lines up with the rail. I won't go through the whole thing again, since it's basically the same thing. Find $M$, draw a line from $M$ to the spot where the target ending position lines up with the rail, and parallel shift the line over to the cue ball to see where it needs to hit the rail.

The two-rail kick

This is a very difficult shot, but if you can make contact on these, then you'll save yourself from giving a good opponent ball-in-hand and running the table on you. Suppose we now have this set-up where there is no way to execute a one-rail kick and hit the 2 first:
Once again, the translucent cue ball is the desired position of the cue ball at contact, and as you can see, the only way to get it there is by hitting the cue ball off of at least two rails. We'll start by drawing a diagram similar to the ones above:
Note that this diagram is not exactly to scale, so some of the lines that should be parallel look a bit off. The same reason it is difficult to draw the diagram is the reason that this shot is trickier to work out than the one-rail kick: we need to work out the unknown contact point $C$, but the trouble is that a change in $C$ changes the angle off the bottom rail and thus also changes the unknown contact point $E$ on the second rail.

The reflection principle will hold at both $C$ and $E$, and we need to work it out so that the cue ball ends up at $G$. The three red triangles are all similar by angle-angle, and this fact gives us a few equations we can use to solve for the location of point $C$.

Define $x_1$ to be the length $L(\overline{BC})$ (unknown), $L_1 = L(\overline{BD})$ (known), $x_2 = L(\overline{DE})$ (unknown), and $L_2 = L(\overline{DF})$ (known). The known quantities can be measured in units of diamonds on the side of the table. Because of the similarity of the red triangles, we have: $$
\begin{align}
\dfrac{x_1}{L_1-x_1} &= \dfrac{L(\overline{AB})}{x_2} \tag{1} \\[3mm]
\dfrac{x_2}{L_2-x_2} &= \dfrac{L_1-x_1}{L(\overline{FG})} \tag{2}
\end{align}
$$ Solving $(1)$ and $(2)$ for $x_1$ gives the location of point $C$. One can do this by solving $(1)$ for $x_2$ and then plugging in the answer every time $x_2$ occurs in $(2)$. This is not very practical while playing, so once again, there is a midpoint/parallel shift method that you can implement easily.

1. Find the midpoint $M$ between $A$, the cue ball starting position, and $G$, the desired cue ball position at contact.

2. Make a line (with your cue stick) from $M$ to $D$, the corner pocket between the two rails off which you're kicking.

3. Move your cue, keeping it parallel, over to the cue ball, and where it intersects the bottom rail is the target contact point $C$.

So there you have it. The two-rail kick.

Now, there's an important point we completely ignored, and that's the effect of spin and, related to that, the fact that the rails are not rigid, but rather have some give and will compress when a ball contacts them and potentially alter the ball's trajectory coming off.

On the two-rail kick, in practice, we want to hit the ball along the lines we calculated above, but with a bit of running English, which means topspin + a bit of spin in the direction the ball will be moving (right English in the example above). I can't find anywhere on the internet why this is the case, but I suspect that it's because when the rail compresses and then launches the cue ball back out, the "launch" makes the angle off the rail greater. Running English compensates for that angle boost and also gives the ball some speed to "run" around the table.

The amount of spin depends on how close the contact point is to the corner pocket (point $D$ in the diagram) and also the speed with which the cue ball is hit.

For the simpler one-rail shots above, especially when the distances involved are relatively small and the cue ball isn't hit too hard (and thus the rail does not compress much), there is a bit of margin for error, and no fancy spin is necessary, but it is important for the two-rail kick where the ball must travel a long distance and with enough speed to go that distance.

I defer to better players to add to the spin point in the comments section or let me know if anything above is off.

Thanks for reading.