Why the mirror trick misses a bank shot, and the math that fixes it

Quick answer: The mirror trick, where you reflect the target across the rail and aim at its image, assumes the ball keeps all of its speed in the bounce. A real rail takes energy out of the part of the motion that heads into the wall, so the ball leaves at a flatter angle and runs on past the target. The fix is to multiply the wall’s share of the aim by the rail’s bounce factor, called the coefficient of restitution, e. On a rail with e = 0.46, the bank point moves from the midpoint to about 31 percent of the way along, and the throw has to cover about a third more distance than the target is away. The rest of this article shows the math and a worked example.

Anyone who has banked a ball off a rail knows the mirror trick. It works on paper and misses on the table, and usually it misses long. That is not bad aim. It is a built in error in the trick, and it is large enough to measure.

The numbers below come from the physics in our game, Bocce, where the rails have a bounce factor of 0.46 and the court is 1.0 wide and 2.2 long, measured in court widths. Real boards, rails and balls differ, and the last sections cover how to estimate your own. The method carries over to any rail you can put a number on.

The mirror trick, and what it assumes

Stand where you throw from, put the rail on one side, and call your distance from the rail a. Your target is a distance b from the same rail. The distance between the two of you along the rail is D. The mirror trick says to aim at the point where a straight line from you to the target’s reflection crosses the rail. That point sits this far along the rail from you:

mirror hit point = D × a ÷ (a + b)

If a and b are equal, you aim at the midpoint. The picture behind it is a ball that arrives at some angle and leaves at the same angle, which is true only if the bounce loses no speed.

What a real bounce does

Split the ball’s motion into two parts: the part running along the rail, and the part running straight into it. A rail that absorbs energy shrinks the straight into part and leaves the along the rail part alone. The ball’s speed into the wall gets multiplied by e, the restitution, while its speed along the wall stays the same.

That changes the exit angle. Measuring angles from the line perpendicular to the rail:

tan(exit angle) = tan(entry angle) ÷ e

Since e is less than one, the exit angle is always bigger than the entry angle. The ball leaves more along the rail and less away from it. Here is how much, for three values of e:

Entry angle	Exit, e = 1.0	Exit, e = 0.8	Exit, e = 0.46	Energy kept, e = 0.46
15°	15.0°	18.5°	30.2°	26%
30°	30.0°	35.8°	51.5°	41%
45°	45.0°	51.3°	65.3°	61%
60°	60.0°	65.2°	75.1°	80%
75°	75.0°	77.9°	83.0°	95%

A 45 degree shot into a rail with e = 0.46 comes off at 65 degrees. Over a long court that difference puts the ball a long way past where the mirror said it would land. The energy column shows the other cost: the ball leaves with less than two thirds of the energy it arrived with, and the shallower the shot, the more it loses.

The corrected aim

Solve for the bank point so that the ball, leaving at the flatter exit angle, still reaches the target. The result is the mirror formula with the near side scaled by e:

corrected hit point = D × a × e ÷ (a × e + b)

With e = 1 it reduces to the mirror trick. With e below one, the hit point slides toward you. If a and b are equal, the hit point is a fraction e ÷ (1 + e) of the way along:

• e = 1.0 gives 50 percent, the midpoint
• e = 0.8 gives 44 percent
• e = 0.46 gives 31.5 percent

In other words, against a soft rail you aim at the rail earlier than the mirror says.

A worked example

Take a game court 1.0 wide. A ball’s center cannot get closer to a rail than one ball radius, which is 0.028 in the game, so the rail’s effective line for the ball’s center is 0.972 from the left side. You throw from the middle of the baseline, at x = 0.5 and y = 0.03. The target is at x = 0.75, y = 1.30, and you bank off the right rail.

• Your distance from the rail line: a = 0.972 − 0.5 = 0.472
• The target’s distance from the rail line: b = 0.972 − 0.75 = 0.222
• Distance between you along the rail: D = 1.30 − 0.03 = 1.27

Mirror trick: hit point = 1.27 × 0.472 ÷ 0.694 = 0.864 along the rail, which is a launch angle of 28.7 degrees from straight up the court.

Corrected, with e = 0.46: hit point = 1.27 × 0.472 × 0.46 ÷ (0.472 × 0.46 + 0.222) = 0.628 along the rail, which is a launch angle of 36.9 degrees.

That is an 8 degree difference in the throw. I ran both shots through a copy of the game’s rail and rolling logic, with the speed chosen correctly for each. The corrected shot stopped exactly on the target. The mirror shot stopped 0.48 past it, a miss of about a fifth of the court’s length.

You also need more speed

A bank costs energy at the rail, so a bank shot has to be thrown as if the target were further away than it is. A ball slowed at a steady rate stops after a distance that depends on the square of its speed, which leads to a clean rule:

effective distance = distance to the rail + (distance from the rail to the target ÷ energy kept)

The energy kept at the rail is the last column of the table above, which is cos²(entry angle) × e² + sin²(entry angle). In the example the ball hits the rail at 53 degrees from the perpendicular, so it keeps 72 percent of its energy. The legs of the shot are 0.786 before the rail and 0.679 after, so the effective distance is 0.786 + 0.679 ÷ 0.72, which is about 1.73. A straight shot at the same target is only 1.29 away. The bank has to be thrown as if the target were about a third further away. That is roughly 16 percent more launch speed, because distance grows with the square of speed, and in the game’s power range it is about 71 percent of full power instead of 53 percent.

If you throw a bank with straight shot speed, it dies early. If you correct the angle and forget the speed, it falls short. Both corrections go together.

How to estimate a real rail’s bounce factor

You do not need instruments. You need a ball that rolls and a wall, and one steady push.

1. Push the ball along an open stretch with your usual smooth push and note how far it rolls before it stops. Call that S. Repeat a few times and average.
2. Now push the ball with the same push straight at the wall from a known distance s, square on to the wall, and measure how far it comes back. Call that r.
3. The bounce factor is: e = √(r ÷ (S − s))

For example, with S = 8 m, a wall 4 m away, and a rebound of 0.85 m, e = √(0.85 ÷ 4), which is about 0.46. The calculation works because a ball with a steady rolling loss stops after a distance proportional to the square of its speed, and the bounce cuts speed by the factor e.

Where the math stops

The correction assumes a rail with no friction and a ball with no spin. A real ball picks up spin from the rail’s surface, and the spin pushes the exit angle further in a direction that depends on speed and the rail’s material. Pool references note that the real rebound depends on the cushion, the speed and the spin, which is why experienced players practice banks rather than trusting a formula. Treat the math as a way to find the right starting point and to understand which direction your misses go. Then adjust from a test throw.

Try it with real rails

Bocce, one of our apps, uses exactly this rebound model, so you can see the 8 degree difference on screen. Aim the mirror way, then aim the corrected way, and watch where the ball ends up.

Frequently asked questions

What is the mirror trick for a bank shot?

You reflect the target across the rail and aim at its image, so the line crosses the rail at the bank point. It assumes the ball keeps all of its speed in the bounce, which a real rail does not allow.

Why does my bank shot miss long?

A real rail takes energy out of the part of the motion heading into the wall. The ball’s speed into the wall is multiplied by the restitution e, so tan(exit angle) = tan(entry angle) ÷ e. The exit angle is flatter and the ball runs on past the target.

What is the corrected bank point?

corrected hit point = D × a × e ÷ (a × e + b). When a and b are equal, you aim a fraction e ÷ (1 + e) of the way along the rail: 50 percent at e = 1.0, 44 percent at e = 0.8 and 31.5 percent at e = 0.46.

Do I need more power on a bank shot?

Yes. The ball loses energy at the rail, so throw as if the target were further away. In the worked example the bank needs about 16 percent more launch speed than a straight shot at the same target.

How can I estimate a rail’s bounce factor?

Push the ball along open ground and measure how far it rolls, S. Then push it the same way at a wall a known distance s away and measure how far it comes back, r. The bounce factor is e = √(r ÷ (S − s)).

Does spin change the answer?

Yes. The correction assumes no friction at the rail and no spin. Real rebounds also depend on the cushion, the speed and the spin, so use the formula as a starting point and adjust from a test throw.