Billiards & Cue Sports formula solvers
Bank Shot Rail Contact Point
Billiards & Cue SportsMechanicsWhere on the rail to send the ball for a one-cushion bank. The schoolbook mirror method reflects the target through the rail and aims at the image — which is this formula with e = 1. Real cushions rebound wider than they receive, so the true contact point sits shifted, and this is the mirror method corrected for restitution.
Cushion Rebound Angle
Billiards & Cue SportsMechanicsWhy a ball does not come off a cushion at the angle it went in. The cushion squashes and pushes back along its own normal, so only the perpendicular part of the velocity is affected — it comes back multiplied by e — while the part parallel to the rail sails through untouched. The result is always a wider angle out than in.
Cushion Rebound Speed
Billiards & Cue SportsMechanicsHow much speed a cushion takes out of a ball. Only the perpendicular component is squeezed and returned at e times its size; the parallel component is untouched. Add the two back as vectors and the answer depends entirely on the angle — a ball hitting the rail square loses the most, and one grazing along it loses almost nothing.
Cut Angle from Ball Fraction
Billiards & Cue SportsMechanicsThe whole of aiming geometry in one line. Two spheres of radius R touch when their centres are 2R apart, so if the cue ball's centre passes a perpendicular distance b to the side of the object ball's centre, the line of centres at contact — and therefore the object ball's departure direction — sits at sin⁻¹(b/2R) from the cue ball's path. A half-ball hit gives exactly 30°.
Rolling Cue Ball Deflection (the real 30° rule)
Billiards & Cue SportsMechanicsWhere a ROLLING cue ball actually ends up after a cut. It leaves along the tangent line at 90° to the object ball, but it is still carrying its forward roll, so friction bends it forward again until it is rolling once more. The final path sits α off the original line of travel — and α hovers near 32° for every cut angle between 20° and 40°, which is the whole reason the 30° rule exists.
Slide Distance Before Natural Roll
Billiards & Cue SportsMechanicsHow far a cue ball struck with no spin slides before cloth friction has spun it up into a roll. It is the range over which a stun shot still behaves like a stun shot, and it is the reason the 90° rule works on a short shot and fails on a long one.
Slide Time Before Natural Roll
Billiards & Cue SportsMechanicsHow long a cue ball struck with no spin slides before it is rolling. The ball decelerates at μg while friction spins it up at 5μg/2R, and the two meet when v = Rω — which happens after 2v₀/7μg, linearly in the starting speed rather than quadratically like the distance.
Speed at Natural Roll
Billiards & Cue SportsMechanicsThe speed a ball settles to once cloth friction has turned its slide into a roll. Friction acts at the contact point, so angular momentum about that point is conserved through the whole slide — and the answer comes out as a fixed weighted average of the starting speed and the starting spin. A ball struck with no spin loses exactly two-sevenths of its speed, always.
Stun Shot — Cue Ball Speed
Billiards & Cue SportsMechanicsWhat the cue ball keeps after a stun shot: the component of its velocity ALONG THE TANGENT LINE, which the collision cannot touch. Together with the object ball's u cos φ this is the 90° rule — two perpendicular velocities whose squares add back to u², so no energy is lost and no momentum goes missing.
Stun Shot — Object Ball Speed
Billiards & Cue SportsMechanicsHow much of the cue ball's speed the object ball actually receives on a cut. The balls are smooth, so the impulse runs along the line of centres and only the component of the cue ball's velocity along that line is transferred — cos φ of it. A thin cut sends the object ball nowhere, which is why thin cuts have to be hit hard.
Tip Offset to Spin Ratio
Billiards & Cue SportsMechanicsHow much spin a given tip offset puts on the cue ball. Strike the ball a distance b off centre and the same impulse that gives it speed also gives it angular momentum about its centre — and because a solid sphere has I = ⅖mR², the ratio of surface spin speed to travel speed comes out as 5b/2R, independent of how hard you hit it.