Cut Angle from Ball Fraction

Every aiming system ever invented for pool, snooker and billiards is a way of estimating one number, and this is that number written down. Two balls of radius $R$ touch when their centres are $2R$ apart. So if your cue ball's centre travels along a line that passes a perpendicular distance $b$ to the side of the object ball's centre, then at the moment of contact the line joining the two centres sits at $\arcsin(b/2R)$ from your cue ball's path — and the object ball leaves along that line of centres, because on a smooth collision that is the only direction an impulse can point.

That is the entire content of it. No mass, no speed, no cloth, no spin. Aiming is pure geometry, and it is the same geometry on a snooker table with 52.5 mm balls as on a pool table with 57.15 mm ones, because $b$ and $2R$ scale together and only their ratio survives.

The fullness fraction is the same number in the language players actually speak. Define $f = 1 - b/(2R)$ and you have the fraction of the object ball's width your cue ball covers as you sight down the shot. Then $\sin\varphi = 1 - f$, and the whole fractional-ball table falls out of one arcsine:

• Full ball, $f = 1$: $\varphi = 0°$. Straight through.
• Three-quarter ball, $f = 0.75$: $\varphi = 14.5°$.
• Half ball, $f = 0.5$: $\varphi = 30°$ exactly.
• Quarter ball, $f = 0.25$: $\varphi = 48.6°$.
• Feather, $f \to 0$: $\varphi \to 90°$.

The half-ball hit is worth committing to memory, and it is the reason the fractional system caught on. At $f = 0.5$ the EDGE of the cue ball lines up with the CENTRE of the object ball — a sighting picture you can actually see, with no estimation in it — and it gives exactly 30°, forever, on any table, with any size of ball. Snooker players build whole break patterns around it.

Notice too how brutally non-linear the table is. The first quarter of fullness, from $f = 1$ down to $f = 0.75$, buys you only 14.5° of cut. The last quarter, from $f = 0.25$ down to zero, spends the remaining 41°. Thin cuts are hypersensitive: a millimetre of aiming error near a feather swings the object ball several degrees, while the same millimetre on a three-quarter ball hit is almost invisible. That is a fact about the arcsine, not about your stroke, and it is why thin cuts feel so much harder than they look.

Now the two mistakes, and the first one is nearly universal.

Aiming at the contact point. The contact point is a spot on the surface of the object ball, and it is the wrong thing to aim at, because your cue tip is not going to arrive there — the cue ball's CENTRE has to pass through a particular place, and that place is one full radius away from the surface. Aim instead at the GHOST BALL: imagine a cue ball already resting against the object ball along the line of centres, and send your cue ball's centre to where that ghost's centre is. This is what $b$ measures. Worse, the contact point is seen at an angle, so your eye foreshortens it, and the error grows exactly as the cut gets thinner — precisely where you could least afford it.

Trusting the geometry to the degree. It will not be right, and the reason is throw. The two balls are not frictionless. During the roughly 200 microseconds they are in contact, the cue ball's surface is sliding across the object ball's surface, and friction pushes the object ball a degree or two off the line of centres — away from the direction the cue ball was travelling on a cut, toward it if there is sidespin working the other way. Throw is largest around a half-ball hit and at slow speeds, and it can reach several degrees, which over the length of a table is easily a missed pot. Experienced players compensate without ever naming it: they aim a fraction thicker on slow cuts.

This site does not publish a throw formula, and that is deliberate rather than an oversight. The naive model — treat the ball-on-ball friction coefficient as a constant and you get $\tan\theta = \mu$, independent of cut angle — is simply wrong, and it is wrong in a way that looks plausible enough to be believed. The honest model needs $\mu$ as a function of the relative surface speed at the contact, which depends on the cut angle, the shot speed and any sidespin, and it has no closed form. A formula nobody has verified is worse than an honest paragraph saying the effect is real and pointing at where it bites.