Slide Distance Before Natural Roll

The previous page established that the SPEED lost during the slide does not depend on the cloth. This page is the other half: how FAR the slide goes, which depends on the cloth completely.

The derivation is two ordinary kinematics lines. Linearly, friction $\mu mg$ decelerates the ball at $\mu g$, so $v = v_0 - \mu g t$. Rotationally, that same friction acts at the contact point with a moment arm $R$, giving torque $\mu mgR$; with $I = \tfrac{2}{5}mR^2$ the angular acceleration is $5\mu g/2R$, so the surface speed climbs as $R\omega = \tfrac{5}{2}\mu g t$. Sliding stops when the two meet:

\[v_0 - \mu g t = \tfrac{5}{2}\mu g t \quad\Longrightarrow\quad t = \frac{2v_0}{7\mu g}\]

Substitute that time back into $d = v_0 t - \tfrac{1}{2}\mu g t^2$ and the distance falls out as $d = 12v_0^2/49\mu g$. The 49 is the 7 squared, and the 7 is the same 7 as the sevenths on the natural-roll page. Everything on this shard traces back to $I = \tfrac{2}{5}mR^2$.

What this distance IS, in playing terms, is the lifetime of the 90° rule. A cue ball struck dead centre is a genuine stun shot for exactly this far and no further. Inside it, a cut really does send the two balls off at right angles. Past it, the ball is rolling, and a rolling cue ball leaves along the tangent line and then bends forward off it by the amount the deflection page computes. Knowing roughly where that boundary sits on your table is more useful than either formula alone.

Rough numbers for a pool table with $\mu \approx 0.2$: a soft 1 m/s shot slides about 125 mm, a medium 2 m/s shot about 500 mm, and a firm 3 m/s shot about 1.1 m. A nine-foot table's diagonal is roughly 2.7 m. So on a soft shot the stun is gone almost immediately, and on a firm one it survives a good fraction of the table.

The square on $v_0$ is the interesting term. Double the stroke speed and the slide lasts twice as long but goes four times as far, because the ball is covering that longer time at higher speed. This is why a hard stun shot holds its stun so much further down the table than a gentle one, and it is the mechanism behind a piece of advice good players give without explaining: if you need stun at distance, hit it firmly.

Two limits on this formula, both worth naming.

It assumes the ball starts with no spin at all. Any follow shortens the slide, any draw lengthens it, and at the natural-roll offset $b = 0.4R$ there is no slide whatever. The full version with initial spin exists but is messier; this is the pure stun case, which is also the case anybody actually asks about.

$\mu$ is the weak link and deserves suspicion. Ball-on-cloth sliding friction is not a material constant. It changes with the nap direction and its wear, with humidity, with how clean and how polished the ball is, and it is measurably different on a napless worsted tournament cloth than on a napped one. Published figures cluster near 0.2 with a wide spread. The right move is to measure your own: strike a ball at a speed you can estimate, watch for the moment the skid ends — it is visible if you look for it, and audible on some cloths — and run this equation backwards. That is what the $\mu$ brain on this page is for.

One more thing this does NOT model: the rolling friction that acts afterwards. Once the ball is rolling it keeps slowing, at a much gentler rate governed by a rolling resistance coefficient near 0.01, and that is what eventually stops it. This page covers only the sliding phase.