The half-ball stun shot, measured

Cue sports · the physics under a position play

A half-ball hit: the aim line runs through the edge of the object ball, which puts the ghost-ball offset at exactly one ball radius — 28.575 mm on a standard 57.15 mm pool ball. The cue ball arrives at 3 m/s with pure stun, no follow and no draw. The cloth's sliding friction runs about 0.20. Find the cut angle a half-ball aim actually is, the speed the cue ball carries off the contact, and how far it slides before friction turns the slide into natural roll.

top viewuφ90°bRd

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • b = 28.575 mmGhost-ball offset (one radius)
  • R = 28.575 mmBall radius
  • u = 3 m/sCue ball speed at impact
  • μ = 0.2 Cloth sliding friction
  • g = 9.81 m/s²Gravitational acceleration
Determine
  1. (a)the cut angle of a half-ball hit
  2. (b)the cue ball's speed after the stun contact
  3. (c)the slide distance before natural roll
Step 1 of 3(a) · solve for Cut angle

The half-ball hit is the game's fixed star: offset equals radius, sin φ equals one half, and the cut is exactly 30° — the one angle a player can aim by geometry alone, which is why every aiming system is anchored to it.

bφR
Rearranged for φ
φ=arcsin ⁣(b2R)\varphi = \arcsin\!\left(\frac{b}{2R}\right)
Your values, in your units
φ=arcsin ⁣((28.575 mm)2(28.575 mm))\varphi = \arcsin\!\left(\frac{\left( 28.575\ \text{mm} \right)}{2 \cdot \left( 28.575\ \text{mm} \right)}\right)
Answer
φ=30 \varphi = 30\ ^{\circ}

Carried onward at full precision, not this rounded figure.

Open the Cut Angle from Ball Fraction solver →

Step 2 of 3(b) · solve for Cue ball speed after contact

In a stun collision the object ball takes the line through centres and the cue ball keeps the perpendicular — the famous 90° rule. Its share of the speed is sin φ: at 30°, half. Three metres a second in, 1.5 m/s of cue ball out, moving at right angles to the object ball's path.

uvCB
Rearranged for vCB
vCB=usinφv_{CB} = u \sin\varphi
30 °carried from step 1
Your values, in your units
vCB=(3 m/s)sin(0.523599 rad)v_{CB} = \left( 3\ \text{m/s} \right) \, \sin \left( 0.523599\ \text{rad} \right)
Converted to base units
vCB=(3 m/s)sin(30 )v_{CB} = \left( 3\ \text{m/s} \right) \, \sin \left( 30\ ^{\circ} \right)
Answer
vCB=1.5 m/sv_{CB} = 1.5\ \text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Stun Shot — Cue Ball Speed solver →

Step 3 of 3(c) · solve for Slide distance

A stunned ball leaves with no spin, so the cloth has to teach it to roll: friction eats speed and feeds rotation until they agree, 12/49ths of v²/μg later. At 1.5 m/s on this cloth that is 281 mm of skid — the window in which the cue ball travels dead straight and position play is geometry rather than judgement.

v0μd
Rearranged for d
d=12v0249μgd = \frac{12 v_0^{2}}{49 \mu g}
1.5 m/scarried from step 2
Your values, in your units
d=12(1.5 m/s)249(0.2)(9.81 m/s2)d = \frac{12 \cdot \left( 1.5\ \text{m/s} \right)^{2}}{49 \cdot \left( 0.2 \right) \cdot \left( 9.81\ \text{m/s}^{2} \right)}
Answer
d=280.85 mmd = 280.85\ \text{mm}

Carried onward at full precision, not this rounded figure.

Open the Slide Distance Before Natural Roll solver →

Answer

Therefore a half-ball aim is a 30° cut exactly, the stunned cue ball leaves at 1.5 m/s along the tangent line, and it skids 281 mm before natural roll takes over — the straight-line window a position player is actually using.

Why this order

Three pieces of real mechanics hide in one routine shot. The half-ball relation sin φ = b/2R is geometry a player can trust under pressure precisely because it has no judgement in it. The stun split is momentum conservation for equal masses with negligible friction at the contact: the balls part at 90°, and the speeds divide as cos φ to the object ball, sin φ to the cue ball — so a thin cut keeps the cue ball hot, and a full hit kills it, which is the entire grammar of position play. The slide relation is the odd one out: the 12/49 comes from a sphere's moment of inertia, 2/5 mR², worked through the friction that simultaneously brakes the ball and spins it up.

The trap is treating the slide as decoration. During those 281 mm the cue ball follows the tangent line exactly; after natural roll begins, any masse, cling or table roll bends the path. Hit the same shot at 1 m/s and the skid collapses to 31 mm — speed is not just how far the cue ball goes, it is how long its path stays honest.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.