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Snookered! Part II (Posted on 2007-02-08) Difficulty: 3 of 5
In “Snookered” there was a realistic value for friction of the rolling cue ball, but the bounce off the rail of the billiards table was “perfect”. A more realistic value for the “coefficient of restitution” of such a bounce is 0.7 (perfect = 1.0). For that case, how much harder would you need to hit the cue ball (compared to Part I) in order to again just make the shot? And where would you place the ball this time?

  Submitted by Kenny M    
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When the coefficient of restitution (COR) is <> 1, then the angle of reflection (Ar) is not equal to the angle of incidence (Ai). Instead, with COR = 0.7:

Tan (Ai) = COR * tan(Ar) .

This I because only the velocity directly in line to the impact is reduced by the COR; the velocity perpendicular to the rail is not affected. Using the notation in Part I:

x1/2s = 0.7 * x2/8s & (again) x1 + x2 = 2s leads to Ai = 8.472 deg, Ar = 12.011 deg, x1=.29787s and x2=1.70213s

The cue ball starting spot is 4 * x1 + x2 = 2.8936 s = 2.8936 rail sights from the target pocket.

From here, realizing that path length is slightly different from part I, and also taking into account that only the perpendicular portion of the cue ball velocity is reduced by the factor of 0.7 when it bounces off the foot rail, an initial speed of 4.8929 * s^.5 is required, or about 22% faster than for Part I.

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  Subject Author Date
Carebrianjn2007-02-11 06:47:33
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