I'd post a followup for those who might be interested.
>>> the left, a rolling cue ball will deflect three times that much to the
>>Bob, can you post the derivation for this?
>I'll start the outline of a derivation, and leave the details as an
>exercise for the reader. The following assumes that the balls are
>First result: A cue ball hit with maximum draw (assumed to be the
>opposite of smooth rolling) will slow to 3/7 of its initial velocity
>when smooth rolling is achieved.
I=(2/5)*m*r^2, then given a ball with velocity V_0 and rotating at
omega_0, the final velocity will be
V = (5/7)V_0 + (2/7)*r*omega_0 Eq. 1
When the cue ball is struck with a cue tip, then both V_0 and omega_0 are
set. V_0 = F*t/m and omega_0 = r.X.F*t/I = 5*h*F*t/(2*m*r^2) where h is
the height of contact above the center of the cue ball. Eliminating those
controversial quantities F and t, the force of the cue tip on the cue ball
and the contact time respectively, gives
V = (5/7) * V_0 * (z/r)
where z=r+h is the height of the contact point measured above the cloth.
If z=(7/5)*r, then V=V_0 and smooth rolling is achieved immediately with
no sliding; this is an above center hit on the cue ball. If z=r, then
V=(5/7)*V_0; this is a center ball hit on the cue ball. If z=(3/5)*r,
then V=(3/7)*V_0; this is the same displacement from center ball, but with
a below center hit instead of above center.
>smoothly rolling cue ball is given by:
> theta = atan(7*tan(alpha)/2)-alpha
Immediately upon contact with the cue ball, it starts off in a direction
V_c with magnitude V_0*sin(alpha). The initial x-component of the
direction is given by V_cx=V_c*cos(alpha), and the initial y-component is
given by V_cy=V_c*sin(alpha). The x direction is perpendicular to the
initial cue ball direction (omega_0=0), so according to Eq. 1, the
velocity in this direction slows down eventually to V_fx=(5/7)*V_cx
=(5/7)*V_0*sin(alpha)*cos(alpha). The velocity in the y direction
includes the component from the original rolling cue ball, and according
to Eq. 1 is given by ((5/7)*sin(alpha)^2 + (2/7))*V_0. The trick here is
that r*omega_0=V_0, and this spin doesn't change upon impact of the cue
ball. The tangent of the final deflection angle is given by the ratio of
these two components. So this gives finally that
tan(theta) = sin(alpha)*cos(alpha)/(sin(alpha)^2 + (2/5))
The last expression comes from some half-angle trig relations. Anyway, I
can't finangle this into BJ's simple result, but I've ploted several
points with both expressions and they are indeed equal to each other.
Anyone see the trig trick?
Now we have to see if knowing this helps us run 9-ball racks any better! ;-)
$.02 -Ron Shepard