>> >Draw on the cue ball transfers negligible follow to ball b.
>> Depending on how sticky are the balls, you can get "b" to follow only
>> 3"-4" by transfering draw.
>I don't believe the effect is that large, except maybe at very high
>speeds. Let's try an experiment. Place the two object balls 5mm apart.
>Place a third object ball near the first two to act as a marker, or use
>some other method of measurement. Place the cue ball in the same line
>as the first two balls. Shoot straight at the two balls with draw.
>Take notes. Report results.
I was demonstrating this very shot last week to some of our league
players. The ball separation might have been more like 1/2" rather than
1/4", but still too close for the cloth friction to impart the roll. The
balls are sticky where we play, so 3"-4" of follow is typical there. That
is, it usually rolls more than one ball but less than two balls forward.
The cue ball does not hit twice; this is just due to transfer of spin.
>Well, the theory says that as long as the cue ball has more draw than
>the object ball has acquired follow, the surface of the cue ball is
>still slipping on the surface of the object ball and it makes no
>difference whether the excess rotation on the cue ball is 1RPM or
>10000RPM. Therefore, the cue ball need draw no further than you want the
>object ball to follow, neglecting whatever follow rubs off on the second
This is an aspect that I hadn't considered, but I think your argument is
correct. I did this two ways, firm with center ball, and soft with draw,
in order to demonstrate that it is the spin that does the trick and not
the force. I was using enough draw to get the cue from the center to the
end of the table.
>> However, I've noticed that when the cue ball is jumped, then it can
>> sometimes impart extra follow to the ball it contacts.
>I haven't noticed this. Could it be that the object ball is jumping which
>can give the illusion of follow?
FINALLY, after a year of reading your posts about tricky shots, I came up
with a shot that you (BJ that is) don't know! ;-) I don't know why this
shot works. Yes, it could be that the first object ball is airborne when
it contacts the second ball. Another possibility is that the cue ball
hits the first object ball more than once -- that is it bounces on the
ball as it rolls off. Perhaps the last bounces occur after impact between
"a" and "b"? However, this effect is obvious when it occurs; the ball can
roll forward a couple of feet! BTW, the cue ball need not be jumped with
much force. This is still a soft shot.
>> It also works when "a" and "b" are frozen together.
>When the two object balls are touching, neither draw nor jumping is
>required. For example, the double spot shot can be made with a stop
>shot or even follow.
>(A double spot shot is when two balls are spotted on the foot spot, and
>the cue ball is in hand behind the line. The common method of making
>the head ball straight into one of the corner pockets is to put the cue
>ball six inches from the centerline of the table and draw it perfectly
>straight back from the head object ball. Draw is not required on the
>shot, which can be demonstrated by placing a third object ball 5mm from
>the head ball and in the line of the cue ball's path.)
I've seen a variation of this in one-pocket. If a second spotted ball is
placed just a tad to side of the spotted ball in front of it (yes, this is
illegal), then the first spotted ball can be made in the corner without
doing anything special by hitting it head on from the center of the
headstring. Your shot, which can be made with legal spotting of the
balls, is a rotation of this one, but requires, I believe, a bit more
accuracy at that sharper throw angle. When you play one-pocket, you
should always casually check that balls are spotted properly; it can make
a big difference.
>>> the "small cut angle follow rule": "If a ball is cut a small angle to
>>> 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.
This is something that you've posted before. 5/7 for a center ball hit,
3/7 for maximal draw, right?
>Third result: The follow angle theta for a cut angle of alpha for a
>smoothly rolling cue ball is given by:
> theta = atan(7*tan(alpha)/2)-alpha
>This is the fairly simple result that should have accompanied diagram
>5-5 in Koehler. The plotted curve there is clearly not the theoretical
>curve. The follow angle has a maximum value for about a half ball cut.
>Question: what is the theortical cut angle for maximum follow angle?
I need to spend some more time thinking about this, but if the above
expression is correct, then the maximum occurs at a cut angle of
(1/2)acos(5/9), or about 28.1255 degrees, right? This is the zero of the
-1 + 28 / (53 - 45 * cos(2*alpha))
which I got by differentiating and simplifying the trig. As you say, this
is close to a half ball hit, which would be 30 degrees. This corresponds
to about 33.749 degrees of follow angle. A question I have about this is
what happens when friction between the balls is included? I would assume
that this changes slightly both the cue ball and the object ball paths.
$.02 -Ron Shepard