Billiards Trig Question

[winglessbuzzard]

Given:
Cue Ball and Target Ball are both perfectly round
Cue Ball's diameter is D and it's center is X1, Y1
Target Ball's diameter is D and it's center is x2, Y2
Table Pocket's center is X3, Y3
There are no obstructions between X2, Y2 and X3, Y3 or X1, Y1 and X2, Y2

Question:
What is the algebraic expression, given the above variables, for the coordinates
X4 =
Y4 =
Where X4, Y4 = the desired cue ball ending position, a distance of D past the Target ball strike point (assuming it does not change direction when hitting Target ball) after it has struck the Target ball directly toward the X3, Y3.

 

[mmm4444bot]

Your description of this senario contains some ambiguities.

I'm thinking that (x1,y1) (x2,y2) (x3,y3) (x4,y4) all lie on the same line. Is this correct?

(Uploading a diagram would be very helpful.)

?

… a distance of D … Measured from where to where ?

… the Target ball strike point … This is the location of impact (i.e., not a ball center) ?

 

[mmm4444bot]

Is my guess (below) correct?

Double-click the image, to expand.

[attachment=0:2ze6jpdu]4Buzzard.JPG[/attachment:2ze6jpdu]

 

[winglessbuzzard]

[attachment=0:201qon97]billiards question diagram.JPG[/attachment:201qon97]

Sorry,

x1,y1
x2,y2
x3,y3
and
x4,y4

do not lie on the same line.

D = the diameter of the ball (both balls with the same dameter)

Diagram attached

I'm working on a 'ghost mouse' application to help on computer based billiard games. Basically, you'd
*put the mouse in the middle of the cue ball, hit a button
*put the mouse in the middle of the target ball, hit a button
*put the mouse in the middle of the pocket, hit a button

then the computer would position the mouse cursor at X4,Y4 ... effectively where you need to aim the cue ball in order to strike the Target ball, driving it into the pocket.

 

[mmm4444bot]

Maybe distance D is this ?

[attachment=0:1b1b8ghr]4Buzzard2.jpg[/attachment:1b1b8ghr]

 

[Deleted member 4993]

Given:
Cue Ball and Target Ball are both perfectly round
Cue Ball's diameter is D and it's center is X1, Y1
Target Ball's diameter is D and it's center is x2, Y2
Table Pocket's center is X3, Y3
There are no obstructions between X2, Y2 and X3, Y3 or X1, Y1 and X2, Y2

Question:
What is the algebraic expression, given the above variables, for the coordinates
X4 =
Y4 =
Where X4, Y4 = the desired cue ball ending position, a distance of D past the Target ball strike point (assuming it does not change direction when hitting Target ball) after it has struck the Target ball directly toward the X3, Y3.

So the target ball does change direction after hitting the ball.

This - as I see it - is going to be quite difficult problem. Solvable - but quite difficult.... and time consuming.

 

[winglessbuzzard]

Yes, technically it would change direction. But we're modifying that rule of physics today and assuming it just continues straight through the Target ball without any change in direction (for simplicity sake). Thanks for your time. If this is too much hassle, then I don't want to pressure you.

 

[Denis]

Ahhh....methinks I see what's going on...

Say the centers of cue ball, target ball and pocket form a straight line;
then the cue ball will simply end up exactly in target ball's original position, right?

To remove ambiguity (like the table's banks not interfering), problem can be looked at
as a fixed circle (the pocket), then a moving circle (the cue ball; straight line path) contacting
a temporarily fixed circle (target ball) such that this circle moves in a straight line towards the
fixed circle, ending up exactly in the fixed circle's position.

After contact, the "cue circle" continues in a straight line (same straight line as before) for
a distance equal to it's diameter. All 3 circles have same diameter.

For all intents and purposes, due to symmetry, all cases can be looked at as happening in
the top right quadrant, pocket circle tangent to x and y axes, target circle to the right on
x axis or higher, cue circle to right of target circle.

With diameters = D, then pocket circle is fixed at center (D/2, D/2).

Centers of circles:
pocket circle: P(x1,y1)
target circle: T(x2, y2)
cue circle: C(x3,y3)
x2 > x1, y2 => x2, x3 > x2

And the problem becomes calculating the slope of the line CM (M on circle T's circumference)
such that the point of contact M creates straight line TP.
Add D to end of line CM (creating straight line CD through M) to get circle C's final position.

Looks like point M is simply end of straight line PM going through point T;
anyhow, that's certainly the case if circle T is on x axis.
Need my 2nd coffee before "trying" all that out...

 

[Denis]

Turns out to be relatively/confusingly easy: similar right triangles....

A(a,b) = original Cue ball, B(g,h) = Cue ball at contact of Target ball, C(i,j) = Cue ball's final position.
D(c,d) = Target ball, E(e,f) = Pocket.

A,B,C on same straight line; B,D,E on same straight line; angle ABD > 90 degrees.

I'll use r = radius instead of D = diameter.

Givens are r,a,b,c,d,e,f. Problem: calculate i,j

SOLUTION:
i = a - [(a - g)(v + 2r)]/v
j = f + [(h - b)(v + 2r)]/v
where:
u = SQRT[(c - e)^2 + (d - f)^2] : that's length of line DE

g = e + [(c - e)(u + 2r)]/u
h = f + [(d - f)(u + 2r)]/u .....coordinates of B

v = SQRT[(a - g)^2 + (h - b)^2] : that's length of line AB

Try it using r = 5, a = 38, b = 13, c = 18, d = 16, e = 6, f = 7.
You'll get full integer case: i = 18, j = 28 ; g = 26, h = 22, u = 15, v = 15.