Object ball squirt?

M

Mitchxout

AzB Silver Member
Silver Member
When breaking a rack of nineball the corner balls aren't dead in the corner pockets. However, it's my understanding that the rack moves back when struck, allowing the corner balls to essentially become wired.

My question is, what happens to the object ball when hit at a high rate of speed? Why does this cause cut shots to be undercut? Does the OB slide forward then change directions?
 
Mitchxout said:
When breaking a rack of nineball the corner balls aren't dead in the corner pockets. However, it's my understanding that the rack moves back when struck, allowing the corner balls to essentially become wired.

My question is, what happens to the object ball when hit at a high rate of speed? Why does this cause cut shots to be undercut? Does the OB slide forward then change directions?
Click to expand...
For high-speed shots, the cut angle is actually closer to the ideal. You can test this for yourself with frozen combinations struck at an angle of about 45 degrees. There is less throw at higher speeds.

Breaking a solid rack is a very complicated event. All of the balls compress slightly during the time that the cue ball is on the one ball. The exit angles of the balls coming out of the rack depend on exactly how much each contact point was compressed.

A similar shot is the "double spot shot" in which two balls are spotted on the foot spot and the cue ball is shot from the kitchen. The front ball can be made straight into a corner pocket even without draw on the cue ball. All three balls are together for an instant (actually, about 200 microseconds) and the outbound angle of the middle ball depends on the details of how the balls compress.

There are more explanations of this on Dr. Dave's site.
 
Object ball

Mitchxout said:
When breaking a rack of nineball the corner balls aren't dead in the corner pockets. However, it's my understanding that the rack moves back when struck, allowing the corner balls to essentially become wired.

My question is, what happens to the object ball when hit at a high rate of speed? Why does this cause cut shots to be undercut? Does the OB slide forward then change directions?
Click to expand...

Most cut shots are undercut for a couple reasons.

1. Contact Induced Throw (aka Collision Induced Throw) - Even though the balls are smooth and polished there is still friction between the 2 balls. On a cut shot the friction action imparts spin on the cue ball and the object ball. The spin on the object ball results in Throw which changes its path slightly (to be undercut). Speed is important because the softer you hit the more throw is caused. Plus any english you have can reduce/elimate this or magnify it (outside english reduces/eliminates this and inside magnifies it.

2. The collision actually causes the balls to slide back a little due to the force. The best example is the 9 ball break. If you break from the left side you will notice the wing ball's tangent line is no to the pocket, but the force of the break actually causes the rack to push back and twist to the right a little. This slight change makes it possible for the wing ball to be made in the corner. Its the same effect on cut shots but not as big. The harder you shoot the more this has an effect.

Most people don't understand these concepts so they have taught themselves that "I have to shoot this shot with outside english to make it". If you understand why this is, then you can shoot center ball and cut the shot slightly thinner.

Learn how to pocket the balls cleanly with center ball hits. After this then learn how to pocket the balls cleanly with center ball hits at the various speeds (soft, medium, hard). After this then learn how to pocket the balls cleanly with Left and right english at different speeds.
 
Bob Jewett said:
For high-speed shots, the cut angle is actually closer to the ideal. You can test this for yourself with frozen combinations struck at an angle of about 45 degrees. There is less throw at higher speeds.
Click to expand...
There's a point of no return where high enough speed causes undercuts. I'll check with DR Dave. Thanks.
 
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Bob Jewett said:
All three balls are together for an instant (actually, about 200 microseconds) and the outbound angle of the middle ball depends on the details of how the balls compress.
Click to expand...
I couldn't find anything on Dr Dave's site about how the balls compress.
 
Bob Jewett said:
For high-speed shots, the cut angle is actually closer to the ideal. You can test this for yourself with frozen combinations struck at an angle of about 45 degrees. There is less throw at higher speeds.
Click to expand...
Ok, take the same shot but with a single OB. Isn't there an optimum speed

that minimizes CIT? I know very soft speed needs more cut and adding

speed gradually reduces CIT.

Only to a point, though. I've found that if hit hard enough, CIT will go back

up increasing the cut again. I've never read this explained before.
 
Mitchxout said:
Ok, take the same shot but with a single OB. Isn't there an optimum speed

that minimizes CIT? I know very soft speed needs more cut and adding

speed gradually reduces CIT.

Only to a point, though. I've found that if hit hard enough, CIT will go back

up increasing the cut again. I've never read this explained before.
Click to expand...

I don't think I've seen this effect. Do you have a way to demonstrate it?
 
RWOJO said:
2. The collision actually causes the balls to slide back a little due to the force. The best example is the 9 ball break. If you break from the left side you will notice the wing ball's tangent line is no to the pocket, but the force of the break actually causes the rack to push back and twist to the right a little. This slight change makes it possible for the wing ball to be made in the corner. Its the same effect on cut shots but not as big. The harder you shoot the more this has an effect.
Click to expand...
Thank you very much, this is what I was looking for.
 
Cit

The effect of CIT is greater with a greater angle of a cut shot. Also the softer you hit it appears to increase the CIT but you are really giving the ball more time to spin.

A way to test this over and over is freeze 2 object balls. 1 on the foot spot and 1 directly above it so they are straight in line with the head spot. Now from near the foot rail try a bunch of different shots.

1. try cutting the ball thats on the footspot as if to make it in the side pocket (even though it wont go). Try this at different speeds and see where the frozen ball goes. If the cut is to the left side pocket (from your view at the foot rail) - this imparts opposite spin (right) and in return that imparts opposite spin to the left on the frozen ball. You will see the frozen ball instead of going straight for the head spot it will change its path a little to the left. At different speeds this varies.

2. Try placing the OB and FB the same as before, but CB over to the left corner pocket. Cut the OB as if to make it in the right side pocket and see where the Frozen ball goes. again hit at different speeds, try different english and watch the results.
 
Bob Jewett said:
I don't think I've seen this effect. Do you have a way to demonstrate it?
Click to expand...

The collision causes the OB to slide back a bit due to the force. It's easy enough to demonstate, just knock the hell out of a 1/2 ball hit and watch where it goes.
 
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RWOJO said:
The effect of CIT is greater with a greater angle of a cut shot. Also the softer you hit it appears to increase the CIT but you are really giving the ball more time to spin.
Click to expand...

Are you saying CIT isn't affected by speed?

A way to test this over and over is freeze 2 object balls.. try cutting the ball thats on the footspot as if to make it in the side pocket (even though it wont go). Try this at different speeds and see where the frozen ball goes. If the cut is to the left side pocket (from your view at the foot rail) - this imparts opposite spin (right) and in return that imparts opposite spin to the left on the frozen ball. You will see the frozen ball instead of going straight for the head spot it will change its path a little to the left. At different speeds this varies.

2. Try placing the OB and FB the same as before, but CB over to the left corner pocket. Cut the OB as if to make it in the right side pocket and see where the Frozen ball goes. again hit at different speeds, try different english and watch the results.
Click to expand...

Are you explaining CIT or saying the OB curves?
 
Freeze two balls in the middle of the table, dead straight to the corner pocket.
Shoot the cue ball from about a 30 degree angle straight at the first ball for pocket speed.
I'm betting you can miss this "dead combination" quite easily.

Steve
 
Here is a plot of some measurements I did a while ago. It shows that for soft shots, the throw increases up to some limit as the cut angle increases. For harder shots, there is a maximum amount of throw at about a half-ball hit (30-degree cut) and then it decreases for larger angles.
throw.gif


Physics predicts something very similar to this, and it's covered somewhere on Dr. Dave's site.

While the above was done for a frozen combination, I'm pretty sure you would get the same result for a stun shot with no spin on the cue ball.
 
Which ball gets thrown more? I think the cue ball moves of line more than the object ball. So if you are striking a set of nine balls, throw has very little to do with the wing balls path towards the corner pockets.
 
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Bob Jewett said:
Here is a plot of some measurements I did a while ago. It shows that for soft shots, the throw increases up to some limit as the cut angle increases. For harder shots, there is a maximum amount of throw at about a half-ball hit (30-degree cut) and then it decreases for larger angles.
throw.gif


Physics predicts something very similar to this, and it's covered somewhere on Dr. Dave's site.

While the above was done for a frozen combination, I'm pretty sure you would get the same result for a stun shot with no spin on the cue ball.
Click to expand...

I've seen this graph as well as Dr Dave's and didn't realize it was with frozen balls.

Question: Is it possible to slide the cueball almost the length of the table with hard centerball? It's possible I'm simply applying stun from a long distance with a hard hit. Anyway, the result is with high speed I have to cut the shot more.

Although, this still doesn't explain how a passing shot works and only works in one direction and only with a hard hit. This goes back to saying the OB slides back before it rolls forward on a hard hit.
 
RWOJO said:
Most cut shots are undercut for a couple reasons.

1. Contact Induced Throw (aka Collision Induced Throw) - The spin on the object ball results in Throw which changes its path slightly (to be undercut).

2. The collision actually causes the balls to slide back a little due to the force.
Click to expand...

Are both throw and slide always present equally on cut shots?

On a hard hit could slide outweigh throw? Would a soft hit throw more but slide less?

Is a thinner hit more throw and less slide?

A thicker hit more slide than throw?
 
RWOJO said:
...
2. The collision actually causes the balls to slide back a little due to the force. The best example is the 9 ball break. If you break from the left side you will notice the wing ball's tangent line is no to the pocket, but the force of the break actually causes the rack to push back and twist to the right a little. This slight change makes it possible for the wing ball to be made in the corner. Its the same effect on cut shots but not as big. The harder you shoot the more this has an effect. ...
Click to expand...
I'm pretty sure that none of the balls in a nine ball rack move more than a millimeter or so before they separate. If that's true, what you describe can't occur because the movement is not large enough to change the kiss lines. I think something more complicated happens in a full, tight rack.
 
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