Throwing Frozen Object Balls Question

[MattPoland]

I know if two object balls are frozen, you can manipulate the path of the second object ball. Typically the advice is to project the line-of-centers between both object balls. Cutting into the left side of that line on the first object ball will throw the second object ball to the right. And vice versa. I'm curious about a similar scenario.

Two object balls are frozen. Your cueball is in line with the line-of-centers. You choose to hit that first ball square for whatever reason. But you still want to manipulate the path of the second object ball. You choose to put right english on the cueball. You adjusted for deflection to hit that first object ball square.

What is the behavior of the second object ball?

1. Throws strongly right because the spin-induced throw is highly dominant?
2. Throws strongly left because the collision-induced throw is highly dominant?
3. Goes dead straight forward because the spin-induced throw and cut-induced throw net each other out?
4. Throws weakly right because both throws net out but spin-induced throw was slightly more dominant?
5. Throws weakly left because both throws net out but collision-induced throw was slightly more dominant?

I plan on playing with it when I get home. I doubt it's anything I'll find useful for an in-game scenario.

I'm sure there's a Dr. Dave page illustrating the details of this. I just can't help finding the thought process interesting. Makes me curious what others would think the answer is from the tops of their heads based off experience or reasoning. At least before a definitive answer is easily determined.

I also wouldn't be surprised to find out shot speed and tip offset may introduce some variability.

 

[Fore Rail]

The second object ball will go to the right.

The object ball will not throw as much as possible because of the full hit.

Speed of course is also a component of the amount of throw. Throw in how much dirt is involved as well.

 

[maha]

two frozen balls want to act as one ball.

 

[jsp]

I'm not sure how CIT plays a part if you're hitting the first ball absolutely full. I would think the scenario would be similar to hitting the first ball left of center with the cue, which means the second ball will throw to the right. I don't see how the second ball experience any force component to the left.

EDIT: Nevermind, I see how CIT can play a part in it...if the first ball can be thought of has being thrown to the left. Good question. I still feel like the net effect would be that the second ball goes right.

 

[Patrick Johnson]

The spin on OB1 is gearing spin on OB2, so no throw.

pj
chgo

 

[Tennesseejoe]

Slightly right.

 

[jsp]

The spin on OB1 is gearing spin on OB2, so no throw.

View attachment 756711
pj
chgo

Can you clarify your answer? Does "no throw" mean that OB2 will go completely straight?

 

[pt109]

I know if two object balls are frozen, you can manipulate the path of the second object ball. Typically the advice is to project the line-of-centers between both object balls. Cutting into the left side of that line on the first object ball will throw the second object ball to the right. And vice versa. I'm curious about a similar scenario.

Two object balls are frozen. Your cueball is in line with the line-of-centers. You choose to hit that first ball square for whatever reason. But you still want to manipulate the path of the second object ball. You choose to put right english on the cueball. You adjusted for deflection to hit that first object ball square.

What is the behavior of the second object ball?

1. Throws strongly right because the spin-induced throw is highly dominant?
2. Throws strongly left because the collision-induced throw is highly dominant?
3. Goes dead straight forward because the spin-induced throw and cut-induced throw net each other out?
4. Throws weakly right because both throws net out but spin-induced throw was slightly more dominant?
5. Throws weakly left because both throws net out but collision-induced throw was slightly more dominant?

View attachment 756678

I plan on playing with it when I get home. I doubt it's anything I'll find useful for an in-game scenario.

I'm sure there's a Dr. Dave page illustrating the details of this. I just can't help finding the thought process interesting. Makes me curious what others would think the answer is from the tops of their heads based off experience or reasoning. At least before a definitive answer is easily determined.

I also wouldn't be surprised to find out shot speed and tip offset may introduce some variability.

Shot speed……if hit hard, the combo will go where ever it‘s lined up.

 

[MattPoland]

two frozen balls want to act as one ball.

When you suggest two frozen balls want to act as one ball, are you suggesting that they'd behave like this? The spin on the cueball throws them both to the left and as a pair they rotate clockwise together around their joint center of mass (momentarily until they separate) and as such each would have slight clockwise spin on them?

 

[MattPoland]

The spin on OB1 is gearing spin on OB2, so no throw.

View attachment 756711
pj
chgo

Would you say that means the "gearing contact" means the CIT and SIT on OB2 are both cancelled out and the final direction is the original path?

Or are you suggesting the "gearing contact" only cancels the CIT throw and the SIT is still active?

 

[Patrick Johnson]

Can you clarify your answer? Does "no throw" mean that OB2 will go completely straight?

It means throw to the left on OB2 from collision with OB1 will be minimal to none, so OB2 will go more straight or right.

pj
chgo

 

[MattPoland]

It means throw to the left on OB2 from collision with OB1 will be minimal to none, so OB2 will go more straight or right.

pj
chgo

So you're in the ballpark of this...

And you'd imagine if it has any throw to the right, it will be weakened. Nowhere near what you'd get from just cutting the CB into the left side of OB1.

 

[Patrick Johnson]

So you're in the ballpark of this...

View attachment 756716

And you'd imagine if it has any throw to the right, it will be weakened. Nowhere near what you'd get from just cutting the CB into the left side of OB1.

Right. Of course, this all depends a lot on the amount of spin, condition of balls, etc.

pj
chgo

 

[maha]

i think because of the total friction involved the throw aspect overcomes the english to the second ball.
using pool balls in the pool room.

 

[Patrick Johnson]

i think because of the total friction involved the throw aspect overcomes the english to the second ball.
using pool balls in the pool room.

For throw to dominate, OB1 must be thrown left faster than its surface spins right against OB2. Is that what you mean?

pj
chgo

 

[justnum]

These measures are only useful on the initial ball to ball contact.

There is rail induced throw in carom billiards.

Softer shots are more common to carom because rail induced throw can be managed easily when using multiple rails.

Learning the optimal amounts of spin for specific type of shots is more appropriate than a general analysis.

Most shooters train for targets that occur most. Frozen balls do not happen often.

A better discussion would have focused on the cue tip and cue ball contact surface interaction and how it influences spin.

 

[Bob Jewett]

I think there will be a little throw due to spin on the middle ball. I think this is a hard effect to measure because it is small and it can be masked by errors in the alignment or hit.

 

[kling&allen]

So you're in the ballpark of this...

View attachment 756717

And you'd imagine if it has any throw to the right, it will be weakened. Nowhere near what you'd get from just cutting the CB into the left side of OB1.

This is an excellent diagram.

 

[Patrick Johnson]

This is an excellent diagram.

Yes. Lots of info in the colors/shades.

pj
chgo

 

[3kushn]

I think there will be a little throw due to spin on the middle ball. I think this is a hard effect to measure because it is small and it can be masked by errors in the alignment or hit.

I'm thinking if there's a better diagram that includes the table, the topic would be more understandable.

If the OP is looking for a lot of throw by hitting ball 2 directly in the center??? Good luck.

Haven't played these shots for a long time but my memory tells me about the most you can get is only 1/4? diamond throw using max spin and cutting ball 2 about 1/4 ball shooting full length of table.