# Throwing Frozen Object Balls Question

#### MattPoland

##### AzB Silver Member
Silver Member
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.
Truth in that. The OB1 gives the cueball a lot of resistance. The right spin sends the cueball right every time. So you can’t count on the cueball acting like a stop shot to know a hit was square.

You can go a little too far left of square too far right of square. Both with right English on the cueball. In all those attempts, OB2 never went left. It pretty much always went dead straight or at most 1-3 degree to the right (if I estimate angles well).

Interestingly enough hitting it with no spin and contacting OB1 as a half ball hit (on the left side) didn’t throw OB2 that much more. Maybe 2-4 degrees to the right. It wasn’t until I did that half ball hit and put right spin on the cueball that the two combined sent the OB2 to the right more like 5-7 degrees (like you’d expect to throw a miss into a make).

#### 3kushn

##### AzB Silver Member
Silver Member
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 1/4? diamond throw using max spin and cutting ball 2 about 1/4 ball shooting full length of table.

#### MattPoland

##### AzB Silver Member
Silver Member
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 1/2 or maybe only 1/4? diamond throw using max spin and cutting ball 2 about 1/4 ball shooting full length of table.

It was one of those FB reels. Where they show hitting it straight won’t make it but hitting the left side of the first object ball throws the second object ball to the right and that will make it.

Someone commented “Left English will do the same thing.” My kneejerk reaction was “It would need to be right English”. But then it got me wondering if it was a bit more complicated than that.

But I do know if I really want to throw that ball, I’d contact about a half ball hit on the left side with maximum right spin. You mentioned a quarter ball hit, I’d say either way is similar thinking.

#### dr_dave

##### Instructional Author
Gold Member
Silver Member
I'm sure there's a Dr. Dave page illustrating the details of this.

This topic has been on "my list" for a long time. Maybe I'll do a video (maybe along with a supporting math/physics analysis) soon; although, I have a very busy month ahead.

Nice job with the illustration. Do you mind if I include it in my video (with a "shout out" to you and your thread) when I can find the time?

bbb

#### MattPoland

##### AzB Silver Member
Silver Member
This topic has been on "my list" for a long time. Maybe I'll do a video (maybe along with a supporting math/physics analysis) soon; although, I have a very busy month ahead.

Nice job with the illustration. Do you mind if I include it in my video (with a "shout out" to you and your thread) when I can find the time?
No problem. Go for it!

#### Mensabum

##### AzB Gold Member
Gold Member
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.

#### VarmintKong

##### Cannonball comin’!
Gold Member
Is it Byrne’s or 99 Shots that outlines the “2nd ball principle”? Can’t remember where I learned it, but I darn sure internalized it.

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#### straightline

##### AzB Silver Member
Silver Member
Synapses get all twisted without proper mind training.

#### mikemosconi

##### AzB Silver Member
Silver Member
In straight pool when you have four balls frozen within the stack it is the side of the second frozen ball closest to the pocket and what side of that ball is contacted by the third ball in the frozen group of four or more balls that determines the path of the frozen ball closest to the object pocket

bbb

Silver Member

#### 7stud

##### AzB Silver Member
Silver Member
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.
Can you explain that to me? The way I see it, CIT is an effect created by cut angle. In the diagrammed shot, there is no cut angle. It doesn't matter how much spin the CB imparts to OB1, it does not create any cut angle between OB1 and OB2.

In reality, there has to be a slight cut angle due to the deflection of the CB then the swerve back to the contact point with OB1 on the line of centers of OB1 and OB2 (the condition stated by the op), but that doesn't seem like what you are pointing out. If you dismiss that slight cut angle, then the effects seem like they are all spin related to me, i.e. SIT.

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#### MattPoland

##### AzB Silver Member
Silver Member
Can you explain that to me? The way I see it, CIT is an effect created by cut angle. In the diagrammed shot, there is no cut angle. It doesn't matter how much spin the CB imparts to OB1, it does not create any cut angle between OB1 and OB2.

In reality, there has to be a slight cut angle due to the deflection of the CB then the swerve back to the contact point with OB1 on the line of centers of OB1 and OB2 (the condition stated by the op), but that doesn't seem like what you are pointing out. If you dismiss that slight cut angle, then the effects seem like they are all spin related to me, i.e. SIT.
You’re right.

Here’s the scenario. OB1 is moving left from OB2. OB1 is spinning clockwise. So what is happening at the point of contacts at the front of OB1?
1. The movement left and the rotation right can be the same speed and no throw is applied to OB2.
2. The movement left is faster than the rotation right, and the net effect is OB2 gets weakened CIT to the left.
3. The movement left is slower than the rotation right, and the net effect is OB2 gets weakened SIT to the right.
I was incorrectly saying CIT and SIT were competing together for a net weakened effect. But that’s not the right terminology. You either get CIT or you get SIT or you get neither in this scenario. Those terms describe the final effect. They don’t describe the forces that build up to that effect.

In this case the result I’ve observed is this shot always yields #3.

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#### MattPoland

##### AzB Silver Member
Silver Member
Now considering a different scenario. What if the CB hits OB1 on the left side and has heavy right spin. Now OB1 is moving right and has clockwise spin. And the point of contacts at the front of OB1 has both factors do combining to create an effect that would be considered both “CIT and SIT” and we are just talking magnitude.

#### Patrick Johnson

##### Fish of the Day
Silver Member
3. The movement left is slower than the rotation right, and the net effect is OB2 gets weakened SIT to the right.

In this case the result I’ve observed is this shot always yields #3.
That's what I'd expect too.

pj
chgo

#### Cue Stick

##### Member
Have you done a cost/benefit analysis on this project All jokes aside, I think the answer would depend on additional factors that you haven't listed...material the balls are made of, cleanliness of the balls, type of cloth, cleanliness of the cloth, relative humidity...

#### Willowbrook Wolfy

##### Going pro
Gold Member
two balls together will cut the throw in half when applying spin because they are double the mass. I think anyway.

#### Mensabum

##### AzB Gold Member
Gold Member
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.
As a 14 to 1 player, you run across this quite often I'm sure, as do I. Frozen throw can make or break a good run. You can include caroms after the fact if you wish. Understanding frozen throw is crucial for any player to know if they truly wish to put up some decent #s.

#### Mensabum

##### AzB Gold Member
Gold Member
My s
As a 14 to 1 player, you run across this quite often I'm sure, as do I. Frozen throw can make or break a good run. You can include caroms after the fact if you wish. Understanding frozen throw is crucial for any player to know if they truly wish to put up some decent #s.

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.
Excellent post by the way. My suggestion would be to put some balls on the table and shoot some Frozen pairs and triples until you're familiar with how they roll afterwards. Use all types of English and speeds. You'll get a feel for it pretty quick. Then you can really start to see how far you can throw a shot that's not online w a pocket by loading up w spin/speed. More or less as needs arise.
Enjoy!! This is where the game starts to get good dude.

#### Baby Huey

##### AzB Silver Member
Silver Member
When discussing throwing balls, be careful because if they are grimmy and dirty they throw great but if they're waxed watch out.............They won't throw at all.