What do you prefer to break with? Heavy or light?

[fastone371]

We all love dirty 8b on beater Valley barboxes!

There are several variables besides velocity that could have been contributing to you getting better breaks. Does your break cue have a wider shaft/tip than your playing cue? (Larger sweet spot). Or a harder tip, or flatter tip? (less inadvertent spin). Any/all of these might be allowing you to get less spin and be more accurate - giving you just a tiny bit more of a square, solid pop into the head ball.

I’m not into break cues, but if i did have one, I think I’d prefer a fatter, conical taper (vs the 12.5 mm pro taper I prefer on my playing cue). I’ve noticed when power slamming on barboxes with dirty/sticky balls and using house cues, I seem to get better results when the cue has fatter, stiffer taper.

This might be what some folks term “more efficient energy transfer”. Or maybe it was a few extra beers and your shoulder was looser - of course that may also be technically considered more efficient energy transfer

Cheers

The break cue shaft is the same size as my playing cue. As far as unintentional spin I am pretty good at parking the cue ball right between the side pockets so I don't think it would be spin. The tips are different, I use a Tiger Everest on my player and I have one of those clear Bulletproofs on my break cue.

 

[DeeDeeCues]

I don't really care. I have cues from about 15.5oz to 19.5oz. I break with whatever I'm playing with. I prefer my lighter cues, normally.

 

[phreaticus]

The break cue shaft is the same size as my playing cue. As far as unintentional spin I am pretty good at parking the cue ball right between the side pockets so I don't think it would be spin. The tips are different, I use a Tiger Everest on my player and I have one of those clear Bulletproofs on my break cue.

Must be the tip, the beers, both, or placebo! The great mystery continues

 

[DeeDeeCues]

you just have to find what your limit factor is. the simplistic formula is F=MA (M=cuestick weight, A= cuestick Acceleration). That being said, going from a 18oz to 20oz break cue would be a 4% increase (EDIT -- its actually a 11.1% increase, Thanks Bob Jewett for pointing that out) in F (which is the force at which the cue ball hits the rack (not accounting for friction of the cloth and etc). So knowing this, can you swing an 18 oz cue 11.1% faster than a 20oz cue???? if you can the answer would be to use a lighter cue, if not the heavier cue is for you. I don't know the answer to that, ,maybe someone with a better physics background then I could answer that. Take that to extreme's of a 16oz break cue vs a 25oz break cue, that is a 36% difference (if I did my quick math right). So the same question, can you accelerate the lighter cue (16oz) 36% faster than the heavier cue (25oz)????

I am sure there is also some confusion on the difference between acceleration and velocity in my simplistic view of this.

F= MA is not really relevant. You need to figure that it is approximately an elastic collision and figure out the kinetic energy of the system before and after contact.

The energy changes directly proportional to mass, but proportional to the SQUARE of the velocity. Thus, a forty percent increase in velocity almost double the kinetic energy.

 

[surffisher2a]

F= MA is not really relevant. You need to figure that it is approximately an elastic collision and figure out the kinetic energy of the system before and after contact.

The energy changes directly proportional to mass, but proportional to the SQUARE of the velocity. Thus, a forty percent increase in velocity almost double the kinetic energy.

I only know very basic physics (think high school level). In the case of the cue stick hitting the cue ball there is still a driving force acting (pushing)on the cue ball during the collision. I think that changes some things with a pure elastic collision (where are rolling ball hits another rolling ball with no other forces involved).

 

[Willowbrook Wolfy]

All this is good info on here. Am I the only one that doesn’t give a crap about a tight rack? I could care less. Even on 9 ball. If one side is loose it’s not a problem to adjust your break accordingly. I know someone’s going to chime in going on about variables, etc. most my breaks come from near the middle. I can get just as good a break with the head ball off one ball 1/16” or 1/32” as with it frozen on both..

 

[DeeDeeCues]

I only know very basic physics (think high school level). In the case of the cue stick hitting the cue ball there is still a driving force acting (pushing)on the cue ball during the collision. I think that changes some things with a pure elastic collision (where are rolling ball hits another rolling ball with no other forces involved).

I never said that a force didn't act on the cueball. The point is that F=MA doesn't directly help figure out what you are trying to figure out. A simple change in mass or velocity does not equate to the same percentage change in the velocity of the cueball.

Conservation of energy and assuming an elastic collision between the cue and the ball is a much better way to figure this. I don't want to go through all the physics here, I know it is on Dr. Dave's website.

 

[KMRUNOUT]

All this is good info on here. Am I the only one that doesn’t give a crap about a tight rack? I could care less. Even on 9 ball. If one side is loose it’s not a problem to adjust your break accordingly. I know someone’s going to chime in going on about variables, etc. most my breaks come from near the middle. I can get just as good a break with the head ball off one ball 1/16” or 1/32” as with it frozen on both..

Absolutely no offense intended, but what you describe sounds like you are differentiating between terrible and a little less terrible, and saying there is little difference.

If I evaluate the grip of the tires on my car, I don’t look at the difference between sliding off the road and crashing hard vs really hard.

If you don’t recognize why a tight rack matters, then it is highly likely that you are not establishing particularly high standards for break quality. Your claim of breaking 9 ball from the center reinforces this idea.

Let me ask...when you break 9 ball, what percentage of the time are you making a ball and getting shape on the 1 ball? Best to try it 10 times or so and see, rather than rely on feelings about it. You might try 10 times with the rack any which way you have it, and 10 times with a properly used template. (10 times would be barely statistically significant, but 1000 breaks would certainly say something)

With respect, there is quite a difference between the ideas of whether “it matters” and whether “you care”.

It matters. Also, I care.

KMRUNOUT



Sent from my iPhone using AzBilliards Forums

 

[KMRUNOUT]

I never said that a force didn't act on the cueball. The point is that F=MA doesn't directly help figure out what you are trying to figure out. A simple change in mass or velocity does not equate to the same percentage change in the velocity of the cueball.

Conservation of energy and assuming an elastic collision between the cue and the ball is a much better way to figure this. I don't want to go through all the physics here, I know it is on Dr. Dave's website.

Well, if all variables are kept constant *other than* the cue mass and resultant velocity of the cue at impact* (given the same input force...meaning the same arm and arm muscles and strength pulling the cue with an equal muscle contraction), then F=MA would precisely describe the difference in cueball velocity. Assuming the identical cue characteristics other than weight, the elasticity of the collision would remain constant regardless of the cue mass.

Can you perhaps elaborate on why you think F=MA is not relevant here?

Thank you,

KMRUNOUT


Sent from my iPhone using AzBilliards Forums

 

[LWW]

18oz shooter, 17oz break.

 

[DeeDeeCues]

Well, if all variables are kept constant *other than* the cue mass and resultant velocity of the cue at impact* (given the same input force...meaning the same arm and arm muscles and strength pulling the cue with an equal muscle contraction), then F=MA would precisely describe the difference in cueball velocity. Assuming the identical cue characteristics other than weight, the elasticity of the collision would remain constant regardless of the cue mass.

Can you perhaps elaborate on why you think F=MA is not relevant here?

Thank you,

KMRUNOUT


Sent from my iPhone using AzBilliards Forums

Look up elastic impacts on wiki. Read Dr. Dave's website. Both of those already show why your assumptions are wrong.

A few things:
1. Not all of the energy of the cue stick is transferred to the ball. The stick is still moving after contact.
2. The ball moves faster than the cue stick does contact.

You can take my word for this and look up the sources I recommend, or you can believe your grasp of hs physics outweighs my engineering degree and professional experience. I don't really care, I'm too busy to explain this further when it exists on wiki, on Dr Dave's site, and many places on a billiards.

 

[Willowbrook Wolfy]

Absolutely no offense intended, but what you describe sounds like you are differentiating between terrible and a little less terrible, and saying there is little difference.

If I evaluate the grip of the tires on my car, I don’t look at the difference between sliding off the road and crashing hard vs really hard.

If you don’t recognize why a tight rack matters, then it is highly likely that you are not establishing particularly high standards for break quality. Your claim of breaking 9 ball from the center reinforces this idea.

Let me ask...when you break 9 ball, what percentage of the time are you making a ball and getting shape on the 1 ball? Best to try it 10 times or so and see, rather than rely on feelings about it. You might try 10 times with the rack any which way you have it, and 10 times with a properly used template. (10 times would be barely statistically significant, but 1000 breaks would certainly say something)

With respect, there is quite a difference between the ideas of whether “it matters” and whether “you care”.

It matters. Also, I care.

KMRUNOUT



Sent from my iPhone using AzBilliards Forums

Well it depends. In 9 ball, I typically make a ball on the break. I can make the 1 ball in 4 different pockets of choice regularly, the other 2 on occasion when trying If I feel like it, and have found a break that usually sets playing field into a “safe” table if I feel like having a safe battle. Well 2. 1 from either side. Therefore if I was breaking for shape it’s on the 2. Not the 1. And thank you for your comment. I’m going to start paying more attention to the 2 from now on. Normally doesn’t matter too much because I get a pretty good spread most of the time.

Same break over and over bores me. On my “safe” break I almost always have great shape on the 1. Giving me a setup for a nice safe on the 2 or 3 behind the 3-4 ball cluster.

The 1 ball off the rack doesn’t have any bearing on break except that I might need to hit from right side or left. If the whole rack is loose just stroke the cue ball with the correct English, hit the 1 ball in the correct spot and you are golden. A ball will almost always fall. I think a wing automatically goes in with a loose one a lot of times. And the 1 still goes where you want it.

Of course my break does sounds like a jet fighter taking off so that could be part of it to. Full force every time. If I was a finesse breaker it might change things up a bit. And I’m by no means a pro in the trade. I don’t mind missing nor do I care about percentage of making a ball on my break. I can usually make the 1 if I want though.

On 8 ball I’ve never even tried to intentionally make the head ball. So I have no input on an 8 ball rack except it can be loose and it’s not an issue for me.(I personally wouldn’t hit it center cause cue ball could go flying)And a ball usually falls.

 

[Willowbrook Wolfy]

Just because most break from the side on 9 ball doesn’t mean it’s the best or correct break for 9ball. It’s just the easiest.

 

[Willowbrook Wolfy]

Now I have to watch pool hall junkies again. this got me in the mood. When dudes talking about the one pro player that makes three balls on his 9 ball break all the time……

 

[garczar]

Just because most break from the side on 9 ball doesn’t mean it’s the best or correct break for 9ball. It’s just the easiest.

Huh?? Its done that way because balls go in far more often than not. You can make the one in four pockets 'by choice'????? Maybe on that GoldCrown-like Centurion. Put the pipe down brother.

 

[The_JV]

Huh?? Its done that way because balls go in far more often than not. You can make the one in four pockets 'by choice'????? Maybe on that GoldCrown-like Centurion. Put the pipe down brother.

My keyboard thanks you

 

[Chili Palmer]

The break location has nothing to do with the OP's original post.

 

[mdavis228]

fwiw… I like my breaker and my player to be very close to the same weight. My 2 cents.
I remember Varner, when he was beating the world, saying he didn’t want something different in his hands when hitting one of the most important shots in the game.
I understand the velocity business. Just lean to the side of hitting the break better vs faster.

 

[Willowbrook Wolfy]

*Deleted**deleted*deleted*deleted*

Huh?? Its done that way because balls go in far more often than not. You can make the one in four pockets 'by choice'????? Maybe on that GoldCrown-like Centurion. Put the pipe down brother.

sry not here to argue so I deleted. To answer your question simply- yes. 2 per side breaking from. And I learned all my 9 ball breaks on a Gold Crown.

 

[garczar]

*Deleted**deleted*deleted*deleted*

sry not here to argue so I deleted. To answer your question simply- yes. 2 per side breaking from. And I learned all my 9 ball breaks on a Gold Crown.

I do not believe that. Sorry. Been around a lot of CHAMPION speed players that couldn't do what you're describing. Not going for it.