# Optimal pool cue weight

1. Aug 3, 2011

### bill duffy

Within an 18-21 gram range, can you produce more force when striking a cue ball with a lighter or heavier cue?

2. Aug 3, 2011

### LostConjugate

Higher weight gives you more stability. Not more or less force.

3. Aug 5, 2011

### bill duffy

Newton 2nd law -- force = mass (grams) * acceleration (miles/sec)**2. (1) I don't see how weight is involved. (2) the body/arm propelling the cue must figure in.

4. Aug 5, 2011

### Drakkith

Staff Emeritus
He means mass not weight.

Anyways, a more massive stick only affects the stability since it takes more force or time to cause the same movements that a less massive stick would get. The key here is that the stick itself isn't the source of the force, but the person behind the stick. You can apply just as much force with a lighter stick.

5. Aug 6, 2011

### DaveC426913

While it is the arm that generates the force, it is the cue that transfers it to the ball.

A heavier cue will transfer more kinetic energy to the cueball, as it will not slow down quite as much in the collision.

Look at the extreme. If you used a sledgehammer (and got it up to the same velocity as a cue), when it hit the ball, the ball would accelerate to 100% of the speed of the sledgehammer, whereas the sledgehammer would decelerate negligibly.

6. Aug 6, 2011

### Phrak

At the risk that we both might be set straight by those that can feel a 1 ounce difference in cue weight and have a feel for this, there's more to it than that.

The arm will not get a cue that has the weight of a 20 lb. sledge to the velocity of a lighter one. Given a particular arm, there is a cue mass that is neither infinite nor zero that delivers maximum cueball velocity.

7. Aug 6, 2011

### S_Happens

An 18-21 gram cue would be pretty much worthless, so obviously you mean ounces.

As for the REAL reason(s) for choosing a certain cue, it's mostly about personal comfort and consistency.

The only time you're really concerned with a max force would be for the break, and many people use a different cue for breaking than for normal play. The "optimum" cue for you is the one that you can produce the best results with, and will almost certainly result more from just being comfortable or what you spend a lot of time using.

8. Aug 6, 2011

### DaveC426913

Who said anything about max force? No matter how hard you're hitting, you'd want the best transfer from your arm to cue to ball.

And it is the cue that should do the work on the ball, not your arm.

You're supposed to just gently grip the cue. If you grip a cue too hard you lose fine control. So gently gripping a too light cue would cause it to stop the moment it hit the ball.

Your ideal cue would be as heavy as possible to give you better transfer, yet not so heavy as to be uncomfortable and uncontrollable.

9. Aug 6, 2011

### S_Happens

The OP. My point is that his actual question and thread title don't necessarily go together.

10. Aug 7, 2011

### Redbelly98

Staff Emeritus
It would actually be close to 200% (or twice) the speed of the sledgehammer. (Assuming an elastic collision, and the sledgehammer mass is a lot larger than the cue ball mass.)

EDIT:
Interesting question. If we consider "optimal" to mean what gives the cue ball it's highest velocity, then ever more massive sticks would have less and less velocity.

Assume your arm exerts a fixed amount of force over a fixed distance, and hence does a fixed amount of work on the stick. I.e., mv2 is a fixed quantity for the stick, hence v decreases with increasing mass. In the high mass limit, $v_{ball} \approx 2v_{stick}$ would decrease as well.

Last edited: Aug 7, 2011
11. Aug 7, 2011

### bill duffy

The range is actually from 18 - 25 oz, although 18-21 is normal, and you can notice that 16.7% increase.
It is safe to assume for present purposes that the coefficient of restitution of the 5 1/2 oz ball is 1.0.
There seems to be a 2 part model. First, the cue acceleration -- a biomechanical model showing how the cue acceleration decreases as cue mass increases. My intuition is that increasing the mass of the cue by 16.7% is accompanied by a less than proportional decrease in v-squared, at least for more muscular players. I think that this is the essential question.
Second, the force acting on the ball produced by the cue. Everyone seems to agree on Newton's 2nd law.

12. Aug 7, 2011

### DaveC426913

Certainly. If this were true:
But it isn't.

The player applies as much force as is necessarily to hit the ball as hard as he needs to. It would be silly for a player to pick up a heavier cue and apply the same force to it as the lighter cue.

13. Aug 10, 2011

### bill duffy

According to R Shepard, Amateur Physics for the QAmateur Pool Player, p22, "for a given for ce on the cue stick and a given stroke length, a light cue stick will acquire the same energy as a heavy cue stick." So that's that...

14. Aug 10, 2011

### DaveC426913

While that may be true, it stops juuuuust short of being the definitive answer. It does not address how much of that momentum is transferred to the ball.

15. Sep 7, 2011

### poolhalljunke

lol ok. So, I play pool everyday every week anytime i can. And first off the cue ball will only stop immediately on contact with the object ball if "English" is used a ball will never just STOP without it. No matter how heavy or light a pool cue is. I use a Lucassi cue myself and it's a 18.5 i break and shoot with it, why? Just for the simple face i have control over the cue with it at all times. The object of Billiards is not to get the 8,9 or 10 ball in (deciding on game type) the actual object is where you put that cue ball, and weight is not an issue my friends. weight is a comfort zone for the player, finesse is key not strength. Any real pool player can and will tell you if you ask them, the Weight, color, brand (custom or production) does not make a difference on the butt of the cue, (Shafts yes buts that's a different story) but it's the player that makes the ball move or react not the stick.