Physics help?

goettlicher

goettlicher

AzB Silver Member
Silver Member
Hey PJ

What is the correct formula to use to calculate:

energy transfer from cue tip to cue ball? In other words, how fast is the cue ball traveling at the time of impact?

I have been lead to understand that with a dead center hit, the cue ball leaves with about an additional 10% of speed.
 
Patrick Johnson

Patrick Johnson

Fish of the Day
Silver Member
goettlicher said:
Hey PJ

What is the correct formula to use to calculate:

energy transfer from cue tip to cue ball? In other words, how fast is the cue ball traveling at the time of impact?

I have been lead to understand that with a dead center hit, the cue ball leaves with about an additional 10% of speed.
Click to expand...
I don't know the math, but I think Bob Jewett has said the CB leaves the tip at about 150% of the tip's speed.

pj
chgo
 
mikepage

mikepage

AzB Silver Member
Silver Member
Patrick Johnson said:
I don't know the math, but I think Bob Jewett has said the CB leaves the tip at about 150% of the tip's speed.

pj
chgo
Click to expand...

The math is straightforward. There are two before and after things that have to be the same, energy (1/2 M V^2), and momentum (MV).

If the stick is 3 times the mass of the ball (i.e., 18oz and 6oz) then to only way this works out is the stick loses half its speed after impact and the ball goes, like you said, at one and a half times the original stick speed.
 
V

vinay

Registered
goettlicher said:
Hey PJ

What is the correct formula to use to calculate:

energy transfer from cue tip to cue ball? In other words, how fast is the cue ball traveling at the time of impact?

I have been lead to understand that with a dead center hit, the cue ball leaves with about an additional 10% of speed.
Click to expand...

So I dusted off parts of my brain I haven't used since college and figured this out. You use conservation of momentum and the coefficient of restitution to solve for the ratio of initial cue speed to final cue velocity. If you want to check my math, it's here.

The final answer is:

v_ball / u_cue = (1 + Cr) / (1 + m_ball / m_cue)

where, v_ball is the final ball velocity, u_cue is the initial cue velocity, Cr is the coefficient of restitution, m_ball is the mass of the ball and m_cue is the mass of the cue.

Plugging in values from Dr. Dave's handy reference, if you have a 6oz ball, 18oz cue, and a leather tip with a tip-ball Cr of 0.75, you get 1.3125. Which is to say the ball will the traveling about 31% faster than the cue was initially.
 
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C

Cron

AzB Silver Member
Silver Member
I must be having a brain fart, I wound up with....

v_ball = u_cue * 1.3125

Where is u_cue to derive a constant? Or do hot assume 1 here for this?
 
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logical

logical

Loose Rack
Silver Member
Two thoughts:

1) you guys should be virtually unstoppable if you figure this out

2) this all makes sense if the cue was thrown at the ball but it isn't, it is attached to your arm and driven thru impact. The mass of the cue needs to include some amount that represents your hand/arm. Whether the cue is lightly held in your finger tips or gripped in a death grip will also matter. I think a light grip approximates a thrown cue but otherwise your hand/arm increase its effective mass.

Sent from my SM-G950U using Tapatalk
 
logical

logical

Loose Rack
Silver Member
mikepage said:
The math is straightforward. There are two before and after things that have to be the same, energy (1/2 M V^2), and momentum (MV).



If the stick is 3 times the mass of the ball (i.e., 18oz and 6oz) then to only way this works out is the stick loses half its speed after impact and the ball goes, like you said, at one and a half times the original stick speed.
Click to expand...
You are ignoring the energy absorption of the tip and incorrectly applying momentum laws.

Sent from my SM-G950U using Tapatalk
 
V

vinay

Registered
Cron said:
I must be having a brain fart, I wound up with....

v_ball = u_cue * 1.3125

Where is u_cue to derive a constant?
Click to expand...

u_cue is the initial velocity of the cue, i.e. how fast the cue is moving at the moment it strikes the ball. So if you strike the ball at 10mph it will be travelling at 13.125 mph, if you strike it at 20mph, it will be 23.25.
 
C

Cron

AzB Silver Member
Silver Member
logical said:
You are ignoring the energy absorption of the tip and incorrectly applying momentum laws.

Sent from my SM-G950U using Tapatalk
Click to expand...

Sure, and gravity, and atmosphere, and resistance.... point is it's a generic base.
 
C

Cron

AzB Silver Member
Silver Member
vinay said:
u_cue is the initial velocity of the cue, i.e. how fast the cue is moving at the moment it strikes the ball. So if you strike the ball at 10mph it will be travelling at 13.125 mph, if you strike it at 20mph, it will be 23.25.
Click to expand...

Yeh, I was just wondering about the 1 you plugged in, makes sense.

But... but... you didn't figure in my shotty hip!!
 
alphadog

alphadog

AzB Silver Member
Silver Member
logical said:
Two thoughts:

1) you guys should be virtually unstoppable if you figure this out

2) this all makes sense if the cue was thrown at the ball but it isn't, it is attached to your arm and driven thru impact. The mass of the cue needs to include some amount that represents your hand/arm. Whether the cue is lightly held in your finger tips or gripped in a death grip will also matter. I think a light grip approximates a thrown cue but otherwise your hand/arm increase its effective mass.

Sent from my SM-G950U using Tapatalk
Click to expand...

Funny I have had this arguement with a pro about having a solid grip at impact while breaking.
 
V

vinay

Registered
logical said:
I think a light grip approximates a thrown cue but otherwise your hand/arm increase its effective mass.
Click to expand...

This is a good point, I'm assuming a light grip so the effective weight is close to the actual weight of the cue. That said, the weight of the cue isn't as big a factor as people think it is. Like if you used a 50oz effective cue weight, the ratio goes up from 1.31 to 1.56. The speed at which you're able to move the cue is a much bigger factor than the weight of the cue.
 
Chopdoc

Chopdoc

AzB Silver Member
Silver Member
goettlicher said:
Hey PJ

In other words, how fast is the cue ball traveling at the time of impact?
Click to expand...


At the time of impact the cue ball is not moving at all.

Every time I see these questions people do lots of calculations ignoring variables.

For example: The cue is being held in your arm, you can't just use the mass of the cue.

Add to that the fact that none of this happens in a vacuum and the environmental variables are as enormous as a golf course.

When I was in college I played occasionally with my physics professor. He was an OK player. He loved to describe some of this stuff. He also said that physics was more useful to describe what just happened than to predict what was about to happen.


A place I used to play had an AC vent that blew straight down on the bar table. In just a few moments the tops of the balls got cooled. You could see the effect quite obviously in a draw shot as the ball would "stutter step" back. In a cut/draw the cue ball would zig-zag. :eek:


Fun stuff. :smile:

.
 
V

vinay

Registered
alphadog said:
Funny I have had this arguement with a pro about having a solid grip at impact while breaking.
Click to expand...

A solid grip would be better IF you were still able to still strike the ball at the same speed and still hit the head ball square.

Let's say you have a super solid grip and the effective cue weight is 100oz. The ball will be traveling at 165% of the initial cue speed instead of the 131% you get with a light grip. But if your speed only drops by 20% because of the death grip you're back to square one.
 
C

Cron

AzB Silver Member
Silver Member
Chopdoc said:
At the time of impact the cue ball is not moving at all

.
Click to expand...

Welllllll, at the time of impact the cueball is moving, as Earth does rotate. You can take this to insane depths, but a general base is all that is needed here.
 
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