Hey Dr. Dave. Can a stick hit a ton?

The Cue Rating Scale:
Sweet Hit = 1 Ton
5 Tons = 1 Mack Truck
5 Mack Trucks = 1 Locomotive
5 Locomotives = 1 Freight Train
:D
 
Houstoer said:
you frickin ruined my day sir !! I was feelin good; now I just feel dumb as sh..t !!
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Sorry about that. I normally try to "hide" this sort of analysis in my TP documents so people don't stumble upon them by mistake. They are obviously of interest only to those with strong math and physics backgrounds. I am sorry I "soiled" the forum with this on such a great day (... but somebody asked for it).

Catch you later,
Dave
 
Bob Jewett said:
So here is a way we could easily exceed a ton.... You shoot the cue ball at Jamie and he shoots the ball back to you. The speed between his stick and the cue ball is about 26MPH (the speed of his cue stick deduced from his break speed) plus your 29MPH. This means that he would have an effective stick speed of 55MPH, which would produce a change in ball speed of about 77MPH. It is the change in speed that's the important thing for calculating the force.

Going from the above calculations, Jamie's tip would reach a force of about one and a quarter tons.

Note that the cue ball would be coming back at you with a speed of 77MPH-29MPH or 48MPH. I recommend full body armor and hockey masks for all participants. Do not try this at home, kids.
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You should put that disclaimer at the beginning of your post... Someone out there is gonna knock their teeth out :thumbup:
 
Bob Jewett said:
So here is a way we could easily exceed a ton.... You shoot the cue ball at Jamie and he shoots the ball back to you. The speed between his stick and the cue ball is about 26MPH (the speed of his cue stick deduced from his break speed) plus your 29MPH. This means that he would have an effective stick speed of 55MPH, which would produce a change in ball speed of about 77MPH. It is the change in speed that's the important thing for calculating the force.

Going from the above calculations, Jamie's tip would reach a force of about one and a quarter tons.

Note that the cue ball would be coming back at you with a speed of 77MPH-29MPH or 48MPH. I recommend full body armor and hockey masks for all participants. Do not try this at home, kids.
Click to expand...

Kind of like these two. :D

https://www.youtube.com/watch?feature=player_detailpage&v=cuogn4JTgn4#t=38s
 
ChopStick said:
Thanks. Now when someone tells me "this stick hits a ton" I can say, no it doesn't. It hits 683 lbs.
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I have certified test reports for all my cues that hit with a force of 1-Ton or greater.
Cue design is critical at 2000-lbs.
The only failures at 683 lbs. were due to tip delamination, possibly caused by the use of
a mud cue ball, or we might of been testing in reverse with a Meucci, I don't remember.
 
Seriously.... I officially feel stupid.

dr_dave said:
To keep things simple, let's use a cue weight of 18 oz and assume a perfect tip with a center-ball hit. For this case, TP A.30 predicts that the outgoing CB speed is about 3/2 (1.5) the incoming cue speed. Let's also assume that the average force during tip contact is about half the peak force. And let's assume the tip is in contact with the ball for 0.001 sec, which is typical.

For any CB speed (vb), given the CB mass (mb), the momentum delivered to the CB is:
mom = mb * vb

For a given duration of contact (dt), this momentum must equal the impulse delivered from the cue:
imp = 1/2 * Fmax * dt

So to find the peak force for a given CB speed:
Fmax = 2*mb*vb/dt

And for a given peak force, the CB speed is:
vb = Fmax*dt/2 / mb

And the cue stick speed required to create this is about:
vs = 2/3 vb = Fmax*dt / 3*mb

For a 20mph break, with a 6oz pool ball, the Fmax equation gives a peak force of:
Fmax = 683 pounds

To achieve a 1 ton (2000 pound) peak force, the vs equation gives a required cue speed of:
vs = 39mph

I hope you can sleep at night now, :grin:
Dave
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YBA said:
dr_dave said:
To keep things simple, let's use a cue weight of 18 oz and assume a perfect tip with a center-ball hit. For this case, TP A.30 predicts that the outgoing CB speed is about 3/2 (1.5) the incoming cue speed. Let's also assume that the average force during tip contact is about half the peak force. And let's assume the tip is in contact with the ball for 0.001 sec, which is typical.

For any CB speed (vb), given the CB mass (mb), the momentum delivered to the CB is:
mom = mb * vb

For a given duration of contact (dt), this momentum must equal the impulse delivered from the cue:
imp = 1/2 * Fmax * dt

So to find the peak force for a given CB speed:
Fmax = 2*mb*vb/dt

And for a given peak force, the CB speed is:
vb = Fmax*dt/2 / mb

And the cue stick speed required to create this is about:
vs = 2/3 vb = Fmax*dt / 3*mb

For a 20mph break, with a 6oz pool ball, the Fmax equation gives a peak force of:
Fmax = 683 pounds

To achieve a 1 ton (2000 pound) peak force, the vs equation gives a required cue speed of:
vs = 39mph

I hope you can sleep at night now, :grin:
Dave
Click to expand...

Seriously.... I officially feel stupid.
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Sorry about that. Again, I normally try to "hide" the math and physics stuff in the TP section of my website so innocent people won't get hurt. :o

Here's the disclaimer on the TP page:

NOTE: this information will be of interest only to people with strong physics and mathematics backgrounds. Others proceed at your own risk.

I probably should have just posted the answers (... but then the math and physics guys out there would have wondered where they came from).

Let me try again:

The maximum force occurring between the tip and CB during a 20mph break is about 680 pounds, and the cue speed necessary to create a "ton of force" is about 40 mph (which would correspond to a CB break speed of about 60 mph).

Is that better? :thumbup:

Regards,
Dave
 
dr_dave said:
Sorry about that. Again, I normally try to "hide" the math and physics stuff in the TP section of my website so innocent people won't get hurt. :o

Here's the disclaimer on the TP page:

NOTE: this information will be of interest only to people with strong physics and mathematics backgrounds. Others proceed at your own risk.

I probably should have just posted the answers (... but then the math and physics guys out there would have wondered where they came from).

Let me try again:

The maximum force occurring between the tip and CB during a 20mph break is about 680 pounds, and the cue speed necessary to create a "ton of force" is about 40 mph (which would correspond to a CB break speed of about 60 mph).

Is that better? :thumbup:

Regards,
Dave
Click to expand...

Dr. Dave, sorry I should have put a smiley after my sentence. :-). I'm actually happy someone was able to answer this question scientifically. This shows the world that pool players are not just drunken bar people :-).

I write communication protocols between satellites and avionics for a living. I often find that my professional background is not really useful in day to day life. I'd be quite happy to apply a touch of scientific toughs in my hobbies. Tis thread is actually a very interesting from my perspective.

Please continue showing the math and physics details. You may catch the attention of a few kids on the board and motivate them to pursue a career in science!

Cheers

Luc
 
dr_dave said:
Sorry about that. Again, I normally try to "hide" the math and physics stuff in the TP section of my website so innocent people won't get hurt. :o

Here's the disclaimer on the TP page:

NOTE: this information will be of interest only to people with strong physics and mathematics backgrounds. Others proceed at your own risk.

I probably should have just posted the answers (... but then the math and physics guys out there would have wondered where they came from).

Let me try again:

The maximum force occurring between the tip and CB during a 20mph break is about 680 pounds, and the cue speed necessary to create a "ton of force" is about 40 mph (which would correspond to a CB break speed of about 60 mph).

Is that better? :thumbup:

Regards,
Dave
Click to expand...

Come on guys. Don't be intimidated by the language. It's basic math. I wanted to compliment Dr. Dave on the way he chose to present the information. It was actually quite clever. Anyone who is accustomed to variable substitution in computer languages, which is a common skill among young people now days, can easily read and understand it. I appreciate Dr. Dave taking the time to post it.

On another note. Mr. Jewett and Dr. Dave have pointed out something that I have long suspected. As we have all observed, even the most accomplished professionals have trouble controlling the erratic behavior of the cue ball during the break shot. I believe this may be due, in part, to the fact that at the pressures involved, the cue ball is no longer round. This may indicate that absolute precision on a break shot is not a reasonable expectation. Only a statistical range of expectations is possible.

Here is a golf ball hitting a steel plate at 175mph. To some degree, however small, some level of this has to happening to the cue ball. It may be so small that it is not even a factor but as we have all seen, the highest possible break speed does not guarantee the best outcome.

https://www.youtube.com/watch?v=aMqM13EUSKw
 
The highest break speed I have ever personally recorded, was 31.8mph. (yes it stayed on the table... just barely...) This was with a leather tip, and a 15.2oz cue. In therory I could possibly do this again with my now 17.1oz break cue, and a g10 tip.

I have two questions.

Can you calculate roughly what the break with a 15.2oz was?

Can you calculate roughly what it will be if I can achieve it with my new 17.1oz will be?

thanks, Dr. Dave!
 
jhanso18 said:
The highest break speed I have ever personally recorded, was 31.8mph. (yes it stayed on the table... just barely...) This was with a leather tip, and a 15.2oz cue. In therory I could possibly do this again with my now 17.1oz break cue, and a g10 tip.

I have two questions.

Can you calculate roughly what the break with a 15.2oz was?

Can you calculate roughly what it will be if I can achieve it with my new 17.1oz will be?

thanks, Dr. Dave!
Click to expand...
Sorry, but I can't, because it is impossible. The answer depends on too many other things. For more info, see the optimal cue weight resource page.

Regards,
Dave
 
ChopStick said:
I believe this may be due, in part, to the fact that at the pressures involved, the cue ball is no longer round. This may indicate that absolute precision on a break shot is not a reasonable expectation. Only a statistical range of expectations is possible.

Here is a golf ball hitting a steel plate at 175mph. To some degree, however small, some level of this has to happening to the cue ball. It may be so small that it is not even a factor but as we have all seen, the highest possible break speed does not guarantee the best outcome.

https://www.youtube.com/watch?v=aMqM13EUSKw
Click to expand...

Good Lord, I love slow motion videos. I might even be addicted to SloMo. ;)

Thanks to your link, I now have a new visualization to use when I'm getting ready to break.

Great thread. Thanks to Dr. Dave and Bob Jewett for your replies!

-Blake

P.S. I think you're right about CB/OB compression.
 
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