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

ChopStick

ChopStick

Unsane Poster
Silver Member
Dr. Dave or someone with the appropriate math background, let's settle the question once and for all. In the back of the Golfsmith golf club construction and repair guide, it states that a 9 oz driver moving at 100 mph, exerts 1 ton of force to a golf ball for 1/10,000th of a second. They also had a formula there that I do not remember.

OK. So, how fast would a 20oz pool cue have to be moving to exert 1 ton of force to a cue ball? Also, given the achievable human break speed of 20 mph, how much would a cue have to weigh to exert 1 ton of force. And lastly, when someone says this cue hits a ton, what percentage exaggeration is it? :D
 
ChopStick said:
Dr. Dave or someone with the appropriate math background, let's settle the question once and for all. In the back of the Golfsmith golf club construction and repair guide, it states that a 9 oz driver moving at 100 mph, exerts 1 ton of force to a golf ball for 1/10,000th of a second. They also had a formula there that I do not remember.

OK. So, how fast would a 20oz pool cue have to be moving to exert 1 ton of force to a cue ball? Also, given the achievable human break speed of 20 mph, how much would a cue have to weigh to exert 1 ton of force. And lastly, when someone says this cue hits a ton, what percentage exaggeration is it? :D
Click to expand...

metric ton or standard ton? also long ton or short ton? 2000 lbs is the short ton 2240 lbs is the long ton. and for which country are we talking?
 
Last edited:
I like to POUND them balls into the pockets.

I guess that makes me a lightweight.:embarrassed2:
 
Since "hitting a ton" is a function of tip impact on the CB, another important question is "what size radius on the tip and what is the tip size?"
 
for a short ton at 100mph a cue could be 20ounces but at a long ton you would either have to apply more force faster than 100 mph but not quite 110mph. so given this i think your golf friends made their side of this up
 
in golf or anything with a 9 ounce weight you would have to be going 248.88888888888888888888888888888888888888888mph not quite 249mph
 
all of this im basing off of this
Work is a transfer of energy so work is done on an object when you transfer energy to that object. The amount of work done on an object depends on the amount of force exerted on the object and the amount of distance the object moves.

Work = Force x Distance



According to Newton's Second Law of Motion, the net force on an object is dependent on the mass of the object, and its acceleration during the movement.

Force = Mass x Acceleration

The common unit of force is the Newton (N). One Newton is the force required to accelerate one kilogram of mass at 1 meter per second per second.

1 N = 1kg m/s2

The amount of work done to push a 10,000 N car a distance of 10 meters would be

10,000 N x 10 m = 100,000 N m or 100,000 J

The Newton-meters are termed joules (J). The joule is named after James Prescott Joule (1818-1889) who first calculated the amount of electrical work needed to produce a unit of heat. In his experiments, Joule discovered that the same amount of heat was produced by the same amount of either electrical or mechanical work ("the mechanical equivalent of heat").
 
and also i was thinking this these answers only apply to tips that don't have give and shafts that don't deflect. because those take away from the force applied.
 
In a collision, the average force is equal to the mass times the change in velocity divided by the time of contact. That works a lot better in MKS units for the calculation than with Imperial units...

average F = 0.17 kilograms * 10 m/s / 0.001 seconds

10 meters / second is 22.4 MPH; 170 grams is 6 ounces

average F = 1700 newtons

What the heck is a newton? It's the metric unit of force. Google will happily do conversions for you if you type in something like "1700 newtons in pound-force". You have to say pound-force because we also have pound-mass, although technically in Imperial units one should give the mass of something in slugs.

If you type in the above you get 382 pounds of force, but that's the average force, which builds up to a peak as the tip is compressed and then drops off. The peak force is about twice that or 764 pounds, roughly. If you had a very hard phenolic tip that cut the contact time to 0.5 milliseconds and hit the ball at 30MPH, then, in fact you would hit the ball a ton, more or less.
 
Bob Jewett said:
In a collision, the average force is equal to the mass times the change in velocity divided by the time of contact. That works a lot better in MKS units for the calculation than with Imperial units...

average F = 0.17 kilograms * 10 m/s / 0.001 seconds

10 meters / second is 22.4 MPH; 170 grams is 6 ounces

average F = 1700 newtons

What the heck is a newton? It's the metric unit of force. Google will happily do conversions for you if you type in something like "1700 newtons in pound-force". You have to say pound-force because we also have pound-mass, although technically in Imperial units one should give the mass of something in slugs.

If you type in the above you get 382 pounds of force, but that's the average force, which builds up to a peak as the tip is compressed and then drops off. The peak force is about twice that or 764 pounds, roughly. If you had a very hard phenolic tip that cut the contact time to 0.5 milliseconds and hit the ball at 30MPH, then, in fact you would hit the ball a ton, more or less.
Click to expand...

i would vote for less. there is no way to get even a short ton of force out of a cue by a human
 
ChopStick said:
Also, given the achievable human break speed of 20 mph, how much would a cue have to weigh to exert 1 ton of force....
Click to expand...
The cue's weight (as a measure of mass) is irrelevant once you specify the speed of the cueball and know the contact time. A leather tip will exert about 300 - 400 lbs to accelerate a cueball to 20 mph in 0.001 seconds. A phenolic tip will exert double that force in about half that time.

Note that we're talking average force, that is, the force averaged over the period of contact. Peak force will be greater and with all the shock waves meandering about, it might reach 2000 lbs for very brief moments.

...Oh, I see Bob has already answered with more detail.

Jim
 
Last edited:
ChopStick said:
Dr. Dave or someone with the appropriate math background, let's settle the question once and for all. In the back of the Golfsmith golf club construction and repair guide, it states that a 9 oz driver moving at 100 mph, exerts 1 ton of force to a golf ball for 1/10,000th of a second. They also had a formula there that I do not remember.

OK. So, how fast would a 20oz pool cue have to be moving to exert 1 ton of force to a cue ball? Also, given the achievable human break speed of 20 mph, how much would a cue have to weigh to exert 1 ton of force. And lastly, when someone says this cue hits a ton, what percentage exaggeration is it? :D
Click to expand...
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
 
Last edited:
The actual force which was measured in ton was never felt by your grip hand anyway.
The peak force only exist in the tip and a small portion of the shaft in form of stress. The
majority of the stress will be buffered by the cue on the way when the shockwave travel
to your grip hand, so the peak force your grip hand feel is much less than a ton.

armoworrior said:
i would vote for less. there is no way to get even a short ton of force out of a cue by a human
Click to expand...
 
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), give the CB mass (mb), the momentum delivered to the CB is:
mom = mb * vb

For a given duration of contact (dt), this momentum has to 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:
Fmax = 683 pounds

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

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

Thanks. Now when someone tells me "this stick hits a ton" I can say, no it doesn't. It hits 683 lbs. :thumbup:

I am surprised how close it comes to actually being possible. There may be someone out there who could do it.
 
ChopStick said:
Thanks. Now when someone tells me "this stick hits a ton" I can say, no it doesn't. It hits 683 lbs. :thumbup:

I am surprised how close it comes to actually being possible. There may be someone out there who could do it.
Click to expand...

I believe Jamie Baracks broke 37mph about 10years ago.. I broke 29mph..

Iowa state tourney was the venue
 
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...

you frickin ruined my day sir !! I was feelin good; now I just feel dumb as sh..t !!
 
btown said:
I believe Jamie Baracks broke 37mph about 10years ago.. I broke 29mph..

Iowa state tourney was the venue
Click to expand...
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.
 
You must log in or register to reply here.
Back
Top