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Shaft
12-17-2007, 05:02 PM
Here is an attempt to settle, once and for all, the question of whether a light cue stick "breaks better" than a heavy cue stick.

The mathematical result surprised me, and it forces me to recant what I have said on previous posts that "it should not make a difference." It does, in fact, make a (small?) difference.

I apologize in advance for the use of metric units, but it makes the math so much easier. I will give the english equivalents wherever they help.

For the purpose of this analysis I will assume that

* The collision is perfectly elastic (energy and momentum will be conserved in the post collision motion of the stick and ball).

* The same shooter exerts a constant force (300 Newtons, about 67.4 lbs) over a 6 inch forward swing to impact. (The actual force and swing distance are not important here so long as they are the same for both cues.)

* The "light" stick weighs 18oz, the "heavy" stick weighs 21oz, and the cue ball weighs 6 oz.

* The stick and ball are free bodies in space. The the stick is accellerated to hit a ball that is initially at rest, and the axis and path of the stick is alligned with the center of the ball. (I know real balls and sticks are not free bodies in space, but this assumption simplifies everything without changing the outcome.)

The trick is to find the post collision speeds of the ball and stick so that energy and momentum are conserved. We have a system of two equations and two unknowns, so it can be solved with some hairy algebra. I used the Excel Solver routine because it was easier and I am LAZY.

A constant force of 300N applied over 6 inches will accelerate an 18oz (0.5103kg) stick to a speed of 13.38614 m/s (29.94mph). The pre-collision momentum of the stick is 6.830947 kg-m/s, and the kinetic energy is 45.72J. Conserving energy and momentum, the post collision speed of the stick is 6.69307m/s (14.97mph) and the post collision speed of the ball is 20.07m/s (44.92mph).

A constant force of 300N applied over 6 inches will accellerate a 21oz (0.5953kg) stick to a speed of 12.39368m/s (27.72mph). The pre-collision momentum of the stick is 7.377956 kg-m/s, and the kinetic energy is 45.72 J. Conserving energy and momentum, the post collision speed of the stick is 6.885376m/s (15.40mph) and the post collision speed of the ball is 19.27905m/s (43.13mph).

DRUM ROLL and TRUMPET FANFARE:::::::::::::
The cue ball has 4.1% more speed and 8.5% more energy when struck by a LIGHTER (18oz) stick than when it is struck by a HEAVIER (21oz) stick, all other things being equal. A 15% reduction in weight results in 4% more speed.

This is true - NOT because the lighter stick moves faster - but because the lighter stick transfers more energy to an object that is more similar in mass.

In fact, I demonstrated with the same spreadsheet that, as the mass of the two colliding objects gets more and more equal, more and more of the energy of the first object is transfered to the second object. When the mass difference is zero, the energy transfer is 100%.

You might find that hard to believe. Remember those annoying clickety-clackety novelty toys that had 5 steel ball bearings suspended on a wooden frame? Take three balls out of the way and try it with just two balls. The first ball will stop dead as the second ball takes 100% of the energy and swings upward.

We can also see this principle on the pool table. A dead-stratight stun shot transfers 100% of the cue ball energy to the object ball. (It is ironic that a clue to the answer has been in front of us all the time.)

In this cas, physics CONFIRMS anecdotal observation.

I suppose we should all go out and buy 6oz cue sticks!!! /ccboard/images/graemlins/grin.gif /ccboard/images/graemlins/grin.gif /ccboard/images/graemlins/grin.gif

Merry Christmas everybody!

dr_dave
12-17-2007, 05:15 PM
See my previous posting on this topic (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&Board=ccb&Number=171855&Foru m=ccb&Words=dr_dave&Match=Username&Searchpage=1&Li mit=25&Old=allposts&Main=125575&Search=true#Post17 1855). It's not just a question of physics. Biomechanics, physiology, and psychology also come into play.

Regards,
Dave

PS: I also have some other comments in <font color="red">red</font color> below.
<blockquote><font class="small">Quote Shaft:</font><hr> Here is an attempt to settle, once and for all, the question of whether a light cue stick "breaks better" than a heavy cue stick.

The mathematical result surprised me, and it forces me to recant what I have said on previous posts that "it should not make a difference." It does, in fact, make a (small?) difference.

I apologize in advance for the use of metric units, but it makes the math so much easier. I will give the english equivalents wherever they help.

For the purpose of this analysis I will assume that

* The collision is perfectly elastic (energy and momentum will be conserved in the post collision motion of the stick and ball).

* The same shooter exerts a constant force (300 Newtons, about 67.4 lbs) over a 6 inch forward swing to impact. (The actual force and swing distance are not important here so long as they are the same for both cues.) <font color="red">I think this assumption is invalid and inappropriate per my previous posting (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&amp;Board=ccb&amp;Number=171855&amp;Foru m=ccb&amp;Words=dr_dave&amp;Match=Username&amp;Searchpage=1&amp;Li mit=25&amp;Old=allposts&amp;Main=125575&amp;Search=true#Post17 1855)</font color>.

* The "light" stick weighs 18oz, the "heavy" stick weighs 21oz, and the cue ball weighs 6 oz.

* The stick and ball are free bodies in space. The the stick is accellerated to hit a ball that is initially at rest, and the axis and path of the stick is alligned with the center of the ball. (I know real balls and sticks are not free bodies in space, but this assumption simplifies everything without changing the outcome.)

The trick is to find the post collision speeds of the ball and stick so that energy and momentum are conserved. We have a system of two equations and two unknowns, so it can be solved with some hairy algebra. I used the Excel Solver routine because it was easier and I am LAZY.

A constant force of 300N applied over 6 inches will accelerate an 18oz (0.5103kg) stick to a speed of 13.38614 m/s (29.94mph). The pre-collision momentum of the stick is 6.830947 kg-m/s, and the kinetic energy is 45.72J. Conserving energy and momentum, the post collision speed of the stick is 6.69307m/s (14.97mph) and the post collision speed of the ball is 20.07m/s (44.92mph).

A constant force of 300N applied over 6 inches will accellerate a 21oz (0.5953kg) stick to a speed of 12.39368m/s (27.72mph). The pre-collision momentum of the stick is 7.377956 kg-m/s, and the kinetic energy is 45.72 J. Conserving energy and momentum, the post collision speed of the stick is 6.885376m/s (15.40mph) and the post collision speed of the ball is 19.27905m/s (43.13mph).

DRUM ROLL and TRUMPET FANFARE:::::::::::::
The cue ball has 4.1% more speed and 8.5% more energy when struck by a LIGHTER (18oz) stick than when it is struck by a HEAVIER (21oz) stick, all other things being equal. A 15% reduction in weight results in 4% more speed. <font color="red">Momentum (the product of mass and speed) is what is important. Cue momentum (not just speed) is what creates cue-ball speed.</font color>

This is true - NOT because the lighter stick moves faster - but because the lighter stick transfers more energy to an object that is more similar in mass. <font color="red">But the amount of energy of the cue also depends on its mass (and not just its speed)</font color>

In fact, I demonstrated with the same spreadsheet that, as the mass of the two colliding objects gets more and more equal, more and more of the energy of the first object is transfered to the second object. When the mass difference is zero, the energy transfer is 100%.

You might find that hard to believe. Remember those annoying clickety-clackety novelty toys that had 5 steel ball bearings suspended on a wooden frame? Take three balls out of the way and try it with just two balls. The first ball will stop dead as the second ball takes 100% of the energy and swings upward.

We can also see this principle on the pool table. A dead-stratight stun shot transfers 100% of the cue ball energy to the object ball. (It is ironic that a clue to the answer has been in front of us all the time.)

In this cas, physics CONFIRMS anecdotal observation.

I suppose we should all go out and buy 6oz cue sticks!!! /ccboard/images/graemlins/grin.gif /ccboard/images/graemlins/grin.gif /ccboard/images/graemlins/grin.gif

Merry Christmas everybody! <font color="red">Ditto!</font color><hr /></blockquote>

Shaft
12-17-2007, 06:38 PM
My comments in green...

<blockquote><font class="small">Quote dr_dave:</font><hr> See my previous posting on this topic (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&amp;Board=ccb&amp;Number=171855&amp;Foru m=ccb&amp;Words=dr_dave&amp;Match=Username&amp;Searchpage=1&amp;Li mit=25&amp;Old=allposts&amp;Main=125575&amp;Search=true#Post17 1855). It's not just a question of physics. Biomechanics, physiology, and psychology also come into play.
<font color="green"> I agree, but when talking about one issue, shaft weight, I am examining only that variable. The difference in weight does not take either stick outside of the bound of the abilities of any player.</font color>
Regards,
Dave

PS: I also have some other comments in <font color="red">red</font color> below.
<blockquote><font class="small">Quote Shaft:</font><hr> Here is an attempt to settle, once and for all, the question of whether a light cue stick "breaks better" than a heavy cue stick.

The mathematical result surprised me, and it forces me to recant what I have said on previous posts that "it should not make a difference." It does, in fact, make a (small?) difference.

I apologize in advance for the use of metric units, but it makes the math so much easier. I will give the english equivalents wherever they help.

For the purpose of this analysis I will assume that

* The collision is perfectly elastic (energy and momentum will be conserved in the post collision motion of the stick and ball).

* The same shooter exerts a constant force (300 Newtons, about 67.4 lbs) over a 6 inch forward swing to impact. (The actual force and swing distance are not important here so long as they are the same for both cues.) <font color="red">I think this assumption is invalid and inappropriate per my previous posting (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&amp;Board=ccb&amp;Number=171855&amp;Foru m=ccb&amp;Words=dr_dave&amp;Match=Username&amp;Searchpage=1&amp;Li mit=25&amp;Old=allposts&amp;Main=125575&amp;Search=true#Post17 1855)</font color>.
<font color="green"> No, this is not an invalid assumption. In any experiment, only one variable should change. We are talking about the effect of shaft weight only. All other factors must remain the same.</font color>
* The "light" stick weighs 18oz, the "heavy" stick weighs 21oz, and the cue ball weighs 6 oz.

* The stick and ball are free bodies in space. The the stick is accellerated to hit a ball that is initially at rest, and the axis and path of the stick is alligned with the center of the ball. (I know real balls and sticks are not free bodies in space, but this assumption simplifies everything without changing the outcome.)

The trick is to find the post collision speeds of the ball and stick so that energy and momentum are conserved. We have a system of two equations and two unknowns, so it can be solved with some hairy algebra. I used the Excel Solver routine because it was easier and I am LAZY.

A constant force of 300N applied over 6 inches will accelerate an 18oz (0.5103kg) stick to a speed of 13.38614 m/s (29.94mph). The pre-collision momentum of the stick is 6.830947 kg-m/s, and the kinetic energy is 45.72J. Conserving energy and momentum, the post collision speed of the stick is 6.69307m/s (14.97mph) and the post collision speed of the ball is 20.07m/s (44.92mph).

A constant force of 300N applied over 6 inches will accellerate a 21oz (0.5953kg) stick to a speed of 12.39368m/s (27.72mph). The pre-collision momentum of the stick is 7.377956 kg-m/s, and the kinetic energy is 45.72 J. Conserving energy and momentum, the post collision speed of the stick is 6.885376m/s (15.40mph) and the post collision speed of the ball is 19.27905m/s (43.13mph).

DRUM ROLL and TRUMPET FANFARE:::::::::::::
The cue ball has 4.1% more speed and 8.5% more energy when struck by a LIGHTER (18oz) stick than when it is struck by a HEAVIER (21oz) stick, all other things being equal. A 15% reduction in weight results in 4% more speed. <font color="red">Momentum (the product of mass and speed) is what is important. Cue momentum (not just speed) is what creates cue-ball speed.</font color> <font color="green"> No, momentum is important, but not determinative. In this case, the heavy stick has more momentum, yet results in a slower cue ball speed.</font color>

This is true - NOT because the lighter stick moves faster - but because the lighter stick transfers more energy to an object that is more similar in mass. <font color="red">But the amount of energy of the cue also depends on its mass (and not just its speed)</font color> <font color="green"> Dave, you should read my post again and check my math. The heavy cue has the same energy, more mass and more momentum. Yet it results in a slower ball speed. Solve the 2-variable, 2-equation system for yourself and you will see that the greater mass and momentum did not win the day.</font color>

In fact, I demonstrated with the same spreadsheet that, as the mass of the two colliding objects gets more and more equal, more and more of the energy of the first object is transfered to the second object. When the mass difference is zero, the energy transfer is 100%.

You might find that hard to believe. Remember those annoying clickety-clackety novelty toys that had 5 steel ball bearings suspended on a wooden frame? Take three balls out of the way and try it with just two balls. The first ball will stop dead as the second ball takes 100% of the energy and swings upward.

We can also see this principle on the pool table. A dead-stratight stun shot transfers 100% of the cue ball energy to the object ball. (It is ironic that a clue to the answer has been in front of us all the time.)

In this cas, physics CONFIRMS anecdotal observation.

I suppose we should all go out and buy 6oz cue sticks!!! /ccboard/images/graemlins/grin.gif /ccboard/images/graemlins/grin.gif /ccboard/images/graemlins/grin.gif

Merry Christmas everybody! <font color="red">Ditto!</font color><hr /></blockquote> <hr /></blockquote>

DeadCrab
12-17-2007, 07:24 PM
Something is askew because your cue ball velocity is about twice that achieved by the world's best break shots.

Bob_Jewett
12-17-2007, 07:32 PM
<blockquote><font class="small">Quote Shaft:</font><hr> ... * The same shooter exerts a constant force (300 Newtons, about 67.4 lbs) over a 6 inch forward swing to impact. ... <hr /></blockquote>
So, would you also conclude that a player will exert a constant force over that six inches for a 0.1-ounce and a 16000-ounce stick? If not, then what is the relation between force during the stroke and stick weight? I doubt that the force is perfectly constant for a 15% change in stick mass.

Also, you said that you are talking only about shaft weight. I assume that you misspoke, since the important weight is the weight of the stick and the individual components are more or less unimportant.

dr_dave
12-17-2007, 11:56 PM
My comments in <font color="purple">purple</font color>...
<blockquote><font class="small">Quote Shaft:</font><hr> My comments in green...
<blockquote><font class="small">Quote dr_dave:</font><hr> See my previous posting on this topic (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&amp;Board=ccb&amp;Number=171855&amp;Foru m=ccb&amp;Words=dr_dave&amp;Match=Username&amp;Searchpage=1&amp;Li mit=25&amp;Old=allposts&amp;Main=125575&amp;Search=true#Post17 1855). It's not just a question of physics. Biomechanics, physiology, and psychology also come into play.
<font color="green"> I agree, but when talking about one issue, shaft weight, I am examining only that variable. The difference in weight does not take either stick outside of the bound of the abilities of any player. </font color> <font color="purple">But the amount of force and energy a player can deliver to cues of different weights can vary significantly.</font color>

PS: I also have some other comments in <font color="red">red</font color> below.
<blockquote><font class="small">Quote Shaft:</font><hr>For the purpose of this analysis I will assume that

* The collision is perfectly elastic (energy and momentum will be conserved in the post collision motion of the stick and ball). <font color="purple">FYI, an easy way to handle the energy conservation is to use the coefficient of restitution relation for an elastic collision (see TP A.30 (http://billiards.colostate.edu/technical_proofs/new/TP_A-30.pdf)). The math can be a lot easier.</font color>

* The same shooter exerts a constant force (300 Newtons, about 67.4 lbs) over a 6 inch forward swing to impact. (The actual force and swing distance are not important here so long as they are the same for both cues.) <font color="red">I think this assumption is invalid and inappropriate per my previous posting (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&amp;Board=ccb&amp;Number=171855&amp;Foru m=ccb&amp;Words=dr_dave&amp;Match=Username&amp;Searchpage=1&amp;Li mit=25&amp;Old=allposts&amp;Main=125575&amp;Search=true#Post17 1855)</font color>.
<font color="green"> No, this is not an invalid assumption. In any experiment, only one variable should change. We are talking about the effect of shaft weight only. All other factors must remain the same.</font color> <font color="purple">But grip force and cue weight are not independent variables, so you can't ignore changes in one while changing the other.</font color>

...

DRUM ROLL and TRUMPET FANFARE:::::::::::::
The cue ball has 4.1% more speed and 8.5% more energy when struck by a LIGHTER (18oz) stick than when it is struck by a HEAVIER (21oz) stick, all other things being equal. A 15% reduction in weight results in 4% more speed. <font color="red">Momentum (the product of mass and speed) is what is important. Cue momentum (not just speed) is what creates cue-ball speed.</font color> <font color="green"> No, momentum is important, but not determinative. In this case, the heavy stick has more momentum, yet results in a slower cue ball speed.</font color> <font color="purple">With your force assumption, you are right here. My bad.</font color>

This is true - NOT because the lighter stick moves faster - but because the lighter stick transfers more energy to an object that is more similar in mass. <font color="red">But the amount of energy of the cue also depends on its mass (and not just its speed)</font color> <font color="green"> Dave, you should read my post again and check my math. The heavy cue has the same energy, more mass and more momentum. Yet it results in a slower ball speed. Solve the 2-variable, 2-equation system for yourself and you will see that the greater mass and momentum did not win the day.</font color> <font color="purple">I just did the math and found (with the help of a little calculus) that the optimal cue weight for maximum cue ball speed, assuming a constant and equal force profile, is half the ball's mass. So with your force assumption, I come up with an optimal break cue weight of 3 oz! This is obviously not a very useful result; otherwise, most of us would probably be using much lighter cues to break. The problem is: it is difficult for a human to generate large forces on something so light moving so fast. That's why I originally suggested that your force assumption was inappropriate.</font color><hr /></blockquote><hr /></blockquote>

Regards,
Dave

Jal
12-18-2007, 01:23 AM
Shaft,

I didn't realize that you were "into" the physics too. There are a couple of problems with your analysis, imo, but it does come to essentially the right conclusion: that differences in stick weight don't mean all that much.

This is really kind of trivial, but the numbers used for the force (300 N) and speeds are much larger than actually seen at the table. Break speeds involve an average force applied to the cue of something like 10-15 lbs, with a peak force of about 15-20 lbs (the force rises and then falls during the forward stroke). But this doesn't really invalidate the analysis.

More critical is the player's forearm/hand mass. If it were zero, then the optimal stick weight for a centerball hit would indeed be 6 oz, the same as the cueball, and assuming you could apply as much force to a 6 oz stick as to a 25 oz stick. Let's assume that's true for the moment and consider the effect of a non-zero forearm/hand mass.

The force or torque that's applied by the muscles has to accelerate both the arm and the stick. If the stick is too light, you lose some potential stick momentum because of the arm's mass, ie, increasing the stick's mass doesn't increase the combined mass of the stick and arm proportionally. On the other hand, if the stick is too heavy, speed (momentum again) suffers.

Thus, there is an optimal stick mass and a way to calculate it IF you know your arm's mass (actually the arm's moment of inertia which can be converted to an equivalent mass). Although not easy to measure, you can make an educated guess at a "typical" value given people's preferences for cue weight. According to this, it should be roughly 18 oz.

Taking this into account, it turns out that the functional differences between an 18 and 21 oz cue are even less than your calculations indicate: more like a fraction of a percent. But this is pretty much in the same ballpark.

However, as Dr. Dave and Bob Jewett point out, when it comes to applying large forces, as with a break or power draw, the assumption that the same force can be generated with differently weighted cues is no longer valid. Physiology now enters into it. If you did happen to know the equivalent mass of your arm, the best that physics could do is report the minimal cue weight that is best suited to you for any particular tip offset.

Jim

Shaft
12-18-2007, 04:11 AM
No, Bob.

Your example is absurd and completely off point.

I agree that there are physiological limits to the acceleration a human can put to the stick. My 6oz closing comment was (I thought) an obvious joke (note the smiley faces).

But the difference between an 18oz and 21oz cue is within the relm of plausibility of any shooter. The context of the question is 17-22oz, and everybody on this post knows that.

Shaft
12-18-2007, 04:14 AM
Reply to Dead Crab:

The force figure was a round number arbitrarily chosen. If you want less force that results in a slower break speed, fine.

The result (a lighter stick results in a faster cue ball speed for same force and stroke distance) does not change for any force you choose.

Shaft
12-18-2007, 05:31 AM
My comments in green again...
<blockquote><font class="small">Quote dr_dave:</font><hr> My comments in <font color="purple">purple</font color>...
<blockquote><font class="small">Quote Shaft:</font><hr> My comments in green...
<blockquote><font class="small">Quote dr_dave:</font><hr> See my previous posting on this topic (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&amp;Board=ccb&amp;Number=171855&amp;Foru m=ccb&amp;Words=dr_dave&amp;Match=Username&amp;Searchpage=1&amp;Li mit=25&amp;Old=allposts&amp;Main=125575&amp;Search=true#Post17 1855). It's not just a question of physics. Biomechanics, physiology, and psychology also come into play.
<font color="green"> I agree, but when talking about one issue, shaft weight, I am examining only that variable. The difference in weight does not take either stick outside of the bound of the abilities of any player. </font color> <font color="purple">But the amount of force and energy a player can deliver to cues of different weights can vary significantly.</font color> <font color="green"> A 200N force over 6 inches is a 200N force over 6 inches, regardless of the final stick speed achieved. We don't normally measure the force applied, except indirectly, and a difference in speed does not prove a difference in force if the cue weights are not the same. Iron Willie or the Meucci robot probably use a weight system that applies constant force (or, more accurately, a consistant force) to a constant stroke.</font color>

PS: I also have some other comments in <font color="red">red</font color> below.
<blockquote><font class="small">Quote Shaft:</font><hr>For the purpose of this analysis I will assume that

* The collision is perfectly elastic (energy and momentum will be conserved in the post collision motion of the stick and ball). <font color="purple">FYI, an easy way to handle the energy conservation is to use the coefficient of restitution relation for an elastic collision (see TP A.30 (http://billiards.colostate.edu/technical_proofs/new/TP_A-30.pdf)). The math can be a lot easier.</font color> <font color="green"> Yes, I know about the coefficient of resitution, but I recognize this post is read by a broad audience of people who are generally intelligent but with different educational backgrounds. I deliberately did not refer to the Coefficient of Restitution in making my point. </font color>

* The same shooter exerts a constant force (300 Newtons, about 67.4 lbs) over a 6 inch forward swing to impact. (The actual force and swing distance are not important here so long as they are the same for both cues.) <font color="red">I think this assumption is invalid and inappropriate per my previous posting (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&amp;Board=ccb&amp;Number=171855&amp;Foru m=ccb&amp;Words=dr_dave&amp;Match=Username&amp;Searchpage=1&amp;Li mit=25&amp;Old=allposts&amp;Main=125575&amp;Search=true#Post17 1855)</font color>.
<font color="green"> No, this is not an invalid assumption. In any experiment, only one variable should change. We are talking about the effect of shaft weight only. All other factors must remain the same.</font color> <font color="purple">But grip force and cue weight are not independent variables, so you can't ignore changes in one while changing the other.</font color> <font color="green"> If my force value is too high for your taste, change it to any value you like. The result does not change. Yes, "grip force" is probably some factor in real life results, along with myriads of other factors you did not list, but it does not change the basic transfer of energy effect I am talking about in the abstract. Cripes, if I follow this logic, there is no mechanical engineering problem in the world that can be solved, because I will always be omitting some real life friction or some other minor dependant variable. The truth is, you use ideal abstrations every day. For example, the assumption of a "constant force" over the length of their swing is obviously an abstraction that simlifies the problem but does no damage to the principle being demonstrated. </font color>

...

DRUM ROLL and TRUMPET FANFARE:::::::::::::
The cue ball has 4.1% more speed and 8.5% more energy when struck by a LIGHTER (18oz) stick than when it is struck by a HEAVIER (21oz) stick, all other things being equal. A 15% reduction in weight results in 4% more speed. <font color="red">Momentum (the product of mass and speed) is what is important. Cue momentum (not just speed) is what creates cue-ball speed.</font color> <font color="green"> No, momentum is important, but not determinative. In this case, the heavy stick has more momentum, yet results in a slower cue ball speed.</font color> <font color="purple">With your force assumption, you are right here. My bad.</font color>

This is true - NOT because the lighter stick moves faster - but because the lighter stick transfers more energy to an object that is more similar in mass. <font color="red">But the amount of energy of the cue also depends on its mass (and not just its speed)</font color> <font color="green"> Dave, you should read my post again and check my math. The heavy cue has the same energy, more mass and more momentum. Yet it results in a slower ball speed. Solve the 2-variable, 2-equation system for yourself and you will see that the greater mass and momentum did not win the day.</font color> <font color="purple">I just did the math and found (with the help of a little calculus) that the optimal cue weight for maximum cue ball speed, assuming a constant and equal force profile, is half the ball's mass. So with your force assumption, I come up with an optimal break cue weight of 3 oz! This is obviously not a very useful result; otherwise, most of us would probably be using much lighter cues to break. The problem is: it is difficult for a human to generate large forces on something so light moving so fast. That's why I originally suggested that your force assumption was inappropriate.</font color><hr /></blockquote><hr /></blockquote> <font color="green"> See my reply to Bob Jewett. I agree there are physiological limits to how fast a human can accelerate a stick, and my closing comment to buy a 6oz cue was an intentional joke I did not think I would have to explain to anyone here. The context of the question is 17-22oz, and I do not think that is an extreme range. But, even if I am wrong: there will always be SOME range of sticks a player has the option to use. For someone who can shoot in the 17-20oz range, 17oz will result in a slightly faster cue ball speed - for the same force and stroke distance. For someone who can shoot in the 19-22oz range, the 19oz will result in a slightly faster cue ball speed - for the same force and stroke distance.

I am proposing a simple pool physics (not biology) axiom: For the same force and stroke distance, regardless of force level, a lighter cue results in a slightly faster cue ball speed.

I think that IS a somewhat useful result.

I am NOT commenting on an optimal cue weight for every player. I am NOT sayng that every person (man, woman, child, chimp, or democrat /ccboard/images/graemlins/wink.gif) can play optimally with a 17oz (or 3oz, or 1oz) cue.

Best to all,
Shaft</font color>

Regards,
Dave <hr /></blockquote>

Artemus
12-18-2007, 06:06 AM
And to think, I did all of this and came to the same conclusions by testing hundreds of cues from 17 oz. to 21.5 and then putting them in the hands of a number of very fine players. My way took a lot longer time and all I came up with was anecdotal evidence that isn't recognized by forum wizards, but it certainly was more entertaining, fun, and an eye opening experience. At least all of those good players and onlookers were convinced. Nothing better than just taking it to the table.

Watch yourself Shaft, they all work as a very close knit gang and you aren't wearing their "colors" on your leather jacket.

Cornerman
12-18-2007, 07:25 AM
<blockquote><font class="small">Quote Artemus:</font><hr> And to think, I did all of this and came to the same conclusions by testing hundreds of cues from 17 oz. to 21.5 and then putting them in the hands of a number of very fine players. My way took a lot longer time and all I came up with was anecdotal evidence that isn't recognized by forum wizards, but it certainly was more entertaining, fun, and an eye opening experience. At least all of those good players and onlookers were convinced. Nothing better than just taking it to the table.

Watch yourself Shaft, they all work as a very close knit gang and you aren't wearing their "colors" on your leather jacket. <hr /></blockquote>Here's the big difference. All of the forum wizards have the same experience of trying out hundreds of cues, hitting thousands of balls, etc. But, the forum wizards know a lot of about the science of the game. I would think that someone like yourself who has been through the trenches (as most of the forum wizards have) would appreciate the added value. If you haven't yet, you might want to read dr dave's post that he linked giving the same conclusion as yours. Isn't that just grand to have the ultimate "forum wizard" come to the same conclusion?

You as well as the forum wizards should know that the science-heavy discussion won't work well with everyone, especially those who aren't versed in some of the topics and theories as these gentlemen are. Likewise, some neanderthal-like discussions won't work well with everyone, especially those who aren't versed in those type either.

Make your choice, but respect the other. We all run on different roads, but are looking at a similar goal.


Fred &lt;~~~ scientifically-minded Neanderthal

Artemus
12-18-2007, 08:22 AM
Fred stated: "Here's the big difference. All of the forum wizards have the same experience of trying out hundreds of cues,

In your case it's true, but it's absolutely not the case for other wizards. I don't think you're in the same class of wizardry. You do it because you go to all of the shows and are a cue addict connoisseur. They don't buy or collect cues, go to shows, or have access to that many cues and it just isn't even their "bag". The only thing that really matters is having a butt, Pamela Anderson or Budweiser, a Predator shaft and some regular shafts to test for deflection and that's it.

Additionally Fred, I don't appreciate forum wizards throwing it up to me that what I did and how I did or whatever doesn't show correctly in the way of a long drawn out formula is only "anectdotal evidence" and should go in the trash can or looked down on with their snitty elitist holier than thou responses. If Dr. Dave came up with the same conclusion and I came up with the same conclusion, I must have done something right. Yet, it was being blasted by other wizards and instructors. Where was Dr. Dave to prove the point then?

Cornerman
12-18-2007, 08:51 AM
<blockquote><font class="small">Quote Artemus:</font><hr> In your case it's true, but it's absolutely not the case for other wizards. <hr /></blockquote>
I can't speak to some of the wizards, but certainly mikepage and Bob Jewett have had as much if not more experience in cue experimentation.


[ QUOTE ]


Additionally Fred, I don't appreciate forum wizards throwing it up to me that what I did and how I did or whatever doesn't show correctly in the way of a long drawn out formula is only "anectdotal evidence" and should go in the trash can or looked down on with their snitty elitist holier than thou responses. If Dr. Dave came up with the same conclusion and I came up with the same conclusion, I must have done something right. Yet, it was being blasted by other wizards and instructors. Where was Dr. Dave to prove the point then? <hr /></blockquote>This is a good and valid point, but just so you know that fair is fair, Bob Jewett and Mike Page have had their share of blasting at over the years. And it doesn't take much of a search to see how many times Dr. Dave has been blasted at on this forum.

It's easy for me to say that coming into a forum, expect posts to be scrutinized. Expect posters to give an opposing viewpoint. Expect posters to give simply a differenet viewpoint that may or may not be opposing at all.

What I'd like to see more of is the science behind the anectdotal evidence, rather than the science trying to dismiss or disprove the anectodal evidence. Sometimes, it's just a terminolgy mix-up. Sometimes, it's too much science in the face of results.

Personally, I don't see much blasting at all. I'd hate to see what would happen if a true flamefest were to happen.

Fred

Artemus
12-18-2007, 08:57 AM
I think you've been around long enough for some of those true flamefests. Either you've erased them from your memory bank or the old memory is starting to fade.
I don't know whether to classify it as a flamefest, but weren't you involved with the mother of heated threads and possibly the longest thread ever recorded right here regarding anectdotal evidence vs. science? And you were the guy coming out entirely from the anecdotal end of it. Rememmmberrrr?!

Cornerman
12-18-2007, 09:29 AM
<blockquote><font class="small">Quote Artemus:</font><hr> I think you've been around long enough for some of those true flamefests. Either you've erased them from your memory bank or the old memory is starting to fade.
I don't know whether to classify it as a flamefest, but weren't you involved with the mother of heated threads and possibly the longest thread ever recorded right here regarding anectdotal evidence vs. science? And you were the guy coming out entirely from the anecdotal end of it. Rememmmberrrr?! <hr /></blockquote>What I meant was that I'd hate to see what your reaction would be if these past couple of threads turned into flamefests. Right now, they seem quite tame and manageable.

I'm no angel. I'd hate to be confused with one. Those white gown things are for pansies.

Fred

dr_dave
12-18-2007, 11:57 AM
<blockquote><font class="small">Quote Shaft:</font><hr>I am proposing a simple pool physics (not biology) axiom: For the same force and stroke distance, regardless of force level, a lighter cue results in a slightly faster cue ball speed.<hr /></blockquote>Agreed.

<blockquote><font class="small">Quote Shaft:</font><hr>I think that IS a somewhat useful result.<hr /></blockquote>Why? I'm not asking to be critical or disrespectful. I just want to know what you and others think. Do you think this result would imply a person should use a light break cue? Just as with baseball bats, some people's biomechanics, body type and size, physiology, and drug usage allow them to generate more ball speed with a heavier cue (bat), and some can do much better with a lighter cue (bat). Physics alone cannot explain this fact. The optimal choice requires experimentation.

I was thinking all along that you were focusing on power shots where extra speed is desirable (e.g., with the break or an extreme power draw shot). If not, what are you suggesting with the analysis and results, as applied to normal (average speed) pool shots?

Thanks,
Dave

dr_dave
12-18-2007, 12:24 PM
<blockquote><font class="small">Quote Shaft:</font><hr>The collision is perfectly elastic (energy and momentum will be conserved in the post collision motion of the stick and ball).<blockquote><font class="small">Quote dr_dave:</font><hr>FYI, an easy way to handle the energy conservation is to use the coefficient of restitution relation for an elastic collision (see TP A.30 (http://billiards.colostate.edu/technical_proofs/new/TP_A-30.pdf)). The math can be a lot easier.<hr /></blockquote> Yes, I know about the coefficient of resitution, but I recognize this post is read by a broad audience of people who are generally intelligent but with different educational backgrounds. I deliberately did not refer to the Coefficient of Restitution in making my point.<hr /></blockquote>
I didn't bring this up to be disrespectful or to scare off the non-physics crowd. I was just suggesting it to you to possibly help you simplify your analysis in case you were using an energy balance equation instead. An energy balance involves squares of terms, and the COR relation does not. I don't expect non-physics readers out there to know or care what this means. I just thought it might help you with this or future analyses.

I'm sorry if I offended you or others.

Regards,
Dave

Artemus
12-18-2007, 01:01 PM
<blockquote><font class="small">Quote Cornerman:</font><hr> <blockquote><font class="small">Quote Artemus:</font><hr> I think you've been around long enough for some of those true flamefests. Either you've erased them from your memory bank or the old memory is starting to fade.
I don't know whether to classify it as a flamefest, but weren't you involved with the mother of heated threads and possibly the longest thread ever recorded right here regarding anectdotal evidence vs. science? And you were the guy coming out entirely from the anecdotal end of it. Rememmmberrrr?! <hr /></blockquote>What I meant was that I'd hate to see what your reaction would be if these past couple of threads turned into flamefests. Right now, they seem quite tame and manageable.

I'm no angel. I'd hate to be confused with one. Those white gown things are for pansies.

Fred <hr /></blockquote>

There are no flamefests or prolonged flame wars any more. The wild west days are over and there are no longer shootouts at high noon.

dr_dave
12-18-2007, 01:07 PM
<blockquote><font class="small">Quote dr_dave:</font><hr>So with your force assumption, I come up with an optimal break cue weight of 3 oz!<hr /></blockquote>Sorry, but I had a small error in my math. Now my math agrees with Jal's. The optimal cue weight, neglecting the effective mass of the grip (related to grip tightness and forearm mass) and using Shaft's fixed energy assumption, is equal to the ball weight (6 oz). This can be derived by substituting the fixed-energy cue speed [vs = sqrt(2Fd/ms)] into Equation 7 of TP A.30 (http://billiards.colostate.edu/technical_proofs/new/TP_A-30.pdf), with x=0. Then take the derivative with respect to ms and set to zero. The answer is ms=mb.

Sorry for the confusion,
Dave

Jal
12-18-2007, 01:09 PM
<blockquote><font class="small">Quote Shaft:</font><hr>...I am proposing a simple pool physics (not biology) axiom: For the same force and stroke distance, regardless of force level, a lighter cue results in a slightly faster cue ball speed.<hr /></blockquote>Not true. If your equivalent arm mass is zero (impossible of course), then 6 oz would be the lower limit. If your equivalent arm mass is 18 oz, then the lower limit is 18 oz (coincidentally). This is for a centerball hit.

If you want to see the physics, let me know. My math could be wrong, but ignoring the inertia of a player's arm definitely gives you an incomplete picture, along with some erroneous conclusions.

Jim

Artemus
12-18-2007, 01:20 PM
<blockquote><font class="small">Quote Cornerman:</font><hr>
I can't speak to some of the wizards, but certainly mikepage and Bob Jewett have had as much if not more experience in cue experimentation.
Fred <hr /></blockquote>

I still have to doubt that statement, but lets assume I just go ahead and accept it at face value, doesn't it stand to reason that they would include some of those experiments and test results from what they did? I've never heard or seen any except for trying to find pivot points in relationship to deflection, but it seems like nobody ever comes up with the same results whether it's Platinum Billiards or each of them individually. Besides, who cares? If a player can't learn what to do or how to handle a little squirt and realize that it exists after his first month of playing, he's in a world of hurt. You have to be blind not to see why you're missing shots every time you apply english. So, you adjust.

A little real life action on the table and the results with all of those cues would go a long way to bolster the science end of it and make believers of people, hold their interest, help them to become better players, and improve participation. It would also help build a bridge over the chasm that exists.

mikepage
12-18-2007, 01:22 PM
<blockquote><font class="small">Quote dr_dave:</font><hr> [....] [vs = sqrt(2Fd/ms)] into Equation 7 [...] <hr /></blockquote>

All right, so here's a geeky puzzler. We all know your shot doesn't start on the break until until the cueball passes the headstring, so there'd be nothing wrong with striking a cueball that's already rolling, right?

If you're going to start with a rolling cueball, and you can swing your 18 OZ stick at 15 mph, would it help to give the cueball a head start by rolling it?


Hint: If you roll the cueball forward at a speed greater than 15 mph, then the stick can't catch up to it. And if you roll the cueball forward at 14.9 mph, then the stick can at best barely catch up and tap the cueball--so that's not the right answer either.


What, if any, initial cueball velocities would help (lead to a higher break speed)?

dr_dave
12-18-2007, 01:31 PM
<blockquote><font class="small">Quote mikepage:</font><hr> <blockquote><font class="small">Quote dr_dave:</font><hr> [....] [vs = sqrt(2Fd/ms)] into Equation 7 [...] <hr /></blockquote>

All right, so here's a geeky puzzler. We all know your shot doesn't start on the break until until the cueball passes the headstring, so there'd be nothing wrong with striking a cueball that's already rolling, right?

If you're going to start with a rolling cueball, and you can swing your 18 OZ stick at 15 mph, would it help to give the cueball a head start by rolling it?


Hint: If you roll the cueball forward at a speed greater than 15 mph, then the stick can't catch up to it. And if you roll the cueball forward at 14.9 mph, then the stick can at best barely catch up and tap the cueball--so that's not the right answer either.

What, if any, initial cueball velocities would help (lead to a higher break speed)? <hr /></blockquote>I'm an engineer. I don't like silly hypothetical physics questions. /ccboard/images/graemlins/wink.gif

If I thought rolling the CB was legal and if someone could maintain hit accuracy with a roll-then-hit approach, then I would be interested. Otherwise, I would rather tackle problems that have useful answers. Engineers don't like science unless it is useful. /ccboard/images/graemlins/cool.gif

I don't mean any disrespect. I just like poking fun at physicists who just study science for the sake of science. I'm not implying you or others here are in this category.

Regards,
Dave

Bob_Jewett
12-18-2007, 01:33 PM
<blockquote><font class="small">Quote Shaft:</font><hr> No, Bob.

Your example is absurd and completely off point.
... <hr /></blockquote>
One absolutely standard way to check reasoning for problems like these is to look at the extremes. If a statement doesn't hold for the extremes, then you need to figure out how close to the extremes it starts to give out. Without such obvious checks, the analysis is suspect.

dr_dave
12-18-2007, 01:38 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote Shaft:</font><hr>...I am proposing a simple pool physics (not biology) axiom: For the same force and stroke distance, regardless of force level, a lighter cue results in a slightly faster cue ball speed.<hr /></blockquote>Not true. If your equivalent arm mass is zero (impossible of course), then 6 oz would be the lower limit. If your equivalent arm mass is 18 oz, then the lower limit is 18 oz (coincidentally). This is for a centerball hit.<hr /></blockquote>Jim,

For fixed energy, the optimal (fastest ball speed) effective cue weight is the ball weight (6 oz). So for heavier cues, the ball speed will be less. But as you get lighter (towards 6 oz), doesn't the resulting ball speed increase?

I'm sorry if I misinterpreted your statement. What do you think?

Regards,
Dave

Bob_Jewett
12-18-2007, 01:41 PM
<blockquote><font class="small">Quote mikepage:</font><hr> ... We all know your shot doesn't start on the break until until the cueball passes the headstring, so there'd be nothing wrong with striking a cueball that's already rolling, right? ... <hr /></blockquote>
With a false premise, the conclusion will be suspect. Some of us believe the rules say otherwise.

There is a standard trick shot to shoot the cue ball up and down the table, and on its return to strike it low so the cue ball goes out a ways and then comes back as if it were hit with a lot of masse.

A related question is: for which pitch is it easier to hit a home run, a fast ball or a soft, arcing lob?

mikepage
12-18-2007, 03:01 PM
<blockquote><font class="small">Quote dr_dave:</font><hr> [...]I'm an engineer. I don't like silly hypothetical physics questions. /ccboard/images/graemlins/wink.gif

If I thought rolling the CB was legal and if someone could maintain hit accuracy with a roll-then-hit approach, then I would be interested. Otherwise, I would rather tackle problems that have useful answers. Engineers don't like science unless it is useful. /ccboard/images/graemlins/cool.gif

I don't mean any disrespect. I just like poking fun at physicists who just study science for the sake of science. I'm not implying you or others here are in this category.

Regards,
Dave <hr /></blockquote>

Oh I'm definitely in that category. I do science, truth be told, for the same reason I play pool: it amuses me.

The making it "useful" part is for me more about getting other people to pay me to ...ahem ... amuse myself.

But let's not forget pool is a GAME. Your "useful" standard here is not about curing disease or reducing hunger or ending conflict; it's about helping people to amuse themselves.

And that's all I'm doing with my puzzler ;-). I thought people might enjoy figuring out the nonintuitive answer that you'd get a better break by rolling a cueball TOWARD the shooter rather than giving it a headstart AWAY from the shooter....

dr_dave
12-18-2007, 03:44 PM
<blockquote><font class="small">Quote mikepage:</font><hr>I thought people might enjoy figuring out the nonintuitive answer that you'd get a better break by rolling a cueball TOWARD the shooter rather than giving it a headstart AWAY from the shooter....<hr /></blockquote>I guess I should have listed baseball in my list of analogies (with golf and bowling) in a previous thread. A fast ball coming towards you is easier to hit out of the park (with an accurate hit).

Happy Holidays,
Dave

Jal
12-19-2007, 12:55 AM
<blockquote><font class="small">Quote dr_dave:</font><hr> <blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote Shaft:</font><hr>...I am proposing a simple pool physics (not biology) axiom: For the same force and stroke distance, regardless of force level, a lighter cue results in a slightly faster cue ball speed.<hr /></blockquote>Not true. If your equivalent arm mass is zero (impossible of course), then 6 oz would be the lower limit. If your equivalent arm mass is 18 oz, then the lower limit is 18 oz (coincidentally). This is for a centerball hit.<hr /></blockquote>Jim,

For fixed energy, the optimal (fastest ball speed) effective cue weight is the ball weight (6 oz). So for heavier cues, the ball speed will be less. But as you get lighter (towards 6 oz), doesn't the resulting ball speed increase?

I'm sorry if I misinterpreted your statement. What do you think?

Regards,
Dave <hr /></blockquote>Hi Dr. Dave,

Thank you for asking my opinion. The only difference between my treatment and yours is that in your relation:

vs = sqrt(2Fd/ms),

I use the combined inertia of the arm and stick instead of just ms:

vs = sqrt(2Fd/(ms+ma)),

where ma is the equivalent mass of the arm.

As I see it, the torque generated by the muscles propels both the arm and the stick. The arm rotates about the elbow while the stick undergoes a more or less linear motion. Relating the angular momentum of both the arm and stick about the elbow to the torque impulse leads to an expression for the arm's equivalent mass: the arm's moment of inertia divided by the square of the moment arm extending from the elbow to the grip. Finding the maxima, as you described earlier, gives the optimal stick to ball mass ratio as:

ms/mb = (1/2u)[1 + Sqrt[1 + 8u(ma/mb)]]

where u is (5/2)(b/R)^2 + 1, and b/R is the fractional tip offset.

Some examples. For a centerball hit, u=1. Then if ma/mb=2, ms=15.4oz. If ma/mb=3, then ms=18oz. And if ma/mb=4, ms=20.2oz. These are minimal values as indicated earlier.

If you get the time to look at this, particularly the equivalent mass formulation, I'd certainly like to have your expert judgement on this.

Jim

Shaft
12-19-2007, 06:19 AM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote Shaft:</font><hr>...I am proposing a simple pool physics (not biology) axiom: For the same force and stroke distance, regardless of force level, a lighter cue results in a slightly faster cue ball speed.<hr /></blockquote>Not true. If your equivalent arm mass is zero (impossible of course), then 6 oz would be the lower limit. If your equivalent arm mass is 18 oz, then the lower limit is 18 oz (coincidentally). This is for a centerball hit.

...
Jim <hr /></blockquote>

<font color="green"> </font color> JAL: Of course I agree with you 100% here. I never seriously proposed that the optimal cue weight was "6oz" (or anything close to it).

Now, we should recognize that your arm mass effect is present with both the heavy and light cues. Therefore, this changes the final numerical values but not the result of the comparison. -Shaft

Artemus
12-19-2007, 06:26 AM
<blockquote><font class="small">Quote mikepage:</font><hr> I thought people might enjoy figuring out the nonintuitive answer that you'd get a better break by rolling a cueball TOWARD the shooter rather than giving it a headstart AWAY from the shooter.... <hr /></blockquote>

Based on the number of responses it doesn't look like people enjoyed anything or gave a shat. It's kinda hilarious to see the motorcycle gang of Ph.D's now quibbling among themselves about nebulous minutiae. LMAO

I don't know about rolling a cueball toward a player, but how about hitting the cueball while it's flying through the air?

http://www.youtube.com/watch?v=BckBIUHx8e8&amp;feature=related

Yeah, I know it's the most hated man on pool forums, but lets see you put THAT into an equation or formula. Just so you don't go jumping to erroneous conclusions, I am not FL and he peeesses me off just as much as you!

You guys need to come up with a name for your Ph.D. motorcycle gang. Maybe there should be a vote or forum participation to come up with the best one. Hells Angels and Pagans have already been taken, how about the GOOGANS!
Btw Shaft, I warned you that they work as a close knit gang if you you haven't gone through the hazing process and proven yourself worthy to wear the jacket colors. Are you ready to be a GOOGAN? Will they accept you? Will you be allowed to be a part of the inner circle? I'm getting excited to find the outcome now that the writer's strike is going to affect regular programming.

DickLeonard
12-19-2007, 08:26 AM
Fred Breaking nineballs has always interested me. As one who never put a ball in on the break I always wondered what I was doing wrong.

In Troy we had a player who put all nineballs in on the break. One day he came into the room I was running in Troy so I asked him to give me a demo of him breaking nineball. He said he hadn't played in 5 years so he wanted to loosen up first. With his finger exercises he looked like he was going to play the piano instead of playing pool. Now he was ready to Break.

Slight of build 5ft 6,125lbs. For the next three hours he gave me a breaking demo. He never put less than two balls in, some racks 3,4 or 5 and the most he put in was 6. He never did anything to explain the action that his cueball hitting one generated.

He never lunged, no wild windup nothing but when the cueball hit the one it was as if a bomb went off. Paul Dayton the cuemaker said if there was a room full of people breaking nineball you would hear his break over the rest of the players.

What I am trying to say is Math doesn't explain everything.####

Artemus
12-19-2007, 08:46 AM
Oh, Oh. You aren't going to be receiving a jacket with the GOOGAN colors either, nor an invitation to the annual Christmas party.

New2Pool
12-19-2007, 09:57 AM
So what is the problem with people who enjoy discussing the minutia of pool physics doing so? I don't really enjoy that but I just skim over the comments and try not to fall asleep before I get to comments of interest.

Individuals sometimes enjoy in-depth discussion of what some of us would consider arcane minutia? One of the threads I enjoyed participating in the most was a discussion 1Time and I had about the academic literature related to expert performance. I suspect everyone else's eyes glazed over but why should a poster have to worry about whether or not their topic is worthy of discussion? If it is not, nobody else will participate.

Artemus
12-19-2007, 10:14 AM
<blockquote><font class="small">Quote New2Pool:</font><hr>I don't really enjoy that but I just skim over the comments and try not to fall asleep <hr /></blockquote>

Thank you for answering your own question with the most astute answer you could have come up with. I concur.

SpiderMan
12-19-2007, 12:24 PM
<blockquote><font class="small">Quote Shaft:</font><hr> Here is an attempt to settle, once and for all, the question of whether a light cue stick "breaks better" than a heavy cue stick.

The mathematical result surprised me, and it forces me to recant what I have said on previous posts that "it should not make a difference." It does, in fact, make a (small?) difference.

I apologize in advance for the use of metric units, but it makes the math so much easier. I will give the english equivalents wherever they help.

For the purpose of this analysis I will assume that

* The collision is perfectly elastic (energy and momentum will be conserved in the post collision motion of the stick and ball).

* The same shooter exerts a constant force (300 Newtons, about 67.4 lbs) over a 6 inch forward swing to impact. (The actual force and swing distance are not important here so long as they are the same for both cues.)

* The "light" stick weighs 18oz, the "heavy" stick weighs 21oz, and the cue ball weighs 6 oz.

* The stick and ball are free bodies in space. The the stick is accellerated to hit a ball that is initially at rest, and the axis and path of the stick is alligned with the center of the ball. (I know real balls and sticks are not free bodies in space, but this assumption simplifies everything without changing the outcome.)

The trick is to find the post collision speeds of the ball and stick so that energy and momentum are conserved. We have a system of two equations and two unknowns, so it can be solved with some hairy algebra. I used the Excel Solver routine because it was easier and I am LAZY.

A constant force of 300N applied over 6 inches will accelerate an 18oz (0.5103kg) stick to a speed of 13.38614 m/s (29.94mph). The pre-collision momentum of the stick is 6.830947 kg-m/s, and the kinetic energy is 45.72J. Conserving energy and momentum, the post collision speed of the stick is 6.69307m/s (14.97mph) and the post collision speed of the ball is 20.07m/s (44.92mph).

A constant force of 300N applied over 6 inches will accellerate a 21oz (0.5953kg) stick to a speed of 12.39368m/s (27.72mph). The pre-collision momentum of the stick is 7.377956 kg-m/s, and the kinetic energy is 45.72 J. Conserving energy and momentum, the post collision speed of the stick is 6.885376m/s (15.40mph) and the post collision speed of the ball is 19.27905m/s (43.13mph).

DRUM ROLL and TRUMPET FANFARE:::::::::::::
The cue ball has 4.1% more speed and 8.5% more energy when struck by a LIGHTER (18oz) stick than when it is struck by a HEAVIER (21oz) stick, all other things being equal. A 15% reduction in weight results in 4% more speed.

This is true - NOT because the lighter stick moves faster - but because the lighter stick transfers more energy to an object that is more similar in mass.

In fact, I demonstrated with the same spreadsheet that, as the mass of the two colliding objects gets more and more equal, more and more of the energy of the first object is transfered to the second object. When the mass difference is zero, the energy transfer is 100%.

You might find that hard to believe. Remember those annoying clickety-clackety novelty toys that had 5 steel ball bearings suspended on a wooden frame? Take three balls out of the way and try it with just two balls. The first ball will stop dead as the second ball takes 100% of the energy and swings upward.

We can also see this principle on the pool table. A dead-stratight stun shot transfers 100% of the cue ball energy to the object ball. (It is ironic that a clue to the answer has been in front of us all the time.)

In this cas, physics CONFIRMS anecdotal observation.

I suppose we should all go out and buy 6oz cue sticks!!! /ccboard/images/graemlins/grin.gif /ccboard/images/graemlins/grin.gif /ccboard/images/graemlins/grin.gif

Merry Christmas everybody!
<hr /></blockquote>

Shaft,

I agree with your math, but not with all your assumptions.

The primary issue I have is with your constant-stick-energy model. This model is accurate if every shot's stick energy is defined by force times distance, where neither of these vary, but that isn't the case.

If there is a constant force applied, then it is more likely at the muscle level, not at the hand-stick interface. In other words, there may a constant force available to accelerate BOTH THE ARM AND THE STICK.

The total kinetic energy imparted to the arm/stick would be constant, but the lighter the stick becomes, the lesser portion of this energy it contains. I've actually demonstrated this concept experimentally by chronographing projectiles from a slingshot. Because the rubber bands must accelerate both themselves and the projectile, the projectile energy, rather than constant, is a monotonically-increasing function of projectile mass. The slingshot is more efficient at imparting it's stored energy to heavier projectiles.

So, using this "modified" constant-energy model, the stick energy increases with stick mass, and (as both you and Dr Dave have shown) the efficiency of energy transfer to the ball decreases with increasing stick mass, over some nominal range. I'm guessing this would yield a single-valued result of stick mass for maximum ball energy.

One thing I'd add, which also seemed a factor in my slingshot experiments, is that there is probably also a diminishing ability of the muscle to apply force, as a function of increasing speed. In other words, the muscle has a maximum possible speed of contraction. The effect of this would be to make the stick energy decrease faster for smaller stick masses than otherwise predicted, so the "real" optimum stick would be a little heavier than predicted by solving the simpler case.

SpiderMan

Jal
12-19-2007, 06:14 PM
<blockquote><font class="small">Quote Shaft:</font><hr> <blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote Shaft:</font><hr>...I am proposing a simple pool physics (not biology) axiom: For the same force and stroke distance, regardless of force level, a lighter cue results in a slightly faster cue ball speed.<hr /></blockquote>Not true. If your equivalent arm mass is zero (impossible of course), then 6 oz would be the lower limit. If your equivalent arm mass is 18 oz, then the lower limit is 18 oz (coincidentally). This is for a centerball hit.

...
Jim <hr /></blockquote>

<font color="green"> </font color> JAL: Of course I agree with you 100% here. I never seriously proposed that the optimal cue weight was "6oz" (or anything close to it). <hr /></blockquote>Nor did I think you did.

<blockquote><font class="small">Quote Shaft:</font><hr>Now, we should recognize that your arm mass effect is present with both the heavy and light cues. Therefore, this changes the final numerical values but not the result of the comparison. -Shaft <hr /></blockquote>It does mean, for instance, that a heavier cue would in fact get you more cueball speed depending on how light a cue we're comparing it to. For someone with an equivalent arm mass of three cueballs, an 18 oz cue would be better than a 15 oz one at centerball hits.

You may already agree with this, if not the exact numbers, but it's not clear from what you just said. I think that if you added various amounts of arm mass to the stick's mass in your program, and then varied the stick mass for each particular arm mass, it would show this. The results should agree with the formula given earlier.

Jim

Jigger
12-24-2007, 01:37 PM
<blockquote><font class="small">Quote Shaft:</font><hr> Here is an attempt to settle, once and for all, the question of whether a light cue stick "breaks better" than a heavy cue stick.

The mathematical result surprised me, and it forces me to recant what I have said on previous posts that "it should not make a difference." It does, in fact, make a (small?) difference.

I apologize in advance for the use of metric units, but it makes the math so much easier. I will give the english equivalents wherever they help.

For the purpose of this analysis I will assume that

* The collision is perfectly elastic (energy and momentum will be conserved in the post collision motion of the stick and ball).

* The same shooter exerts a constant force (300 Newtons, about 67.4 lbs) over a 6 inch forward swing to impact. (The actual force and swing distance are not important here so long as they are the same for both cues.)

* The "light" stick weighs 18oz, the "heavy" stick weighs 21oz, and the cue ball weighs 6 oz.

* The stick and ball are free bodies in space. The the stick is accellerated to hit a ball that is initially at rest, and the axis and path of the stick is alligned with the center of the ball. (I know real balls and sticks are not free bodies in space, but this assumption simplifies everything without changing the outcome.)

The trick is to find the post collision speeds of the ball and stick so that energy and momentum are conserved. We have a system of two equations and two unknowns, so it can be solved with some hairy algebra. I used the Excel Solver routine because it was easier and I am LAZY.

A constant force of 300N applied over 6 inches will accelerate an 18oz (0.5103kg) stick to a speed of 13.38614 m/s (29.94mph). The pre-collision momentum of the stick is 6.830947 kg-m/s, and the kinetic energy is 45.72J. Conserving energy and momentum, the post collision speed of the stick is 6.69307m/s (14.97mph) and the post collision speed of the ball is 20.07m/s (44.92mph).

A constant force of 300N applied over 6 inches will accellerate a 21oz (0.5953kg) stick to a speed of 12.39368m/s (27.72mph). The pre-collision momentum of the stick is 7.377956 kg-m/s, and the kinetic energy is 45.72 J. Conserving energy and momentum, the post collision speed of the stick is 6.885376m/s (15.40mph) and the post collision speed of the ball is 19.27905m/s (43.13mph).

DRUM ROLL and TRUMPET FANFARE:::::::::::::
The cue ball has 4.1% more speed and 8.5% more energy when struck by a LIGHTER (18oz) stick than when it is struck by a HEAVIER (21oz) stick, all other things being equal. A 15% reduction in weight results in 4% more speed.

This is true - NOT because the lighter stick moves faster - but because the lighter stick transfers more energy to an object that is more similar in mass.

In fact, I demonstrated with the same spreadsheet that, as the mass of the two colliding objects gets more and more equal, more and more of the energy of the first object is transfered to the second object. When the mass difference is zero, the energy transfer is 100%.

You might find that hard to believe. Remember those annoying clickety-clackety novelty toys that had 5 steel ball bearings suspended on a wooden frame? Take three balls out of the way and try it with just two balls. The first ball will stop dead as the second ball takes 100% of the energy and swings upward.

We can also see this principle on the pool table. A dead-stratight stun shot transfers 100% of the cue ball energy to the object ball. (It is ironic that a clue to the answer has been in front of us all the time.)

In this cas, physics CONFIRMS anecdotal observation.

I suppose we should all go out and buy 6oz cue sticks!!! /ccboard/images/graemlins/grin.gif /ccboard/images/graemlins/grin.gif /ccboard/images/graemlins/grin.gif

Merry Christmas everybody!
<hr /></blockquote>

Are you serious? I'm impressed....

Why not try different weights until you find what makes you most comfortable. Most cues can be adjusted later as you refine your preference. <font color="brown"> </font color> <font color="blue"> </font color>