I know weight is a matter of preference, but I got a question

[Ssonerai]

I started out in the basic physics camp, F = ma.

Which seems to have become simplified in this discussion by most to mean mass x velocity at impact. (The quote "energy in = energy out"). The working assumption has become "similar acceleration of the CB to absorb similar energy of the cue".

But does it?
A draw shot is not a perfect collision. (like a break shot is, or an ideal break shot would be).

It is a glancing collision, and some of the energy continues with the forward motion and deflection of the cue, unconsciously absorbed by the shooter until the cue stops.

Does a massive cue deflect less? With the same energy (lower speed) as the lighter cue (higher speed) or does the velocity + cancel the mass advantage? Deeper than that, :grin: Does the shooter unconsciously make a miniscule adjustment, hitting the CB somewhat closer, or somewhat lower than center because of the mass (s)he perceives?

smt, supposed to be solving real problems in the shop but this is somehow more compelling.

 

[3kushn]

If you put the same energy into your stoke you'll move a lighter cue faster, a heavier cue slower, and they'll produce the same CB speed/spin.

Energy in = energy out. Cue weight is just the delivery method.

pj <- I think
chgo

I'm not arguing and all I have is anecdotal evidence. This has all been studied by smarter people than myself.

My normal playing cue is around 16.5oz. I'm playing with a heavier ball if that matters one way or another. I don't have any problem sending the ball 9 rails. 10 is a monster stroke. But 9 is enough IMO.

The last few days I switched to a 14oz cue and distances haven't changed. That said, I had Jamison Neu's masse cue for about a week. I never achieved 11 rails but 10 seemed easier.

I just bring this up since I assume everyone is thinking between 18 & 21 or 22 oz. The masse cue is guessing between 28 & 30oz

Years ago at the USBA Nationals at the DCC, Dieckman handed a guy a 14oz cue, who proceeded to hit 11 rails with it I think on the 2nd or 3rd attempt.

 

[Patrick Johnson]

I'm not arguing and all I have is anecdotal evidence. This has all been studied by smarter people than myself.

I don't see anything to disagree with in your post.

My normal playing cue is around 16.5oz. I'm playing with a heavier ball if that matters one way or another.

It matters to the CB's speed - the same energy in your stroke will send a heavier CB at a slower speed...

...but not to its kinetic energy - with the same stroke energy a heavier CB moving at a slower speed and a lighter CB moving at a faster speed both have the same kinetic energy (they'll both move an OB the same distance).

pj <- I think
chgo

 

[3kushn]

I

pj <- I think
chgo

^^^^^^^^^^^^
The crux of the matter.

I have to think technique is a major factor with someone trying to determine this stuff.

I have to defer to the science guys on it and just let it go.

I'll just continue to play with my ultra light cues.

Love the feel.

Get all the action I want.

 

[dr_dave]

Physics question:

Let's say an example, on a draw shot.

We have the same player, with the same exact stroke every single time, shooting one time with 19oz cue, then another time with the exact same cue but reduced to 17oz.

Let's assume he shot the same shot with both, and also same emotion, backstroke, everything to the teeth, also he hit down on the cueball on the same exact point.

Now my question is, with which cue does the cueball travel further to the back?

If cue speed, cue elevation, and effective tip contact point (taking squirt into account) are the same in the comparison, the heavier cue will create more draw; although the stroke will require more effort. FYI, This and many other related questions are answered here:

Draw Shot Physics-Based Advice

Now, if you instead ask what weight cue can create the most draw, that depends on the person per the info here:

Pool Cue Optimal Weight

You might also find the following interesting:

How to get maximum spin

Power draw advice

Enjoy,
Dave

 

[straightline]

If cue speed, cue elevation, and effective tip contact point (taking squirt into account) are the same in the comparison, the heavier cue will create more draw; although the stroke will require more effort. FYI, This and many other related questions are answered here:

Draw Shot Physics-Based Advice

Now, if you instead ask what weight cue can create the most draw, that depends on the person per the info here:

Pool Cue Optimal Weight

You might also find the following interesting:

How to get maximum spin

Power draw advice

Enjoy,
Dave

Thank you Dr. Dave. I will accept this as definitive;

I think...

 

[straightline]

dblpstdltok

 

[Cron]

If cue speed, cue elevation, and effective tip contact point (taking squirt into account) are the same in the comparison, the heavier cue will create more draw; although the stroke will require more effort. FYI, This and many other related questions are answered here:

Draw Shot Physics-Based Advice

Now, if you instead ask what weight cue can create the most draw, that depends on the person per the info here:

Pool Cue Optimal Weight

You might also find the following interesting:

How to get maximum spin

Power draw advice

Enjoy,
Dave

WOW! Thank you very much for these...

https://billiards.colostate.edu/technical-proof/

I had no idea you had these. This looks like everything, is there something you're missing on that page? Were any of them trickier to proof than you first thought? I will now donate (sadly, it can only be $20... I'm poor :-/).

Thank you for your appreciation of billiards.

 

[dr_dave]

WOW! Thank you very much for these...

https://billiards.colostate.edu/technical-proof/

You're welcome. I'm glad somebody appreciates them ... not many do because they require strong backgrounds in math and physics (not strong suits for most pool players, understandably so).

I had no idea you had these. This looks like everything, is there something you're missing on that page? Were any of them trickier to proof than you first thought?

The most useful in terms of insights gained were:

TP A.14 – The effects of cut angle, speed, and spin on object ball throw​

and

TP A.20 – The effect of spin, speed, and cut angle on draw shots​

The most challenging in terms of physics complexity was:

TP B.2 – Rolling resistance, spin resistance, and “ball turn”​

The most time consuming was the effective pocket size and center stuff, especially:

TP 3.6 – Effective target sizes for slow shots into a corner pocket at different angles​

I will now donate (sadly, it can only be $20... I'm poor :-/).

Thank you for your appreciation of billiards.

Thank you ... and you're welcome. I aim to swerve. :grin-square:

Best regards,
Dave

 

[straightline]

@ Dr. Dave. I had a question got deleted in my double post. Why is it gravity acts equally on all objects when every other transfer of motion is subject to the rest of Newtonian physics?

 

[dr_dave]

@ Dr. Dave. I had a question got deleted in my double post. Why is it gravity acts equally on all objects when every other transfer of motion is subject to the rest of Newtonian physics?

The force of gravity is equal and opposite on each object (e.g., an apple and the earth) just like any other "Newtonian physics" force; although, it doesn't "act equally on all objects" ... the magnitude of the force depends on the mass of each object and the distance between their centers of mass. For example, an apple "weighs" more at low altitude than it does at high altitude.

Now, are you asking how gravity actually works? If so, it is due to the fact that mass warps the time-space continuum (I'm not kidding). Don't ask me to explain further because I am a pool player (and an engineer), not a theoretical physicist.

Now, are you asking why the "acceleration due to gravity" (g) is the same for all objects close to the surface of the earth (i.e., they drop at the same rate, neglecting air resistance), regardless of their weight? That's because gravitational force is proportional to the object's mass, so g (= F/m) is constant as a result of the Newtonian equation: F = ma.

Regards,
Dave

 

[straightline]

The force of gravity is equal and opposite on each object (e.g., an apple and the earth) just like any other "Newtonian physics" force; although, it doesn't "act equally on all objects" ... the magnitude of the force depends on the mass of each object and the distance between their centers of mass. For example, an apple "weighs" more at low altitude than it does at high altitude.

Now, are you asking how gravity actually works? If so, it is due to the fact that mass warps the time-space continuum (I'm not kidding). Don't ask me to explain further because I am a pool player (and an engineer), not a theoretical physicist.

Now, are you asking why the "acceleration due to gravity" (g) is the same for all objects close to the surface of the earth (i.e., they drop at the same rate, neglecting air resistance), regardless of their weight? That's because gravitational force is proportional to the object's mass, so g (= F/m) is constant as a result of the Newtonian equation: F = ma.

Regards,
Dave

I'm superficially familiar with the notion of curved space - don't know enough to commit but yes, I was referring to the identical acceleration of falling bodies and wondering why every other motive force, stroking cues for instance, requires more force to move a heavy one than a light one.

 

[pvc lou]

Now, are you asking why the "acceleration due to gravity" (g) is the same for all objects close to the surface of the earth (i.e., they drop at the same rate, neglecting air resistance), regardless of their weight? That's because gravitational force is proportional to the object's mass, so g (= F/m) is constant as a result of the Newtonian equation: F = ma.

Regards,
Dave

hmm...that answer is a bit circular, isn't it Doc? the logical followup question might be, what's the difference between g and a ?
you got close to the answer with "g is the same for all objects close to the surface of the earth", but i guess u didn't want to explain Newton's law of gravitation...or the difference between a gravitational force and an inertial force. anyway, i'm surprised at your lack of rigor, Doc.

 

[dr_dave]

Now, are you asking why the "acceleration due to gravity" (g) is the same for all objects close to the surface of the earth (i.e., they drop at the same rate, neglecting air resistance), regardless of their weight? That's because gravitational force is proportional to the object's mass, so g (= F/m) is constant as a result of the Newtonian equation: F = ma.

yes, I was referring to the identical acceleration of falling bodies and wondering why every other motive force, stroking cues for instance, requires more force to move a heavy one than a light one.

The force of gravity works the same way as the force acting on a cue. It takes more force to accelerate a heavier object at the same rate (W = mg, F = ma).

Regards,
Dave

 

[dr_dave]

Now, are you asking why the "acceleration due to gravity" (g) is the same for all objects close to the surface of the earth (i.e., they drop at the same rate, neglecting air resistance), regardless of their weight? That's because gravitational force is proportional to the object's mass, so g (= F/m) is constant as a result of the Newtonian equation: F = ma.

hmm...that answer is a bit circular, isn't it Doc? the logical followup question might be, what's the difference between g and a ?
you got close to the answer with "g is the same for all objects close to the surface of the earth", but i guess u didn't want to explain Newton's law of gravitation...or the difference between a gravitational force and an inertial force. anyway, i'm surprised at your lack of rigor, Doc.

Nothing was circular. I didn't think more detail was wanted or needed. But since you brought it up. The force due to gravity is:

F = G*m*M/(R^2)​

where G is the universal gravitational constant. So the gravitational force (F) or weight (W) acting on an object at the surface of the earth (distance R from the earth's center, where the earth has mass M) is proportional to the mass of the object (m):

F = m * [G*M/(R^2)] = m * g = W​

Since F = ma (one of Newton's Laws of Motion), g is the constant acceleration due to gravity.

Is that enough "rigor" for you?

Regards,
Dave

 

[straightline]

The force of gravity works the same way as the force acting on a cue. It takes more force to accelerate a heavier object at the same rate (W = mg, F = ma).

Regards,
Dave

Ok but gravity seems not to expend any energy doing it; everything else; not nearly so efficient.

 

[pvc lou]

Nothing was circular. I didn't think more detail was wanted or needed. But since you brought it up. The force due to gravity is:

F = G*m*M/(R^2)​

where G is the universal gravitational constant. So the gravitational force (F) or weight (W) acting on an object at the surface of the earth (distance R from the earth's center, where the earth has mass M) is proportional to the mass of the object (m):

F = m * [G*M/(R^2)] = m * g = W​

Since F = ma (one of Newton's Laws of Motion), g is the constant acceleration due to gravity.

Is that enough "rigor" for you?

Regards,
Dave

That's a little better, Doc. Second question : what does that have to do with the OP's question? ...which I happen to think is an excellent question : all else being equal, how does a cue's weight affect the draw action of a cueball?

I looked through some of the links to your website that you shared earlier, and I still haven't found an analysis of that question...I hope you will take the time to address that question explicitly, Doc.

Love,

Lou(cifer)

 

[dr_dave]

That's a little better, Doc. Second question : what does that have to do with the OP's question? ...which I happen to think is an excellent question : all else being equal, how does a cue's weight affect the draw action of a cueball?

I looked through some of the links to your website that you shared earlier, and I still haven't found an analysis of that question...I hope you will take the time to address that question explicitly, Doc.

If you want the physics and math, it is here:

TP A.30 – The effects of cue tip offset, cue weight, and cue speed on cue ball speed and spin

TP B.8 – Draw shot physics​

They covers how everything affects draw, including cue weight.

Enjoy,
Dave

 

[dr_dave]

Ok but gravity seems not to expend any energy doing it; everything else; not nearly so efficient.

When an object falls due to gravity, energy changes are definitely involved. Potential energy (PE = mgh) is being converted to kinetic energy (KE = 1/2mv^2). If you want to learn more about this, you need to read a basic physics book or read about it online (maybe in Wikipedia). AZB is probably not the best place to learn general physics principles.

Regards,
Dave

 

[pvc lou]

If you want the physics and math, it is here:

TP A.30 – The effects of cue tip offset, cue weight, and cue speed on cue ball speed and spin

TP B.8 – Draw shot physics​

They covers how everything affects draw, including cue weight.

Enjoy,
Dave

Thanks, Doc, I'll check it out right now...but it may take me a while to get through the derivations.

In the meantime, what's the answer, Doc ? Heavy or light cue gives better draw?