Back arm perpendicular- why?

[crosseyedjoe]

Actually, it does apply to all real mechanical systems. The assumption of course is that the acceleration doesn't stay positive indefinitely, but eventually reaches zero at some time.

If acceleration never reaches zero, then it's obvious that the velocity is forever increasing and the peak velocity is whatever it is at the present moment.

Your mechanical system is actually the description you gave; "the assumption of course is that the acceleration doesn't stay positive indefinitely, but eventually reaches zero at some time." It's not all "real" mechanical systems. Unless otherwise you assume all "real" mechanical systems behaves that way.

Just one thing "NET acceleration doesn't stay positive."

An object in the system can still have its acceleration even when net acceleration of the whole system is zero. That's why I'm too concern about people using the word "peak" acceleration most specially when applied to just one part of the whole system.

 

[Patrick Johnson]

Ah, how is that attractive force normally referred to and what does it imply?

The system YOU and EARTH, has a net zero acceleration. But when you isolate the body YOU and treat it as a separate system, YOU is accelerating by 9.8m/s^2. I was just hoping for people to use terms and the associated system consistently most specially the modifier NET.

Well, maybe so, but it still doesn't seem to have anything to do with the difference in how much we weigh on Earth vs. the Moon. I believe that difference is entirely explained by the difference in their masses/gravitational forces.

And this is even less relevant to pool than it was before.

pj
go

 

[jsp]

Well, maybe so, but it still doesn't seem to have anything to do with the difference in how much we weigh on Earth vs. the Moon. I believe that difference is entirely explained by the difference in their masses/gravitational forces.

And this is even less relevant to pool than it was before.

What are you talking about? You mean you don't factor in the moon's gravitational pull when lagging a shot?

 

[softshot]

This is true only if you don't care how fast the cue ball moves. The speed of the cue ball will be directly proportional to the speed of the stick at the moment of tip-to-ball contact.

And to the extent that you control the acceleration of the stick you control its velocity to the same extent.

I control acceleration directly.. I control velocity indirectly.. I believe I should be focused on what I can directly control and let the results be what they are.

 

[crosseyedjoe]

Well, maybe so, but it still doesn't seem to have anything to do with the difference in how much we weigh on Earth vs. the Moon. I believe that difference is entirely explained by the difference in their masses/gravitational forces.

And this is even less relevant to pool than it was before.

pj
go

The acceleration due to gravity is what changing since your mass remains constant.

 

[dr_dave]

enough with the planets

You're talking about the total combined gravitational force of the two bodies. I'm talking about the fraction of the total supplied by the Moon, which is the comparison that explains why we weigh less on the Moon than on the Earth.

Sorry if I misunderstood your original point.

Regards,
Dave

PS: There is only one force that acts between the earth and moon, and it is equal and opposite on both (i.e., the force the earth exerts on the moon is the same as the force the moon exerts on the earth). This doesn't have a thing to do with pool, but it is an often-misunderstood concept.

 

[sfleinen]

I control acceleration directly.. I control velocity indirectly.. I believe I should be focused on what I can directly control and let the results be what they are.

Smorgie's DuneWorm says:

"I controls the spice. I controls the universe. I focuses on what I can directly control (everything) and the results are what they are."

:-D

 

[Patrick Johnson]

...There is only one force that acts between the earth and moon, and it is equal and opposite on both (i.e., the force the earth exerts on the moon is the same as the force the moon exerts on the earth). This doesn't have a thing to do with pool, but it is an often-misunderstood concept.

I'd like to understand it better. Can you point me to something?

Thanks,

pj
chgo

 

[dr_dave]

graviational physics lesson

I'd like to understand it better. Can you point me to something?

Any introductory college physics book covers this. It's called the universal law of gravitation (another one of Newton's contributions to the world). Here's the equation:

F = G * m1 * m2 / d^2
​

where:

F = equal and opposite force acting between the bodies
G = universal gravitation constant
m1 = mass of body 1
m2 = mass of body 2
d = distance between the CGs of the two bodies
​

On the surface of the earth:

g = G*M / d^2​

where:

g = "acceleration due to gravity"
M = mass of the earth
d = average radius of the earth​

To calculate the force acting between the earth and an object (e.g., you) sitting on the surface of the earth:

W = m*g​

where:

W = "weight" of the object
m = mass of the object​

I hope you find this useful.

Regards,
Dave

PS: ... and this is why the forearm should be perpendicular to the cue at CB impact!!!

 

[Patrick Johnson]

Dr Dave:
graviational physics lesson...

Thanks, but I'm interested specifically in a conceptual explanation of the notion that the gravitational attraction between two masses is a unitary force that's not the sum of discrete gravitational forces* generated by each.

pj
chgo

*I'm familiar with the General Relativity concept of gravity as spacetime distortion, so I may be taking liberties with the word "force" here.

 

[mikepage]

I'd like to understand it better. Can you point me to something?

Thanks,

pj
chgo

I only skimmed this thread, but I think you were comparing the Moon-Patrick force to the Earth-Patrick force, and stating the former is less because the mass of the moon is less.

I think Dave is referring (and thinks you are referring?) only to the Moon-Patrick force. Some people say, incorrectly, that the Moon pulls pretty hard on Patrick because it is pretty massive--while Patrick pulls back only weakly on the Moon. This error (that I don't know whether anybody made here) is one I referred to many moons ago in this letter I had in the Washington Post.



A Weighty Physics Problem
[FINAL Edition]
The Washington Post - Washington, D.C.
Date: Nov 20, 1990
Start Page: a.22
Section: OP/ED
Text Word Count: 294

Three cheers for the new column "Why Things Are" {Style Plus, Nov. 2}. While I enjoyed most of the questions and answers, particularly the two on physics, I am compelled to note that one statement violates Newton's third law of motion.

Why do objects fall at the same rate toward the Earth regardless of their weight? Why does Marlon Brando fall at the same rate as a paper clip when dropped from the Empire State Building? It is because "heaviness is a two-sided coin. As you get heavier, gravity pulls harder, but it is also that much harder to budge you. So weight doesn't make you fall faster or slower." Bravo! Well done. The column should have stopped there.

Instead, it goes on to point out an ever-so-slight complication: because these objects pull on the Earth as they fall, the Earth comes up to meet them, and because Mr. Brando pulls harder on the Earth than a paper clip does, the Earth comes a little faster and a little farther toward him and, lo and behold, Mr. Brando wins, providing the experiments are done one at a time.

All right, but when the column then says that Mr. Brando exerts only an "infinitesimally slight pull" on the Earth, it destroys a fundamental and beautiful symmetry in nature: that forces come in equal and opposite pairs; that every action has an equal and opposite reaction; that no matter how hard you try, you can't lift yourself and the chair you're sitting in by pulling on the sides of the chair. Indeed, by using a bathroom scale and Newton's third law, we can easily measure the strength of Brando's gravitational pull on the Earth. It isn't infinitesimal; it's about 300 pounds!
MICHAEL PAGE Manassas

 

[Cameron Smith]

A small technical nit.... It's peak velocity you want, not peak acceleration. Velocity and acceleration are very different in physics, but many people confuse the two. If you hit at peak velocity, you have the most efficient and powerful stroke and likely the most consistent, as well.

Your absolutely right. I got them mixed up. I don't know much about physics, and even less about them in relation to anything other than pool.

 

[JoeyA]

with your grip hand forward, you are pushing the cueball not stroking it.
Players, insecure with their stroke, do this trying to guide or control the cueball more. It is called overcompensation.

Bet I could watch you shoot and tell you what shots that are a problem for you.

I like what you write about pool. I think you really know how to play pool. Hope one day to meet you and have my mettle tested.

FWIW, I have seen MANY professional pool players move their hand forward on "cinch shots" and just push the cue ball instead of stroking it. This forward movement of the grip hand cuts down on the speed of the cue ball as JAL points out in one of his posts. This reduced speed is a good thing especially on tight pocket tables and "cinch shots" IMO.

JoeyA

 

[crosseyedjoe]

Any introductory college physics book covers this. It's called the universal law of gravitation (another one of Newton's contributions to the world). Here's the equation:

F = G * m1 * m2 / d^2
​

where:

F = equal and opposite force acting between the bodies
G = universal gravitation constant
m1 = mass of body 1
m2 = mass of body 2
d = distance between the CGs of the two bodies
​

On the surface of the earth:

g = G*M / d^2​

where:

g = "acceleration due to gravity"
M = mass of the earth
d = average radius of the earth​

To calculate the force acting between the earth and an object (e.g., you) sitting on the surface of the earth:

W = m*g​

where:

W = "weight" of the object
m = mass of the object​

I hope you find this useful.

Regards,
Dave

PS: ... and this is why the forearm should be perpendicular to the cue at CB impact!!!

Dr. Dave,

I think he is asking if you can express this equation

F = G * m1 * m2 / d^2

to something like this;

F = G * m1 * m2 / d^2 = m1g1(moon's force experience at any point equal to the distance of Earth and moon) + m2g2(Earth's force experience at any point equal to the distance of Earth and moon)

 

[Patrick Johnson]

Some people say, incorrectly, that the Moon pulls pretty hard on Patrick because it is pretty massive--while Patrick pulls back only weakly on the Moon. This error (that I don't know whether anybody made here) is one I referred to many moons ago in this letter I had in the Washington Post.

So...

Every time I fall onto the surface of the Earth from 20 feet up (I hate that!) the Earth also falls onto the surface of me. I move 20 feet times the ratio of the Earth's mass to my mass (very close to every bit of 20 feet), the Earth moves 20 feet times the ratio of my mass to its mass (very close to none of 20 feet), we each attain speeds in the same ratio, and the end result is that we slam into each other with the same force (but I'm always the one that goes to the hospital).

Is that what this equal force thing means?

pj
chgo

 

[ShootingArts]

Aha!

Thanks a bunch PJ, one more excuse to add to my arsenal.

"That danged PJ fell again and jiggled the earth just enough my ball jawed the pocket!"

Spoons is sitting somewhere with a bemused and befuddled look on his face, "I just asked a simple question about pool, that is really all I did . . . '

Hu

So...

Every time I fall onto the surface of the Earth from 20 feet up (I hate that!) the Earth also falls onto the surface of me. I move 20 feet times the ratio of the Earth's mass to my mass (very close to every bit of 20 feet), the Earth moves 20 feet times the ratio of my mass to its mass (very close to none of 20 feet), we each attain speeds in the same ratio, and the end result is that we slam into each other with the same force (but I'm always the one that goes to the hospital).

Is that what this equal force thing means?

pj
chgo

 

[arcticmonkey]

Bridge hand placement

I spent years trying to get my arm vertical in this plane. It was largely a waste of time. I would have gotten 10 times the benefit from paying attention to where I was placing my bridge hand I believe. Colin

Colin, I read the entire thread and although I got a fuzzy head from the physics discussion I was disappointed not to see more discussion about your statement above.

I would be very interested to learn more about how you went about correcting your bridge hand placement. Are there threads out there discussing that very topic?

Thanks

arctic

 

[softshot]

Muscle acceleration is tricky because it is under our control if we work at it. My normal tendency is to want to get up to the speed I want to hit the cue ball quickly and basically maintain that speed. However when I work on starting my forward stroke slowly and increasing that speed much more gradually I seem to miss fewer shots. One of the things I need to practice and do practice now and then. I think it should be an ingrained part of my game but it isn't yet.

Hu

I have no proof but my guess would be that if you could somehow pinpoint acceleration and track it through the stroke... and graph it... you would see two distinct outcomes..

one graph is a 45 degree angle ....up until impact.. the right way... the accelerating stroke.. and most decent players once taught could repeat that... semi constantly...


if you graph the player who reaches peak acceleration.. and then attempts to control velocity... you will find a graph that looks more like an EKG graph of a heartbeat.. accelerate stop.. accelerate stop...as the muscle attempts to maintain a velocity.... you are asking a muscle to do something its not designed to do and getting inconsistent results...

when you ask a muscle to do what it does... you get consistent results...

JMO

 

[Patrick Johnson]

if you graph the player who reaches peak acceleration.. and then attempts to control velocity... you will find a graph that looks more like an EKG graph of a heartbeat.. accelerate stop.. accelerate stop...as the muscle attempts to maintain a velocity...

I doubt that anybody does this or imagines doing it.

We control acceleration by trying to do it smoothly and evenly throughout the stroke. We control velocity by deciding how fast we want the stick to be moving at the end of that smooth, even acceleration and scaling the whole thing up or down accordingly. It's not either/or. We do both.

The distinction between acceleration and velocity doesn't come into the action; that's just a topic for academic discussion.

pj
chgo

 

[mikepage]

So...

Every time I fall onto the surface of the Earth from 20 feet up (I hate that!) the Earth also falls onto the surface of me. I move 20 feet times the ratio of the Earth's mass to my mass (very close to every bit of 20 feet), the Earth moves 20 feet times the ratio of my mass to its mass (very close to none of 20 feet), we each attain speeds in the same ratio, and the end result is that we slam into each other with the same force (but I'm always the one that goes to the hospital).

Is that what this equal force thing means?

pj
chgo

Bingo......