Secrets!

[Jal]

Hey Jim,

No, I wasn't really trying to get into that. I agreed with PJ's analogy about a barrell rolling on its edge. My contention :smile: is that an object ball will pick up a small amount of speed from the extreme side spin of the cue ball on a thin cut.

Thanks for the clarification, Mike.

For what it's worth, I don't there's much doubt that, all else being equal, the throw component adds to an object ball's speed, but extremely little.

Let's say you execute a cut shot with just the right amount of outside english to eliminate throw (or maybe put some spit on the contact area). The object ball takes off with a certain speed. You then do the same shot again, but change the amount of outside english so that there is some throw. As Patrick pointed out, the friction acts perpendicular to the direction of the object ball in the first shot. Nevertheless, it now has an additional velocity component which adds, vectorally (Pythagorean addition in this case) to its total velocity. That is, it'll have more speed along its new direction of travel. We're assuming the original forward speed component (the one in the first shot) isn't affected by the friction. This is probably near enough true, but it is an assumption. As indicated earlier, though, it's next to nothing. With 5 degrees of throw, it works out to about a 0.3% increase.

However, if you produced the throw by increasing the outside english, then compensated by hitting the OB thicker to drive it in the same direction as the first shot, then its speed will also be increased because of the thicker hit, as a few posters have pointed out. This particular addition to its velocity is greater at larger cut angles for the same throw compensation (it's almost nothing at very modest cut angles).

This small transfer of speed is possibly proportional to the decreasing amount of force from the impulse of the cue ball and object ball. I seem to see it most when I cut balls 80+ degrees.

In a sense, yes. Let's say you cut a ball 83 degrees without any throw (or very little). Then you repeat the shot with lots of outside english, attempting to send the OB in the same direction. And say you expect about 3 degrees of throw (and possibly get it), so you hit it thicker, equivalent to an 80-degree cut. Because of that second addition to the OB's speed mentioned above, its speed will be increased by a whopping 40% (given the same cueball speed, of course)! Since travel distance is proportional to the square of a ball's speed, it should go about twice as far (1.4 x 1.4 = 1.96). The throw itself contributes virtually nil, as per above.

Comparing this to a 30-degree versus a 33-degree cut, the increase is a mere 3%. The moral is that OB speed is extremely sensitive to changes in contact point at really severe cuts.

I started by saying "in a sense," because it's actually 'how' the impulse drops off with cut angle (cosine of the cut angle), not merely that it drops off.

Jim

 

[mrmusicmr]

With the same setup on the second shot, I used extreme right center spin and lagged the one ball over to the rail again. I hit the same contact point and the one ball travelled the same distance as on the previous shot. Notice the difference of cue ball travel on the second shot.

Best,
Mike

So, If you hit with alot of Right spin, why does the cueball come off the cushion to the left, instead of back to the line it came from or cuts further to the right?
Did the object ball have any change in it's direction with the same thin cut as was previously with the centre ball hit?
Thanks, Neil[/QUOTE]
I'm curious about that too.

 

[Patrick Johnson]

Swerve adds to the cue ball direction.

I doubt that swerve could make nearly so much difference, but we can't really know from this test.

The difference in contact points is so small, it was insignificant. I contacted the extreme edge of the object ball on both shots.

How can you possibly know this? Here's what we can know: if the sidespin's effect on the OB is enough to (theoretically) affect the OB's speed, then it's enough to affect the OB's direction, so in order to send the OB on the same path you must hit a different contact point.

Good for you that you're trying to test your theory, but the critical variables in the tests you've shown (exact contact point, CB speed and amount of spin) are impossible to control precisely, so the tests can't really tell us anything. I haven't thought of any more definitive test, so I guess we're stuck with theory and conjecture.

pj
chgo

 

[Patrick Johnson]

jal:
Let's say you execute a cut shot with just the right amount of outside english to eliminate throw (or maybe put some spit on the contact area). The object ball takes off with a certain speed. You then do the same shot again, but change the amount of outside english so that there is some throw. As Patrick pointed out, the friction acts perpendicular to the direction of the object ball in the first shot. Nevertheless, it now has an additional velocity component which adds, vectorally (Pythagorean addition in this case) to its total velocity.

Does "Pythagorean addition" mean the OB travels the distance of a right triangle's leg without sidespin, but the distance of its hypotenuse with sidespin?

pj
chg

 

[Mikjary]

With the same setup on the second shot, I used extreme right center spin and lagged the one ball over to the rail again. I hit the same contact point and the one ball travelled the same distance as on the previous shot. Notice the difference of cue ball travel on the second shot.

Best,
Mike

So, If you hit with alot of Right spin, why does the cueball come off the cushion to the left, instead of back to the line it came from or cuts further to the right?
Did the object ball have any change in it's direction with the same thin cut as was previously with the centre ball hit?
Thanks, Neil[/QUOTE]

You're right. I didn't diagram the second shot as well as I should have. The cue ball came straight back off of the long rail because of the spin. The object ball cut slightly thinner. I will edit the cue ball track. The object ball track shows more cut, by a couple of inches. Hard to show on a diagram.

Best,
Mike

 

[Mikjary]

Thanks for the clarification, Mike.

For what it's worth, I don't there's much doubt that, all else being equal, the throw component adds to an object ball's speed, but extremely little.

Let's say you execute a cut shot with just the right amount of outside english to eliminate throw (or maybe put some spit on the contact area). The object ball takes off with a certain speed. You then do the same shot again, but change the amount of outside english so that there is some throw. As Patrick pointed out, the friction acts perpendicular to the direction of the object ball in the first shot. Nevertheless, it now has an additional velocity component which adds, vectorally (Pythagorean addition in this case) to its total velocity. That is, it'll have more speed along its new direction of travel. We're assuming the original forward speed component (the one in the first shot) isn't affected by the friction. This is probably near enough true, but it is an assumption. As indicated earlier, though, it's next to nothing. With 5 degrees of throw, it works out to about a 0.3% increase.

However, if you produced the throw by increasing the outside english, then compensated by hitting the OB thicker to drive it in the same direction as the first shot, then its speed will also be increased because of the thicker hit, as a few posters have pointed out. This particular addition to its velocity is greater at larger cut angles for the same throw compensation (it's almost nothing at very modest cut angles).


In a sense, yes. Let's say you cut a ball 83 degrees without any throw (or very little). Then you repeat the shot with lots of outside english, attempting to send the OB in the same direction. And say you expect about 3 degrees of throw (and possibly get it), so you hit it thicker, equivalent to an 80-degree cut. Because of that second addition to the OB's speed mentioned above, its speed will be increased by a whopping 40% (given the same cueball speed, of course)! Since travel distance is proportional to the square of a ball's speed, it should go about twice as far (1.4 x 1.4 = 1.96). The throw itself contributes virtually nil, as per above.

Comparing this to a 30-degree versus a 33-degree cut, the increase is a mere 3%. The moral is that OB speed is extremely sensitive to changes in contact point at really severe cuts.

I started by saying "in a sense," because it's actually 'how' the impulse drops off with cut angle (cosine of the cut angle), not merely that it drops off.

Jim

Thanks, Jim. I will look at this again and come up with more questions. :grin:

Best,
Mike

I sent some green your way! Great post!

 

[Jimbojim]

here is another interesting shot. Little quizz


You are on the short rail CB and OB are 1/8' away from the rail and your dead straight in. Without going rail-first and you can't cheat the pocket, how do you make the CB deflect on the side with major spin to bring back to the other end of the table....if someone can post a cuetable layout in picture format it would be great.

 

[Mikjary]

Swerve adds to the cue ball direction.

I doubt that swerve could make nearly so much difference, but we can't really know from this test.

How can you possibly know this? Here's what we can know: if the sidespin's effect on the OB is enough to (theoretically) affect the OB's speed, then it's enough to affect the OB's direction, so in order to send the OB on the same path you must hit a different contact point.

Good for you that you're trying to test your theory, but the critical variables in the tests you've shown (exact contact point, CB speed and amount of spin) are impossible to control precisely, so the tests can't really tell us anything. I haven't thought of any more definitive test, so I guess we're stuck with theory and conjecture.

pj
chgo

I know this is a drawing of something from a guy on a pool table, but I've discussed this before this thread with notable players and they had no answer. It is an assumption from the results and those results are interesting. I've been trying to come up with a solid test, but it's hard with just a diagram of a table.

This stroke adds a new dimension to your game in cue ball speed control. That is why I posted it originally here. This fact is starting to get lost in all the research. All comments are welcome.

Best,
Mike

 

[dr_dave]

Thanks for the clarification, Mike.

For what it's worth, I don't there's much doubt that, all else being equal, the throw component adds to an object ball's speed, but extremely little.

Let's say you execute a cut shot with just the right amount of outside english to eliminate throw (or maybe put some spit on the contact area). The object ball takes off with a certain speed. You then do the same shot again, but change the amount of outside english so that there is some throw. As Patrick pointed out, the friction acts perpendicular to the direction of the object ball in the first shot. Nevertheless, it now has an additional velocity component which adds, vectorally (Pythagorean addition in this case) to its total velocity. That is, it'll have more speed along its new direction of travel. We're assuming the original forward speed component (the one in the first shot) isn't affected by the friction. This is probably near enough true, but it is an assumption. As indicated earlier, though, it's next to nothing. With 5 degrees of throw, it works out to about a 0.3% increase.

However, if you produced the throw by increasing the outside english, then compensated by hitting the OB thicker to drive it in the same direction as the first shot, then its speed will also be increased because of the thicker hit, as a few posters have pointed out. This particular addition to its velocity is greater at larger cut angles for the same throw compensation (it's almost nothing at very modest cut angles).


In a sense, yes. Let's say you cut a ball 83 degrees without any throw (or very little). Then you repeat the shot with lots of outside english, attempting to send the OB in the same direction. And say you expect about 3 degrees of throw (and possibly get it), so you hit it thicker, equivalent to an 80-degree cut. Because of that second addition to the OB's speed mentioned above, its speed will be increased by a whopping 40% (given the same cueball speed, of course)! Since travel distance is proportional to the square of a ball's speed, it should go about twice as far (1.4 x 1.4 = 1.96). The throw itself contributes virtually nil, as per above.

Comparing this to a 30-degree versus a 33-degree cut, the increase is a mere 3%. The moral is that OB speed is extremely sensitive to changes in contact point at really severe cuts.

I started by saying "in a sense," because it's actually 'how' the impulse drops off with cut angle (cosine of the cut angle), not merely that it drops off.

Jim

Excellent post, Jim! Well thought out, and well written.

Regards,
Dave

 

[Patrick Johnson]

This stroke adds a new dimension to your game in cue ball speed control. That is why I posted it originally here. This fact is starting to get lost in all the research. All comments are welcome.

Best,
Mike

Let's get back to it. What stroke? How is it performed differently from other similar strokes? How does the result compare with other similar strokes?

pj
chgo

 

[Jal]

Does "Pythagorean addition" mean the OB travels the distance of a right triangle's leg without sidespin, but the distance of its hypotenuse with sidespin?

pj
chg

Maybe.

I mean I'm not sure whether by "sidespin" you're referring to the throw component of velocity added to the OB by the sidepin on the CB, or the induced spin on the OB itself. What I meant was, if Vn is the OB's speed in the forward direction (line of centers direction), and Vt is the throw component in the perpendicular direction (tangent line direction), per your earlier diagram, then its total speed is the square root of the sum of the squares of Vn and Vt:

V = Sqrt(Vn^2 + Vt^2) = Sqrt(Vn X Vn + Vt X Vt)

And yes, Vn and Vt are the legs of a right triangle where V is the hypotenuse!

The induced spin on the OB is irrelevant as far as its initial speed after the collision is concerned. It may be irrelevant to total travel distance as well (i.e., how fast it loses speed), but I can't say I'm really sure about that.

Jim

 

[Jal]

Excellent post, Jim! Well thought out, and well written.

Regards,
Dave

Thanks much, Dr. Dave.

Jim

 

[duckie]

The distance between the CB and OB is also a factor, actually it is the major factor.

The closer the CB and OB are, the more speed control you have of the shot. And speed control is the real secret to shot making period.

 

[JoeyA]

I think another factor here is that it's so difficult to describe verbally things that you learn in a physical activity such as pool billiards. There are things that you could say, but you realize that they might sound banal and trivial when said out loud.

I think a good example is the concept of relaxed arm. I think it's safe to say that most people agree that it's best to keep arm relaxed when shooting. But term itself is pretty much meaningless. If your hand is completely relaxed, the cue stick will fall to ground. Of course, saying that the hand has to be relaxed, can give some guidance to a newcomer, but eventually everyone has to find just the right level of relaxedness themselves (or, everyone's brain has to figure that out).

The deeper you go, the more difficult it becomes to verbally describe the things you've learned.

I agree that it is very difficult to verbally explain not only the concept of a relaxed arm but many other things. I spend lots of time with a friend who does not have much time to play pool and I attempt to articulate what I see and feel. It is easy for me to explain myself in words but my explanation sometimes falls short.

Besides my own failings to communicate my thoughts, it is possible that the other person must be or must have experienced some of what you are discussing. If they haven't experienced or aren't experiencing what you are discussing, you can only hope they will remember what you attempted to communicate when they do cross that particular bridge.

 

[(((Satori)))]

I believe there are two kinds of secrets out there: some are shot specific (like a 1pocket bank Freddy the Beard showed me when I took a lesson with him and made my peepers open up wide). And then there are those secrets that, while they might have value to others, are more idiosyncratic to the player. These secrets revolve around their PSR and being able to always produce an accurate stroke, day-in-day-out, for all kinds of shots, and under all kinds of pressure. These would be the toughest to pry loose

Lou Figueroa

Once again, Lou produces a great post.

I'll add that I doubt there are any secrets that are hidden from the recreational player. The information on how to play this game and at a high level can be found. The real secret is practice and utilizing what clicks with you.

 

[JoeyA]

Practice is GREAT but..............

The real secret is practice and utilizing what clicks with you.

The "REAL SECRET" is being exposed to what you don't know.

If you think you can just practice and wait for something to click for you, you will be waiting a LONG, LONG time to reach a high level of proficiency.

 

[(((Satori)))]

The "REAL SECRET" is being exposed to what you don't know.

If you think you can just practice and wait for something to click for you, you will be waiting a LONG, LONG time to reach a high level of proficiency.

I agree.

I just think that the information can be found if you seek it and it will not come in the form of something magical that will instantly raise your game to a whole new level.


Although I have had moments when something clicked and the game seemed easier all of a sudden, I wouldn't say that I had discovered a secret, more like the questioning of other players, the studying, and the practice finally paid off.



Here is an example of questioning other players. The questioning process usually goes like this.
Me: I'm having trouble doing________ recently can you check this out and see what you think?


Player: Try doing ____________. or You are doing ______________.


The answer is not something I have never heard of or some piece of knowledge that could not be found by a recreational player. Just something I needed reminded of or perhaps more awareness of what it meant to me and my game.

I think all to often people look for a majic bullet that the pros use, a secret aiming system or something like that, and imo there is no such thing.

 

[keilperry]

Heres a Secret

Top Player = Have the opportunity to learn/practice/compete for several hours a day +
Have the desire to learn/practice/compete for several hours a day

There is no secret to becoming a top player. Its Just HARD because it takes time and effort.

"Every good worth possessing must be paid for in strokes of daily effort." - William James

 

[prewarhero]

Top Player = Have the opportunity to learn/practice/compete for several hours a day +
Have the desire to learn/practice/compete for several hours a day

There is no secret to becoming a top player. Its Just HARD because it takes time and effort.

"Every good worth possessing must be paid for in strokes of daily effort." - William James

practice is key, but there are secrets. I am not an A player but I do know a couple of "secrets" I have learned. Some of these secrets have been put on video over the years and some are more along the lines of "tips". TBS, these tips are necessary to learn and no amount of practice would overcome not knowing these tips. You would just keeping practicing knowing the same stuff you already new. Same reason they don't have you reread the same textbook ever year at school.

 

[JoeyA]

practice is key, but there are secrets. I am not an A player but I do know a couple of "secrets" I have learned. Some of these secrets have been put on video over the years and some are more along the lines of "tips". TBS, these tips are necessary to learn and no amount of practice would overcome not knowing these tips. You would just keeping practicing knowing the same stuff you already new. Same reason they don't have you reread the same textbook ever year at school.

Exactly!

I think some people get all fidgety when someone talks about "SECRETS" in pool. Utilizing the word "SECRETS" is just one way to garner someone's attention so that you can share the information that has so far, been "hidden" from them because they haven't been exposed to it.

Everyone, well almost everyone enjoys learning a new SECRET every now and then.

I don't want to waste my time with those that wail and whine that the information isn't a secret and that the information is readily available here or there.

I enjoy sharing the secrets of the professional players when I do commentary and I have to believe that they too, enjoy knowing that someone out there enjoys learning what they believe is information that is not common knowledge.

Sharing that knowledge is fun and if you don't know the information then I am going to call it a SECRET and share it with you. :smile:

JoeyA