Seems to me unless there's a miscue the no-spin path should be in the direction the tip is moving...?...the no-spin path is in the direction of the cue's long-axis.
pj
chgo
Seems to me unless there's a miscue the no-spin path should be in the direction the tip is moving...?...the no-spin path is in the direction of the cue's long-axis.
The pics below show the same cue hitting the CB on the same spot while traveling in the same direction.
The cue on the left is stroked normally, straight ahead parallel with the CB's no-squirt path (the dotted white arrow).
The cue on the right is also stroked parallel with the CB's no-squirt path, but with the cue angled as shown (don't ask me how to do that; this is a thought experiment).
Which one produces more squirt (compared with the no-squirt CB path)?
pj <- not a quiz - I don't know the answer
chgo
View attachment 63025
Differs from shaft shaft end weight will vary squirt amount
The pics below show the same cue hitting the CB on the same spot while traveling in the same direction.
The best answer I can come up with - and I'm not completely happy with it myself - is that in the transverse direction, the cue acts like a very weak spring. With a weak spring, it takes much more deformation of some sort (compression, tension, bending...) to cause the spring to resist the deformation by generating a certain specific amount of force, as compared to a strong spring. With a cue, there's a large difference in its transverse versus longitudinal "springiness."Seems to me unless there's a miscue the no-spin path should be in the direction the tip is moving...?
pj
chgo
The pics below show the same cue hitting the CB on the same spot while traveling in the same direction.
The cue on the left is stroked normally, straight ahead parallel with the CB's no-squirt path (the dotted white arrow).
The cue on the right is also stroked parallel with the CB's no-squirt path, but with the cue angled as shown (don't ask me how to do that; this is a thought experiment).
Which one produces more squirt?
pj <- not a quiz - I don't know the answer
chgo
View attachment 63025
PJ,
I'm trying to simulate a swoop stroke at the same angle of attack as a straight stroke, to explore how the squirt for both might affect the outcome. [NOTE: I've replaced the pic of the angled cue with a more realistic one that has a smaller angle.]I wouldn't call the CB motion resulting from the 2nd hit "squirt" or "cue ball deflection." I would call it a hit in a totally different direction with a simulated swoop or swipe, which isn't as big of a possible effect as some people might think.
The white arrows are the intended no-spin path of the CB (I edited the OP to be clear about that), so it's assumed that the CB will actually go somewhere to the right of that - will it go different directions each way? Why?
My assumption is that the angled cue moving as shown above is a fair simulation of a swoop stroke at the same point-of-contact. Do you agree?
It is when viewed parallel with the cue.If the intended path of CB is the dotted line then the angled cue should be coming across the CCB point. For your pictured shot the angled cue should be coming from the right side into the point you are hitting.
Yeah, as I said in the OP.Looks impossible to move an angled cue that way if one had an anchored bridge hand.
So I was wrong when I said in the swoop thread that the stick's orientation at contact is mostly irrelevant to the CB's reaction...?The best answer I can come up with - and I'm not completely happy with it myself - is that in the transverse direction, the cue acts like a very weak spring. With a weak spring, it takes much more deformation of some sort (compression, tension, bending...) to cause the spring to resist the deformation by generating a certain specific amount of force, as compared to a strong spring. With a cue, there's a large difference in its transverse versus longitudinal "springiness."
Thus for something traveling due North that encounters something that opposes its motion, the net force on that obstruction will be the vector sum of the reaction forces.
View attachment 409184
Jim
I would also add the following:The best answer I can come up with - and I'm not completely happy with it myself - is that in the transverse direction, the cue acts like a very weak spring. With a weak spring, it takes much more deformation of some sort (compression, tension, bending...) to cause the spring to resist the deformation by generating a certain specific amount of force, as compared to a strong spring. With a cue, there's a large difference in its transverse versus longitudinal "springiness."
Thus for something traveling due North that encounters something that opposes its motion, the net force on that obstruction will be the vector sum of the reaction forces.
View attachment 409184
Jim
I think this concept finally brought the dynamics home for me. Thanks again!Creating a sideways force on the CB with the tip requires friction, and friction requires "normal" force (in the direction of the cue, perpendicular to the CB surface).
If the intended path of CB is the dotted line then the angled cue should be coming across the CCB point. For your pictured shot the angled cue should be coming from the right side into the point you are hitting.
It is when viewed parallel with the cue.
pj
chgo
Thanks. FYI, much more info (and additional demonstrations) can be found on the stroke swoop resource page.Nice experiment on swoop stroke Dr Dave! I just saw the video you made linked in your post above.
agree........... you can dissect pool to the Nth degree using formulas and tests by robots.............
in the end, it won't help.... you still have to shoot the shot yourself............
Kim
If the intended path of CB is the dotted line then the angled cue should be coming across the CCB point. For your pictured shot the angled cue should be coming from the right side into the point you are hitting.
It is when viewed parallel with the cue.
pj
chgo
I mean when viewed like this.Guess i'm missing something, can you elaborate farther.
So you're saying that if you want to swoop, the tip should follow a sinusoidal-like path on the way to the CB. Is that the moral?![]()
Thank you kindly.
Just a small token payment for all the stuff I've learned from you....Thanks, as always, for your patient guidance, Jim and Dave!