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Andrew Manning
10-24-2008, 09:20 AM
A question for those brave enough to attempt the physics, even though I know there are too many variables to arrive at a 100% reliable answer:

I have a standard maple hardwood shaft with a tip diameter of 13.25mm. How much will the squirt, or more relevantly the pivot point, change if I have it turned down to 12.5mm? I'm looking for a quantitative answer, so spare me any answers that consist of "a lot" or "not much".

If it helps, assume the pivot point is currently at about 9.5 inches.

-Andrew

Black-Balled
10-24-2008, 10:30 AM
How about...

That is way too much change. I would'nt go 1/2 that far and would advise you to first goto hi 12mm, then another cut (or even 2) months later.

Thinking (zzzzzzzzzzzzzzz)...you were threatened with snooker recently. Is that your impetus here?

ftgokie
10-24-2008, 10:57 AM
I bet there will be lots of good replies to this thread...I will be checking to see what wonderful answers are posted

Andrew Manning
10-24-2008, 11:08 AM
How about...

That is way too much change. I would'nt go 1/2 that far and would advise you to first goto hi 12mm, then another cut (or even 2) months later.

Thinking (zzzzzzzzzzzzzzz)...you were threatened with snooker recently. Is that your impetus here?

Actually, I failed to follow up on the threatened snooker outing. Thanks for the reminder, I'll try to make that happen.

But for the cue, I think you misunderstand my intentions. I have a Joss plain-jane cue, and a predator 314 shaft for it. For the past 4 years, I have exclusively shot with the 314. I haven't used the original shaft on the cue except to experiment with it now and then since the first two months I had the cue.

When I do use the original shaft, the tip seems so large that I feel like I'm shooting with a broom handle, and the difference in squirt between the Joss and predator shafts is night and day.

Lately, however, I've been a little dissatisfied with the low-deflection shaft, because when I'm using follow and english, the swerve on the shot more than counteracts the squirt; when using high right english, I actually have to compensate my aim to the left a little, and it always proves very difficult to judge.

But last night I couldn't play with the predator because the 3-year-old talisman pro tip finally popped off it, and so I played for several hours with a one-piece house cue with about a 12.5mm tip diameter, and I was playing really well (for me) with it, using back-hand english (aim and pivot).

So that made me want to try turning down the Joss shaft to a diameter I'm comfortable with, and actually playing with it instead of letting it slowly fossilize in my case.

So does anyone know, quantitatively, how much the squirt would change?

-Andrew

Patrick Johnson
10-24-2008, 11:14 AM
A question for those brave enough to attempt the physics, even though I know there are too many variables to arrive at a 100% reliable answer:

I have a standard maple hardwood shaft with a tip diameter of 13.25mm. How much will the squirt, or more relevantly the pivot point, change if I have it turned down to 12.5mm? I'm looking for a quantitative answer, so spare me any answers that consist of "a lot" or "not much".

If it helps, assume the pivot point is currently at about 9.5 inches.

-Andrew

Assuming the first 8 inches or so are cylindrical (a pro taper) and remain cylindrical after being turned down, the percentage of "end mass" removed is:

((pi x (13.25/2)^2) - (pi x (12.5/2)^2)) / (pi x (13.25/2)^2) = 11.0%

Maybe that's also the amount of squirt reduction...?

If you increase the taper (reduce the diameter at the tip more than you reduce it at 8 inches from the tip), you should be able to get squirt reduction without sacrificing much stiffness (you might even increase stiffness). For reducing squirt, I think the amount of material removed is most important right at the tip and less important as you get farther from the tip (even within the 8-inch "end mass" area).

pj
chgo

ftgokie
10-24-2008, 11:19 AM
Assuming the first 8 inches or so are cylindrical (a pro taper) and remain cylindrical after being turned down, the percentage of "end mass" removed is:

((pi x (13.25/2)^2) - (pi x (12.5/2)^2)) / (pi x (13.25/2)^2) = 11.0%

Maybe that's also the amount of squirt reduction...?

If you increase the taper (reduce the diameter at the tip more than you reduce it at 8 inches from the tip), you should be able to get squirt reduction without sacrificing much stiffness (you might even increase stiffness). For reducing squirt, I think the amount of material removed is most important right at the tip and less important as you get farther from the tip (even within the 8-inch "end mass" area).

pj
chgo

WOW, Pj beat me to that answer:thumbup:

Andrew Manning
10-24-2008, 11:28 AM
Assuming the first 8 inches or so are cylindrical (a pro taper) and remain cylindrical after being turned down, the percentage of "end mass" removed is:

((pi x (13.25/2)^2) - (pi x (12.5/2)^2)) / (pi x (13.25/2)^2) = 11.0%

Maybe that's also the amount of squirt reduction...?

If you increase the taper (reduce the diameter at the tip more than you reduce it at 8 inches from the tip), you should be able to get squirt reduction without sacrificing much stiffness (you might even increase stiffness). For reducing squirt, I think the amount of material removed is most important right at the tip and less important as you get farther from the tip (even within the 8-inch "end mass" area).

pj
chgo

So would squirt be linearly proportional to end-mass? I would guess the relevant property is impulse, which would imply that the relationship is linear.

And is pivot point proportional to the inverse of the squirt?

-Andrew

Patrick Johnson
10-24-2008, 12:50 PM
So would squirt be linearly proportional to end-mass?

I don't know. My guess assumes so.

And is pivot point proportional to the inverse of the squirt?

Yes, if I understand your question. That's simple geometry.

pj
chgo

Shaft
10-24-2008, 01:41 PM
It is impossible to give an accurate, quantitative answer --- even with much more information, such as the specifics of the intended taper.

The modulus of elasticity of your particular shaft is an important factor and that is unknowable without testing. Every shaft made of natural materials will be different.

This may sound counter-intuitive to some of you but, based on the vectors I have sketched out, I believe the pivot point will get longer if you turn down your shaft. (I could be wrong.) How much longer is impossible to calculate.

This assumes the same stroke force, the same tip-to-ball friction and the same side spin moment arm.

The easiest way to get your answer is to turn the shaft and re-calibrate your pivot point. A few minutes time at the table is all you need.

Patrick Johnson
10-24-2008, 04:04 PM
This may sound counter-intuitive to some of you but, based on the vectors I have sketched out, I believe the pivot point will get longer if you turn down your shaft.

I don't know why you think it would be counter-intuitive. Pretty much everybody agrees with this. Less end mass = less squirt = longer pivot point.

I bet it's also fairly easy to calculate, although I'm not sure I know how.

pj
chgo

Bob Jewett
10-24-2008, 04:18 PM
So would squirt be linearly proportional to end-mass? I would guess the relevant property is impulse, which would imply that the relationship is linear.

And is pivot point proportional to the inverse of the squirt?

-Andrew
I think Dr. Dave has done an experiment with added mass and you could estimate the result for reduced mass. The write-up should be on his web site. Mike Page has also described an added-mass experiment.

A different experiment that you could try is to remove the ferrule and just have a fiber pad between the wood and the tip. Most ferrules are denser than wood, so removing it will make the front of your stick lighter by quite a bit. Unfortunately, it's hard to get the length back if you change your mind.

From the data I've seen, the relationships don't seem to be linear, but that could be due to the measurements being polluted by swerve and such. The simple theory says that the squirt angle should be proportional to the end mass to first order.

Besides Dr. Dave's site, you may want to look at Ron Shepard's paper on squirt at http://www.sfbilliards.com/Shepard_squirt.pdf

mikepage
10-24-2008, 04:27 PM
One test that doesn't destroy your shaft in the process is to ADD the same amount of mass in the same places that you would like to REMOVE--say with tape of some sort. Then assume (reasonable for small changes) that magnitude of the squirt change would be the same in both directions.

A question for those brave enough to attempt the physics, even though I know there are too many variables to arrive at a 100% reliable answer:

I have a standard maple hardwood shaft with a tip diameter of 13.25mm. How much will the squirt, or more relevantly the pivot point, change if I have it turned down to 12.5mm? I'm looking for a quantitative answer, so spare me any answers that consist of "a lot" or "not much".

If it helps, assume the pivot point is currently at about 9.5 inches.

-Andrew

Bigkahuna
10-24-2008, 05:18 PM
One test that doesn't destroy your shaft in the process is to ADD the same amount of mass in the same places that you would like to REMOVE--say with tape of some sort. Then assume (reasonable for small changes) that magnitude of the squirt change would be the same in both directions.

I don't know if he still does but Mike AKA Fargo on youtube.com had a video explaining squirt. It is pretty good and Mike, hats off, funny as heck. You should probably start off with "you should not do this at home"!

GordonRamsay
10-24-2008, 05:51 PM
if your shaft is squirting that much.... i'd go see a doctor right away...:D

Shaft
10-24-2008, 07:52 PM
I agree with Bob Jewett and others that the "added mass" experiment is the most practical way. This will yield an approximate result that may be "close enough" for the owner to make a decision on whether or not to reduce the shaft.

This would not be "simple to calculate," except by using simplified empirical formulas derived from testing the shaft itself, and even that would yield an approximate answer. The mass is not being removed from a specific point, but in an ???undefined??? way along the last ??? inches of the tip.