I’ve found the opposite to be true for me, from 8 to straight and one pocket. I find I get the best action out of about an 18oz. a 21 just feels too weighty for me.
Do you still play?
That’s a pretty interesting thoughts. I do believe though that with the carom and Pyramid games that much heavier cues aren’t uncommon, but that’s really not at the heart of this discussion. Just something to keep in mind.
FYI - Weigh those cues on an accurate scale. You will probably find that a good many of them are mislabeled as to weight! (It may say 21 oz, but is actually 19 oz. Or say 19 oz and is actually 20 oz., etc.)
A pool player will know instantly if his cue is 1 oz off in weight. It would be interesting to audit the weights on house sticks but I know I can tell without ever reading the marked weight.
I play English pool/snooker (though I haven’t played snooker in a while) and I prefer a slightly heavier cure on the break/in general as I have very long arms and I find that little bit of extra stability helps. Though feeling comfortable with the cue is much more important.
I went from a 21 to a 20 to a 19 when I was playing regular, I may have gone even lower if I would have kept playing
Nice thread.
Glad it wasn’t bounced to Games forum.
One method I use for breaking might benefit from a heavier cue after reading these posts. I put my weight on the left foot and step into the break with my right arm much closer to straight than it would be on a normal stroke.
Thanks.
I think there’s a very obvious way to see if this is true: Just look at the cues pro 9-ball players use to break with. If you could train yourself to break harder with a 25 oz cue, then that’s what you’d see on the 9-ball circuit since the pros look for any advantage they can get and there’s no rule I know of restricting the weight of a breaking cue. Some amateur leagues have a 25oz weight restriction, but I’ve played a lot of league pool and I’ve played with a few pro players, and no one I know uses a break cue more than 21oz in weight. And the trend I’ve seen over the years is that players have slowly shifted to slightly lighter break cues. I’ll bet the common weight for a break cue now is between 19 and 20 oz - probably the same as the player’s regular cue for consistency.
Something that needs to be said is that power isn’t everything. If you are trying to come up with some extreme technique to give yourself a slightly harder break, and you are giving up some cueball control to do it, you’re probably hurting yourself.
The ideal break in 9-ball or 8-ball will be done with enough draw and good enough aim that the cueball will strike the head ball and then pop back roughly into the middle of the table without touching a rail. If you’re using a second-ball break from the side in 8-ball, you want the cueball to rebound straight across into the side rail and then back to the middle of the table.
It still might get hit in a collision, but controlling the cueball this way maximizes the probability that it will stay out of a pocket and that it will be placed to give you the best chance of having a shot at another ball should you make one on the break.
So if you’re trying to improve your breaking skills and your cueball tends to go wildly random after impact, the first place to start is to learn to control the cueball after it impacts the rack. Once you can do that consistently, then you can start working on adding power.
I get my best hard brakes with a little draw, if I break hard with follow it will often jump the table.
I would be interested. I don’t really care about the numbers per se but am interested in the formulas for the mechanics.
Insightful & I agree completely about actual playing. A player won’t be successful hitting any shot harder than he/she can control what happens at the first contact and after. Some real amateurs think the break is all power all the time & point of contact, draw/follow, and English mean nothing. Not so. As you clearly know.
Way back when I was competing locally I was pretty sure the rules limited cue weight to 21oz. I’ve been out of it for 30 years now. I just checked and these guys say 25oz: http://www.wpa-pool.com/web/WPA_Tournament_Table_Equipment_Specifications although obviously they’re not the only sanctioning body on the planet and other rules may differ.
For sure the best practical answer is look at what the champions of the various leagues use and how that compares to their equipment rules.
As I read it, the OP started with more of a physics question: “Assuming I’m *really *strong, which cue weight (including ridiculously light & heavy ones) produces the hardest break?” Which is an interesting question in its own right, but isn’t relevant to winning games.
That makes it more difficult :). Would you be satisfied if I used some plausible-sounding numbers and let you extrapolate from there? Otherwise there are a lot of constants to keep track of and the post editor isn’t great for formulas.
Let’s look at the stick-ball collision first. We’ll model this as a perfect elastic collision (it’s not, but it’s pretty close). The formula for the final velocity of the impacted object (which you can look up on Wikipedia) is:
v[sub]2[/sub] = (u[sub]2[/sub](m[sub]2[/sub] - m[sub]1[/sub]) + 2m[sub]1[/sub]u[sub]1[/sub]) / (m[sub]1[/sub] + m[sub]2[/sub])
u[sub]2[/sub] is the initial velocity of the ball, which is 0. I’ll also plug in 0.16 kg for m[sub]2[/sub].
v[sub]2[/sub] = 2m[sub]1[/sub]u[sub]1[/sub] / (m[sub]1[/sub] + 0.16 kg)
All right, so we have v[sub]2[/sub] in terms of the mass of the stick and stick velocity. We want to reduce this to just the stick mass.
The stick is accelerated by the arm, which can accelerate the arm+stick through a fixed distance. Let’s say that the arm can exert 150 N of force through 0.5 m distance and has 2 kg of effective swinging arm mass. F=ma, so:
150 N = m[sub]1[/sub]a
a = 150 / (m[sub]1[/sub] + 2)
We also have d = 0.5at[sup]2[/sup]:
0.5 m = 0.5at[sup]2[/sup]
a = 1/t[sup]2[/sup]
Setting equal:
1/t[sup]2[/sup] = 150 / (m[sub]1[/sub] + 2)
t = sqrt((m[sub]1[/sub] + 2) / 150)
Finally, v=at:
u[sub]1[/sub] = (150 / (m[sub]1[/sub] + 2)) * sqrt((m[sub]1[/sub] + 2) / 150)
u[sub]1[/sub] = sqrt(150 / (m[sub]1[/sub] + 2))
Substituting into the earlier formula (and change m[sub]1[/sub]->m and v[sub]2[/sub]->v):
v = 2m*sqrt(150 / (m + 2)) / (m + 0.16)
At this point I’m going to cheat and use Wolfram Alpha to find the maximum (yeah, I could do the derivative, but it’s painful). It reports a positive maximum at:
m = 0.884 kg (v = 12.21 m/s)
All right; that’s 31 oz, which is rather higher than the (apparent) actual optimal number. So there’s obviously other stuff going on, or my estimates are off. But the OP asked about a 1 oz and a 50 oz stick, and I think we can see from the physics that “somewhere in the middle” is the right answer. The empiricists can deal with any remaining factors of two :).