Help understanding the faults

bratnikotin

Registered
Hello. Please help me understand my mistakes, and/find a path to correct.

Problem 1:
- I seem to not being able to grasp the difference between "shoot fast" and "shoot hard"
To me the two concept seem the same, but I can't a way to shoot "not hard" but fast, or the other way around. Can anyone explain?

Problem 2:
- I am sure it has been asked before, but .. I can't keep "consistent stroke". Someone who teaches me, just a local "good" player, the "house specialist" so to speak. He says he would help me to improve, but he can't "understand" why I don't keep consistent stroke. So he says that I "shoot 11 times, and he sees 11 different strokes". Again, some question as above. I can practice and I know I am slowly getting better, but I need to "understand" what is it I am doing that is wrong.
 

Buzzard II

AzB Silver Member
Silver Member
You need a quality stroke instructor. "Quality" is the key here. Scott, Randy or whoever they recommend.
SPF.
 

Patrick Johnson

Fish of the Day
Silver Member
Hello. Please help me understand my mistakes, and/find a path to correct.

Problem 1:
- I seem to not being able to grasp the difference between "shoot fast" and "shoot hard"
To me the two concept seem the same, but I can't a way to shoot "not hard" but fast, or the other way around. Can anyone explain?
They're the same.

Problem 2:
- I am sure it has been asked before, but .. I can't keep "consistent stroke". Someone who teaches me, just a local "good" player, the "house specialist" so to speak. He says he would help me to improve, but he can't "understand" why I don't keep consistent stroke. So he says that I "shoot 11 times, and he sees 11 different strokes". Again, some question as above. I can practice and I know I am slowly getting better, but I need to "understand" what is it I am doing that is wrong.
Nobody can help with that unless they see your stance/stroke.

pj
chgo
 

goettlicher

AzB Silver Member
Silver Member
Hello. Please help me understand my mistakes, and/find a path to correct.

Problem 1:
- I seem to not being able to grasp the difference between "shoot fast" and "shoot hard"
To me the two concept seem the same, but I can't a way to shoot "not hard" but fast, or the other way around. Can anyone explain?

Problem 2:
- I am sure it has been asked before, but .. I can't keep "consistent stroke". Someone who teaches me, just a local "good" player, the "house specialist" so to speak. He says he would help me to improve, but he can't "understand" why I don't keep consistent stroke. So he says that I "shoot 11 times, and he sees 11 different strokes". Again, some question as above. I can practice and I know I am slowly getting better, but I need to "understand" what is it I am doing that is wrong.


There's a great PBIA Instructor in your area, Mr Mark Finkelstien.
Mark teaches the SPF Stroke method.
Please look him up at:

www.playbetterbilliards.com
 

Bob Jewett

AZB Osmium Member
Staff member
Gold Member
Silver Member
... Problem 1:
- I seem to not being able to grasp the difference between "shoot fast" and "shoot hard"
...
One use of shooting "fast" is to say a player takes very little time between shots. Find a YouTube video of Tony Drago for someone who shoots fast in that sense.

Shooting "hard" usually refers to the speed of the cue ball which is directly related to the speed of the cue stick. Many beginners shoot too hard. They lose accuracy and consistency.

Many beginners simultaneously shoot hard but not fast.:grin:
 

Bob Jewett

AZB Osmium Member
Staff member
Gold Member
Silver Member
... Problem 2:
- I am sure it has been asked before, but .. I can't keep "consistent stroke". ...
No one here has any idea how your fundamentals are. Maybe you have both feet in line with the cue stick. Maybe your bridge hand is not even on the table. Maybe your bridge formation is somewhere between ludicrous and "get outta here!" We have no idea. A good instructor can help with your fundamentals which include how you stand and how you move your arm.

In the absence of a good instructor, take a video of yourself and compare it to top players. It helps to know what to look for, but you might see something obvious, like jumping up in the middle of the shot.

Also in the absence of an instructor, you can practice simple shots with very precise goals. Try to play the cue ball (no object ball on the table) up and down the center line of the table to leave it on the rail where you shoot from or as close as possible. (This is called a lag shot and is often used to decide who will take the first shot.) Next, try to hit it a little harder to leave the cue ball exactly on the far rail.

Simply trying to do something accurately like that can get you to be more consistent and improve your fundamentals if you are a quick learner.
 

bratnikotin

Registered
Apologies, for not providing more details

I am above novice level .. so stance, bridge .. I am sort of OK on that front.
And, yes, I meant "shooting fast" as a speed of a cue ball, directly related to "speed" of the cue.
Now, I do not understand what is it I do, when I want to shoot "fast" and NOT hard, and instead, I end up shooting hard. So, the object ball swiftly goes to the pocket or to chosen target, but ... at the same time, cue ball makes few rounds around the table, and I would want it to, maybe just, run q quarter table off, keeping the swiftness of the object ball.
 
Last edited:

Patrick Johnson

Fish of the Day
Silver Member
Apologies, for not providing more details

I am above novice level .. so stance, bridge .. I am sort of OK on that front.
And, yes, I meant "shooting fast" as a speed of a cue ball, directly related to "speed" of the cue.
Now, I do not understand what is it I do, when I want to shoot "fast" and NOT hard, and instead, I end up shooting hard. So, the object ball swiftly goes to the pocket or to chosen target, but ... at the same time, cue ball makes few rounds around the table, and I would want it to, maybe just, run q quarter table off, keeping the swiftness of the object ball.
Fast and hard are the same thing. How fast/hard you hit the shot is about how far you want the CB to go. There's usually no reason to want the OB to go fast.

pj
chgo
 

BC21

https://www.playpoolbetter.com
Gold Member
Silver Member
Keep in mind that the cb basically leaves the ob with whatever speed is left over from the collision. For example, if you shoot the cb at a medium speed, say 8 feet per second, sending it straight into the ob (a full ball hit), the ob will take off at 8ft per second and the cb will stop moving. If you hit a halfball shot at 8ft/sec the ob will take half of the cb speed and move away at 4ft/sec while the cb glances off the ob and retains the other 4ft/sec of the total speed. If you only hit an eighth of the ob then the ob will only take 1/8 of the cb's speed, departing the impact with a speed of only about 1ft/sec, and the cb will keep most of its speed, 7/8 of it, or about 7ft/sec.

I probably made this sound much more complicated than it is. All I'm getting at is that if you're sending the cb around the table on most shots, you are leaving yourself a lot of thin cuts. Or your stroke is so just too inconsistent right now to develop a good sense of speed control. Speed is something that requires practice, but also requires a consistent cue delivery.

I think it was Tor Lowry or Billy Incardona that said in one of their videos, "I don't care who you are, you shoot to hard." I think that's true for a lot of players.
 

Patrick Johnson

Fish of the Day
Silver Member
Keep in mind that the cb basically leaves the ob with whatever speed is left over from the collision. For example, if you shoot the cb at a medium speed, say 8 feet per second, sending it straight into the ob (a full ball hit), the ob will take off at 8ft per second and the cb will stop moving. If you hit a halfball shot at 8ft/sec the ob will take half of the cb speed and move away at 4ft/sec while the cb glances off the ob and retains the other 4ft/sec of the total speed. If you only hit an eighth of the ob then the ob will only take 1/8 of the cb's speed, departing the impact with a speed of only about 1ft/sec, and the cb will keep most of its speed, 7/8 of it, or about 7ft/sec.

I probably made this sound much more complicated than it is. All I'm getting at is that if you're sending the cb around the table on most shots, you are leaving yourself a lot of thin cuts. Or your stroke is so just too inconsistent right now to develop a good sense of speed control. Speed is something that requires practice, but also requires a consistent cue delivery.

I think it was Tor Lowry or Billy Incardona that said in one of their videos, "I don't care who you are, you shoot to hard." I think that's true for a lot of players.
I believe the CB/OB speed ratios are affected by the CB’s vertical rotation. For example a rolling CB travels the same distance as the OB after a half ball hit (30 degree cut), but a stunned CB doesn’t. For a stunned CB to travel the same distance the cut must be 45 degrees.

pj
chgo
 

BilliardsAbout

BondFanEvents.com
Silver Member
Hello. Please help me understand my mistakes, and/find a path to correct.

Problem 1:
- I seem to not being able to grasp the difference between "shoot fast" and "shoot hard"
To me the two concept seem the same, but I can't a way to shoot "not hard" but fast, or the other way around. Can anyone explain?

Problem 2:
- I am sure it has been asked before, but .. I can't keep "consistent stroke". Someone who teaches me, just a local "good" player, the "house specialist" so to speak. He says he would help me to improve, but he can't "understand" why I don't keep consistent stroke. So he says that I "shoot 11 times, and he sees 11 different strokes". Again, some question as above. I can practice and I know I am slowly getting better, but I need to "understand" what is it I am doing that is wrong.

I teach a progressive bridge drill, to gain feel and skill in cue ball speed (later you can use one bridge length for all open shots if you wish, in actual play):

The bridge is the distance between the cue stick through the loop of your fingers and the cue ball. Set up some practice shots, then shoot each shot at three different bridge lengths, 3", 5" and 7", focused on two things:

1) A SMOOTH, unhurried backstroke and forward stroke

2) Stroke ALL the way back to the loop at every bridge length (harder than it sounds, most amateurs stroke back part way then jab forward, hitting the ball too hard/offline)

The longer bridges allow more time to accelerate the cue, giving you a baseline for smooth soft strokes (3"), smooth medium strokes of 5" and smooth hard strokes of 7", and this with all the strokes taking the same, unhurried, smooth pace and feel.

Try it and report back your results here.

Thank you.
 

BC21

https://www.playpoolbetter.com
Gold Member
Silver Member
I believe the CB/OB speed ratios are affected by the CB’s vertical rotation. For example a rolling CB travels the same distance as the OB after a half ball hit (30 degree cut), but a stunned CB doesn’t. For a stunned CB to travel the same distance the cut must be 45 degrees.

pj
chgo

Of course cb rotation affects the speed at which the cb departs the ob. The actual "equal" speed for a rolling cb occurs on a cut a little thinner than a halfball hit. But for general purposes, and simplicity, a good method for estimating speed is to use the method I described, where you estimate the cb/ob speed based on simple impact ratios/proportions.
 
Last edited:

Patrick Johnson

Fish of the Day
Silver Member
...the cb basically leaves the ob with whatever speed is left over from the collision.
Dr. Dave's calculations confirm what you say about a stunned CB: its after-collision speed is 1 minus the overlap fraction times its incoming speed (1/4 ball overlap = 3/4 of incoming speed).

But the after-collision speed of the OB is greater than the remainder of the CB's incoming speed - they always (except for full hits and 90 degree hits) add up to more than 100%. That surprised me, but Dave explains it this way: "Energy (proportional to speed squared) is conserved, but speed is not. Momentum (and therefore vector velocity) is conserved, but not speed magnitude."

Here's Dave's graph of the CB/OB after-collision speeds (as percentages of CB incoming speed) and some examples of the after-collision speeds as percentages of the CB's incoming speed (based on Dave's formulas)...

pj
chgo

P.S. Sorry for the detour in this thread...

cb-ob speeds 3.jpg
 
Last edited:

dr_dave

Instructional Author
Gold Member
Silver Member
Dr. Dave's calculations confirm what you say about a stunned CB: its after-collision speed is 1 minus the overlap fraction times its incoming speed (1/4 ball overlap = 3/4 of incoming speed).

But the after-collision speed of the OB is greater than the remainder of the CB's speed - they always (except for full hits and 90 degree hits) add up to more than 100%. That surprised me, but Dave explains it this way: "Energy (proportional to speed squared) is conserved, but speed is not. Momentum (and therefore vector velocity) is conserved, but not speed magnitude."

Here's Dave's graph of the CB/OB after-collision speeds (as percentages of CB incoming speed) and some examples of the after-collision speeds as percentages of the CB's incoming speed (based on Dave's formulas)...

pj
chgo

P.S. Sorry for the detour in this thread...

View attachment 539630
Good post, Pat. For those interested, a lot more information about CB and OB speeds can be found here:

CB and OB speed and travel distances for different types of shots

Enjoy,
Dave
 

BC21

https://www.playpoolbetter.com
Gold Member
Silver Member
Dr. Dave's calculations confirm what you say about a stunned CB: its after-collision speed is 1 minus the overlap fraction times its incoming speed (1/4 ball overlap = 3/4 of incoming speed).

But the after-collision speed of the OB is greater than the remainder of the CB's incoming speed - they always (except for full hits and 90 degree hits) add up to more than 100%. That surprised me, but Dave explains it this way: "Energy (proportional to speed squared) is conserved, but speed is not. Momentum (and therefore vector velocity) is conserved, but not speed magnitude."

Here's Dave's graph of the CB/OB after-collision speeds (as percentages of CB incoming speed) and some examples of the after-collision speeds as percentages of the CB's incoming speed (based on Dave's formulas)...

pj
chgo

P.S. Sorry for the detour in this thread...

Notice how post collision cb speed corresponds with the fractional proportion of the cut shot. In other words, sending the cb into a 3/4 hit with the ob will cause the cb to lose 3/4 of its speed. So an 8ft/sec cb will leave the ob with a speed of only about 2ft/sec. Only hitting 1/4 of the ob will result in the cb only losing 1/4 of its speed, retaining 3/4.

This is what I was referring to about a simple way to estimate how fast the cb is going to travel after it hits the ob, taking in account how fast/hard you initially struck the cb. If you don't have a good sense of cb speed, exactly how much is needed on any given shot, then it's a good method to work with until you build up the experience of just knowing what needs to happen, and knowing how hard to shoot to allow it to happen. It ends up being a feel thing, not something you calculate, but something that can be estimated pretty accurately by paying attention to the cb/ob relationship, fractional speaking, that's going to occur.
 
Last edited:

Patrick Johnson

Fish of the Day
Silver Member
Notice how post collision cb speed corresponds with the fractional proportion of the cut shot. In other words, sending the cb into a 3/4 hit with the ob will cause the cb to lose 3/4 of its speed. So an 8ft/sec cb will leave the ob with a speed of only about about 2ft/sec. Only hitting 1/4 of the ob will result in the cb retaining 3/4 of it's speed.
Yes, I agreed with that (first sentence in my post) - only clarifying the OB's speed (which isn't as important to your point).

This is what I was referring to about a simple way to estimate how fast the cb is going to travel after it hits the ob, taking in account how fast/hard you initially struck the cb. If you don't have a good sense of cb speed, exactly how much is needed on any given shot, then it's a good method to work with until you build up the experience of just knowing what needs to happen.
Agreed again.

pj
chgo
 

BilliardsAbout

BondFanEvents.com
Silver Member
Notice how post collision cb speed corresponds with the fractional proportion of the cut shot. In other words, sending the cb into a 3/4 hit with the ob will cause the cb to lose 3/4 of its speed. So an 8ft/sec cb will leave the ob with a speed of only about 2ft/sec. Only hitting 1/4 of the ob will result in the cb only losing 1/4 of its speed, retaining 3/4.

This is what I was referring to about a simple way to estimate how fast the cb is going to travel after it hits the ob, taking in account how fast/hard you initially struck the cb. If you don't have a good sense of cb speed, exactly how much is needed on any given shot, then it's a good method to work with until you build up the experience of just knowing what needs to happen, and knowing how hard to shoot to allow it to happen. It ends up being a feel thing, not something you calculate, but something that can be estimated pretty accurately by paying attention to the cb/ob relationship, fractional speaking, that's going to occur.

Yes, CB+OB = 1 is a great rule, and leads toward three easy safety options for a given ob/cb for an amateur--thin hit sending the cb far from the ob, thick or full hit doing vice versa, half-ball hit sending both balls an equal distance (like to the rails from the table's midline, for a painful bank on the key ball).
 

Patrick Johnson

Fish of the Day
Silver Member
CB + OB = 1 is a great rule
Except that's not the rule. The rule is CB remaining fraction of original speed + CB/OB overlap fraction (not OB speed) = 1.

As described above, CB outgoing speed + OB outgoing speed = more than CB incoming speed.

pj
chgo
 
Last edited:

BilliardsAbout

BondFanEvents.com
Silver Member
Except that's not the rule. The rule is CB remaining fraction of original speed + CB/OB overlap fraction (not OB speed) = 1.

As described above, CB outgoing speed + OB outgoing speed = more than CB incoming speed.

pj
chgo

"Rule of thumb". For those unable to memorize your chart, an inverse ratio that totals 1 is more than adequate for placing both balls for safeties.
 

Patrick Johnson

Fish of the Day
Silver Member
"Rule of thumb". For those unable to memorize your chart, an inverse ratio that totals 1 is more than adequate for placing both balls for safeties.
No, it's mostly not.

For a stunned CB (as we've been discussing), the OB travels anywhere from 1/10 more than your rule of thumb (for very thin hits) to 7 times farther (for very thick hits). For any hit over 1/2 ball thick the OB travels at least twice as far as your rule of thumb predicts.

Your rule of thumb comes closer for a rolling CB, but still mostly too far off to be useful: the OB travels about 50% farther than your rule predicts for hits over about 1/2 ball thick, up to 5 times as far for very thick hits.

P.S. There's no need to memorize my chart for stun hits - I summarized it in a few simple words. Unfortunately, I don't think the real rule for a rolling CB can be described as simply.

pj
chgo
 
Top