Back arm perpendicular- why?

[softshot]

Human muscles don't "maintain" peak velocity - it's "maintained" naturally by the fact that your muscles have stopped accelerating your arm.



They're naturally good at doing nothing during "peak velocity" - it's what they naturally do.

pj
chgo

you don't maintain peak anything... thats why it's the peak...

a muscle doing nothing... isn't moving

a muscle in mid contraction is accelerating.. at some rate.

muscles accelerate towards their own center it is all they CAN do.. or other muscles are pulling and stretching them back..

that's why a pause is a good thing it makes every other muscle stop

then allows you to concentrate on the accelerating contraction of the bicep muscle.. because at the end of the day that is all a stroke is... if you are doing it right..

 

[Patrick Johnson]

Me:
"Better speed control" means creating a better chance for the correct stick velocity and impact with the cue ball to occur simultaneously. How could "accelerating through" (accelerating continuously up to and past impact) do this?

When accelerating through to obtain some particular speed, the curve of acceleration plotted against time is generally flatter than when not accelerating or decelerating at impact. This makes it less sensitive to timing errors, i.e., shifting it, or some portion of it, right or left on the time scale.

But the objective is to have a flatter curve of velocity around the moment of impact. How is "accelerating through" conducive to this?

pj
chgo

 

[softshot]

That's not implausible, but what sort of study determined this? Do you have a reference?

As for what we can sense of the motion, I think we can sense at least three factors in motion and controlling motion: position, velocity and acceleration. It's not clear to me that separating them out somehow gives either a better feeling of what's going on or more control. Also, I suspect that we sense force far more than acceleration, but if you use the same cue stick all the time they are directly proportional.

I think we may also be able to sense -- have a feel for -- jerk, which is the rate of change of acceleration. Minimizing jerk also seems to be important in developing a consistent stroke.

currently I cannot site the specific source of that information but it does exist...

I was at the library a month or so ago with some time to kill and on a whim started looking at books on sports medicine and things like physics and the human body.. and I learned quite a bit..... I can't get you the specific passage.. off hand right now.. but the next time I am at the library I will attempt to find it..

the passage on acceleration as a natural muscle function .. was in relation to a baseball pitcher throwing an off speed pitch .. and how they have to remember to accelerate the arm consistently even though it was at a slower speed.. and how it is easier than you think .. because consistent acceleration is a natural ability of muscles.

 

[Jal]

But the objective is to have a flatter curve of velocity around the moment of impact. How is "accelerating through" conducive to this?

It might help to imagine two extreme cases.

Suppose the acceleration was absolutely constant throughout. A graph of it against time (or distance) would then be a horizontal line - as flat as flat can be. The final speed would be the area under the curve, that is, the product of that constant magnitude of acceleration times the length of time it took for the cue to reach the cueball. If you were to attempt to stretch or contract this acceleration line over time, it would be exactly the same and you would end up with the same final speed, assuming a fixed bridge length. In other words, under the assumption that the acceleration is constant, stretching or contracting it really has no meaning.

The opposite extreme would be an acceleration curve that shot up to some peak value in a very short time, then back down to zero just as fast. The cue would acquire its velocity in almost an instant, then coast at this speed the rest of the way, assuming no decelerating force was subsequently applied. As with the flat line just described, this speed would be the area under that spike-like curve. But now if you were to stretch it out over time, i.e., change the timing, it would have a very different area underneath it. Generally speaking, the steeper the rise and fall, the more sensitive the area is to timing.

I think (hope) that illustrates the general principle. With more realistic curves which do rise and fall but at a slower rate, the same applies; it's just not as obvious and the benefit of making them flatter, to whatever extent that can be done, isn't as dramatic.

Jim

 

[mikepage]

Good question.

During the forward stroke the cue is being accelerated up to some speed (duh). Take the time from the start of the stroke up to impact and divide it into small intervals (ideally infinitesimal). During each of these intervals the magnitude of the acceleration will have a particular value. Multiply the magnitude of the acceleration by the length of the time interval. The cue's final speed at the commencement of impact will be the sum of all these products, one for each interval. So its final speed depends on its acceleration at each interval throughout the entire stroke, not just at the end. In that sense, none (or any group) of the time intervals is more important than any other.

When accelerating through to obtain some particular speed, the curve of acceleration plotted against time is generally flatter than when not accelerating or decelerating at impact. This makes it less sensitive to timing errors, i.e., shifting it, or some portion of it, right or left on the time scale. It's kind of hard to explain why this is true without a diagram, which I can't offer at this time.

It has to be said, however, that small variations in bridge length favor coasting through if no adjustments are made for them. It then becomes a question of how well these variations are noticed by the player and what adjustments are made. This element adds considerable fuzziness to the whole issue.

Jim

Jim - I'm not SURE I understand your argument, but I think I do. Tell me if the jist is right here.

My understanding of Jim's argument
**************
Assume we can generate a desired peak force (acceleration) accurately, and assume we ramp up to this peak acceleration and then back down like a hump of a sine wave. The area under this acceleration versus time "hump" up to any point in time is the speed the stick has achieved up to that time. If we strike the ball when the stick is coasting, like Patrick is endorsing in this thread, that means we have generated the speed corresponding to the area under the entire hump.

Now instead of our sine hump lasting for 1.0 seconds, suppose the whole thing is wider in time and lasts for 1.1 seconds (but has the same height). This new stick will be moving faster at the end of the hump (more area under this wider hump)

Or if the whole hump lasted 0.9 seconds instead of 1.0 seconds--that stick would be moving more slowly at the end.

But if you look at a 0.9-second hump right on top of a 1.0 second hump (starting at the same instant of time), then things are more interesting. We know the area under the 0.9 is less, but for the early parts of the stroke--for the first few tenths of a second, the area is actually more because the 0.9 hump is rising faster. Then on the back half the area under the 0.9 becomes less by a greater degree. That's how it is overall less.

Jim's argument is if you only go about halfway down the far side the areas (and thus speed of the stroke) are the same.

Thus at a particular point on the downhill, the stick speed is insensitive to this "timing variable."

I will say what I think of this argument later, but I'd first like to know whether I basically have it right.

 

[Jal]

Jim - I'm not SURE I understand your argument, but I think I do. Tell me if the jist is right here.

My understanding of Jim's argument
**************
Assume we can generate a desired peak force (acceleration) accurately, and assume we ramp up to this peak acceleration and then back down like a hump of a sine wave. The area under this acceleration versus time "hump" up to any point in time is the speed the stick has achieved up to that time. If we strike the ball when the stick is coasting, like Patrick is endorsing in this thread, that means we have generated the speed corresponding to the area under the entire hump.

Now instead of our sine hump lasting for 1.0 seconds, suppose the whole thing is wider in time and lasts for 1.1 seconds (but has the same height). This new stick will be moving faster at the end of the hump (more area under this wider hump)

Or if the whole hump lasted 0.9 seconds instead of 1.0 seconds--that stick would be moving more slowly at the end.

But if you look at a 0.9-second hump right on top of a 1.0 second hump (starting at the same instant of time), then things are more interesting. We know the area under the 0.9 is less, but for the early parts of the stroke--for the first few tenths of a second, the area is actually more because the 0.9 hump is rising faster. Then on the back half the area under the 0.9 becomes less by a greater degree. That's how it is overall less.

Jim's argument is if you only go about halfway down the far side the areas (and thus speed of the stroke) are the same.

Thus at a particular point on the downhill, the stick speed is insensitive to this "timing variable."

I will say what I think of this argument later, but I'd first like to know whether I basically have it right.

Dr. Page,

I'm still confident in my argument, but ain't arrogant or dumb enough (close) to not realize that when a professor gets on your case, you've got trouble.

My argument does not really resemble your take on it. It's based entirely on the relative flatness of the curves. Actually, when the computer spit out the numbers (this was years ago), it surprised me. I then made up a diagram to see why this is true, and just concocted the response to Patrick's question in lieu of the diagram (that computer isn't working).

I've been looking over the math to see if I had an expression for the derivative of the final velocity with respect to phase frequency (w in sin(wt), where the force function is Asin(wt), or with harmonics, A1sin(wt)+A2(sin(2wt)...), but don't. It's complicated by the fact that I use an iterative process to find the phase angle (wt) at impact. I don't know if a closed analytic expression exists, but I'll see if I can derive one which will make things more obvious, one way or the other. (Obviously, you could do it too, and with greater ease, but may have better things occupying your time).

I agree with your general statements about the .9, 1.0, and 1.1 second curves, but didn't really follow the conclusion about the speed sensitivity to timing as drawn from them, which methinks you're prepared to demolish anyway. Yeah, I seem to be evasive, but really didn't look at it in the way you presented it...at least I don't think so.

(Got my hardhat on, hammer as needed.)

Jim

 

[Jal]

deleted by Jal

 

[Jal]

...I've been looking over the math to see if I had an expression for the derivative of the final velocity with respect to phase frequency...

As a double check of arguments given earlier, I calculated the sensitivity of the velocity at impact to changes in overall timing for the simple force function F=Asin(wt). This is the derivative dVi/dw, where "Vi" is the cue's speed at impact. If "ti" is the time it takes for the cue to reach the cueball, and m its mass, the derivative is:

dVi/dw = [A/(mw^2)][2(wti - sin(wti))/(1 - cos(wti)) - (1 - cos(wt))]

When coasting at impact, wti is equal to pi. When "accelerating through", wti is something less than pi. To make a comparison, the peak force amplitude "A" has to be adjusted to equalize the velocity and bridge distance traveled. For some different w, w', the adjusted amplitude A' is given by:

A'/A = (w'^2/w^2)[(w'ti' - sin(w'ti'))/(wti - sin(wti))

Using 0.9pi and 0.8pi for w'ti' as examples, the sensitivity (derivative) is reduced to 0.69 and 0.46 of its value when wti=pi (coasting), respectively. These are significant reductions, imo.

Jim

 

[Jal]

...
dVi/dw = [A/(mw^2)][2(wti - sin(wti))/(1 - cos(wti)) - (1 - cos(wt))]

Using 0.9pi and 0.8pi for w'ti' as examples, the sensitivity (derivative) is reduced to 0.69 and 0.46 of its value when wti=pi (coasting), respectively.

In case someone is actually interested enough to verify the math, I should note that a factor of sin(wti) was accidentally dropped from one of the terms for the derivative. It should read:

dVi/dw = [A/(mw^2)][2(wti - sin(wti))sin(wti)/(1 - cos(wti)) - (1 - cos(wt))]

Then instead of the cited reductions of 0.69 and 0.46, they are 0.72 and 0.46, respectively. (The second one doesn't change within the 2-digit precision.)

Jim

 

[mikepage]

What I was describing and could describe better with a diagram was a graphical way to compute this derivative, dVi/dw --or, to be clear, the partial derivative of Vi with respect to w at fixed A.

Imagine a hump corresponding to frequency w. Then imagine another hump starting at the same time of frequency w + dw that is superimposed over the first one. Though the two humps almost fall on top of one another, the second hump is a little shorter and overall has a little less area under it (higher w means quicker stroke). This difference in area is dVi. Divide that by dw and you have the derivative dVi/dw evaluated at the end of the hump, i.e., at wt=pi. This derivative is negative.

Now think a little about what that area difference looks like. The shorter hump (the w + dw) actually rises faster in the first half. So if you were to evaluate dVi/dw at, say, wt=pi/2, the derivative would be positive instead of negative (positive area change divided by dw). Then when you start falling down the back side, the shorter hump drops more quickly and has less area under it.

So the left-side area difference is positive (say, +3) and the right side area difference is negative (say, -5) such that the total area difference is -2 (negative as it should be.)

If we want to find the point on the downhill for which dVi/dw is zero, then we just need to go down the hill until the right-side area difference is -3, i.e., just enough to cancel the left-side area difference.

At this point, dVi/dw is zero and the speed is insensitive to the "timing" of the stroke for fixed A.

In case someone is actually interested enough to verify the math, I should note that a factor of sin(wti) was accidentally dropped from one of the terms for the derivative. It should read:

dVi/dw = [A/(mw^2)][2(wti - sin(wti))sin(wti)/(1 - cos(wti)) - (1 - cos(wt))]

Then instead of the cited reductions of 0.69 and 0.46, they are 0.72 and 0.46, respectively. (The second one doesn't change within the 2-digit precision.)

Jim

 

[mikepage]

Dr. Page,

I'm still confident in my argument, but ain't arrogant or dumb enough (close) to not realize that when a professor gets on your case, you've got trouble.[...]

Hey I'm just a B-player and I haven't been shooting much in the last month and my eyes aren't what they used to be and I just changed tips and I'm coming in cold off the street and ....

 

[Patrick Johnson]

What I was describing and could describe better with a diagram was a graphical way to compute this derivative, dVi/dw --or, to be clear, the partial derivative of Vi with respect to w at fixed A.

Imagine a hump corresponding to frequency w. Then imagine another hump starting at the same time of frequency w + dw that is superimposed over the first one. Though the two humps almost fall on top of one another, the second hump is a little shorter and overall has a little less area under it (higher w means quicker stroke). This difference in area is dVi. Divide that by dw and you have the derivative dVi/dw evaluated at the end of the hump, i.e., at wt=pi. This derivative is negative.

Now think a little about what that area difference looks like. The shorter hump (the w + dw) actually rises faster in the first half. So if you were to evaluate dVi/dw at, say, wt=pi/2, the derivative would be positive instead of negative (positive area change divided by dw). Then when you start falling down the back side, the shorter hump drops more quickly and has less area under it.

So the left-side area difference is positive (say, +3) and the right side area difference is negative (say, -5) such that the total area difference is -2 (negative as it should be.)

If we want to find the point on the downhill for which dVi/dw is zero, then we just need to go down the hill until the right-side area difference is -3, i.e., just enough to cancel the left-side area difference.

At this point, dVi/dw is zero and the speed is insensitive to the "timing" of the stroke for fixed A.

Will this be made available with English subtitles?

Alternatively, can you say whether you've concluded anything about acceleration's vs. velocity's importance to hitting the cue ball at the right speed?

pj
chgo

 

[mikepage]

Will this be made available with English subtitles?

Ok... when I have time ;-)

Alternatively, can you say whether you've concluded anything about acceleration's vs. velocity's importance to hitting the cue ball at the right speed?

pj
chgo

Both arguments are mathematically correct, and both are of the form

[All else being equal,]
Ball speed is least sensitive to
[type of error in stroke]
when
[particular condition]
is satisfied.

I think your argument is the more reasonable one because I think your [all else being equal] is better satisfied and your [type of error in stroke] is imo more closely related to the kinds of errors that plague our strokes. But these judgments are qualitative. So I'll explain my reasoning when I can.

 

[Jal]

What I was describing and could describe better with a diagram was a graphical way to compute this derivative, dVi/dw --or, to be clear, the partial derivative of Vi with respect to w at fixed A.

Imagine a hump corresponding to frequency w. Then imagine another hump starting at the same time of frequency w + dw that is superimposed over the first one. Though the two humps almost fall on top of one another, the second hump is a little shorter and overall has a little less area under it (higher w means quicker stroke). This difference in area is dVi. Divide that by dw and you have the derivative dVi/dw evaluated at the end of the hump, i.e., at wt=pi. This derivative is negative.

Now think a little about what that area difference looks like. The shorter hump (the w + dw) actually rises faster in the first half. So if you were to evaluate dVi/dw at, say, wt=pi/2, the derivative would be positive instead of negative (positive area change divided by dw). Then when you start falling down the back side, the shorter hump drops more quickly and has less area under it.

So the left-side area difference is positive (say, +3) and the right side area difference is negative (say, -5) such that the total area difference is -2 (negative as it should be.)

If we want to find the point on the downhill for which dVi/dw is zero, then we just need to go down the hill until the right-side area difference is -3, i.e., just enough to cancel the left-side area difference.

At this point, dVi/dw is zero and the speed is insensitive to the "timing" of the stroke for fixed A.

Thanks for the clarification. I have reservations about drawing any conclusions from it since w+dw also causes a change in ti, ti--->ti+dti. It's difficult (for me) to see how this will affect the area without some hindsight from the math. The graphical arguments I've used were to illustrate what the math was already indicating. If there are doubts as to whether dVi/dw decreases as wti becomes less than pi, I would prefer that either you did the math, or looked over my derivation to check for errors. However, from your response to Patrick, it looks like you don't disagree.

Jim

 

[mikepage]

Thanks for the clarification. I have reservations about drawing any conclusions from it since w+dw also causes a change in ti, ti--->ti+dti. It's difficult (for me) to see how this will affect the area without some hindsight from the math. The graphical arguments I've used were to illustrate what the math was already indicating. If there are doubts as to whether dVi/dw decreases as wti becomes less than pi, I would prefer that either you did the math, or looked over my derivation to check for errors. However, from your response to Patrick, it looks like you don't disagree.

Jim

Hi Jim. I don't disagree. I get a little different expression for the derivative though

dV/dw = A/mw^2 [ wt*sin(wt) + cos(wt) -1]

This is zero when the stuff in square brackets is zero which occurs when wt = 0.742*pi, in other words about halfway down the hill. Maybe I made an error here. I don't know. But I think my graphical argument is rigorous and proves that your conclusion about dV/dw being zero somewhere on the downhill is right.

 

[mikepage]

[...]
Alternatively, can you say whether you've concluded anything about acceleration's vs. velocity's importance to hitting the cue ball at the right speed?

pj
chgo

Think about a person walking.

On every step, the person places his foot RIGHT below his center of mass. Now ask the person to walk faster. The center of mass (person) moves faster, the legs swing faster and/or longer, and still those feet get planted right below the person's center of mass on each step. Ask the person to start trotting. Same thing. This is something we're really good at.

What would happen if the person started trotting and the leg didn't swing around quite fast enough, or didn't swing out quite far enough? The person would plant a foot BEHIND his center of mass and fall forward.

My point is we are remarkably good at COUPLING the speed we are moving to the way in which we swing our legs to achieve the desired result --planting a foot below our center of mass.

It's not to the same degree, but I think we are remarkably good at coupling the length of our stroke to the peak force to achieve a desired stick speed. That is, if a one-pocket player is hitting an object ball three rails to the vicinity of his pocket, he can do it with a short stroke or a medium stroke or a long stroke. If he shortens the length of his backstroke by half an inch, he is good at generating the right amount of extra force on the forward stroke to get the desired outcome.

So if the force versus time curve for his stroke looks like the hump of a sine curve

F(t) = A sin(wt)

then I'm saying players are remarkably good at coupling together w and A to achieve the desired speed of the stroke. w can vary over a pretty wide range and the player will naturally also change A over a wide range to compensate.

This makes consideration of how the quality of the outcome varies with respect to w holding A fixed (like Jal does) or considering how the quality of the outcome varies with respect to A holding w fixed [which nobody's talked about yet and favors decelerating at impact ;-)] just not very well related to the nature of our actual errors.

Think about the hot and cold water controls in a shower--two degrees of freedom, two variables. But like A and w, H and C are two degrees of freedom that are strongly coupled together. What we care about is the temperature of the water. If we increase the intensity of H and of C together, we can keep the temperature of the water the same. This suggests a coordinate transformation to more natural variables (ones that aren't so coupled)

The new coordinates are I (intensity) and T (temperature), and they're obtained from the old ones by

I = H + C

T = H - C

and these are the better controls in a shower (pull it out further to get more water and twist it to change the temperature)

 

[shovelmd]

wow, that's crazy if i had to think about all of that i would never be able to drop a ball. don't forget to add top or bottom english to your formula.:smile:

 

[Jal]

wow, that's crazy if i had to think about all of that i would never be able to drop a ball. don't forget to add top or bottom english to your formula.:smile:

It's essential for the next level of position play. Efren was kind enough to send the formula along, however, I find that I'm still limited to +/- half a table length of precision....maybe he's playing games with us.

Jim