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cushioncrawler
08-03-2012, 03:42 PM
I used to hav trouble understanding momentum and energy. Still do.
Momentum iz eezyest.
Momentum = mv , ie mass by speed.
And change in momentum = Ft , ie average force by time.

In an impakt overall momentum stays the same, ie momentum iz konserved.
If a qball hits a red the total speed stays the same.
If qball speed before impakt iz 5 and red's speed after impakt iz 2 then i know that the qball speed iz 3 (if balls hav same wt).

This iz eezyer to understand when u konsider that change in momentum iz Ft.
In an impakt the qball feels the same forces az the red, for the same times.
Therefore the momentum lost by the qball must be equal to the momentum gained by the red.
This iz allways so, for all impakts, no matter what sort of impakt.

What the law aktually means iz that action and reaction forces are allways equal -- its one of Newton's laws.

Its even simpler. Total V stays the same.
If the qball speeding at 5 hits 3 stationary balls then u know that immediately after impakt the 4 speeds add to 5 (a bit less than 5 if the impakts are not simultaeneous, ie the balls slow down over time).

Getting back to the kamui hitting the qball, u shoodnt say that the qtip absorbs some of the momentum of the kamui-qball impakt.
In a way it duzz, temporaryly at least (ie a bit of vibration during impakt kan be sayd to be stored momentum, but after impakt none of the heat or vibration iz what we call momentum).

But it iz ok to say that the kamui absorbs energy.
It absorbs some energy temporaryly during impakt and some of this remains in the kamue after impakt (ie heat and vibration).
Energy iz lost during all impakts.
Or putting it another way, energy iz never lost or gained, it just changes form.

But overall momentum iz never lost or gained, and it never changes form.

Rotational momentum iz eezy. There iz no such thing.
Rotational momentum iz purely a math trick. Dont be fooled -- mathematicianologysts are.
mac.

SHITSU-TONKA FRIED WHALE -- FINGERLICKEN GOOD.

LoneWolf
08-24-2012, 04:09 AM
It's really not that complicated. Do you think efren reyes does a geometry equation before he shoots? It's about feel and experience and watching good playeres.

cushioncrawler
08-24-2012, 04:52 AM
Yes, but that speed stuff haz got nothing to do with aktual play, its all to do with skoolkid math and skoolkid fizzix.

On the other hand, math and fizzix kan help our understanding of the game, and hencely our play of the game, so i shoodnt say "nothing to do with aktual play".

We will never know whether Efren would be a better player had he had an understanding of math and fizzix.
mac.

LoneWolf
08-24-2012, 06:04 PM
Why don't you use the table as guide to the mysteries of pool. It is more accurate.

cushioncrawler
08-25-2012, 04:45 AM
Yes, but there are lots of things that happen on the table that are hard to understand or accept and end up costing u praps years, whereaz once u understand the fizzix u realize that it aint a flaw in your action or something and u dont waste time fixing something that dont exist.
mac.

Jal
08-25-2012, 03:56 PM
Mac, with the exceptions of a straight-in shot and a 90-degree cut (a true geometric 90-degree cut, i.e., the object ball doesn't move), the sum of the cueball's and object ball's speeds after impact are always greater than the cueball's speed before impact. Nevertheless, momentum is conserved.

Momentum is a vector entity, meaning that it's conserved in any <u>given direction</u>. For instance, if you label the cueball's pre-impact direction as the y-direction, then the sum of the cueball's and object ball's velocity components in <u>that</u> direction do indeed add up to the cueball's pre-impact speed (assuming equal masses). Momentum, therefore, is conserved in that direction.

In addition, though, for any cut other than straight-in or 90-degrees, both the cueball and object ball also acquire non-zero velocity components perpendicular to the y-direction. Call that the x-direction. These components in the x-direction exactly cancel each other - they're equal and opposite (have opposite signs). Thus, momentum is conserved in that direction as well (which was zero before the collision). But when you "add" (pythagorean-wise: square root of the sum of the squares) the cueball's and object ball's x-direction post-impact components to their y-direction post-impact components, the total is actually greater than the cueball's pre-impact speed.

Jim

cushioncrawler
08-25-2012, 09:16 PM
Jim. Yes u are korrekt again & again i aint. Ok lemmesee.

1BALL. If a qball goze up the table (north) at 1.0000 m/s and hits the 1ball such that the 1ball heads rightish (northeast) at 45dg & if all friktions are zero and if the coefficient of balltoball impakt iz 1.0 then the 1ball will be speeding at 1/root2 m/s [= (root2)/2] which iz 0.70707070707 m/s. This equates to an easterly komponent of 0.5000 m/s & a northerly komponent of 0.5000 m/s.

Likewize the qball will head leftish (northwest) at 45dg at 0.70707070707 m/s (ie 0.5000 m/s west & 0.5000 m/s north).

The kombined speed northwards iz 1.0000 m/s, ie az before impakt. The kombined speed eastwards iz 0.0000 m/s, ie az before impakt (if u say that east & west cancel).

The kombined speed of the two balls iz 1.4141414141 m/s (ignoring direktion), which iz an inkreec of 41.41414141%.

So, eech time the qball hits another ball it passes 0.7 of its speed to that ball, & it retains 0.7, & the kombined speed gain iz 41%. This iz for a 45dg impakt.

1BALL. The qball & 1ball will eech hav a speed of 0.7070, & a kombined speed of 1.4141.

2BALL. So, if the qball then hits the 2ball (at 45dg), the qball & 2ball will eech hav a speed of 0.7 times 0.7070, which iz 0.5000, & a kombined speed of 1.0000. The grand total will be 1.7070 inklooding the 1ball.

3BALL. If the qball then hits the 3ball (at 45dg), the qball & 3ball will hav a speed of 0.7 times 0.5000 which iz 0.3535, & a kombined speed of 0.7070. The grand total will be 0.7070 plus 0.5000 (2ball) plus 0.7070 (1ball), which iz 1.9141.

4BALL. If the qball then hits the 4ball (at 45dg), the qball & 4ball will hav a speed of 0.7 times 0.3535 which iz 0.2500, & a kombined speed of 0.5000. The grand total will be 0.5000 plus 0.3535 (3ball) plus 0.5000 (2ball) plus 0.7070 (1ball), which iz 2.0606.

So, here the qball haz suffered 4 impakts eech deflekting the qball 45dg, & now the qball iz heading south, whereaz it woz originally heading north, & the total speed of the balls haz rizen to 2.0606 m/s whereaz it woz originally 1.0000 m/s.

Jim, how duzz this look????
mac.

Jal
08-25-2012, 11:10 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">Jim. Yes u are korrekt again & again i aint.</div></div>
I don't know what that "again" business is about. You've described a number of things in the past that I was pretty sure were wrong at first sight, but after much labor, found out otherwise.

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">1BALL. If a qball goze up the table (north) at 1.0000 m/s and hits the 1ball such that the 1ball heads rightish (northeast) at 45dg & if all friktions are zero and if the coefficient of balltoball impakt iz 1.0 then the 1ball will be speeding at 1/root2 m/s [= (root2)/2] which iz 0.70707070707 m/s. This equates to an easterly komponent of 0.5000 m/s & a northerly komponent of 0.5000 m/s.

Likewize the qball will head leftish (northwest) at 45dg at 0.70707070707 m/s (ie 0.5000 m/s west & 0.5000 m/s north).

The kombined speed northwards iz 1.0000 m/s, ie az before impakt. The kombined speed eastwards iz 0.0000 m/s, ie az before impakt (if u say that east & west cancel).

The kombined speed of the two balls iz 1.4141414141 m/s (ignoring direktion), which iz an inkreec of 41.41414141%.

So, eech time the qball hits another ball it passes 0.7 of its speed to that ball, & it retains 0.7, & the kombined speed gain iz 41%. This iz for a 45dg impakt.

1BALL. The qball & 1ball will eech hav a speed of 0.7070, & a kombined speed of 1.4141.

2BALL. So, if the qball then hits the 2ball (at 45dg), the qball & 2ball will eech hav a speed of 0.7 times 0.7070, which iz 0.5000, & a kombined speed of 1.0000. The grand total will be 1.7070 inklooding the 1ball.

3BALL. If the qball then hits the 3ball (at 45dg), the qball & 3ball will hav a speed of 0.7 times 0.5000 which iz 0.3535, & a kombined speed of 0.7070. The grand total will be 0.7070 plus 0.5000 (2ball) plus 0.7070 (1ball), which iz 1.9141.

4BALL. If the qball then hits the 4ball (at 45dg), the qball & 4ball will hav a speed of 0.7 times 0.3535 which iz 0.2500, & a kombined speed of 0.5000. The grand total will be 0.5000 plus 0.3535 (3ball) plus 0.5000 (2ball) plus 0.7070 (1ball), which iz 2.0606.

So, here the qball haz suffered 4 impakts eech deflekting the qball 45dg, & now the qball iz heading south, whereaz it woz originally heading north, & the total speed of the balls haz rizen to 2.0606 m/s whereaz it woz originally 1.0000 m/s.

Jim, how duzz this look????
mac. </div></div>
Looks good to me. It's interesting too in that despite the conservation of momentum and energy, we seem to come out like a thief after the four collisions. Never thought about it that way.

Jim

cushioncrawler
08-26-2012, 12:37 AM
Jim -- I am going to do an excel table to see what iz what for varyus impakts with up to say 15 objektballs.

For instance, what iz the qball speed u might expekt if the qball duzz a 360dg circuit with 15 equal impakts????
I am thinking that this might work out to be say 0.3m/s (if initially 1.0 m/s).
In which case i guess that this allso means that if no friktion or rezistance of any kind then the qball kood go round and round for ever (allbeit slower and slower), if u kood set the balls up in an infinite spiral (which of course aint possible).
mac.

cushioncrawler
08-26-2012, 12:46 AM
And duzz it mean that if i sent the qball off inside a 360dg circuit made up of millions of balls then the qball would go around and still hav allmost all of its 1.0 m/s remaining.
mac.

Jal
08-26-2012, 01:26 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">Jim -- I am going to do an excel table to see what iz what for varyus impakts with up to say 15 objektballs.

For instance, what iz the qball speed u might expekt if the qball duzz a 360dg circuit with 15 equal impakts????
I am thinking that this might work out to be say 0.3m/s (if initially 1.0 m/s).
In which case i guess that this allso means that if no friktion or rezistance of any kind then the qball kood go round and round for ever (allbeit slower and slower), if u kood set the balls up in an infinite spiral (which of course aint possible).
mac. </div></div>
Mac, I was wondering about the same things and got some results for an infinite number of collisions (the formula is slightly simpler for an unending series of them). But I suspect you'd like to work out some of the above cases yourself, so I'll desist.

The general trend is that the thinner the cut, the greater the sum of all the balls speeds (at least for the infinite case where the cueball's energy is completely depleted). If you consider ridiculously thin cuts such as 89.9999 degrees over very large numbers of collisions, the sum gets quite large. So large in fact that it almost seems absurd that the total kinetic energy can be no more than the cueball's original amount, which of course must still be true.

(The infinite case ignores quantum limitations of the energies).

Jim

cushioncrawler
08-26-2012, 03:15 PM
Jim -- Are u saying that that absurd rezult duzz in fakt go over the original kinetik energy.
What appears to be the best speed of the qball after one 360dg circuit with 15 balls, with 360 balls, with squillions.
mac.

Hmmmmmm -- I will havta set up my snooker balls (15 reds say) and see what happens. What iz the best rezult that i kan get, ie iz 360dg possible. Perhaps it iz, uzing outside english. Perhaps it iz but only if uzing 20 balls or 30 balls.
mac.

Jal
08-26-2012, 10:49 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">Jim -- Are u saying that that absurd rezult duzz in fakt go over the original kinetik energy.</div></div>
If I did say that, I'd have had to have messed up somewhere. Our assumption is ideal balls (elastic, frictionless) and the numerical speeds of the balls after each collision are derived from that assumption - energy conservation is built into those numbers. Nonetheless, in the case of an infinite number of collisions at 89.9999 degrees, the sum of all the object balls' speeds turns out to be over a million times the cueball's initial speed! My gut (not my head) reads that as a gross violation of energy conservation. Of course, it's no more a violation than for the single collision case, but my innards don't understand all that.

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">What appears to be the best speed of the qball after one 360dg circuit with 15 balls, with 360 balls, with squillions.

Hmmmmmm -- I will havta set up my snooker balls (15 reds say) and see what happens. What iz the best rezult that i kan get, ie iz 360dg possible. Perhaps it iz, uzing outside english. Perhaps it iz but only if uzing 20 balls or 30 balls.
mac. </div></div>
With 15 balls (66-degree cuts), after each collision the cueball's speed is reduced by a factor of 0.9135454... After 15 impacts, it'll be 0.258 of its initial. After 30 impacts (two 360-deg, circuits), 0.066 of its initial.

With 30 balls instead of 15 (78 deg cuts), the reduction/impact is 0.9781476. After 30 collisions (one 360-deg. circuit), it'll still have over 50% of its original speed (0.515). The more balls the merrier.

(We're talking ideally.)

Jim

cushioncrawler
08-27-2012, 04:14 AM
I am going to make a spiral template and see if i kan make a qball go 360dg.
I set up 22 balls making about 200dg and the qball went around ok, so 360dg shood be possibe.
I notice that the qball only touched about 1 ball in 3 out of that 22 balls, and one or two of them were spat out a long way, so it looks like it iz going to take a lot of attempts and some luck.
mac.

cushioncrawler
08-27-2012, 04:38 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">Jim -- Are u saying that that absurd rezult duzz in fakt go over the original kinetik energy.</div></div><div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Jal</div><div class="ubbcode-body">If I did say that, I'd have had to have messed up somewhere. Our assumption is ideal balls (elastic, frictionless) and the numerical speeds of the balls after each collision are derived from that assumption - energy conservation is built into those numbers. Nonetheless, in the case of an infinite number of collisions at 89.9999 degrees, the sum of all the object balls' speeds turns out to be over a million times the cueball's initial speed! My gut (not my head) reads that as a gross violation of energy conservation. Of course, it's no more a violation than for the single collision case, but my innards don't understand all that........ Jim</div></div>Jim -- I suspekt that the qball goze on and on for ever, but that duznt meen that it suffers an infinite number of kollizions. It just meens that it goze slower and slower. So your series there will head for its natural limit, ie the total energy wont exceed the original kinetik energy of the qball.
I failed pythagorus, and i missed the class on series.
mac.

Anyhow, in the long run, we all end up swallowed by a blackhole, balls and all.

Jal
08-27-2012, 05:36 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">Jim -- I suspekt that the qball goze on and on for ever, but that duznt meen that it suffers an infinite number of kollizions. It just meens that it goze slower and slower. So your series there will head for its natural limit, ie the total energy wont exceed the original kinetik energy of the qball.
I failed pythagorus, and i missed the class on series.
mac.</div></div>
No argument here. (You keep saying that you don't know this or that math, yet you come up with numerical results at times that seem to belie those statements. How does one do that?)

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">Anyhow, in the long run, we all end up swallowed by a blackhole, balls and all. </div></div>I hope I'm not around when that happens. However, if you watch pop expositions of the the latest in cosmology, then you might know that there is a hypothesis, taken somewhat seriously, that our universe is a holographic projection of information on a two dimensional sphere which surrounds it. This is analogous to what happens with black holes (the holographic information regarding whatever it swallows up remains at the event horizon). Now I'm not sure whether they think our universe is in fact a black hole, or it's just an analogous situation. But it might be that we've already been gulped down, so to speak.

How's the project coming along? Have you achieved 360 degrees? If I get the time, I'd like to do some math on that tonight. I want to see if some positioning of the object balls results in auto-correcting of the cueball's path. Maybe you've already thought it through?

Jim

cushioncrawler
08-27-2012, 07:31 PM
I am going to havta make a solidish heavyish circle vortex torus to help re-set the balls eech time. Might take a while.

I like aether theory, where matter iz made from nothing, and returns to nothing in blackholes, in an infinite number of discreet cells, and there twernt no bigbang.

But math iz eezy. Instead of wrestling with calculas, u uze arithmetik.
All i do iz uze excel, uzing simple equations, slicing the impakt or whatever into 1000 slices per second or whatever. Uzing delta ft = delta mv and delta fd = delta mv^2/2, untill the cows kum home.
mac.

cushioncrawler
08-27-2012, 07:33 PM
Hell, i uzed excel to help proov that uzing money management a gambler kan inkreece hiz/her takings (or reduce losses) over the house by 0.666%. Thus solving probly the biggest question in gambling theory in my lifetime.
mac.

Aktually, now that i think of it, my theory/rezults seem to break the law of maturing odds, ie the law saying that just koz u throw 2 heads in a row it shoodnt affekt the chances of throwing another head. But that law iz a small scale law, and it haz a similar larger scale brother, which iz the aktual law i break (proov wrong). And the funny thing iz that to do it my system aktually uzes the reverse of the small-scale law, ie i bet that the next throw iz more likely to be a head, but in aktuallity none of that iz kritikal at that level.
mac.

Jal
08-28-2012, 09:32 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">I am going to havta make a solidish heavyish circle vortex torus to help re-set the balls eech time. Might take a while.</div></div>
Mac, I'm not trying to invite myself onto your project, but out of curiosity I did work out the positions of the balls in a general way. As you indicate, a spiral pattern is called for. If you're satisfied with what you've come up with, great, but if you want to compare notes....

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">I like aether theory, where matter iz made from nothing, and returns to nothing in blackholes, in an infinite number of discreet cells, and there twernt no bigbang.</div></div>
I couldn't defend or critique the standard view, thus making a comparison with yours pointless....unless your theory happens to be derived from pure logic and unassailable premises? Given the utterly unintuitive nature of the discoveries of 20'th century physics, it'd be pretty hard to have any confidence in axioms that aren't already well-established.

Jim

Jal
08-28-2012, 09:51 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: cushioncrawler</div><div class="ubbcode-body">Hell, i uzed excel to help proov that uzing money management a gambler kan inkreece hiz/her takings (or reduce losses) over the house by 0.666%. Thus solving probly the biggest question in gambling theory in my lifetime.
mac.

Aktually, now that i think of it, my theory/rezults seem to break the law of maturing odds, ie the law saying that just koz u throw 2 heads in a row it shoodnt affekt the chances of throwing another head. But that law iz a small scale law, and it haz a similar larger scale brother, which iz the aktual law i break (proov wrong). And the funny thing iz that to do it my system aktually uzes the reverse of the small-scale law, ie i bet that the next throw iz more likely to be a head, but in aktuallity none of that iz kritikal at that level.
mac. </div></div>
I don't have any experience with probability and statistics except at the most elementary of elementary levels. If you brought it up because you're looking to bounce it off someone, anyone that might, against all odds, have a glimmer of understanding, I wouldn't be uninterested in getting a pm about it. On the other hand, if your idea is so dear that it would be devastating to have it proved wrong, unlikely though that may be, particularly by me, maybe it wouldn't be such a good idea to pass it along. (I don't have Excel.)

Just saying, Mac. /forums/images/%%GRAEMLIN_URL%%/smile.gif

Jim