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bradb
02-10-2007, 06:29 PM
Hi Jim, this is the best way to reach you with all these threads.

We were discussing a set throw before and I just wanted to get your take on my thinking. I know this has been discussed much especially in Daves material but I want to get to the very essence of this shot.

There are two touching balls with the QB behind. The set is not quite "on" to the pocket. The QB strikes the first ball at 45 dg on the right pushing it into the second ball. Does not the strike on ball 1 generate some rotation by friction as a result of the side swipe by the QB? Its tiny of course (like the slight turning of a doorknob) As it pushes ball 2 away the turn "throws" it slightly to the left beginning an altered path to the pocket. Also of course the push is a sideways motion again helping to add further to the throw on ball 2.
I realize that once launched, ball 2 is rolling and has no spin it was thrown left by the turn of ball 1.

In conclusion Directional energy supplies the propulsion, rotational energy supplies the throw. Yipes, do I know what I'm talking about?

/ccboard/images/graemlins/smile.gif

1hit1der
02-11-2007, 08:20 AM
While I would agree that both rotation and "directional energy" affect the throw, I wouldn't necessarily break it down exclusively the way you did in your conclusion. I only say this because if the cue ball was 45 degrees off of the two frozen balls and you hit the first ball full, there is ideally no rotation transferred between any of the balls, but you can still get throw.

bradb
02-11-2007, 08:35 AM
I should have said cut, sending the OB off 45 dg. If a full ball hit then there would be only the rotation to the 2 ball.
Incidently I said there would only be roll to ball 2 there would have to be some axis rotation. But not effecting the path.

Jal
02-11-2007, 11:25 AM
Brad, there are plenty of people here who could answer your question, but I'm deeply humbled that you picked me....though you couldn't have chosen better. /ccboard/images/graemlins/smile.gif

As 1hit1dir indicated, since ball 1 is being driven at an angle to the line between centers of 1 and 2, especially at 45 degrees, the surface speed (rubbing action) at their mutual point of contact is also due to this motion. In fact, most of it is due to this and not to spin induced on the first ball. Except for this, I believe you have it right.

Having multiple balls in contact at the same time is a whole level of complexity beyond that of just two balls colliding with each other . Without either experiments or an accurate numerical model (having the computer calculate forces and resulting motions step by small step), not much more can be said about it.

Jim

cushioncrawler
02-11-2007, 04:04 PM
<blockquote><font class="small">Quote bradb:</font><hr> ....In conclusion Directional energy supplies the propulsion, rotational energy supplies the throw. Yipes, do I know what I'm talking about? /ccboard/images/graemlins/smile.gif <hr /></blockquote> bradb -- I koodnt rezist saying that i reckon that "energy" duznt exist, it is a helpfull math concept only, and that energy never "did" anything. "Forces" do do things (everything actually). "Momentum" i like, koz in fact momentum theoryz are just force theoryz anyhow, so i am happy to say that momentum did this or did that, but not "energy".

Here i acknowledge that nobody knows "what a force iz", so its all bullkrap anyhow, but there is "bullkrap" and there is "BullKrap". madMac.

bradb
02-11-2007, 05:32 PM
Crawler, yeah lets leave energy to exxon! /ccboard/images/graemlins/cool.gif

Jim, thanks for your analysis I'll get back to you when I absorb it. Brad

PS, beer offer is still good!! /ccboard/images/graemlins/smile.gif

cushioncrawler
02-11-2007, 06:37 PM
bradb -- I think that some of this stuff is touched upon by Jim and others in a thread (that i started) called "how to maximize throw", but duznt have a real lot of comment re physics. I did write down some thorts in an article, and i can send u a copy if u PM me with an email address. That thread that i mentioned started with the following questions....

"With 2 balls touching, ie a set (which some call a plant) -- u want to pot the "2nd" ball, by hitting the 1st ball -- but u can see that the "line of centers" is pointing a long way off the pocket. Ok, how would u maximize the throw, to get the pot -- i think that there are 4 things u shood do.

Q1... Where would u place the qball (in hand) to throw the 2nd ball to the max??
Q2... What contact would u make on the 1st ball??
Q3... What spin or screw would u put on the qball??
Q4... How hard would u hit (ie to get max throw)??"....
madMac.

bradb
02-12-2007, 10:43 AM
<blockquote><font class="small">Quote Jal:</font><hr>
As 1hit1dir indicated, since ball 1 is being driven at an angle to the line between centers of 1 and 2, especially at 45 degrees, the surface speed (rubbing action) at their mutual point of contact is also due to this motion. In fact, most of it is due to this and not to spin induced on the first ball. Except for this, I believe you have it right.

Jim <hr /></blockquote>

Jim, are you saying that since ball 1-2 are touching (or near nuff as Crawler explained) then the rotation on ball 1- QB strike is negated and only the directional force and rotation of 1-2 is applied? Bear with me, I'm a total layman here.

If true, then if the set should be say 1/16" apart are we in a whole new ballgame?

bradb
02-12-2007, 11:02 AM
<blockquote><font class="small">Quote cushioncrawler:</font><hr>
Q1... Where would u place the qball (in hand) to throw the 2nd ball to the max??
Q2... What contact would u make on the 1st ball??
Q3... What spin or screw would u put on the qball??
Q4... How hard would u hit (ie to get max throw)??"....
madMac.


<hr /></blockquote>

(Crawler) You can get my mailbox with this reply, but I'm a old pool rat so if it gets into physics it might as well be in Ferengie! /ccboard/images/graemlins/smile.gif

Snapshot9
02-12-2007, 11:22 AM
<blockquote><font class="small">Quote bradb:</font><hr> <blockquote><font class="small">Quote cushioncrawler:</font><hr>
Q1... Where would u place the qball (in hand) to throw the 2nd ball to the max??
Q2... What contact would u make on the 1st ball??
Q3... What spin or screw would u put on the qball??
Q4... How hard would u hit (ie to get max throw)??"....
madMac.


<hr /></blockquote>

(Crawler) You can get my mailbox with this reply, but I'm a old pool rat so if it gets into physics it might as well be in Frengie! /ccboard/images/graemlins/smile.gif <hr /></blockquote>

For example, if 1st and 2nd frozen balls were lined up to the left of the corner pocket, and you would use low left to normally cut the 1st ball into the pocket, then you use low left hitting the 1st ball on right side with a soft medium stroke to throw the 2nd ball into the corner pocket.

This holds true for all frozen ball throws - another example, 2 frozen balls aimed to the right of side pocket, normally would use low right to cut 1st ball into the pocket, then use low right hitting 1st object ball on opposite side (left side) to throw the 2nd ball in.

I find that soft to soft medium speed works best for throw, and to control the cue ball.

bradb
02-12-2007, 11:31 AM
Agreed, if to much pace then the directional force seems to overpower any rotational action and the ball follows its original line more.

dr_dave
02-12-2007, 11:46 AM
FYI, Section 7.03 ("Frozen-ball throw shots") in my book ("The Illustrated Principles of Pool and Billiards") has lots of good illustrations, examples, and explanations.

Dave

<blockquote><font class="small">Quote bradb:</font><hr> Hi Jim, this is the best way to reach you with all these threads.

We were discussing a set throw before and I just wanted to get your take on my thinking. I know this has been discussed much especially in Daves material but I want to get to the very essence of this shot.

There are two touching balls with the QB behind. The set is not quite "on" to the pocket. The QB strikes the first ball at 45 dg on the right pushing it into the second ball. Does not the strike on ball 1 generate some rotation by friction as a result of the side swipe by the QB? Its tiny of course (like the slight turning of a doorknob) As it pushes ball 2 away the turn "throws" it slightly to the left beginning an altered path to the pocket. Also of course the push is a sideways motion again helping to add further to the throw on ball 2.
I realize that once launched, ball 2 is rolling and has no spin it was thrown left by the turn of ball 1.

In conclusion Directional energy supplies the propulsion, rotational energy supplies the throw. Yipes, do I know what I'm talking about?

/ccboard/images/graemlins/smile.gif <hr /></blockquote>

Jal
02-12-2007, 02:07 PM
<blockquote><font class="small">Quote bradb:</font><hr> Jim, are you saying that since ball 1-2 are touching (or near nuff as Crawler explained) then the rotation on ball 1- QB strike is negated and only the directional force and rotation of 1-2 is applied?<hr /></blockquote>Hi Brad,

I'm not sure to what extent the rotation of the first ball is negated because of contact with the second ball. The only way the second object ball can affect it is through friction, which would of course contribute to the throw of the second ball, more or less.

But that aside, let's assume, rightly or wrongly, that the first ball can spin pretty freely. The amount of surface speed at the contact point that develops because of this is still small compared to the surface speed that results from being hit at a 45 degree angle. I think this might be the case for any cut angle, but I've only checked it out for 45 degrees.

<blockquote><font class="small">Quote bradb:</font><hr>Bear with me, I'm a total layman here.<hr /></blockquote>No bearing necessary. It's these kind of questions that get some of us thinking (sometimes even correctly). /ccboard/images/graemlins/smile.gif Your questions and insights are most welcome.

<blockquote><font class="small">Quote bradb:</font><hr>If true, then if the set should be say 1/16" apart are we in a whole new ballgame? <hr /></blockquote>Some very casual tests I did (read sloppy) seemed to indicate the amount of throw was about the same from frozen to a separation of about 3/16 of an inch (5 mm). But I'll quickly defer to Mac or anyone else that has done this more carefully.

Jim

cushioncrawler
02-12-2007, 03:17 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote bradb:</font><hr>If true, then if the set should be say 1/16" apart are we in a whole new ballgame? <hr /></blockquote>Some very casual tests I did (read sloppy) seemed to indicate the amount of throw was about the same from frozen to a separation of about 3/16 of an inch (5 mm). But I'll quickly defer to Mac or anyone else that has done this more carefully. Jim <hr /></blockquote> I recall that in my tests i only got "super-throw" (allmost double normal throw) when the balls were well frozen. Any seperation (eg 1/16") and i just got "normal-throw". And, when the balls were say 4/16" to 5/16" apart, i got "zero-throw", ie the No2 ball allwayz went dead straight (unless i hit the No1 ball much finer than say 1/2 ball). madMac.

cushioncrawler
02-12-2007, 03:26 PM
<blockquote><font class="small">Quote bradb:</font><hr> (Crawler) You can get my mailbox with this reply, but I'm a old pool rat so if it gets into physics it might as well be in Ferengie! /ccboard/images/graemlins/smile.gif <hr /></blockquote> bradb -- I sent a PM. madMac.

cushioncrawler
02-12-2007, 03:49 PM
<blockquote><font class="small">Quote Snapshot9:</font><hr> ....For example, if 1st and 2nd frozen balls were lined up to the left of the corner pocket, and you would use low left to normally cut the 1st ball into the pocket, then you use low left hitting the 1st ball on right side with a soft medium stroke to throw the 2nd ball into the corner pocket.

This holds true for all frozen ball throws - another example, 2 frozen balls aimed to the right of side pocket, normally would use low right to cut 1st ball into the pocket, then use low right hitting 1st object ball on opposite side (left side) to throw the 2nd ball in.

I find that soft to soft medium speed works best for throw, and to control the cue ball. <hr /></blockquote> Snapshot -- Here above do u meen that if u hit the No1 ball on its left side (koz the frozen line points left of the pocket) then u havta use right-hand english (inside english) on the qball. If so, then that sounds ok to me.

I found that uzing "outside english" tended to minimize the throw -- this surprised me -- i used to think that outside english would help the throw. Anyhow, i reckon that this factoid would be good to know, not just to maximize throw, but allso to minimize throw.

For example, u might wish to uze (outside) english on the qball plus hit the No1 ball thinnish to help get good pozzy for yor next shot, but the frozen-line is straight to the center of the pocket, so u know that hitting the No1 ball anywhere off center would be fraught. Ok, u go ahead and hit the No1 ball the way u wished, ie thinnish, knowing that the outside english will help quieten the throw enuff to still pocket the No2 ball. madMac.

bradb
02-12-2007, 04:24 PM
Jim <hr /></blockquote> I recall that in my tests i only got "super-throw" (allmost double normal throw) when the balls were well frozen. Any seperation (eg 1/16") and i just got "normal-throw". And, when the balls were say 4/16" to 5/16" apart, i got "zero-throw", ie the No2 ball allwayz went dead straight (unless i hit the No1 ball much finer than say 1/2 ball). madMac. <hr /></blockquote>

Max, could this be because the collision is now more of a hit instead of a complete push so some of the friction is lost?

cushioncrawler
02-12-2007, 05:59 PM
<blockquote><font class="small">Quote bradb:</font><hr> <blockquote><font class="small">Quote madMac:</font><hr> I recall that in my tests i only got "super-throw" (allmost double normal throw) when the balls were well frozen. Any seperation (eg 1/16") and i just got "normal-throw". And, when the balls were say 4/16" to 5/16" apart, i got "zero-throw", ie the No2 ball allwayz went dead straight (unless i hit the No1 ball much finer than say 1/2 ball). madMac.<hr /></blockquote> <hr /></blockquote> Max, could this be because the collision is now more of a hit instead of a complete push so some of the friction is lost? <hr /></blockquote>bradb -- Slippage during a simple balltoball impact leaves a skuffmark on both balls -- the skuffmark is a rough-patch. If u align the objectball so that this skuffmark gets involved in a new impact, then u can expect to see a larger throw angle. The new skuffmark is now rougher, a "super-skuff". If u try to polish a super-skuff away u will see that it takes more effort, compared to polishing away an ordinary skuffmark. Super-skuff givz "super-friction", ie a (potentially) larger throw angle.

Now, in a frozen set-shot, a very similar super-skuff mark is obtained, due not to a double-hit, but due to the extra long impact time, and the extra long balltoball slippage distance. The "trapping" of the No1 ball givz a longer impact time between the qball and No1 ball, and an even longer impact time between the No1 ball and No2 ball.

An ordinary "impact-mark" duznt necessaryly do the trick. A balltoball impact-mark duznt do much skuffing, and, the residue is largely still "sitting there". This residue (burnt bakelite) acts like a mixture of ball-bearings and oil -- it lessens friction. But, if u carefully wipe some of the residue away, without polishing the ball, an impact-mark starts to give similar (friction) rezults to a skuffmark. All balltoball friction is like that, it is allways a battle between roughness and residue.

For example, if u put a bit of chalk on an objectball, u can get a big chalk-kick (and super-super-throw). Now, if u get that ball and align it so that the kick-skuffmark gets into a new impact, u can get another giant kick (throw) allmost az big az the chalk-kick, and this time there is allmost zero chalk involved.

This same "residue" mechanism can be used to explain the lower friction at higher impact speeds, and the lower friction at higher spins. Its in the residue.

The more force during slippage, the more roughness, and the more residue. Somehow the residue wins.

The more slippage during impact, then the more fresh non-rough surface that enters the impact flatspot during impact -- here i am looking at the qball's surface. The skuffmark on the objectball will be allmost circular, and the skuffmark on the qball will be oval, the more the slippage the longer the oval. The skuffmark on the objectball will be rougher, but the oval skuffmark on the qball will be smoother -- somehow, smoother wins. Food for thort. madMac.

bradb
02-12-2007, 07:12 PM
Max, very interesting as a guide for a layman, (right now I'm prostrate) those analogies actually makes some sense. You lost me with the ball bearings though I was rolling along nicely till I got to that part. I think I going to need some laymans terms for you laymans terms Max but I'll work with the scuff analogy I think I got that. Gimme a month and I'll get back to ya.
-Brad /ccboard/images/graemlins/tongue.gif /ccboard/images/graemlins/smile.gif

PS I owe Jim a round of beers, whadoya drink down there?

cushioncrawler
02-12-2007, 08:32 PM
<blockquote><font class="small">Quote bradb:</font><hr> Max, very interesting as a guide for a layman, (right now I'm prostrate) those analogies actually makes some sense. You lost me with the ball bearings though I was rolling along nicely till I got to that part. I think I going to need some laymans terms for you laymans terms Max but I'll work with the scuff analogy I think I got that. Gimme a month and I'll get back to ya. Brad /ccboard/images/graemlins/tongue.gif /ccboard/images/graemlins/smile.gif PS I owe Jim a round of beers, whadoya drink down there? <hr /></blockquote> I think Jim is near u. Me, down in Ozz, i just sip dry (cask) red from 6pm till bedtime, while thinking about ball impacts etc, then i go to bed and dream about cue vibration. But friction is a funny thing. Dr Dave &amp; others often say "a perfectly smooth surface" or somesuch, meening zero friction, but, for some materials, the smoother the surface the "more" the friction. Many would disagree with my little "rough-residue war" theory. One guy that i know is convinced that the kraps have an oil in them as a part of the bakelite process, and sez that this oil is constantly coming to the surface, and thusly making the friction enigma even more complicated. Some believe that a sonic vibration can somehow keep the surfaces apart for a time at a microscopic level, and thusly reduce friction. For instance, the annoying metallic squeak of a bulldozer's tracks iz said to be deliberately designed to reduce friction within the tracks. Moisture can reduces the friction of some sorts of plastics by az much az 30%. I uzually keep an eye out for any stuff on this etc. madMac.

Stretch
02-12-2007, 10:29 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> <blockquote><font class="small">Quote bradb:</font><hr> Max, very interesting as a guide for a layman, (right now I'm prostrate) those analogies actually makes some sense. You lost me with the ball bearings though I was rolling along nicely till I got to that part. I think I going to need some laymans terms for you laymans terms Max but I'll work with the scuff analogy I think I got that. Gimme a month and I'll get back to ya. Brad /ccboard/images/graemlins/tongue.gif /ccboard/images/graemlins/smile.gif PS I owe Jim a round of beers, whadoya drink down there? <hr /></blockquote> I think Jim is near u. Me, down in Ozz, i just sip dry (cask) red from 6pm till bedtime, while thinking about ball impacts etc, then i go to bed and dream about cue vibration. But friction is a funny thing. Dr Dave &amp; others often say "a perfectly smooth surface" or somesuch, meening zero friction, but, for some materials, the smoother the surface the "more" the friction. Many would disagree with my little "rough-residue war" theory. One guy that i know is convinced that the kraps have an oil in them as a part of the bakelite process, and sez that this oil is constantly coming to the surface, and thusly making the friction enigma even more complicated. Some believe that a sonic vibration can somehow keep the surfaces apart for a time at a microscopic level, and thusly reduce friction. For instance, the annoying metallic squeak of a bulldozer's tracks iz said to be deliberately designed to reduce friction within the tracks. Moisture can reduces the friction of some sorts of plastics by az much az 30%. I uzually keep an eye out for any stuff on this etc. madMac. <hr /></blockquote>

Then you must be aware that tapping the bottle with your knuckles changes the molecular structure of the Ketchup allowing it to run out the neck of the bottle more quickly. /ccboard/images/graemlins/smile.gif St.

bradb
02-12-2007, 11:23 PM
I knew it, that doo wop album I downloaded was interfering with the harmonics of my cut shots.... Maybe the smooth silky tones of Coletrain will lay down some good vibrating sonics for set transfer!
With the smooth surface therory I see another practical usage...think about how much undrank beer is left on the inside of a glass...It adds up!
Max, you and stretch are on to something here!

Enjoy that red, gotta Moosehead pale here! (life could be worse) ... Cheers. Brad:) /ccboard/images/graemlins/cool.gif /ccboard/images/graemlins/smile.gif

cushioncrawler
02-13-2007, 02:35 PM
<blockquote><font class="small">Quote Stretch:</font><hr> ....Then you must be aware that tapping the bottle with your knuckles changes the molecular structure of the Ketchup allowing it to run out the neck of the bottle more quickly. /ccboard/images/graemlins/smile.gif St. <hr /></blockquote> Stretch n brad -- I remember doing lots of frozen throw tests, and things werent working out (ie rezults were wrong). I woz halfway throo fudging the data when i looked out the window -- bloody bulldozer working -- phew, that explained it. madMac.

Jal
02-13-2007, 06:22 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> <blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote bradb:</font><hr>If true, then if the set should be say 1/16" apart are we in a whole new ballgame? <hr /></blockquote>Some very casual tests I did (read sloppy) seemed to indicate the amount of throw was about the same from frozen to a separation of about 3/16 of an inch (5 mm). But I'll quickly defer to Mac or anyone else that has done this more carefully. Jim <hr /></blockquote> I recall that in my tests i only got "super-throw" (allmost double normal throw) when the balls were well frozen. Any seperation (eg 1/16") and i just got "normal-throw". And, when the balls were say 4/16" to 5/16" apart, i got "zero-throw", ie the No2 ball allwayz went dead straight (unless i hit the No1 ball much finer than say 1/2 ball). madMac. <hr /></blockquote>Thanks Mac.

I interpret it as a case of static friction getting into the act, which you may not agree with (from reading your your earlier response to Brad), but I can't see anything else causing the abnormal throw for the frozen ball case?

Jim

cushioncrawler
02-13-2007, 06:38 PM
<blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote cushioncrawler:</font><hr> <blockquote><font class="small">Quote Jal:</font><hr> <blockquote><font class="small">Quote bradb:</font><hr>If true, then if the set should be say 1/16" apart are we in a whole new ballgame? <hr /></blockquote>Some very casual tests I did (read sloppy) seemed to indicate the amount of throw was about the same from frozen to a separation of about 3/16 of an inch (5 mm). But I'll quickly defer to Mac or anyone else that has done this more carefully. Jim <hr /></blockquote> I recall that in my tests i only got "super-throw" (allmost double normal throw) when the balls were well frozen. Any seperation (eg 1/16") and i just got "normal-throw". And, when the balls were say 4/16" to 5/16" apart, i got "zero-throw", ie the No2 ball allwayz went dead straight (unless i hit the No1 ball much finer than say 1/2 ball). madMac. <hr /></blockquote>Thanks Mac.
I interpret it as a case of static friction getting into the act, which you may not agree with (from reading your your earlier response to Brad), but I can't see anything else causing the abnormal throw for the frozen ball case?jim <hr /></blockquote>Jim -- I see what u mean about static-friction, at the least the slow speed givz the highest values, static or not. How duz your (Dr Dave's) equation for Marlow's friction/speed data check out for very small values of slippage -- what sort of (ave) speed reduction iz needed to double the friction??? madMac.

Jal
02-13-2007, 11:55 PM
Mac, for small initial surface speeds, given some Vs1 which yields a certain value for the coefficient of friction, to double it the surface speed has to be reduced to (approximately):

Vs2 = 2(Vs1) - 1.004

For instance, if Vs1 = 0.75 m/s, then Vs2 = 0.50 m/s. As you get to larger surface speeds, the above equation doesn't apply very well. In fact, it's not even close. In my version, which treats u as variable during impact, the constant 1.004 used above is replaced with 0.818 - but both produce similar values for throw.)

You might get a better idea by looking directly at his plot of u vs surface speed (my labels):

http://ww2.netnitco.net/users/gtech/cof.jpg

The red circles are the three data points from Marlow. I don't know what his testing procedure was like, ie, whether he used frozen balls (hope not). And neither Dr. Dave's nor mine factor out the shifting of the tangent line during compression.

To get an idea of typical surface speeds, consider a shot at a moderate cueball speed of 7 mph = 3.13 m/s, without spin. For the following cut angles (theta), the surface speed at the contact point, Vsin(theta), is :

5 deg - 0.27 m/s

15 deg - 0.81 m/s

30 deg - 1.57 m/s

45 deg - 2.21 m/s

60 deg - 2.71 m/s

For any other initial cueball speed, just multiply by the ratio of the speeds. To figure in sidespin, expressed as a fraction of maximum tip offset (0.5R), add or subtract (1.25)(3.13 m/s)f, where f is the fraction of maximum offset.

Jim

cushioncrawler
02-14-2007, 12:55 AM
<blockquote><font class="small">Quote Jal:</font><hr> .....The red circles are the three data points from Marlow. I don't know what his testing procedure was like, ie, whether he used frozen balls (hope not). And neither Dr. Dave's nor mine factor out the shifting of the tangent line during compression....<hr /></blockquote> Jim -- Thanks for that, i will look into it all. Offhand i would say that the shifting of the tangent line, and/or flatspot-sqeez (and friction-sqeez), are allready inherent in (simple) throw-angle measurements anyhow. But, if Marlow uzed frozen-balls to help measure simple-throw, Hmmmmmm, yes, that would be a bummer. We would need the help of the Hubble-Lens Team, back-up by the FBI computer-image enhancement team, after softening it all up with a raid from the WhiteHouse's WMD storm-troopers.

Pity i karnt get my newton-shmoos to work on a frozen impact, it would take i reckon 100Megabytes the way i do it, perhaps when the next generation of computers kum out.

I just thort of a simple analogy re frozen-throw, ie why it can be double the simple throw angle. To the No2 ball, the No1 ball will feel like it has double the rotational inertia (ie double the mass), and of course allmost double the linear inertia (or mass). It would be simple for me to double the (effective) mass of my "qball", in my program for a simple 2-ball impact, and the results would be somewhat like what we (can) get for a 3-ball frozen impact. Probably duznt proov anything, but.... Hmmmmmmm. I'll be back. madMac.