PDA

View Full Version : Object Ball Deflection

paulfr
01-15-2005, 01:30 AM
I am having a hard time understanding this concept of Object Ball Deflection. That is that a hard hit CB will cause the OB to deflect inside the line of centers of the CB-OB. This is exactly opposite of contact throw which causes an OB direction to be outside the line of centers.
Here is an article on the subject, but it offers no explanations for why it occurs or any way to demonstrate that it occurs ....
http://www.easypooltutor.com/article239.html

Anyone understand this principle ??

Thanks

paulfr
01-17-2005, 02:02 AM
Do you compensate for it ?
Anyone notice missing hard cuts with an unexplained overcut ??
Thank you.

pooltchr
01-17-2005, 05:59 AM
I don't believe there is any way possible for the opject ball to move in any more than a 90 degree angle. I can demonstrate how it can be LESS than 90 degrees, but in my limited knowledge of physics, more than 90 just ain't happening!
Fred?????????????? Randy????????????
steve

SpiderMan
01-17-2005, 07:13 AM
That's one of the worst-written articles I've ever seen.

My interpretation:

The author is making an observation that shots played with hard draw will tend to be overcut, hanging a name on the observation, and noting you should compensate your aim.

His attempt at explaining why this happens is completely without insight, and he doesn't even address the draw as a factor.

SpiderMan

HallofFame
01-17-2005, 07:56 AM
<blockquote><font class="small">Quote SpiderMan:</font><hr> That's one of the worst-written articles I've ever seen.

My interpretation:

The author is making an observation that shots played with hard draw will tend to be overcut, hanging a name on the observation, and noting you should compensate your aim.

His attempt at explaining why this happens is completely without insight, and he doesn't even address the draw as a factor.

SpiderMan

<hr /></blockquote>

Didn't read the article but "SpiderMan" seems to have uncovered an error.

According to what I learned from a top pro, putting draw on the cue ball eliminates skid and throw on an object ball; NOW, according to Don "The Preacher" Feeney, cut shots with hard draw are missed because we "naturally" tend to correct for throw with our follow shots and when we shoot a hard draw cut shot many players miss because "throw" now does not effect the object ball because draw has eliminated it. What Don is saying is: Players are naturally correcting for throw on draw shots when they DO NOT have to, that is why the shot is missed.

I don't know, it makes sense to me /ccboard/images/graemlins/tongue.gif

dr_dave
01-18-2005, 08:31 AM
I think the effect they are describing is "collision induced throw" (AKA cling, drag, push), but I'm not sure. See my posting on the types of throw (http://www.billiardsdigest.com/ccboard/showthreaded.php?Cat=&amp;Board=ccb&amp;Number=172771&amp;page =&amp;view=&amp;sb=&amp;o=&amp;vc=1) for more information and demonstrations.

SpiderMan
01-18-2005, 09:25 AM
"Collision-Induced Throw" would result in undercut, the phenomenon being described results in the OB being overcut or at least the CIT undercut being reduced. Sounds like another series of experiments in the making, if we can't find a reference to it being done before.

SpiderMan

paulfr
01-22-2005, 06:16 PM

There are two good reasons why it is possible for an OB to take off INSIDE the line of CB-OB centers.
1/ If the balls are very smooth and the CB is hit hard, the CB can slide a bit on the OB before friction takes hold and the ball takes off.
2/ A hard hit CB will leave the table easily, and it will bounce (inperceptibly some times). If it is airborne when it hits the OB, the reduced diameter effect described by Koehler in Science of Pocket Billiards, page 153, will cause an overcut INSIDE the line of centers.

Anyone experience this hard hit OB overcut ?
Anyone agree or disagree with these possible explanations ?

Thanks

pooltchr
01-23-2005, 05:20 AM
It would still take off at 90 degrees, and only alter it's course after is lands from that small jump.
Steve

SpiderMan
01-24-2005, 08:18 AM

There are two good reasons why it is possible for an OB to take off INSIDE the line of CB-OB centers.
1/ If the balls are very smooth and the CB is hit hard, the CB can slide a bit on the OB before friction takes hold and the ball takes off.
2/ A hard hit CB will leave the table easily, and it will bounce (inperceptibly some times). If it is airborne when it hits the OB, the reduced diameter effect described by Koehler in Science of Pocket Billiards, page 153, will cause an overcut INSIDE the line of centers.

Anyone experience this hard hit OB overcut ?
Anyone agree or disagree with these possible explanations ?

Thanks
<hr /></blockquote>

Paul,

I think #1 is unlikely, because if the balls are completely friction-free the OB/CB paths diverge at 90 degrees (centerline/tangentline). Any additional surface friction leads to undercutting, not overcutting.

#2 is clearly possible. I've read several sources (I believe one was a Robert Byrne book) claiming that an object ball can be "overcut" beyond 90 degrees by jumping the cueball.

What about the possibility that, with substantial draw on the shot, the object ball may be subject to a slight masse' because the contact point on the CB has a vertical component of motion? If this is the cause, then the effect would be more noticable for dirty (high-friction) ball surfaces than for slick, clean ones.

SpiderMan

jordan
01-28-2005, 09:12 AM
I believe there's one way of making a cutshot over 90* (without using a rail): play a jumpshot and land on the backside of the objectbal (it doesn't have to be easy, has it?).

01-28-2005, 03:46 PM
Here is an easy way to demonstrate to yourself the principle of deflection. Take an object that is lighter than a cue ball. I crumble up a few pieces of white paper into a ball. Take the crumbled piece of paper and set it on a table. Hit the piece of paper with your cue stick and hit the paper with left or right english. Notice how the paper does not go straight forward. It spins to the opposite side of the english applied. This also an excellent training aid for people who do not own a home table.

paulfr
01-29-2005, 06:03 AM
Christopher
I think I was not clear when I said the CB-OB line of centers. Of course the OB takes off opposite the side of the English. But I meant inside the line of centers of the
GHOST BALL-OB, or the CB-OB at impact. Throw will cause the OB to move OUTSIDE this line.
My previous post explains my theory about why the OB can move INSIDE this line between centers of the CB at impact and the OB at impact.

cushioncrawler
10-23-2005, 10:41 PM
Temporary impakt flatspots are prezent during all impakts, and they allwayz rezult in a finer cut angle for the objekt ball. The softer the ball, and the faster the impakt, the larger the flatspot and the greater this added angle. I like to call it Flatspot Sqeez.
My MathLand calculationz suggest that the cut angle for a half-ball contact iz inkreeced by 0.1 deg for each 1m/s of qball speed. Thusly u would get an extra 0.5 deg of cut for a speed of 5m/s, instead of say 30 deg u would get 30.5 degreez.
The calculationz suggest that the extra cut angle for an 1/8th ball contact iz double the above, ie 0.2 deg per m/s of qball speed. And much more at much finer contacts --- in fact the graph bekumz allmost exponential.
My MathLand calculationz allso suggest that friktion can inkreec or dekreec the cut angle by allmost 3.0 deg for any and all contacts.
Hence, in theory, uzing running sidespin, u can get a total of say 4 or 5 deg of overcut, in which case u can cut a ball more than 90 deg. But there are at least 2 problemz here.
Firstly, in putting running side on the Qball with the cue, u karnt avoid swerv --- swerv iz going to rob more degreez than it givz.
Secondly, for very fine contacts, say below 1/16th ball, the objekt ball goez nowhere --- the world'z hardest hitter would be struggling to make it go one foot.
Most of this stuff iz mentioned i think by Bob &amp; Koehler &amp; Doc &amp; otherz elsewhere.

Jal
10-24-2005, 03:57 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> Temporary impakt flatspots are prezent during all impakts, and they allwayz rezult in a finer cut angle for the objekt ball. The softer the ball, and the faster the impakt, the larger the flatspot and the greater this added angle. I like to call it Flatspot Sqeez.
My MathLand calculationz suggest that the cut angle for a half-ball contact iz inkreeced by 0.1 deg for each 1m/s of qball speed. Thusly u would get an extra 0.5 deg of cut for a speed of 5m/s, instead of say 30 deg u would get 30.5 degreez.
The calculationz suggest that the extra cut angle for an 1/8th ball contact iz double the above, ie 0.2 deg per m/s of qball speed. And much more at much finer contacts --- in fact the graph bekumz allmost exponential.<hr /></blockquote>Can you provide a little more explanation as to why the flatspot causes an increase in cut angle?

Jim

cushioncrawler
10-24-2005, 06:32 PM
Hi Jim -- Az we all know, for a half-ball contact, when the Qball first contacts the Objectball the line from center to center iz at 30.0dg to the Qball line of travel. At midimpakt, due to temporary but serious ball deformation at the contact, the line from center to center iz at say 30.5dg. At the end of impakt perhaps 31.0dg. Hence, the object ball will depart at 30.5dg (altho ball-to-ball friktion can make up to plus or minus say 3.0dg difference to this). In MathLand, the ball iz assumed to be infinitely hard (ie zero flatspot), and the impakt time to be infinitely small, hence the obove figurez are uzually taken to be 30.0dg, 30.0dg, and 30.0dg -- hence in MathLand the red departs at 30.0dg (ignoring friktion).
I notice on chalkboard that one player sez that he allowz for the extra cut-angle that he gets for higher speedz, but he must be very accurate, i reckon that the extra angle between slow and fast speedz would havtabe less than 5mm at the pocket.
I myself only make such an allowance when i am putting draw (bottomspin) on the Qball, koz here the cut-angle iz inkreeced by a lot (ie u get a finer cut-angle) -- but this iz due to throw, not flatspot sqeez.
However, when i say that flatspot sqeez allwayz inkreecez the cut-angle, i am really only referring to MathLand. In the RealWorld we all play by feel (even thoze who dont), and the sqeez and throw etc are all in our familiar equation for feel. This feel iz probably based on medium speed play -- hence we dont really havta make any/much cut allowance for shots that are slower or faster. Regardz... MadMac.

Jal
10-24-2005, 08:25 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> Hi Jim -- Az we all know, for a half-ball contact, when the Qball first contacts the Objectball the line from center to center iz at 30.0dg to the Qball line of travel. At midimpakt, due to temporary but serious ball deformation at the contact, the line from center to center iz at say 30.5dg. At the end of impakt perhaps 31.0dg. Hence, the object ball will depart at 30.5dg (altho ball-to-ball friktion can make up to plus or minus say 3.0dg difference to this). In MathLand, the ball iz assumed to be infinitely hard (ie zero flatspot), and the impakt time to be infinitely small, hence the obove figurez are uzually taken to be 30.0dg, 30.0dg, and 30.0dg -- hence in MathLand the red departs at 30.0dg (ignoring friktion).
I notice on chalkboard that one player sez that he allowz for the extra cut-angle that he gets for higher speedz, but he must be very accurate, i reckon that the extra angle between slow and fast speedz would havtabe less than 5mm at the pocket.
I myself only make such an allowance when i am putting draw (bottomspin) on the Qball, koz here the cut-angle iz inkreeced by a lot (ie u get a finer cut-angle) -- but this iz due to throw, not flatspot sqeez.
However, when i say that flatspot sqeez allwayz inkreecez the cut-angle, i am really only referring to MathLand. In the RealWorld we all play by feel (even thoze who dont), and the sqeez and throw etc are all in our familiar equation for feel. This feel iz probably based on medium speed play -- hence we dont really havta make any/much cut allowance for shots that are slower or faster. Regardz... MadMac. <hr /></blockquote>Thanks MadMac. I get the same numbers for the half-ball hit and essentially the same for the 1/8 ball hit. But Bob Jewett posted on this subject on a different forum and figured the maximum increase to be less than a quarter of a degree at all cut angles: Here is a link to it:

I'm wondering where the discrepancy lies?

Also, if you assume the contact time to be nearly the same at all cut angles and speeds, I don't see the added angle increasing nearly exponentially as you approach 90 degrees. The speed along the tangent line simply approaches v ( sine(theta) --&gt; 1) which would limit the added angle to not much more than you get for a 1/8 ball hit.

I'm not saying this is right, but would like to know why you figure the added angle to be much more than this for shots near 90 degrees? Thanks.

Jim

cushioncrawler
10-25-2005, 05:37 AM
Hi Jim -- My equation for impakt time derived from my ball load tests &amp; my impakt tests givz T(sec)=0.003979*V^(-0.0531) where V iz the full-ball impakt speed in m/s. Hence T iz 0.00035sec @ 5.5m/s, and 0.00045sec @ 0.1m/s, and in theory goes thru the roof @ smaller speedz. This iz for the smaller (52.5mm) English Billiardz ball. Timez for the larger (&amp; softer??) pool ballz should be longer still, but if Marlow'z electronik mezurements show 0.00030sec then i wouldnt argue. But i would argue that impakt timez do get longer at lower speedz, even tho this iz hardly 50% in the sensible range.
Az u say, the Qball'z tangential speed duznt inkreec much az the cut angle nearz 90dg -- but my calculationz showing the inkreecing flatspot sqeez angle are based mainly on impakt time.
I hav checked my calcs &amp; they are ok in the sensible speed range, which woz the original goal. But the equationz showing the geometry of 2 spherez colliding are not accurate enough at say 85dg, 86dg, 87dg, 88dg, 89dg &amp; 90dg cut anglez. This bekumz glaringly obvious when u look at 90dg -- the flatspot at a 90dg cut angle iz obviously zero, koz this iz a near miss, hence the flatspot sqeez should be zero -- which in my calculationz it woz. So, i expect that if i amend my computer program, the "allmost exponential" rize in flatspot sqeez will peak at 0.70dg at say 85dg, and then glide down to zero at 90dg.
A cut angle of 85dg involvz a contact a bit thinner than 1/200th ball -- so my original calcs are ok up to that. By the way, an 1/8th ball givz a cut angle of only about 61.5dg -- this surprizez me every time i see it -- it's a thicker contact than we think.
Re Bob'z lowish figure for sqeez -- this probably arizez from hiz smaller impakt time, and might be correct -- but i would bet that it should rize and rize az u approach 85dg, and then (in line with yor query) fall to zero at 90dg.
Most of my stuff goes back to 1999 -- have u ever tryd to decipher old computer programz??? Regardz... MadMac.

Bob_Jewett
10-25-2005, 04:17 PM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> Hi Jim -- My equation for impakt time derived from my ball load tests &amp; my impakt tests givz T(sec)=0.003979*V^(-0.0531) where V iz the full-ball impakt speed in m/s. ...
<hr /></blockquote>
I assume you dropped a zero from 0.0003979. I think that Hertz' law predicts that the time goes as the fifth root of the inverse of the speed or V^(-0.2). Why do you get a different result?

cushioncrawler
10-25-2005, 05:23 PM
Hi Bob -- yes, it iz 0.0003979, thanks. I derived the equation for imakt time from 2 of my rough homemade tests -- hence it iz only az accurate az theze tests.
Flatspot/Impakt-Speed test, done on verandah, givz D(flatspot dia mm)=2.4561 V^(0.3885) where V iz m/s impakt speed.
Flatspot/Load test, done in driveway, givz D=0.5476 F^(0.3698) where F iz kg. Knowing the mass of the ball (0.138kg) and uzing Newton, one can derive my very impressive looking equation for time.
The equation(z) might stink for 3 reezonz -- my driveway and verandah are still waiting for accreditation -- the load test woz a statik test (ball-ball impakts are anything but statik) -- az all of my stuff woz for my own satisfaction and never meant to be public, i woznt worryd about which ball(z) i grabbed out of my trunk, i woz happy to uze pool ballz or billiard ballz etc, super crystalate (polyester they say) or aramith (phenolic rezin they say) etc, and woznt worryd about exakt weights and sizez. I satisfyd myself whether Impakt Time woz 0.01sec or 0.001sec or 0.0001 sec and i woz happy. So i dont expect to get a Noble Prize for billiardz.
But i could never understand Hertzian stuff (did he get a Noble Prize???). Over the yearz i remember seeing a few research paperz where Hertz had to be uzed -- but in the paperz where Hertz didnt havtabeuzed, but where the researcher woz interested enuff to check what Hertz predikted, no test rezults ever got within a faktor of 2 (or 0.5) of what woz predikted. Regardz.... MadMac.

Jal
10-26-2005, 10:22 AM
<blockquote><font class="small">Quote cushioncrawler:</font><hr> Hi Jim -- My equation for impakt time derived from my ball load tests &amp; my impakt tests givz T(sec)=0.003979*V^(-0.0531) where V iz the full-ball impakt speed in m/s. Hence T iz 0.00035sec @ 5.5m/s, and 0.00045sec @ 0.1m/s, and in theory goes thru the roof @ smaller speedz. This iz for the smaller (52.5mm) English Billiardz ball. Timez for the larger (&amp; softer??) pool ballz should be longer still, but if Marlow'z electronik mezurements show 0.00030sec then i wouldnt argue. But i would argue that impakt timez do get longer at lower speedz, even tho this iz hardly 50% in the sensible range.
Az u say, the Qball'z tangential speed duznt inkreec much az the cut angle nearz 90dg -- but my calculationz showing the inkreecing flatspot sqeez angle are based mainly on impakt time.
I hav checked my calcs &amp; they are ok in the sensible speed range, which woz the original goal. But the equationz showing the geometry of 2 spherez colliding are not accurate enough at say 85dg, 86dg, 87dg, 88dg, 89dg &amp; 90dg cut anglez. This bekumz glaringly obvious when u look at 90dg -- the flatspot at a 90dg cut angle iz obviously zero, koz this iz a near miss, hence the flatspot sqeez should be zero -- which in my calculationz it woz. So, i expect that if i amend my computer program, the "allmost exponential" rize in flatspot sqeez will peak at 0.70dg at say 85dg, and then glide down to zero at 90dg.
A cut angle of 85dg involvz a contact a bit thinner than 1/200th ball -- so my original calcs are ok up to that. By the way, an 1/8th ball givz a cut angle of only about 61.5dg -- this surprizez me every time i see it -- it's a thicker contact than we think.
Re Bob'z lowish figure for sqeez -- this probably arizez from hiz smaller impakt time, and might be correct -- but i would bet that it should rize and rize az u approach 85dg, and then (in line with yor query) fall to zero at 90dg.<hr /></blockquote>Thanks again for your generous explanations. I appreciate you sharing this. I have nothing to add and need to do more studying of the geometry and Hertz's Law (which may or may not be a good thing /ccboard/images/graemlins/smile.gif ).<blockquote><font class="small">Quote cushioncrawler:</font><hr>
Most of my stuff goes back to 1999 -- have u ever tryd to decipher old computer programz??? Regardz... MadMac. <hr /></blockquote>I feel your pain!

Jim