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ras314
09-15-2005, 10:09 AM
Now that I can actually see the edges of a ball I'm playing with some old semi-trick shots I used to like.

Anybody care to comment on this one and whether or not it can be done? The balls should be set up so that it takes a 90 deg or better angle between the cb path and the ob path to drop.

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%KJ5P7%LJ5N2%MK6Q4%NJ5R0%OJ5M0%Pa1U4%WB5V2%Xu8V0%Y Z3Z7%ZZ2C3
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Jal
09-15-2005, 04:33 PM
<blockquote><font class="small">Quote ras314:</font><hr>
Anybody care to comment ...<hr /></blockquote>

I'll bite.

If you take the cueball direction as shown literally (90 degrees), I would say no. If not, you could jump it while cutting it at less than 90 degrees.

It is possible in principle, I think, to cut the ball more than 90 degrees by using more than "natural roll" outside english. For average balls, hitting the cueball halfway between the center and edge (if you could manage to do it without miscuing), and cutting the object ball very near 90 degrees, should throw it more than 90 degrees. But the throw diminishes with speed and you need a lot of speed to get the object moving at those angles. According to results I got, you couldn't get the object ball to move the one diamond distance required. A few inches maybe, but not a foot.

And, of course, you could masse it (you could, I couldn't).

Jim

Candyman
09-15-2005, 05:09 PM
The shot you discribed is not a true 90 degree shot. Move the 1 ball up 1 1/8 inches and it will be a 90 degree shot and "no" I don't think you can cut it.

ras314
09-15-2005, 05:22 PM
<blockquote><font class="small">Quote Jal:</font><hr> If you take the cueball direction as shown literally (90 degrees), I would say no. If not, you could jump it while cutting it at less than 90 degrees.

Jim <hr /></blockquote>
I think that is the way it is made, some would argue the point. A masse may be possible, but would be a very low percentage shot, at least for me. I don't think you could get enough throw even with dirty balls. I have made similar shots to this without really understanding what was going on, stupid as that sounds. One clue is the shot doesn't have to be fast and english is not necessary.

BTW what I call a "semi-trick" shot is one I would try in a game.

ras314
09-15-2005, 05:41 PM
You are correct,thanks. The way I set it up for practice is to use a board at right angles to the cushion and move it down the rail until the ob won't drop rolling along the board. And try to line the cb up by using the cue to point in line with the short rail spots. Then the ob distance from the long rail is set one ball radius inside a line between the short rail diamonds.

This looks more like it.

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%KJ5P7%LJ5N2%MK6Q4%NJ5R0%OJ5M0%Pa1U4%WB5V2%Xu8V0%Y Z3Z7%ZZ2C3
%[[1[1%\Z4W3%][5U5%^a4U3
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Barbara
09-15-2005, 05:48 PM
<blockquote><font class="small">Quote ras314:</font><hr> You are correct,thanks. The way I set it up for practice is to use a board at right angles to the cushion and move it down the rail until the cb won't drop rolling along the board.

This looks more like it.

START(
%AZ3V7%BL7P8%CJ5O4%DL7N1%EM7P1%FK6P1%GK6N8%HM7N8%I L7O4%JK6M5
%KJ5P7%LJ5N2%MK6Q4%NJ5R0%OJ5M0%Pa1U4%WB5V2%Xu8V0%Y Z3Z7%ZZ2C3
%[[1[1%\Z4W3%][5U5%^a4U3
)END <hr /></blockquote>

It looks like you have more than a 90 degree cut to the OB.

In any case, I don't believe a 90 degree cut is possible because of CIT (collision induced throw) and the fact that you'd have to hit that sucker about 90 miles an hour because of the angle to get any speed on the OB.

This is a situation where it's theoretically possible, but physically impossible.

Barbara

ras314
09-15-2005, 06:21 PM
<blockquote><font class="small">Quote Barbara:</font><hr>
It looks like you have more than a 90 degree cut to the OB.

In any case, I don't believe a 90 degree cut is possible because of CIT (collision induced throw) and the fact that you'd have to hit that sucker about 90 miles an hour because of the angle to get any speed on the OB.

This is a situation where it's theoretically possible, but physically impossible.

Barbara <hr /></blockquote>
Hi Barbara,

I had to edit the description several times to try to get it right. The idea is to set up a shot that does take more than a 90 deg cut to make.

MacGyver
09-23-2005, 10:52 PM
I hate to get all math but every shot with an object ball has 3 values basically:

The vector of the object ball, and then the vector of the cueball as broken up into an X and Y vector in terms of the object ball vector.

So basically any cue ball has an X and a Y component.

The Y component is what transfers completely(in a stun shot) to the object ball and X is what is left over(tanget to the object ball)

The point is that you can get Y's power in terms of original hit by taking cosine of angle between X+Y(path of cue ball prehit) and Y(path of object).

At 90 this is 0.
At 89 this is 1.7%.
88 = 3.5%
87 = 5.2%
86 = 7%
ect

So while you may be getting some spin induced throw, there is no way you are cutting at 90 or more, and even at 89 you would need to hit the cueball 100 feet for every foot that the object ball would move.(really it would be 1%, which might not work out because of roll, skid, friction, but the idea is that you need to hit it 100x harder than you want it to move and that is at 89 degree's)

So other than hoping for some awesome throw, I don't think a cut angle of 90 degrees is possible.

ras314
09-24-2005, 11:09 AM
<blockquote><font class="small">Quote MacGyver:</font><hr> So basically any cue ball has an X and a Y component..........
...So other than hoping for some awesome throw, I don't think a cut angle of 90 degrees is possible. <hr /></blockquote>
Don't forget the Z component.

Bob_Jewett
09-24-2005, 03:38 PM
&gt; The vector of the object ball, and then the vector of the
&gt; cueball as broken up into an X and Y vector in terms of the
&gt; object ball vector.

Without a diagram, or at least an explanation, this doesn't make much sense.

&gt; So basically any cue ball has an X and a Y component.

I guess you mean the velocity of the cue ball has those components.

&gt; The Y component is what transfers completely(in a stun shot)
&gt; to the object ball and X is what is left over(tanget to the object ball)

This leads me to believe that you are always shooting the object ball along the Y axis. While that's one way to arrange the diagram, lots of people do it other ways. You might want to check out Dr. Dave's pages such as http://www.engr.colostate.edu/~dga/pool/technical_proofs/TP_3-1.pdf where the cue ball's path begins along the Y axis.

&gt; The point is that you can get Y's power in terms of original hit
&gt; by taking cosine of angle between X+Y(path of cue ball prehit)
&gt; and Y(path of object).

That will give the velocity of the object ball rather than the power. The distance that the object ball rolls is proportional to the square of the initial velocity.

&gt; ... , but the idea is that you need to hit it 100x harder than you
&gt; want it to move and that is at 89 degree's)

If you phrase it in terms of the energy of the two balls, I'd agree. A typical player can hit the ball fairly easily at 5 meters per second, but not 10 meters per second and still keep the cue ball on the table. An 89-degree cut with such a very fast cue ball would get the object ball to travel only about a ball's diameter.

MacGyver
09-24-2005, 03:50 PM
Well I was inserting math into a non-mathematical discussion so I wasn't double checkin each term or value but trying to give an overview of why a 89-90 degree cut is practically impossible.

Regardless of the label you put on it, anytime a cueball strikes a ball, There is a vector going in the direction of the object ball, a vector going in the tangent of the object ball, and the cueball which would be traveling in the (direction of objectball + direction of tangent of object ball).

So if Y is where the objectball is going, X is tangent to that and where the cueball is going after hit, then before the hit the cueball is X+Y... it only transfers the Y component in a hit, while X remains the same.

Well anyway, yes power and velocity (and even distance of roll) are not equal terms but again it was a non-scientific discussion and I was just trying to get across that a 90 degree cut is impossibily and a 89 degree cut is practically impossibily, due to the low transfer of energy.

Snapshot9
10-12-2005, 07:58 AM
That's a pretty good argument except for one
thing you both left out, and that is english
or spin on the cue ball, which creates force
on the object, and therefore moves the object.
I have played Pool 44 years, and I say a 90 degree
cut is possible and makeable. And, yes, I know
about using inside english on the cue to gently
masse the cue into the 90 degree cut on an object
ball, but I have done it before without masse-ing
the cue ball.

Sid_Vicious
10-12-2005, 11:11 AM
If it is not beyond 90 degrees, I too have made it without any masse. I use so much outside english on that stroke for max spin that most players wouldn't trust their tip to hold. A local money player taunted me into it after using my cue and seeing it's playing ability. A true 90 is a tester in all regards though...sid

Stretch
10-12-2005, 11:57 AM
<blockquote><font class="small">Quote Snapshot9:</font><hr> That's a pretty good argument except for one
thing you both left out, and that is english
or spin on the cue ball, which creates force
on the object, and therefore moves the object.
I have played Pool 44 years, and I say a 90 degree
cut is possible and makeable. And, yes, I know
about using inside english on the cue to gently
masse the cue into the 90 degree cut on an object
ball, but I have done it before without masse-ing
the cue ball. <hr /></blockquote>

Darn i can't see how you've set it up. If the ob is against the rail you've a better chance at making it. 90 deg. angles then become very possible by spinning the cb into the shot rail first. Plus pocket slop, i've even double kissed them down that way, it's like pushing it down. If the ob is out on it's own at 90%................a bank or safe is lookin real good. Actually the closer the cb is to the ob the less likely you'd ever make one i'd imagine. I don't know about the exact angle but i've cut the paint off close in shots by aiming to miss useing outside english to throw it down, and i can connect with the long ones with low inside. You just aim to miss and let the low inside swerve the difference. I luv shots where you're aiming up to miss. Since i discovered at an early age that i was real good at missing lol.

Actually small swerves are not hard to do or control (to me anyway) and i believe it's an esential tool in manufacturing angles where there's none to get at......all part of the game /ccboard/images/graemlins/smile.gif St.

Bob_Jewett
10-12-2005, 06:16 PM
<blockquote><font class="small">Quote Sid_Vicious:</font><hr> If it is not beyond 90 degrees, I too have made it without any masse. I use so much outside english on that stroke for max spin that most players wouldn't trust their tip to hold. A local money player taunted me into it after using my cue and seeing it's playing ability. A true 90 is a tester in all regards though...sid <hr /></blockquote>
Lots of people will look at an 80-degree cut and think it's 90.

Here is a 90-degree shot: put the cue ball on the head spot. Place the object ball on the center of the foot rail, and then move it 1/2 ball off the cushion. You should be able to cut the object ball into either corner pocket as they are both 90-degree cuts.

Is that the sort of shot you meant?

If that shot is longer than you want to try, place the cue ball on the head spot and the object ball on the head string one ball off the side cushion. Cut the ball into the head pocket.

Vagabond
10-12-2005, 08:10 PM
Hi ras314,
I shall reach sandia Casino, Albuquerque by 8 pm tomorrow.Introduce youself to me.Ask Sarah Rousey or Carol or Jenifer Baretta to point the way towards me.I will show u how to pot those `so called can not be done` cut shots.I do them with INSIDE ENGLISH.I do not know anything about wei tables and hence did not respond on WEI

Bob_Jewett
10-13-2005, 09:47 AM
<blockquote><font class="small">Quote Vagabond:</font><hr> ... I do them with INSIDE ENGLISH.I do not know anything about wei tables and hence did not respond on WEI <hr /></blockquote>
Have you tried the shot I described? I'd pay to see it done with inside.

Jal
10-13-2005, 05:47 PM
<blockquote><font class="small">Quote Snapshot9:</font><hr> That's a pretty good argument except for one
thing you both left out, and that is english
or spin on the cue ball, which creates force
on the object, and therefore moves the object.
I have played Pool 44 years, and I say a 90 degree
cut is possible and makeable....<hr /></blockquote>It is possible to make this shot, and over quite a distance with less than superhuman speed, but a new layer of precision is required for an already difficult shot. It can be done if you get the outside english within some very tight tolerances, so as to generate a great deal of throw. The throw itself doesn't generate much object ball speed, and most of it is in the wrong direction, but it allows you to hit it at a more moderate cut angle than, say, 89 degrees.

For instance, an 86 degree cut shot hit at 12 mph can throw it more than 90 degrees and move the object ball over a foot. An 85 degree cut shot at 14 mph can also accomplish the same thing and move the object ball around 30".

But the trick is to get the spin just right. For the 86 degree shot you have to make tip contact with the cueball within a range of offsets about 0.03" wide. For the 85 degree shot, it's only about 0.005" wide.

In a way, this level of precision is possible, and in a way it's not. An ordinary spot shot requires you to hit an area of the cueball within a tolerance of 0.005". But here you have a clear target (the object ball) to help you with the alignment. Trying to locate such a small band on the cueball, for the purpose of applying english, is much more difficult.

If only for amusement, here is a table which lists the results of various shots near 90 degrees.

http://ww2.netnitco.net/users/gtech/90cut.txt

It shows the smallest and largest spin (as a fraction of cueball speed) required to throw the object ball at an angle greater than 90 degrees and the size of the span of english offsets required to get the spin within this range. It also shows the maximum throw you can expect, and the corresponding cueball spin/tip offset and distance the object ball will travel.

The numbers aren't to be taken too literally since conditions vary, but I think they provide an idea of what can be done, and how hard it is to do it!

Jim

Colin
10-14-2005, 01:42 AM
Jal,
In reference to this:
http://ww2.netnitco.net/users/gtech/90cut.txt

I had assumed that the outside contact edge would not be turning faster than the CB. That it might be stationary at best. But I haven't looked deeply into that.

What you are suggesting is that the outside edge is actually moving backward and can push the OB back at a wider angle.

How was this table generated, what assumptions, and has there been any testing to confirm this?

Bob_Jewett
10-14-2005, 09:38 AM
<blockquote><font class="small">Quote Jal:</font><hr> ... For instance, an 86 degree cut shot hit at 12 mph can throw it more than 90 degrees and move the object ball over a foot. An 85 degree cut shot at 14 mph can also accomplish the same thing and move the object ball around 30".
... <hr /></blockquote>
Well, yes, but... To get the cue ball to move at 14MPH with retrograde side requires the player to swing the stick faster than most players do on break shots. Retrograde side is when the side of the cue ball is actually moving backwards relative to the table. Retrograde spin of about 20% (rw/v of 1.200 in Jim's table) was observed in the Jacksonville Project, so it is possible at some speeds, but I'm not sure it's possible at 14MPH of cue ball speed.

At that stick speed, it is unlikely that the cue ball will be on the cloth when (if) it hits the object ball.

Has anyone tried my proposed 90-degree cut yet? To recap: put the cue ball on the head spot. Put the object ball on the headstring one ball ball off the side cushion. Cut the object ball into the corner pocket. (I suppose you shouldn't let the object ball hit the side rail on the shot.)

Bob_Jewett
10-14-2005, 10:14 AM
<blockquote><font class="small">Quote Colin:</font><hr> ... I had assumed that the outside contact edge would not be turning faster than the CB. That it might be stationary at best. But I haven't looked deeply into that.

What you are suggesting is that the outside edge is actually moving backward and can push the OB back at a wider angle. <hr /></blockquote>

Measurements from the Jacksonville project showed that part of the cue ball can actually move backwards on a shot if you use extreme spin. An article describing that is available on-line at:

http://www.sfbilliards.com/articles/1999-02.pdf

Jal
10-14-2005, 05:15 PM
<blockquote><font class="small">Quote Colin:</font><hr>...I had assumed that the outside contact edge would not be turning faster than the CB. That it might be stationary at best. But I haven't looked deeply into that.

What you are suggesting is that the outside edge is actually moving backward and can push the OB back at a wider angle.<hr /></blockquote>I think Bob Jewett has answered your question along with providing experimental verification (whew!...I must confess I was a little worried while reading the article).

<blockquote><font class="small">Quote Colin:</font><hr>How was this table generated, what assumptions, and has there been any testing to confirm this? <hr /></blockquote>I don't know how much detail you want so I'll just give a brief summary.

For any of the given cut angles and cueball speeds the program increments the spin/speed ratio in steps of .0001. At each ratio it calculates the amount of throw and adds it to the cut angle. It records the lowest and highest spin in which this is equal to or greater than 90 degrees. The spin ratio it finds that yields maximum throw does agree exactly with more analytical ways of arriving at this.

The corresponding english offsets are back calculated using:

rw/v=(5/2)(b/r) or b=(2/5)r(rw/v) where b is the offset, r is the cueball's radius and w is the spin in radians/sec.

The critical thing then is the throw calculation. Dr. Dave Alciatore who posts here often has a great article on this here:

http://www.engr.colostate.edu/~dga/pool/technical_proofs/new/TP_A-14.pdf

I, however, used a different version of his throw equation because it produces, I think, slightly more favorable results regarding the proposition at hand. His version uses an average value of the coefficient of friction over the impact period, whereas mine tries to account for the changes in it as the relative surface speeds decrease. Both give results which are very close to each other. The math behind mine is here in "ThrowMath.doc"

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

But I'm not sure how accurate either of them are because they both rely on Wayland Marlow's experimental results ("The Physics of Pocket Billiards") to determine the constants involved in calculating the friction coefficient. If he used balls that were initially frozen to each other to measure how much throw is generated, the fact that they were frozen might skew things a bit. (I don't have the book so I don't know.)

And, of course, throw will vary (sort of) with the composition and polish/cleanliness of the balls.

So I listed things to four decimal places not because I have much faith that they are anywhere near this accurate a representation of reality, but because the margins are so small between cutting a ball +90 degrees or not. Whatever the actual values are, they should at least look something like those given in the table.

If you want more details let me know. I hope that helped.

Jim

Bob_Jewett
10-14-2005, 06:11 PM
<blockquote><font class="small">Quote Jal:</font><hr> ... But I'm not sure how accurate either of them are because they both rely on Wayland Marlow's experimental results ("The Physics of Pocket Billiards") to determine the constants involved in calculating the friction coefficient. If he used balls that were initially frozen to each other to measure how much throw is generated, the fact that they were frozen might skew things a bit. (I don't have the book so I don't know.) ... <hr /></blockquote>
I have copies of the book available at \$30 including shipping to the US.

tateuts
10-14-2005, 10:18 PM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr> <blockquote><font class="small">Quote Jal:</font><hr> Has anyone tried my proposed 90-degree cut yet? To recap: put the cue ball on the head spot. Put the object ball on the headstring one ball ball off the side cushion. Cut the object ball into the corner pocket. (I suppose you shouldn't let the object ball hit the side rail on the shot.) <hr /></blockquote>

It was easier for me to shoot from the head spot to the foot rail. If I nick the object ball with lots of outside, I can get it going weakly at the pocket but maybe only half or two thirds of the way there. I made 4 or 5 shots clean at it and made some really good thin hits - I don't think I could do any better than this. The spin definitely helps. I didn't try slamming it hard because I can't see where that would work.

Chris

Jal
10-14-2005, 11:53 PM
<blockquote><font class="small">Quote Bob_Jewett:</font><hr> <blockquote><font class="small">Quote Jal:</font><hr> ... For instance, an 86 degree cut shot hit at 12 mph can throw it more than 90 degrees and move the object ball over a foot. An 85 degree cut shot at 14 mph can also accomplish the same thing and move the object ball around 30".
... <hr /></blockquote>
Well, yes, but... To get the cue ball to move at 14MPH with retrograde side requires the player to swing the stick faster than most players do on break shots. Retrograde side is when the side of the cue ball is actually moving backwards relative to the table. Retrograde spin of about 20% (rw/v of 1.200 in Jim's table) was observed in the Jacksonville Project, so it is possible at some speeds, but I'm not sure it's possible at 14MPH of cue ball speed.<hr /></blockquote>

A cueball speed of 14 mph with a spin ratio of rw/v ~ 1 corresponds, I think, to a speed of about 18 mph with a centerball hit. I'm not sure why you think this, or even a little faster speed, wouldn't be feasible?

<blockquote><font class="small">Quote Bob_Jewett:</font><hr>At that stick speed, it is unlikely that the cue ball will be on the cloth when (if) it hits the object ball.<hr /></blockquote>Yes, this is another problem. I'll admit I wasn't thinking about the higher stick speed required to generate these cueball speeds, and the tendency of the cueball to jump.

Jim

Colin
10-15-2005, 01:27 AM
Thanks Bob and Jal!

I should have recalled the points made regarding overspin in Byrne's book. It should not be surprising to achive +1 spin / speed ratio with extreme side english.

It sure would be cool to capture a 90+ degree cut on high speed video for final proof.

Bob_Jewett
10-17-2005, 02:14 PM
<blockquote><font class="small">Quote Jal:</font><hr> ... A cueball speed of 14 mph with a spin ratio of rw/v ~ 1 corresponds, I think, to a speed of about 18 mph with a centerball hit. I'm not sure why you think this, or even a little faster speed, wouldn't be feasible?
... <hr /></blockquote>
I'm pretty sure the ratio is higher than that. Also, you need more than 1.0 for the spin ratio.

I also believe that at faster speeds, the tip rides around on the cue ball more during tip-to-ball contact, so you have to start closer to the center for extreme power+spin shots, but this needs more theory.