I've seen Larry Nevel cut a ball using inside English that seems impossible. I haven't gotten a chance to ask him how he does it, but it appears he curves the ball slightly using inside. Maybe he'll share his secret here.
What's the lowest cut shot angle possible??
- Thread starter Maniac
- Start date
> There used to be a player from Kentucky named Mike Patton that shot these 3 shots during his warm-up session,and he rarely missed them.
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In the first one he's using low right,he's using low left on the second,on the 3rd he's using a touch of high right or outside,with more overspin than sidespin. He might miss any one of these 3 shots twice in 10 tries,and have seen him make each 10 times in a row. The first one is really tough at the speed he shoots it,drilling the shot. He slows it down some on the second but still whacks it,and the 3rd one is with just enough speed to make it 2 rails out of the corner and back to the nameplate,the tricky part here is overcutting it enough to not scratch. He admitted to me that with the spin he uses,he's having to aim to miss the shot and using the curvature of the cue ball to adjust the actual contact point. Larry can probably make all 3 of these. Tommy D.
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In the first one he's using low right,he's using low left on the second,on the 3rd he's using a touch of high right or outside,with more overspin than sidespin. He might miss any one of these 3 shots twice in 10 tries,and have seen him make each 10 times in a row. The first one is really tough at the speed he shoots it,drilling the shot. He slows it down some on the second but still whacks it,and the 3rd one is with just enough speed to make it 2 rails out of the corner and back to the nameplate,the tricky part here is overcutting it enough to not scratch. He admitted to me that with the spin he uses,he's having to aim to miss the shot and using the curvature of the cue ball to adjust the actual contact point. Larry can probably make all 3 of these. Tommy D.
Maniac said:
The lowest cut shot angle is 0 degrees, a straight in. I know he is actually asking what is the greatest possible angle for a cut shot. Just wanted to make sure he understood the terminology. It might help in presenting this elsewhere.
My opinion is that a ball can be cut as much as 120 degrees, if it is frozen to the rail. Yes, you must go rail first with extreme inside english, but it is a cut shot. As for cutting balls out on the table, a 90 degree cut may actually be possible, by using extreme throw english. That is, by using excessive right hand english, and hitting the object ball wafer thin, you may succeed in cutting it to the left a full 90 degrees.
I'm not going to bet on this shot....yet.
Jude Rosenstock said:
Jude, sounds like you're talking about the other side of the angle. His 86 or 88 are your 92 or 94.
You're right, though, it seems like it would be exceedingly difficult to use spin to cut a ball thinner than perpendicular. I would think that if you're looking for a shot you can make with any consistency, you can probably cut a ball thinner without throw.
-Andrew
To be a nitpicker, if you go rail first on a cut shot, the true cut angle is not the incoming angle to the rail, since the CB is cutting the OB from the other side (rail side). The true cut angle the CB makes with the OB on contact, if you go rail first, should be close to (but less than) 90 degrees, even if the incoming angle is greater than 90 degrees.jay helfert said:
jay helfert said:
You're right. From the untrained eye, it would appear to be a cut-shot exceeding the 90 degree mark but it is not a cut-shot. It's a kick-shot and the cut-angle is wider than 90 degrees.
We are simply talking about cut-shots where the only elements involved are the cue-ball and object-ball. You are including the rail which is only confusing the issue. Let's keep it simple. No kick shots please.
Yes, it's possible to cut a ball 180 degrees this way. The shot, however, is not practical in any reasonable situation. This was discussed in the previous thread that was more or less about this topic:td873 said:
http://forums.azbilliards.com/showthread.php?t=38109
I think it's reasonable to restrict the discussion to shots that can be made at least 10%-20% of the time. Here's the open-table setup I tried: put the object ball on the center spot (exactly between the two side pockets, not usually marked on tables). Put the cue ball on the head string. See how close you can get the cue ball to the head spot and still cut the object ball in the side. My best for a reasonable percentage (about 15%) was 80 degrees.
If anyone wants to try this and report the result, just let us know how close the cue ball was to a ball spotted on the head spot and I'll do the geometry to find the angle.
Oh, and with a masse shot, it's possible to cut a ball more than 180 degrees. The cue ball can go out on the left side of the object ball and reverse and come back on the right side. I'd say that counts as more than 180 degrees of cut.
Bob Jewett said:
Just asking for a clarification (not trying to split hairs)???
If I put the object ball on the center spot and the cue ball on the head spot and cut the object ball in the side pocket:
1. Would that be a 90 degree cut or more than a 90 degree cut?
2. Should the cue ball be placed one ball width to the side of the head spot to be 90 degrees?
If the object ball goes into the center of the side pocket, shooting with the cue ball on the head spot, it is clearly more than a 90 degree cut. (About 3 degrees more, I think.)Tennesseejoe said:
If you put an object ball on the head spot, and then freeze the cue ball to that ball (and have it on the headstring), and then remove the ball from the head spot, cutting the object ball into the center of the side pocket will be a 90-degree shot.
I cut (2) balls backwords on a 5x10' snooker table with a curve shot as
follows:
Frozen on the rail the cue ball & object ball, you have seen the cue ball
at one end of the table on the middle diamond & the object ball also in the middle of the table, people shoot this one all the time at 90 degrees and make the object ball.
Here is what I did: Move the object ball to the left diamond on the back rail and the cue ball at the opposite end of the table frozen on the right
diamond, I cut (2) object balls backwords to the right side of the table
into the right corner pocket.
"POOL TABLE"
__________________
Object ball is here """o"" (Frozen on left diamond)
"o" (Cue Ball is here, frozen on right diamond)
__________________ ... I made (2) balls in this corner pocket on a 5x10 snooker table
follows:
Frozen on the rail the cue ball & object ball, you have seen the cue ball
at one end of the table on the middle diamond & the object ball also in the middle of the table, people shoot this one all the time at 90 degrees and make the object ball.
Here is what I did: Move the object ball to the left diamond on the back rail and the cue ball at the opposite end of the table frozen on the right
diamond, I cut (2) object balls backwords to the right side of the table
into the right corner pocket.
"POOL TABLE"
__________________
Object ball is here """o"" (Frozen on left diamond)
"o" (Cue Ball is here, frozen on right diamond)
__________________ ... I made (2) balls in this corner pocket on a 5x10 snooker table
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Jal said:
Actually, no. It's impossible. Part of the reason WHY it's impossible has to do with the maximum amount of rotation you can place on a cue-ball. The other is the degree of influence it will have because you're skimming the ball. A cue-ball cannot spin faster than it is shot. In otherwords, the force shown on the vertical axis cannot be greater than the force shown on the horizontal axis, at least, not when you are shooting horizontally. Imagine trying to spin a coin by simply flicking it with your finger. It's impossible unless you're holding the quarter in place when you do it.
With that said, when the cue-ball contacts an object ball on a very thin cut, you get a roll-off effect. There's no force that would bring the object-ball backwards and it's so minimal that the line is almost identical to what would be achieved with zero english. The thow can only be seen in angles that are ahead of the point of contact. The more they are ahead of the point of contact, the more you can throw. This is a simple element of transfer of force.
roncook said:
I'm too tired to try and imagine what you are saying. Please use the Wei table. You can find it under "Quick Links" on the top-menu.
Bob Jewett said:
Have you also considered pocket-width? Perhaps you should place the object-ball an inch closer to the foot-spot so that you can be assured the ball can only be pocketed if it's cut at a true 90 degrees. Otherwise, you may be simply hitting the footside of the side pocket (they are often more than 2-balls width).
The bottom line is, pocketing a ball is not always proof of a physical theory. The pooltable is, by far, not a perfect test environment. There are shots that create illusions, even to the most trained eye. If you send the cue-ball on a straight path and next come in contact with an object ball, that object ball MUST travel forward. This is a plain and simple truth. Ask any physics professor and they will agree. This fact alone makes a 90 degree cut impossible.
If you think of it in two dimensions, everything becomes clear. A 90 degree cut will send the object ball neither forward nor backward. That would give it zero velocity which, in physics means, it didn't move. For it to move backwards would mean that a negative amount of force would need to be applied. If someone can figure out a way to achieve that, they should call the Nobel committee.
If you think of it in two dimensions, everything becomes clear. A 90 degree cut will send the object ball neither forward nor backward. That would give it zero velocity which, in physics means, it didn't move. For it to move backwards would mean that a negative amount of force would need to be applied. If someone can figure out a way to achieve that, they should call the Nobel committee.
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Hi Jude,Jude Rosenstock said:
Simple physics says that if you use a tip offset of two-fifths of the ball's radius (2/5 R), the "inside" surface speed at the equator will be zero. I'm using "inside" in the sense that for right english, the inside surface would be on the left side of the ball . The "outside" surface speed at the equator (on the right side of the ball for right english) will be twice the ball's speed. If you hit with a greater offset, say 1/2 R, the inside surface speed will be negative, or retrograde: a point on the surface at the equator will actually move backward with respect to the table's surface, or the object ball it may contact. In reality things are a little more complicated, but I think it's been verified that you can get the inside surface to move backwards (Jacksonville Project).
I'll admit that what I'm about to say is theory and, as far as I know, hasn't been confirmed by tests. But it isn't atomic physics, and until proven wrong I'll believe it's right.Jude Rosenstock said:
It predicts that you will get more throw with cut angle (if you get the spin right) exactly because of the skimming or grazing contact. This reduces the compression impulse (average force times the duration of the contact) acting along the line between the centers of the balls. For reasons that are difficult to put in way that would sound intelligible (I'll try if you like), this must increase the coefficient of friction between the balls. The throw angle (tangent of) is equal to this coefficient, so the throw must increase.
We agree that the average compression force between the balls is less as the cut angle increases. And as a direct consequence, so is the average friction force. But because of what was just said (coefficient stuff), the friction force does not diminish quite as much as the compression force. As a result, the ball gets thrown sideways by the friction with a little more force in proportion to how much it gets propelled forward by the compression. The effect is more throw.
That's what the theory says anyway. If correct, you should be able to make the initial geometric cut at, say 86 degrees, and then throw it backward more than 4 degrees to achieve a greater than 90 degree cut. But the spin tolerances are insanely tight and it would probably be a matter of sheer luck to pull it off.
I doubt that I've convinced you since I've skimmed over a few details, but if you want to look at it closer, I'm willing. Maybe you see something I don't.
Jim
The coefficient of friction can vary depending on the surface speeds. Have you ever heard of static and kinetic friction? Well, there can be two different coefficients of friction for the same materials.Boro Nut said:
I wouldn't try to mock Jal if I were you, he probably knows more physics on this board than anyone.
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Jal said:
However, there is MUCH LESS force getting transferred as you increase the cut (which is something I think you just fail to realize). Look, the amount of spin that would be necessary to throw an object-ball backwards is going to be significantly greater than the speed of the cue-ball. It's gotta be spinning like a top in order to do it. You can't do that, at least, not with a cue alone.
I have to ask: WHAT IS THE JACKSONVILLE PROJECT??? You're referencing something as the basis of your proof yet you provide no links. Does this project apply to pool? I want to see this exact report.
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