No throwing a ball in...

[MacGyver]

"Sorry for my ignorance but what are SIT and CIT?"

SIT is "Spin Induced Throw" This means if you have english on the cue, it will transfer(like a gear though it has been suggested this model is wrong, but when at the pool table it does make sense) to the object ball.

CIT is "Cut Induced Throw" This means anytime you hit a ball non-straight(cutting it), it will impart some english in the direction of the scraping of the Cueball(ie cutting the ball to the right will put left english on the object ball).

SIT has a much greater effect than CIT, meaning you can throw a ball with english a lot more than you can just by cutting it.

but make no mistake, they definatly DO affect it.

Just learning simple banks this becomes apparent, because if you learn to naturally change the angle based on cut(not using a system using english) you'll have to move the rail contact point towards the Cueball(if cueball to left move aim left, if cueball to right, move contact point right)...

Or else you won't be making those banks...

 

[mikepage]

SIT is "Spin Induced Throw" This means if you have english on the cue, it will transfer(like a gear though it has been suggested this model is wrong, but when at the pool table it does make sense) to the object ball.

Here's one reason it doesn't make sense. With gears, the two surfaces are connected. That means if one moves faster, the other will move faster too. If the gear analogy worked, then increasing the spin on the cueball would increase the throw on the object ball. But it doesn't.

CIT is "Cut Induced Throw" This means anytime you hit a ball non-straight(cutting it), it will impart some english in the direction of the scraping of the Cueball(ie cutting the ball to the right will put left english on the object ball).

SIT has a much greater effect than CIT, meaning you can throw a ball with english a lot more than you can just by cutting it.

No, it's not a greater effect; it's the same effect. You can throw an object ball the same amount with one as with the other. The object ball doesn't know *why* the cueball is sliding across it (angle or spin or some combination of these), just that it is.


mike page
fargo

 

[MacGyver]

No, it's not a greater effect; it's the same effect. You can throw an object ball the same amount with one as with the other. The object ball doesn't know *why* the cueball is sliding across it (angle or spin or some combination of these), just that it is.

it may be the same "effect", ie "Throw", but there are differnt methods of INDUCING throw, thus SIT and CIT terms.

I don't see how you can disagree that SIT isnt greater than CIT, look at colin's video or just sit at a pool table, you can get way more angle of throw from a ball using english than just cutting it.

 

[mikepage]

it may be the same "effect", ie "Throw", but there are differnt methods of INDUCING throw, thus SIT and CIT terms.

I see where you're coming from. But I think pretending there are two different effects has its downsides. For instance, if you cut a ball to the left, you get some throw -- CIT in your terminology. Now if you hit the same shot but this time with some inside english, you should get some SIT as well, right? And that CIT should act in the same direction as the CIT. But the resulting throw ( the CIT plus the SIT) is about the same as the CIT alone. In fact it is sometimes a little *less* than the CIT alone.

I don't see how you can disagree that SIT isnt greater than CIT, look at colin's video or just sit at a pool table, you can get way more angle of throw from a ball using english than just cutting it.

I think I've done this experiment. But I'll try it again. Here's the plan

For the "SIT"

Place an object ball on the headspot.
Aim the cueball for a full hit aiming the object ball over the footspot.
See how far (on the bottom rail) you can throw it with spin.


For the "CIT"

Freeze two balls at the headspot aiming toward the footspot. Drive the cueball at an angle into the first of these. See how far (on the foot rail) you can throw the object ball.

mike page
fargo

 

[Jal]

...I think I've done this experiment. But I'll try it again. Here's the plan

For the "SIT"

Place an object ball on the headspot.
Aim the cueball for a full hit aiming the object ball over the footspot.
See how far (on the bottom rail) you can throw it with spin.


For the "CIT"

Freeze two balls at the headspot aiming toward the footspot. Drive the cueball at an angle into the first of these. See how far (on the foot rail) you can throw the object ball.

Mike, I agree with your overall point, but there is something else to consider. On a cut shot the 'apparent' throw is less than the frictional effect because of the added cut angle, due to the finite compression time - the line between centers changes during contact.

A rough estimate is that the added angle is:

del_theta = Arcsin[(1/2)(v)(sin(theta))T/R]

where T is the contact time and v is the cueball's speed and theta is the cut angle.

For theta = 60 deg, v = 10 mph, and T = .0002 sec, I get .78 deg. This is more than the predicted throw, .66 deg, for a stun shot with no english!

I don't know if you agree with the above equation (I think it underestimates del_theta somewhat because of the different tangent line that's really in effect). I should mention that Bob Jewett calculated this once and gotten much smaller numbers - I don't know the details.

Jim

 

[pooltchr]

Better yet, try freezing the cue ball to the object ball straight down the table...shoot into the cue ball straight down the table with side spin and see how much "SIT" you get.
Then freeze two object balls the same way, and shoot the cue ball from an angle into the first ob, and see how much CIT you can get.
Steve

 

[mikepage]

Mike, I agree with your overall point, but there is something else to consider. On a cut shot the 'apparent' throw is less than the frictional effect because of the added cut angle, due to the finite compression time - the line between centers changes during contact.

A rough estimate is that the added angle is:

del_theta = Arcsin[(1/2)(v)(sin(theta))T/R]

where T is the contact time and v is the cueball's speed and theta is the cut angle.

For theta = 60 deg, v = 10 mph, and T = .0002 sec, I get .78 deg. This is more than the predicted throw, .66 deg, for a stun shot with no english!

I don't know if you agree with the above equation (I think it underestimates del_theta somewhat because of the different tangent line that's really in effect). I should mention that Bob Jewett calculated this once and gotten much smaller numbers - I don't know the details.

Jim

I haven't looked at the details yet. But 10 mph is close to Allison's break speed! Let's try 2 meters per second and a 45 degree cut. So the sideways speed is something like 1.5 meters/second. The contact time is something like .0002 s, but for the effect you're talking about, we should look at where the cueball is after half the contact time to get an averaged result. At 1.5 meters/second, the cueball only travels .15 mm in .0001 s, right? That's pretty small. I'll have to think about this...

mike page
fargo

 

[juanbond]

Great vids, Colin. Great discussion, everyone, I'm really enjoying it.

I'm surprised that there has been little or no mention to the factor of cue/object ball cleanliness/dirtyness within this discussion of throw... This makes a relatively large difference in the amount of throw applied to an object ball. In other words, the impulse of the collision is changed dramatically depending on how smooth or dirty the ball surfaces are. If your balls were perfectly smooth (least friction possible), you would find it very hard to induce any kind of throw, due to the extremely short duration of the impulse during collision. On the other hand, lots of friction makes for a longer impulse, and thus more throw.

 

[Jal]

I haven't looked at the details yet. But 10 mph is close to Allison's break speed! Let's try 2 meters per second and a 45 degree cut. So the sideways speed is something like 1.5 meters/second. The contact time is something like .0002 s, but for the effect you're talking about, we should look at where the cueball is after half the contact time to get an averaged result. At 1.5 meters/second, the cueball only travels .15 mm in .0001 s, right? That's pretty small. I'll have to think about this...

Okay, I did choose a case which would highlight the effect. The equation does have a factor of 1/2 to yield, as you indicate it should, 1/2 the contact time.

The .15mm gives an added angle of:

Arcsin(.15mm/1.125") = Arcsin(.15/(25.4*1.125)) = .30 deg.

The throw at this speed should be around 1.8 deg (stun without sidespin). So the effect is much smaller - you might say borderline in significance.

I wonder how figuring this based on the adjusted tangent line affects it? Since you have compression acting along this direction, I'm beginning to see where the numbers should be smaller. I have to think about it more too.

Jim

 

[BRKNRUN]

All you really need to do to show that CIT exists is to set this shot up. I promise on a 9-foot table you will miss all day with any angle cue. And I also "bet" I can make this shot all day long....(Darn spots!!)...

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SIT is also easy to prove exists.......Just go ask the best 9-baller on the planet...Earl Strickland....