SIT and cut angle

[7stud]

I've read some of Dr. Daves stuff on Spin Induced Throw (SIT), and I've watched the video where he shows that paradoxically maximum SIT occurs with only 50% of maximum side spin--for straight in shots. You might think maximum throw would occur with maximum side spin, but that is not the case.

Periodically, when reading the stuff on SIT, I've wondered how the amount of SIT varies with cut angle--but I could not find that information anywhere, until today. I read in one place that maximum possible SIT occurs with straight in shots. However, I now believe that statement was a misinterpretation of what Dr. Dave said. Dr. Dave only said that maximum SIT for straight in shots occurs with 50% of maximum side spin--he did not state that maximum SIT occurs with straight in shots. So at what cut angle does maximum SIT occur? According to Dr. Dave's research, the maximum possible SIT *increases* with cut angle. In the graph below, the labels on the axes are a little confusing, but as far as I can tell, the amount of SIT is shown on the vertical axis, and the cut angle is shown on the horizontal axis. There is a graph for both slow speed shots(red line) and high speed shots(dashed blue line):

https://billiards.colostate.edu/technical_proofs/new/TP_B-25.pdf

However, as the cut angle increases, you need to hit the CB with more and more side spin to achieve the maximum SIT:

A 1/4 ball hit is 49 degrees, and for a thinner, 60 degree cut, you need to hit the CB with almost 100% of maximum possible side spin. For cut angles greater than 60 degrees, the amount of side spin necessary for maximum throw tails off a bit, but you still need roughly 90% of maximum side spin to achieve maximum throw.

How significant is the increase in throw as the cut angle increases? It looks to me like a straight in shot (0 degrees), has throw of about 5.2, and a 1/4 ball cut (roughly 50 degrees) has a throw of about 5.6, so the increase in maximum throw between a straight in shot and a 1/4 ball cut is about 0.4/5.2 = .08 or 8 percent. So if you can throw a straight in shot 4", then you can throw a 1/4 ball cut about 4 1/4 - 4 3/8" over the same distance.

Maximum throw, under typical conditions, is about 1 inch per foot of OB travel, or 1/2 a ball per diamond on a 9′ table.

Maximum Throw - Dr. Dave Pool Info

Answers to frequently-asked questions about the maximum amount pool balls can be thrown.

billiards.colostate.edu

 

[Patrick Johnson]

As the cut angle increases, you need to hit the CB with more and more side spin to achieve the maximum SIT

Actually the opposite is true.

If half of maximum spin produces maximum throw with no cut angle, and no spin produces maximum throw with a half ball hit, then the range of spin to produce maximum throw is none (at half ball) to 50% (at full ball). More than 50% side spin produces less than maximum throw with every cut angle. For cut angles greater than half ball (30°) you actually need to add some outside spin to get max throw.

pj
chgo

 

[7stud]

and no spin produces maximum throw with a half ball hit,

Where did you get that from? A half ball hit is about 30 degrees, and according to the graph on the bottom of my post, you need 85-90% max spin for max throw at that cut angle.

I also think you can look at it this way: the amount of side spin needed for gearing English increases as the cut angle increases(https://billiards.colostate.edu/bd_articles/2015/jan15.pdf), so to achieve more side spin than gearing English in order to produce Spin Induced Throw, you have to hit the CB with more and more side spin as the cut angle increases. Am I wrong?

 

[Patrick Johnson]

no spin produces maximum throw with a half ball hit

Where did you get that from?

The usual place...

pj
chgo

Maximum cut-induced throw (CIT), with no english, occurs with a slow-speed stun shot at about a 1/2-ball hit.

Maximum Throw - Dr. Dave Pool Info

Answers to frequently-asked questions about the maximum amount pool balls can be thrown.

billiards.colostate.edu

 

[7stud]

Cut Induced Throw(CIT)??? This thread is about SIT.

 

[7stud]

As far as I can tell, the spin percentage for maximum SIT is higher than gearing English at every cut angle, so there is no CIT.

 

[Patrick Johnson]

Cut Induced Throw(CIT)??? Check the title of this thread.

The important thing is the rubbing speed between the balls' surfaces - it's the same at full ball/half spin and at half ball/no spin - and that rubbing speed produces maximum throw. Any amount of rubbing speed (with inside, outside or no spin) that exceeds that will produce less than maximum throw, like a car's tire peeling out from over-acceleration.

pj
chgo

 

[7stud]

The important thing is the rubbing speed between the balls' surfaces - it's the same at full ball/half spin and at half ball/no spin - and that rubbing speed produces maximum throw. Any amount of rubbing speed (with inside, outside or no spin) that exceeds that will produce less than maximum throw, like a car's tire peeling out from over-acceleration.

pj
chgo

You seem to be implying that I am misinterpreting the first graph in my post:

How significant is the increase in throw as the cut angle increases? It looks to me like a straight in shot (0 degrees), has throw of about 5.2, and a 1/4 ball cut (roughly 50 degrees) has a throw of about 5.6, so the increase in maximum throw between a straight in shot and a 1/4 ball cut is about 0.4/5.2 = .08 or 8 percent. So if you can throw a straight in shot 4", then you can throw a 1/4 ball cut about 4 1/4 - 4 3/8" over the same distance.

How do you interpret that graph?

 

[Patrick Johnson]

You seem to be implying that I am misinterpreting the first graph in my post:

I think we're just miscommunicating. I'm responding to your words, not your graphs, and don't understand where/how you're making the distinction between inside and outside spin.

pj
chgo

 

[7stud]

and don't understand where/how you're making the distinction between inside and outside spin.

I never mentioned inside or outside spin. I assumed the graphs were dealing with outside spin.

 

[Scratch85]

Does it help to clarify, if we refer to SIT as throw and CIT as push?


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[Bob Jewett]

Does it help to clarify, if we refer to SIT as throw and CIT as push?

I don't think so.

 

[jsp]

For cut angles greater than 60 degrees, the amount of side spin necessary for maximum throw tails off a bit, but you still need roughly 90% of maximum side spin to achieve maximum throw.

Everything seems intuitive to me except for the above (shown graphically below). Why is it that you actually need less side spin for cut angles greater than 60 degrees to maximize throw? I would have thought the curves would asymptotically approach 100% spin as the cut angle increases, but they don't. The curves actually turn around at some point.

 

[7stud]

I don't think so.

Can you tell me what that first graph is saying? Can you really get more throw with a 1/4 ball hit (49 degrees) than a 1/2 ball hit (30 degrees)?

 

[7stud]

The important thing is the rubbing speed between the balls' surfaces - it's the same at full ball/half spin and at half ball/no spin - and that rubbing speed produces maximum throw. Any amount of rubbing speed (with inside, outside or no spin) that exceeds that will produce less than maximum throw, like a car's tire peeling out from over-acceleration.

pj
chgo

By definition, a 1/2 ball hit with no side spin produces zero Spin Induced Throw(SIT). We also know that maximum SIT for a straight in shot (0 degrees) occurs with 50% of maximum sidespin (confirmed by the first graph). Yet, according to the first graph, a 1/2 ball hit (30 degrees) has a maximum SIT that is greater than a straight in shot( 0 degrees). Therefore, if a 1/2 ball hit with no side spin produces the same amount of throw as a straight in shot with 50% maximum side spin, then a 1/2 ball hit with no side spin does not produce the maximum throw.

I guess that's a tortured way of saying, according to the first graph, it's possible to throw a ball more than a straight in shot with 50% max sidespin.

 

[Bob Jewett]

Can you tell me what that first graph is saying? Can you really get more throw with a 1/4 ball hit (49 degrees) than a 1/2 ball hit (30 degrees)?

I assume it is because for a thinner cut shot the contact between the balls is lighter and that will in general get more throw. (Assumes the cue ball speed is fixed.)

 

[Patrick Johnson]

I assume it is because for a thinner cut shot the contact between the balls is lighter and that will in general get more throw.

Really? I thought the opposite: heavier contact = more friction/throw.

pj
chgo

 

[Bob Jewett]

Really? I thought the opposite: heavier contact = more friction/throw.

pj
chgo

In general hard shots -- shots at high speed with a lot of force at the contact point -- have less throw than the same shot at slower speed. There are exceptions to this so for some shots the amount of throw is the same for fast or slow shots and even the same if you chalk the contact point. There are lots of details. Friction is not as simple as they teach in high school.

When you play a thin cut shot, the force between the balls is relatively small so under some conditions you may see larger maximum throw than for a harder shot.

 

[dr_dave]

I've read some of Dr. Daves stuff on Spin Induced Throw (SIT), and I've watched the video where he shows that paradoxically maximum SIT occurs with only 50% of maximum side spin--for straight in shots. You might think maximum throw would occur with maximum side spin, but that is not the case.

The reason why is explained here:

throw speed effects​

Periodically, when reading the stuff on SIT, I've wondered how the amount of SIT varies with cut angle--but I could not find that information anywhere, until today. I read in one place that maximum possible SIT occurs with straight in shots.

Where did you see that? If it is on my site, I would like to correct it.

However, I now believe that statement was a misinterpretation of what Dr. Dave said. Dr. Dave only said that maximum SIT for straight in shots occurs with 50% of maximum side spin--he did not state that maximum SIT occurs with straight in shots. So at what cut angle does maximum SIT occur? According to Dr. Dave's research, the maximum possible SIT *increases* with cut angle. In the graph below, the labels on the axes are a little confusing, but as far as I can tell, the amount of SIT is shown on the vertical axis, and the cut angle is shown on the horizontal axis. There is a graph for both slow speed shots(red line) and high speed shots(dashed blue line):

View attachment 617495

https://billiards.colostate.edu/technical_proofs/new/TP_B-25.pdf

As the cut angle increases, you need to hit the CB with more and more side spin to achieve the maximum SIT:

View attachment 617497
A 1/4 ball hit is 49 degrees, and for a thinner, 60 degree cut, you need to hit the CB with almost 100% of maximum possible side spin. For cut angles greater than 60 degrees, the amount of side spin necessary for maximum throw tails off a bit, but you still need roughly 90% of maximum side spin to achieve maximum throw.

How significant is the increase in throw as the cut angle increases? It looks to me like a straight in shot (0 degrees), has throw of about 5.2, and a 1/4 ball cut (roughly 50 degrees) has a throw of about 5.6, so the increase in maximum throw between a straight in shot and a 1/4 ball cut is about 0.4/5.2 = .08 or 8 percent. So if you can throw a straight in shot 4", then you can throw a 1/4 ball cut about 4 1/4 - 4 3/8" over the same distance.

Agreed. The difference is not very significant, especially since it can be more difficult to judge the required amount of spin for maximum throw with larger cut angles. 50% is easy to judge for most people.

Maximum Throw - Dr. Dave Pool Info

Answers to frequently-asked questions about the maximum amount pool balls can be thrown.

billiards.colostate.edu

 

[dr_dave]

I never mentioned inside or outside spin. I assumed the graphs were dealing with outside spin.

That is correct. The graphs you posted deal with spin-induced throw (SIT) created by outside spin greater than the gearing amount of spin. Cut induced throw (CIT) is a totally different matter.