Close ball separation angles, half distance passing point, and the secant function

[Vorpal Cue]

Close ball separation angles, half distance passing point, and the secant function

It's a bit difficult to make shots where the balls are close together and I got curious about the maximum cut angle available. I broke out the graphing stuff and discovered some interesting things about the relationship between the separation of the balls and the maximum angle you can get. I found that the secant function will give the MAX cut angle and the point half way between the balls centers can be used to fine tune the shot line for those close in shots. (DD)? I'll explain the trig behind the ball geometry later in the post for those that are interested in the math.

In the first column, S is the separation in ball diameters. The second column is the MAXIMUM cut angle available. Since the data is in ball diameters it's valid for any two balls of the same size. (snooker, bowling, bb, marbles, etc.) The last column is the ball separation in inches and it's for standard 2 1/4" balls only.


For balls with less than 1 diameter separation

Separation: Max

S = 0 : 0 , Frozen Balls

S = 1/8 : 27 , .28" (> 1/4")

S = 1/4 : 37 , .56" (> 1/2")

S = 3/8 : 43 , .84" (< 7/8")

S = 1/2 : 48 , 1.125" (1 1/8")

S = 5/8 : 52 , 1.40" (> 1 3/8")

S = 3/4 : 55 , 1.69" (< 1 3/4")

S = 7/8 : 58 , 1.97" (< 2")

For balls above 1 diameter separation

S=1 : 60

S=2 : 70

S=3 : 75

S=4 : 78

S=5 : 80 This is about one diamond separation on an 8' table ( 1' per diamond is 5 1/3 balls)

S=6 : 81

S=7 : 83

S=8 : 84

S=9 : 84

S=10: 85 About 2 diamonds separation

S=15: 86

S=20: 87 About 1/2 table

S=25: 88

S=43: 89 (MAX=88.7) This is the length of 8' table ( 42 2/3 diameters)

S=48: 89 (88.8, 9' table)

S=53: 89 (88.9, 10' table)

S=70: 89 (89.2, 12' snooker table with 2 1/15 balls)


The table shows that the MAX angle grows very quickly from no separation out to one ball diameter. Then it slowly becomes more even (linear) as the separation gets larger. You should remember that the MAX angle is the thinnest cut possible, a '90 degree' cut at the extreme edge of the object ball. On the table you'll probably only be able to get within 5 - 10 degrees of the MAX angle due to the difficulty of the cut.

I got the data by looking at the internal tangent lines between two balls. I examined a lot of diagrams and found the secant function made all the pieces fit together. Looking at the point where the tangents intersect notice that it's at the midpoint on the center to center line. The line segment from the midpoint to the ball center is the secant line. Notice the secant line is the same for both balls. The balls are separated by two radii and two equal line segments that 'jut out' of the balls. Since two radii are one diameter and two equal extra pieces of radii are equal to one extra diameter piece, the separation can be expressed in diameter units.



The MAX angle is found by using the internal tangent lines as shown in the diagrams. The internal tangent lines will define the part of the object ball which can be hit due to the separation of the balls. The angle the tangent line forms when it crosses the center line between the balls is the compliment (90*-x) of the secant angle. The value of the secant angle will give the distance between balls in 'diameter' units. Here's a few more diagrams with 1/4, 1/2, and 2 ball separations.








Using the above information a simple equation can be written to find the MAX angle as a function of ball diameters or the distance between the balls. The equation for Max angle if the separation S is in ball diameters : MAX = arcsecant [S + 1].

The equation for finding MAX if you know the distance between the balls is : MAX = arcsecant [1 + (d/D)] where d is the distance between the balls and D is the ball diameter.

As an example, for a distance of 1 inch using 2 1/15" snooker balls (31/15") : MAX = arcsecant [1 + (1 / (31/15))] = arcsecant [1 + .484] = 47.6 rounded down to 47 degrees.


-----------------------------------------------------

The half distance passing point may be related to double distance aiming or it may BE DD. I don't know enough about the system to be sure. In the diagrams notice that the edge of the cue ball will roll over the midpoint ( call it MP ) to make a 90* cut. Also the center of the cue ball will roll over MP to make a 0* cut. This implies that the point of the cue ball that rolls over MP will be the point that makes contact with the object ball. The MP is pretty easy to find when the balls are tight and it provides an extra check on the alignment. Once the contact point on the CB is found it can be aimed at MP. If it rolls over the MP it will hit the contact point on the OB and the ball will drop. It's a handy aid and easy to use. It works well for those close ball edge to edge shots.

 

[AtLarge]

... I got curious about the maximum cut angle available.
...

Just for clarity here for other readers, I'd point out that the "cut angle" you are talking about is the acute angle between these two lines: (1) the line through the centers of the CB and the OB and (2) the line of travel of the OB after the collision.

This is not the standard definition of a cut angle for a pool shot.

The actual cut angle achieved with a shot, by standard definition, is the number of degrees the OB is knocked off the line of travel (prior to the collision) of the CB. That is, it is the acute angle between these two lines: (a) the line of travel of the CB prior to the collision and (b) the line of travel of the OB after the collision.

You are using a 90-degree cut angle (by standard definition) in all cases, and calculating the resulting angle of departure of the OB from the line through the ball centers.

Edit -- I guess I might also note that the "separation" between balls you are using is the distance between their closest two points. Sometimes when people talk about the separation between two balls they mean the distance between the bases of the balls.

 

[Vorpal Cue]

Just for clarity here for other readers, I'd point out that the "cut angle" you are talking about is the acute angle between these two lines: (1) the line through the centers of the CB and the OB and (2) the line of travel of the OB after the collision.

This is not the standard definition of a cut angle for a pool shot.

The actual cut angle achieved with a shot, by standard definition, is the number of degrees the OB is knocked off the line of travel (prior to the collision) of the CB. That is, it is the acute angle between these two lines: (a) the line of travel of the CB prior to the collision and (b) the line of travel of the OB after the collision.

You are using a 90-degree cut angle (by standard definition) in all cases, and calculating the resulting angle of departure of the OB from the line through the ball centers.

Edit -- I guess I might also note that the "separation" between balls you are using is the distance between their closest two points. Sometimes when people talk about the separation between two balls they mean the distance between the bases of the balls.

Guess I should have boned up on pool terms along with my trig definitions. Looks like I learned something new today.

Hitting edge to edge is how you get the maximum 'deflection' if you address the balls center to center.

I think if you told most people that the balls were separated by one ball that they wouldn't expect the balls to be frozen. Instead they would assume you could fit a ball between the two other balls. Mathematically it's probably the correct definition but I think it would cause confusion to state that frozen balls have a one ball separation.

 

[paultex]

Ok thanks, this is interesting numbers.

FACT: I just woke up and skimmed the writing and also looked secant function.

QUESTION: Are these numbers based on striking center cueball? I assume these numbers are based on a static coeffiency?

FACT: Static in this case can be skewed outside the vacuum with what I call "spin lubricant" or "spin warp".

Thanks

 

[paultex]

One more question sir if you please. I believe I will express this clearly enough, but before you answer (if possible), please ASK BACK the same question as YOU understand it, therefore, I can concur and possibly save you from a wasted effort and your time, which I greatly appreciate either way. I assume you will agree to these terms for the sake of efficiency but if my question is clear to you or anyone else and don't need to ask back, then go for it and answer away. Thanks.

QUESTION: What is maximum side spin rpm from a cueball forward delivery velocity of 1 ball length?

PARAMETERS: Typical atmospheric playing conditions in a standard air conditioned pool hall, simonis average wear, level cue relative to level surface slate, 3'oclock or 9 o'clock on cb at miscue limit and shaft/tip contact angle is parallel to 90 degrees of cb center.

FACT: I am striking cb under these perameters and delivering the cb to a static rest exactly 1 ball diameter forward from original launch point.

I hope this is clear enough.

CONJECTURE: Feel free to expand on this subject in relation to forward distances and side spin rpm capabilities.

SPECULATION: If 1 side spin revolution is maximum, over the stated distance of 1 ball forward travel, then a forward travel of 3 ball lengths, could exponentially yield much more side spin revolution due to forward velocity?

In other words, side spin is relative to forward velocity? Or irrelevant?

ADDITIONAL PERAMETER: Shaft/tip contact angle relativity can be altered if maximum side spin conclusion is dependent BUT forward vertical travel of cb must still reach static rest on original intended destination as if struck straight, despite obvious swerve or curve along the path.

Me thinks this is too complex but I have no idea. If too complex to explain or bother, then state it as so but be clear as to expertise and not just uninformed opinion. If bob jewett said it is explainable but too complex, then that would be very clear to me and ironically, a great answer that helps me tremendously.

Thanks and good luck.

 

[Vorpal Cue]

Ok thanks, this is interesting numbers.

FACT: I just woke up and skimmed the writing and also looked secant function.

QUESTION: Are these numbers based on striking center cueball? I assume these numbers are based on a static coeffiency?

FACT: Static in this case can be skewed outside the vacuum with what I call "spin lubricant" or "spin warp".

Thanks

Yes the cue ball is struck in the center. I imagine a snappy masse shot could be used instead of a normal stroke when the balls are tight for extreme spin. I'll have to try it out on the table.

 

[Vorpal Cue]

One more question sir if you please. I believe I will express this clearly enough, but before you answer (if possible), please ASK BACK the same question as YOU understand it, therefore, I can concur and possibly save you from a wasted effort and your time, which I greatly appreciate either way. I assume you will agree to these terms for the sake of efficiency but if my question is clear to you or anyone else and don't need to ask back, then go for it and answer away. Thanks.

QUESTION: What is maximum side spin rpm from a cueball forward delivery velocity of 1 ball length?

PARAMETERS: Typical atmospheric playing conditions in a standard air conditioned pool hall, simonis average wear, level cue relative to level surface slate, 3'oclock or 9 o'clock on cb at miscue limit and shaft/tip contact angle is parallel to 90 degrees of cb center.

FACT: I am striking cb under these perameters and delivering the cb to a static rest exactly 1 ball diameter forward from original launch point.

I hope this is clear enough.

CONJECTURE: Feel free to expand on this subject in relation to forward distances and side spin rpm capabilities.

SPECULATION: If 1 side spin revolution is maximum, over the stated distance of 1 ball forward travel, then a forward travel of 3 ball lengths, could exponentially yield much more side spin revolution due to forward velocity?

In other words, side spin is relative to forward velocity? Or irrelevant?

ADDITIONAL PERAMETER: Shaft/tip contact angle relativity can be altered if maximum side spin conclusion is dependent BUT forward vertical travel of cb must still reach static rest on original intended destination as if struck straight, despite obvious swerve or curve along the path.

Me thinks this is too complex but I have no idea. If too complex to explain or bother, then state it as so but be clear as to expertise and not just uninformed opinion. If bob jewett said it is explainable but too complex, then that would be very clear to me and ironically, a great answer that helps me tremendously.

Thanks and good luck.

Sorry, but I don't have the knowledge to answer your questions. The topic is an interesting one and I'd like to hear some answers too.

I think you would be better served if you start a new thread with your post. Not too many people are going to see it when it's buried in a math post. You'll get more responses in a new thread imho.

 

[paultex]

Sorry, but I don't have the knowledge to answer your questions. The topic is an interesting one and I'd like to hear some answers too.

I think you would be better served if you start a new thread with your post. Not too many people are going to see it when it's buried in a math post. You'll get more responses in a new thread imho.

Funny, at the moment, I don't feel compelled to because I believe I got my answer from your statement, and common sense.

However my friend, if you feel compelled, and you think my question was stated clear enough, then copy paste it as if it were your own question and do what you suggested. I'm in agreement that it is still an interesting subject and very pertinent to your number chart as well.

If you take the leap from compelled to execution, please provide a link to new thread, as I'm not sure if it would be in "aiming conversations".....thanks.

.....and I'm off to another 10-12 on the table. Ttyl.

 

[paultex]

Yes the cue ball is struck in the center. I imagine a snappy masse shot could be used instead of a normal stroke when the balls are tight for extreme spin. I'll have to try it out on the table.

For my gratitude, I feel compelled to take this additional time to say something here. I think it's extremely important.

I tried yesterday, what you stated you will try.

From cb to ob at 1 diameter separation, I was able to apply 9'oclock spin warp to relieve the thickness.

"Thickness" itself though, is relative to perception. The nature of the static rest of closeness, like a tractor beam, pulls you in thick from perception, thus why most, what I consider, amateur players, put a twist in their stroke during stick delivery execution and "spin out".

The end result is going multiple rails for position, or just to simply pocket the ob. It works.

However, the more advanced player recognizes, the only thing that is "owning" the amateur player in this case, is FRICTION.

If the balls are close enough, perhaps half a ball distance and of course, distance energy requirement for ob to reach the "bottom of the pocket", then velocity itself can and will take over, no matter what lubricant or warp is applied, thus, lubricant is neutralized......unless possibly a jacked up masse (maximum "warp") can over come this, and I'm not at liberty to say yes or no at this time, because distance parameters become relative as well etc etc.

Here's the important thing and I'm obviously preaching to the choir for some I'm sure.

Fact: I cannot go or wish to spin out in a certain situation.

Fact: A different perception can and will alter what appears to be unachievable.

Fact: A proper perception will allow a proper shaft angle as well, in conjunction with knowledge of friction and velocity requirements, to formulate the proper delivery and achieve the only thing that matters in the game of pool:

"Launch point of cueball to it's static destination."

Everything else is just a formulation.

These close together shots can be manipulated to a very high degree with proper perception, stick angle, slight bounce in the air and contact point frictional grab knowledge. The key is to trust the "over cut" if that's the way you fall into perception or it can be made to look "perfect" as well if your formulations are of a high degree.

These ultra refined nuances are so various, but there is a lot of spill over commonality that can be applied. Still though........1000's of hours imo to achieve the familiarity and programming.

It's a beautiful game.....I'm outta here.

 

[AtLarge]

... I think if you told most people that the balls were separated by one ball that they wouldn't expect the balls to be frozen. Instead they would assume you could fit a ball between the two other balls. Mathematically it's probably the correct definition but I think it would cause confusion to state that frozen balls have a one ball separation.

True; I think your word "separation" is fine for what you did. I was just pointing out that ball-to-ball distances are often described in terms of the bases of the balls.

Example -- OB on foot spot, CB on head spot. I think many people would say these two balls are four diamonds apart (50" on a 9-footer, which is just over 22 ball diameters), or that the CB is four diamonds away from the OB. But the balls have 47¾" of space between them, so, in that sense, they are "separated" by about 21 ball diameters of space.

Ah. pool lingo. :smile:

By the way -- many hand calculators do not have an arcsecant function. So you could use the arccosine function instead:

arcsec[1+S] = arccos[1/(1+S)]

 

[garczar]

Close ball separation angles, half distance passing point, and the secant function

It's a bit difficult to make shots where the balls are close together and I got curious about the maximum cut angle available. I broke out the graphing stuff and discovered some interesting things about the relationship between the separation of the balls and the maximum angle you can get. I found that the secant function will give the MAX cut angle and the point half way between the balls centers can be used to fine tune the shot line for those close in shots. (DD)? I'll explain the trig behind the ball geometry later in the post for those that are interested in the math.

In the first column, S is the separation in ball diameters. The second column is the MAXIMUM cut angle available. Since the data is in ball diameters it's valid for any two balls of the same size. (snooker, bowling, bb, marbles, etc.) The last column is the ball separation in inches and it's for standard 2 1/4" balls only.


For balls with less than 1 diameter separation

Separation: Max

S = 0 : 0 , Frozen Balls

S = 1/8 : 27 , .28" (> 1/4")

S = 1/4 : 37 , .56" (> 1/2")

S = 3/8 : 43 , .84" (< 7/8")

S = 1/2 : 48 , 1.125" (1 1/8")

S = 5/8 : 52 , 1.40" (> 1 3/8")

S = 3/4 : 55 , 1.69" (< 1 3/4")

S = 7/8 : 58 , 1.97" (< 2")

For balls above 1 diameter separation

S=1 : 60

S=2 : 70

S=3 : 75

S=4 : 78

S=5 : 80 This is about one diamond separation on an 8' table ( 1' per diamond is 5 1/3 balls)

S=6 : 81

S=7 : 83

S=8 : 84

S=9 : 84

S=10: 85 About 2 diamonds separation

S=15: 86

S=20: 87 About 1/2 table

S=25: 88

S=43: 89 (MAX=88.7) This is the length of 8' table ( 42 2/3 diameters)

S=48: 89 (88.8, 9' table)

S=53: 89 (88.9, 10' table)

S=70: 89 (89.2, 12' snooker table with 2 1/15 balls)


The table shows that the MAX angle grows very quickly from no separation out to one ball diameter. Then it slowly becomes more even (linear) as the separation gets larger. You should remember that the MAX angle is the thinnest cut possible, a '90 degree' cut at the extreme edge of the object ball. On the table you'll probably only be able to get within 5 - 10 degrees of the MAX angle due to the difficulty of the cut.

I got the data by looking at the internal tangent lines between two balls. I examined a lot of diagrams and found the secant function made all the pieces fit together. Looking at the point where the tangents intersect notice that it's at the midpoint on the center to center line. The line segment from the midpoint to the ball center is the secant line. Notice the secant line is the same for both balls. The balls are separated by two radii and two equal line segments that 'jut out' of the balls. Since two radii are one diameter and two equal extra pieces of radii are equal to one extra diameter piece, the separation can be expressed in diameter units.

View attachment 461343

The MAX angle is found by using the internal tangent lines as shown in the diagrams. The internal tangent lines will define the part of the object ball which can be hit due to the separation of the balls. The angle the tangent line forms when it crosses the center line between the balls is the compliment (90*-x) of the secant angle. The value of the secant angle will give the distance between balls in 'diameter' units. Here's a few more diagrams with 1/4, 1/2, and 2 ball separations.

View attachment 461344

View attachment 461345

View attachment 461346


Using the above information a simple equation can be written to find the MAX angle as a function of ball diameters or the distance between the balls. The equation for Max angle if the separation S is in ball diameters : MAX = arcsecant [S + 1].

The equation for finding MAX if you know the distance between the balls is : MAX = arcsecant [1 + (d/D)] where d is the distance between the balls and D is the ball diameter.

As an example, for a distance of 1 inch using 2 1/15" snooker balls (31/15") : MAX = arcsecant [1 + (1 / (31/15))] = arcsecant [1 + .484] = 47.6 rounded down to 47 degrees.


-----------------------------------------------------

The half distance passing point may be related to double distance aiming or it may BE DD. I don't know enough about the system to be sure. In the diagrams notice that the edge of the cue ball will roll over the midpoint ( call it MP ) to make a 90* cut. Also the center of the cue ball will roll over MP to make a 0* cut. This implies that the point of the cue ball that rolls over MP will be the point that makes contact with the object ball. The MP is pretty easy to find when the balls are tight and it provides an extra check on the alignment. Once the contact point on the CB is found it can be aimed at MP. If it rolls over the MP it will hit the contact point on the OB and the ball will drop. It's a handy aid and easy to use. It works well for those close ball edge to edge shots.

Seriously, how can you play with all that going on in your head? Just go hit balls and don't worry about all that stuff. Making it WAY more complicated than it is.

 

[Vorpal Cue]

Seriously, how can you play with all that going on in your head? Just go hit balls and don't worry about all that stuff. Making it WAY more complicated than it is.

I don't play with calipers and a slide rule either. All that's not in my head when I'm playing. I'm just curious about the math behind the ball geometry. There's a few members that like to 'tinker' with the physics and come up with novel methods for potting balls and I like to see how they work. If you didn't enjoy seeing the naked secant pop out of the cake, I guess this thread isn't your cup of tea. :boring:

 

[BC21]

One more question sir if you please. I believe I will express this clearly enough, but before you answer (if possible), please ASK BACK the same question as YOU understand it, therefore, I can concur and possibly save you from a wasted effort and your time, which I greatly appreciate either way. I assume you will agree to these terms for the sake of efficiency but if my question is clear to you or anyone else and don't need to ask back, then go for it and answer away. Thanks.

QUESTION: What is maximum side spin rpm from a cueball forward delivery velocity of 1 ball length?

PARAMETERS: Typical atmospheric playing conditions in a standard air conditioned pool hall, simonis average wear, level cue relative to level surface slate, 3'oclock or 9 o'clock on cb at miscue limit and shaft/tip contact angle is parallel to 90 degrees of cb center.

FACT: I am striking cb under these perameters and delivering the cb to a static rest exactly 1 ball diameter forward from original launch point.

I hope this is clear enough.

CONJECTURE: Feel free to expand on this subject in relation to forward distances and side spin rpm capabilities.

SPECULATION: If 1 side spin revolution is maximum, over the stated distance of 1 ball forward travel, then a forward travel of 3 ball lengths, could exponentially yield much more side spin revolution due to forward velocity?

In other words, side spin is relative to forward velocity? Or irrelevant?

ADDITIONAL PERAMETER: Shaft/tip contact angle relativity can be altered if maximum side spin conclusion is dependent BUT forward vertical travel of cb must still reach static rest on original intended destination as if struck straight, despite obvious swerve or curve along the path.

Me thinks this is too complex but I have no idea. If too complex to explain or bother, then state it as so but be clear as to expertise and not just uninformed opinion. If bob jewett said it is explainable but too complex, then that would be very clear to me and ironically, a great answer that helps me tremendously.

Thanks and good luck.

Wow. What a complicated question! I'm​ not sure how to calculate it, but the side spin should have a greater rpm in the first few inches. Wouldn't the forward velocity decrease the spin due to friction between the cloth and the sliding cb?

 

[paultex]

Wow. What a complicated question! I'm​ not sure how to calculate it, but the side spin should have a greater rpm in the first few inches. Wouldn't the forward velocity decrease the spin due to friction between the cloth and the sliding cb?

I'm not sure either but it makes my head hurt.

I believe you would be right if swerve or curve was involved but not nearly as much or at all, dependent on more truer vertical center axis rotation.

Thats why I believe in the importance of a parallel alignment, shot/stroke line. It produces a truer axis roll and from my experience, if done properly, does not swerve the cue ball nearly as much as stroking inside the parallel line or "corridor" as I like to call it.

This also causes much less deflection at point of contact between cb and ob after collison.

This also produces a much less aggressive spin angle after rail contact. I call it a soft spin. Nothing has changed in physics in the after effect, it's just that this type of delivery holds a more realistic tangent line that exists between angle of cb/ob before delivery.

I would also say from experience, because of spin throw of ob, but much less collision deflection because of much less swerve, when struck really pure, it can even defy the tangent line toward the unnatural side of the angle to a small degree.

A parallel spin, as well as a swerve spin, must be mastered imo.

This is also why I am curious as to how much actual side spin can be applied with varying velocities.

One thing I do know, a twist stroke does not produce more side spin vs velocity. It feels like it does, but it doesn't and that's easy to prove.

You have miscue limit and stick angle. A twist stroke, produces a outside angle delivery, and that can easily be produced with a straight stroke with same angle.

I believe its fair to say that a twist stroke ultimately produces less spin imo because there is a tension that is created in human aspect of physical delivery and there's just something about it that isn't user friendly.

Most if not all in general, very good players grip light and some even throw the cue forward. I believe the official terminology is called "stroke slip".

Even the players who don't throw the cue, simulate a throw with a very nice and fluid wrist and hand action.

Earl Strikland says he GRIPS the stick and I believe it, but to what degree would be nice to know. I speculate, when he pulls the trigger in most cases, his grip lightens.

So, there are exceptions to the rules in any case of life, but the fact remains, good players stroke straight and grip very light.

Ive also found that getting your alignment straight, automatically straightens your stroke significantly. A crooked or wobbly stroke is due in most cases from a constant visual shift between both eyes, because novice or uninformed players don't line up under or close enough to their dominant eye.

This is my opinion from my own personal experience and visual observations. That is also why I say alignment is the system. Conventional "systems" are just a method.

Once again, my opinion.

 

[Vorpal Cue]

True; I think your word "separation" is fine for what you did. I was just pointing out that ball-to-ball distances are often described in terms of the bases of the balls.

Example -- OB on foot spot, CB on head spot. I think many people would say these two balls are four diamonds apart (50" on a 9-footer, which is just over 22 ball diameters), or that the CB is four diamonds away from the OB. But the balls have 47¾" of space between them, so, in that sense, they are "separated" by about 21 ball diameters of space.

Ah. pool lingo. :smile:

By the way -- many hand calculators do not have an arcsecant function. So you could use the arccosine function instead:

arcsec[1+S] = arccos[1/(1+S)]

Sure, When spotting balls, the balls centers are assumed to be put on the spots. I think everyone agrees with that.

The secant function isn't on a lot of calculators and your tip of using the inverse of the arc cosine is a good workaround. When I was getting the data for the table my old calculator (TI-84) wouldn't power up. I used a handy web graphing calculator that I'll recommend. It made the data gathering a snap.

https://web2.0calc.com/

 

[paultex]

Secant is rarely used now from what I understand but im not a trig guy....or calculus lol. But I like numbers and believe even word descriptors should be replaced with a numerical value that is more concise. Words like "usually" for instance, can be abused or miscontrued. Language is a maleable weapon, numbers are definitive and final to a conclusive extent.

 

[Vorpal Cue]

Secant is rarely used now from what I understand but im not a trig guy....or calculus lol. But I like numbers and believe even word descriptors should be replaced with a numerical value that is more concise. Words like "usually" for instance, can be abused or miscontrued. Language is a maleable weapon, numbers are definitive and final to a conclusive extent.

I was surprised to see the secant defined the angle vs.separation ratio too. Now I'm scratching my head and looking at the cosecant. Maybe something's there, maybe not.

 

[bbb]

I wish I could understand your diagrams and description.....:embarrassed2::embarrassed2:

 

[Vorpal Cue]

I wish I could understand your diagrams and description.....:embarrassed2::embarrassed2:

The top ball in the pix are the object ball. The bottom ball is the cue ball. The ball(s) between them are only to show the separation. Think of them as 'ghost ball(s)'. I guess I should have used dotted lines to show them but I'm not that artistic.

 

[Bob Jewett]

I wish I could understand your diagrams and description.....:embarrassed2::embarrassed2:

What he describes is dead simple geometry with a little trig thrown in so that you can get numbers out. If you are interested in learning geometry and trig, it's likely that your local school system offers adult education courses. While it is satisfying to understand that stuff so you can do accurate analyses of some parts of pool, it would mostly be an exercise in personal development and prevention of cerebral calcification.

To find the maximum you can cut the object ball for a close position, just find the line of the thinnest possible hit. The object ball can be driven at 90 degrees to that line, in theory.

In practice, with outside english, you can cut such a close ball more than 90 degrees.