Cue Ball and Object Ball Travel Distances Plot

[Colin Colenso]

Colin, thanks for the valuable tool. Can you think of an easy way to do one for a sliding CB? If I'm not mistaken, a sliding CB travels the same distance as the OB at a 45-degree cut rather than a 30-degree (half ball) cut.

Maybe Dr. Dave or Ron Shepard have already plotted this...

pj
chgo

Dr. Dave has some charts for that in a paper he linked to on the thread I linked to up top.

As it's very hard to apply pure stun to most shots, I don't think it is very useful but I'd like to see the two graphs compared in the format I used above.

The calculations are considerably easier for the sliding CB.

Colin

 

[Colin Colenso]

Dick, you're certainly right that a skilled player's "feel" for distance and speed can't be approached by an unskilled player with a chart...
pj
chgo

Patrick,

While it's a valuable tool for teaching low skilled players I wouldn't underestimate the value of these rule of thumb ratios for advanced players.

While I'm no Efren Reyes, and not intending to brag, just pointing out that I'm not a beginner, I used the 1/6th ratio maybe a dozen or more times during matches in last years national championships where I made the semis. I find it helps me to pin-point CB position on quite a few shots.

For example, it's sometimes very good for determining the following:
1. If you'll get just past, or just short of straight in position on the next ball.
2. If you'll stop just short of or just past a snooker.

I think many players can feel the speed they will roll the OB more than they can feel the speed they will roll the CB through the OB on straightish shots, especially when they move onto different speed tables. Not inch for inch, but inch per half foot. The ratio kind of leverages the degree of accuracy, much like how a player would have better speed control with a light cue as compared to a 30 ounce cue.

So if they can judge the OB travel distance to within a foot, they are able to calculate the CB travel distance to within 2 inches.

I think it also has the advantage of just being able to focus on one variable.

btw: I had a very good snooker player trying to advise me to play stun through shots after noticing I was dribbling a few balls in, not realizing there was some method to my madness. Judging stun through shots to within 2 inches is a heck of a task, especially when moving between tables of varying cloth speeds.

The dribble, or roll in shot is often considered a beginner method, but I think it is under utilized because it is a shot that can be developed with quite high precision.

As an interesting aside: In snooker and other smaller ball and smaller pocket games, kicks (skids) often make slow rolling shots a risky choice, but it doesn't occur as often, nor have such disastrous consequences in US pool (assuming the tables are pretty level).

Colin

 

[Jal]

Colin, thanks for the valuable tool. Can you think of an easy way to do one for a sliding CB? If I'm not mistaken, a sliding CB travels the same distance as the OB at a 45-degree cut rather than a 30-degree (half ball) cut..

A quick and dirty way is to go here:

http://www.walterzorn.com/grapher/grapher_e.htm

Type in 0 for xmin, 90 for xmax, 0 for ymin, and 5 for ymax (the 5 is arbitrary as the following function goes to infinity as x, the cut angle, approaches 90 degrees). For the function, which is the ratio of the distance traveled by the cueball to the distance traveled by the object ball, type (or copy and paste):

tan(x*pi/180)^2

This is the square of the trig tangent function. The x*pi/180 stuff converts x from degrees to radians. The function describes the ratio of distances for a pure stun shot and ignores slight modifications due to inelasticity and mutual throw, ie, ball/ball friction.

Check out what happens at 45 degrees.

Jim

 

[Colin Colenso]

Estimating Angles for Using the Rules of Thumb

This system is most useful for shots under 20 degree cut angle I believe.

Here are the basic Rules of Thumb
Full Ball = 1/6
5 degrees = 1/5
10 degrees = 1/4
15 degrees = 1/3
20 degrees = 1/2​

So how to recognize these angles? Here's a quick and pretty accurate way!

Project a line out 3 feet (36 inches) or just under 3 diamonds past the OB along the line with the CB. Estimate the distance from this line to the line from the OB to the pocket. You might want to take note of a mark on your cue that is 36 inches from the tip for easy measuring at the table.

Remember these estimates:
5 degrees = 3 inches
10 degrees = 6 inches
15 degrees = 9 inches
20 degrees = 13 inches

So just remember 3,6,9,13 and you have a guide to estimating all straightish angles.

The more precise figures are actually 3.2, 6.3, 9.3 and 13.1. And the values for 30, 45 and 60 degrees are 20.8, 36 and 62 inches in case you are interested.

 

[Colin Colenso]

Sliding CB Plot

A quick and dirty way is to go here:

http://www.walterzorn.com/grapher/grapher_e.htm

Jim

If we're gonna do things quick and dirty, here's my attempt ;-)

The lower blue line, starting at 0,0 represents a sliding CB, with the same translational velocity of the rolling CB that would travel 12 units.

btw: JAL, do my calculations seem ok to you?

Colin

 

[Dead Crab]

***********
A much easier method of angle estimation is doubling the number of inches that the joint line of the cue is from the CB-OB line, as this is equal to the number of degrees of the angle. Allowing for an inch of the OB, and 29 inches from the cue tip to the joint line, the hypotenuse side is 30", which has some favorable properties.

So if the cue joint to (CB-OB line) distance is 10 inches, the cut angle is 20 degrees, 12 inches ---> 24 degrees, ect..

I have a mark on my cue shaft at 14", so that I can get the angle by multiplying the distance from this point to the (CB-OB) by four, which is easier when the CB-OB distance is small. This mark also allows me to find my pivot point easily when using BHE.

With the method above, no memorization is needed.

DC

 

[Colin Colenso]

***********
A much easier method of angle estimation is doubling the number of inches that the joint line of the cue is from the CB-OB line, as this is equal to the number of degrees of the angle. Allowing for an inch of the OB, and 29 inches from the cue tip to the joint line, the hypotenuse side is 30", which has some favorable properties.

So if the cue joint to (CB-OB line) distance is 10 inches, the cut angle is 20 degrees, 12 inches ---> 24 degrees, ect..

I have a mark on my cue shaft at 14", so that I can get the angle by multiplying the distance from this point to the (CB-OB) by four, which is easier when the CB-OB distance is small. This mark also allows me to find my pivot point easily when using BHE.

With the method above, no memorization is needed.

DC

Great tip DC!

Colin

 

[Patrick Johnson]

***********
A much easier method of angle estimation is doubling the number of inches that the joint line of the cue is from the CB-OB line, as this is equal to the number of degrees of the angle. Allowing for an inch of the OB, and 29 inches from the cue tip to the joint line, the hypotenuse side is 30", which has some favorable properties.

So if the cue joint to (CB-OB line) distance is 10 inches, the cut angle is 20 degrees, 12 inches ---> 24 degrees, ect..

I have a mark on my cue shaft at 14", so that I can get the angle by multiplying the distance from this point to the (CB-OB) by four, which is easier when the CB-OB distance is small. This mark also allows me to find my pivot point easily when using BHE.

With the method above, no memorization is needed.

DC

So if you point your cue through the OB at the pocket (with your tip at the OB), the cut angle in degrees is twice the distance in inches from your cue's joint to the OB/CB line? Cool.

Do you measure perpendicular to the cue, perpendicular to the OB/CB line, or neither (to form an isosceles triangle)?

pj
chgo

 

[Patrick Johnson]

So if you point your cue through the OB at the pocket (with your tip at the OB), the cut angle in degrees is twice the distance in inches from your cue's joint to the OB/CB line? Cool.

Do you measure perpendicular to the cue, perpendicular to the OB/CB line, or neither (to form an isosceles triangle)?

pj
chgo

A nit: shouldn't the ghost cue ball be substituted for the object ball in this measurement?

pj
chgo

 

[Dead Crab]

Tip of the cue is placed under the edge of the OB, pointed at the pocket (not the ghost ball point). so the cue tip is about an inch from OB center.

Right triangle. Hypotenuse is the OB center to joint, 90 degree angle formed by dropping perpendicular to CB-OB line.

The perpendicular distance to the CB-OB line (in inches) is then estimated, and multiplied by 2 if perp origin taken at the joint, 4 if taken at my 14" shaft mark, and by 1 if taken from 1" past the end of my 58" cue (making for a 60" distance to the OB center).

I can accurately estimate the cut angle to within a degree or two. Still working on a stroke that can deliver that accurately.

 

[CueAndMe]

Can somebody draw what DeadCrab is describing? I'm trying to understand, but I'm getting lost. Sounds very valuable.

 

[Patrick Johnson]

Can somebody draw what DeadCrab is describing? I'm trying to understand, but I'm getting lost. Sounds very valuable.

Here's what I think he's describing. The red line is the measurement from the cue's joint to the Ghost Ball-CB line. In this diagram the cut angle is 45 degrees, so the length of the red line should be 22.5 inches, which looks about right.

CueTable Help

Dead Crab says to measure from the OB, but I show the measurement being made from the ghost ball instead, because I think that's where the cut angle is really measured from. It's probably not necessary to be that finicky for this, but I drew it this way to show Dead Crab what I mean.

pj
chgo

 

[CueAndMe]

Thanks Patrick. And thanks DeadCrab. Very interesting method.

 

[Jal]

If we're gonna do things quick and dirty, here's my attempt ;-)

The lower blue line, starting at 0,0 represents a sliding CB, with the same translational velocity of the rolling CB that would travel 12 units.

btw: JAL, do my calculations seem ok to you?

Colin

Yes they do Colin. But I've been wondering why you're relating them to a cueball that rolls 12 units, since you're really interested (I think) in the ratio of the distances. In the case of a stun shot (your latest graph), dividing the distances eliminates any error (relative to the 12 units of distance), which comes from ignoring the distance each ball travels while sliding before reaching natural roll. Not a biggy, but since your primary interest seems to be in the ratio...

I've sort of double checked your earlier graph (rolling CB) against Dr. Dave's, which includes modifications for throw and energy loss (but also not sliding distances), and they do agree pretty well, which you've likely already confirmed. (The math behind it looks fine.)

I don't usually do any calculations or use established references (with one exception) while playing, but I think the kind of stuff you (and Dr. Dave) presented here can be very useful. Thanks for taking the time.

Jim

 

[Solartje]

Tip of the cue is placed under the edge of the OB, pointed at the pocket (not the ghost ball point). so the cue tip is about an inch from OB center.

Right triangle. Hypotenuse is the OB center to joint, 90 degree angle formed by dropping perpendicular to CB-OB line.

The perpendicular distance to the CB-OB line (in inches) is then estimated, and multiplied by 2 if perp origin taken at the joint, 4 if taken at my 14" shaft mark, and by 1 if taken from 1" past the end of my 58" cue (making for a 60" distance to the OB center).

I can accurately estimate the cut angle to within a degree or two. Still working on a stroke that can deliver that accurately.

What do you do if u live in europe and only count in cm's
i'll guess i need to find a point where it works in cm...

Colin: i really missed posts like this from you !!! I'm your #1 azb fan

 

[Colin Colenso]

Yes they do Colin. But I've been wondering why you're relating them to a cueball that rolls 12 units, since you're really interested (I think) in the ratio of the distances. In the case of a stun shot (your latest graph), dividing the distances eliminates any error (relative to the 12 units of distance), which comes from ignoring the distance each ball travels while sliding before reaching natural roll. Not a biggy, but since your primary interest seems to be in the ratio...

I've sort of double checked your earlier graph (rolling CB) against Dr. Dave's, which includes modifications for throw and energy loss (but also not sliding distances), and they do agree pretty well, which you've likely already confirmed. (The math behind it looks fine.)

I don't usually do any calculations or use established references (with one exception) while playing, but I think the kind of stuff you (and Dr. Dave) presented here can be very useful. Thanks for taking the time.

Jim

Jim,
You're right, I could have used any unit to get the same ratio, but I think using the unit of 12 makes it visually more understandable.

I've found this type of ratio most useful in shots where the OB is played 4 to 6 diamonds. There may be a better way to illustrate this, but this seems like a way that makes sense to me and I hope makes sense to others who may not grasp the ratios presented in other forms.

I had to guess the ratios before based on a little intuition, but I think I can use the 1/5th, 1/4th and 1/3rd ratios with considerably more accuracy now I've seen the plots shown this way.

Colin

 

[Colin Colenso]

What do you do if u live in europe and only count in cm's
i'll guess i need to find a point where it works in cm...

Colin: i really missed posts like this from you !!! I'm your #1 azb fan

Hi Solly,

Yeah, the cm v inches or feet is a battle for some. In Australia we grew up with both so can easily adapt.

Perhaps think of a foot as a diamond, though a diamond is about 1 inch or 2.5cm more. A yard (3 feet) is about 8 cm less than a meter.

Anyway, the ratios are the main thing. Use a converter for distances if you want to get the figures into metric.

Cheers mate, and thanks for the compliments. I'll be coming out with a video and instruction manual on BHE and Pivot Points soon that I think you'll enjoy.

Oh, and I'm getting a 300 frames per second video camera to get some action shots demonstrating BHE and breaking

Colin

 

[Jal]

Here's what I think he's describing. The red line is the measurement from the cue's joint to the Ghost Ball-CB line. In this diagram the cut angle is 45 degrees, so the length of the red line should be 22.5 inches, which looks about right.

CueTable Help

Dead Crab says to measure from the OB, but I show the measurement being made from the ghost ball instead, because I think that's where the cut angle is really measured from. It's probably not necessary to be that finicky for this, but I drew it this way to show Dead Crab what I mean.

pj
chgo

If you place the tip, or whatever happens to be 29" from the joint, at the center of the ghostball and draw an isosceles triangle (as you mentioned earlier) instead of a right triangle, doubling the base of the triangle will give you a more accurate measurement over a greater range of angles.

For instance, with a 10-degree cut, using Dead Crab's method modified for the ghostball, you would measure 5.2" and double it to 10.4", while the above method would yield 10.1". With 30, 45 and 60-degree cuts, the comparative values are 30.00" vs 30.02", 42.4" vs 44.4", and 52.0" vs 58.0", respectively.

Jim

 

[Solartje]

Hi Solly,

Yeah, the cm v inches or feet is a battle for some. In Australia we grew up with both so can easily adapt.

Perhaps think of a foot as a diamond, though a diamond is about 1 inch or 2.5cm more. A yard (3 feet) is about 8 cm less than a meter.

Anyway, the ratios are the main thing. Use a converter for distances if you want to get the figures into metric.

Cheers mate, and thanks for the compliments. I'll be coming out with a video and instruction manual on BHE and Pivot Points soon that I think you'll enjoy.

Oh, and I'm getting a 300 frames per second video camera to get some action shots demonstrating BHE and breaking

Colin

300 fps ! sounds like someone will have some fun

The BHE is now imbedded. I have 3 lines on my shaft that give me enough reference for the moment. i'm not using it alot in 14-1, unless on the breakshot!!! by puting my bridge just 10cm closer i aint missing those shots as much anymore. (i still continu to miss EVERY single breakball to get to the third rack... but thats probably more mental )

For 9ball it has improved my game ALOT. The only time i'm in trouble is a slow shot, with much english and cb against the rail. Impossible to use BHE on those :grin-square:

looking forward to the video's, and the compliments are free. You deserve each and everyone of them.

 

[Dead Crab]

I place the vertex of the triangle at the OB center, as I define the cutangle as the intersection between the CB-OB line and the OB-Pocket line.

let x be the perp distance between the joint and CB-OB line, then

Since cutangle = arcsine(x"/30")

it is noted that units cancel, so unit of measurement is irrevelent.

x cutangle by above eqn 2x (which is the estimate used)

3 5.7 6 degrees
8 15.5 degrees 16 degrees
10 19.5 degrees 20 degrees
15 30 30
18 36.9 36 degrees

So, for practical purposes, errors are generally a fraction of a mm.