Cue Ball and Object Ball Travel Distances Plot

[Jal]

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.

In the context of determining how far the cueball will roll, your method should generally give about as much accuracy as you can reasonably expect, except perhaps when the balls are fairly close together. The plots that Colin and Dr. Dave produced are geared toward the more standard definition of cut angle: the angle between the cueball's pre-impact direction and the object ball's post-impact direction (sans throw). This definition diverges from yours more severely as the CB-OB distance diminishes.

Not knocking your method, just pointing out where some problems might occur. Visualizing your right triangle has some advantages over trying to set up the isosceles version.

Jim

 

[CueAndMe]

Why does Dead Crab's method work at 30"?

 

[Bob Jewett]

Following up on the discussion in this thread, I've completed what I believe to be an accurate plot of travel distances for the Cue Ball and Object Ball for the full range of cut angles.
...

Over in CCB, a poster named "cushioncrawler" or some such, has shown a diagram that presents this data in a very interesting form. His experiment (I don't think he did the math) shows the final locations of the cue ball and object ball for a constant rolling speed on the cue ball (down a ramp) with a varying fullness of hit on the object ball. The two sets of ending points form more or less (but not quite) circular arcs. It is a good way of combining the angle information with the distance information.

Similar diagrams can be done for stun and "full draw" shots, but the draw case is much harder to do experimentally.

 

[Jal]

Why does Dead Crab's method work at 30"?

I don't know how deep of a reason you're looking for, but here is a simple, if somewhat superficial, explanation.

In a triangle where one of the angles is 90 degrees, as in Patrick's diagram, the longest side is called the hypotenuse. For either of the other two angles in the triangle, the ratio of the length of the side opposite to it (one of the "legs" of the triangle) to the length of the hypotenuse is defined as the "sine" of that angle. Every angle from 0 to 90 degrees has a unique sine value, which ranges from 0 to 1. In Patrick's diagram, the sine of the angle formed by the black line and shaft would be the length of the red line divided by the length of the shaft (that is, if he were to move the cue forward a little so that the tip met with the black line thus forming a close triangle).

The sine of 30 degrees happens to be 1/2. Since Dead Crab uses a hypotenuse of 30 inches (shaft + ball radius, approx.), the length of the perpendicular he drops to the CB-OB line is thus 1/2 of 30 inches when his "cut angle" is 30 degrees. So doubling it would return the number 30 exactly if he were to set up so that the hypotenuse was exactly 30 inches.

Edit: changed "In Patrick's diagram, the sine of the angle formed by the red line and shaft... to "In Patrick's diagram, the sine of the angle formed by the black line and shaft...".

Jim

 

[CueAndMe]

I don't know how deep of a reason you're looking for, but here is a simple, if somewhat superficial, explanation.

In a triangle where one of the angles is 90 degrees, as in Patrick's diagram, the longest side is called the hypotenuse. For either of the other two angles in the triangle, the ratio of the length of the side opposite to it (one of the "legs" of the triangle) to the length of the hypotenuse is defined as the "sine" of that angle. Every angle from 0 to 90 degrees has a unique sine value, which ranges from 0 to 1. In Patrick's diagram, the sine of the angle formed by the red line and shaft would be the length of the red line divided by the length of the shaft (that is, if he were to move the cue forward a little so that the tip met with the black line thus forming a close triangle).

The sine of 30 degrees happens to be 1/2. Since Dead Crab uses a hypotenuse of 30 inches (shaft + ball radius, approx.), the length of the perpendicular he drops to the CB-OB line is thus 1/2 of 30 inches when his "cut angle" is 30 degrees. So doubling it would return the number 30 exactly if he were to set up so that the hypotenuse was exactly 30 inches.

Jim

Thanks for taking the time to explain. I see it now.

 

[Colin Colenso]

Over in CCB, a poster named "cushioncrawler" or some such, has shown a diagram that presents this data in a very interesting form. His experiment (I don't think he did the math) shows the final locations of the cue ball and object ball for a constant rolling speed on the cue ball (down a ramp) with a varying fullness of hit on the object ball. The two sets of ending points form more or less (but not quite) circular arcs. It is a good way of combining the angle information with the distance information.

Similar diagrams can be done for stun and "full draw" shots, but the draw case is much harder to do experimentally.

Thanks for the heads up Bob.

I scoured cushioncrawler's posts but couldn't find it, so I messaged him.

I've met Max actually. He lives less than an hour's drive from my place

Must be something in the water around here

Colin

 

[Jal]

Thanks for taking the time to explain. I see it now.

Except I goofed. Please see the edit, Jeff. Sorry about that.

Jim

 

[CueAndMe]

Except I goofed. Please see the edit, Jeff. Sorry about that.

Jim

You're allowed. I actually assumed that was the angle you were referring to anyway.
Thanks again.

 

[dr_dave]

distance graphs and ratios including inelasticity and friction

Actually, these charts do not take into account frictional and restitutional losses between the balls on impact.

FYI, I just added graphs and distance proportions to TP A.16, taking into consideration ball inelasticity and friction effects (see pages 10 and 11). I already had all of the equations ... I just needed to add the distance stuff.

Where thee straight shot ratio is calculated at 6.1 to 0.98, frictional forces would probably make this closer to 5.8 to 1.0 in reality. The friction of the OB tends to pull the CB with it a little and the OB will lose some additional velocity due to imperfect restitution. (Maybe some collision experts can explain this more clearly than I.)

Actually, I calculate a ratio closer to 7 to 1 (see page 11 in TP A.16). To better understand and quantify the effects of friction (mu) and the coefficient of restitution (e), I calculated the ratio for different cases (this isn't in my TP document). Here's what I got:

ideal (e=1, mu=0): OB/CB = 6.3
inelasticity only (e=0.94, mu=0): OB/CB = 5.1
friction only (e=1, mu=0.06): OB/CB = 8.7
inelasticity and friction (e=0.94, mu=0.06): OB/CB = 6.8
​

The final case is what is documented at the end of TP A.16. I think the simplest explanation is that the effect of friction to slow the CB's forward roll during impact is a big factor.

Regards,
Dave

 

[Colin Colenso]

I think the simplest explanation is that the effect of friction to slow the CB's forward roll during impact is a big factor.

Regards,
Dave

That makes some sense Dave, but what about the effect of the OB pulling the CB forward?

I used to play around with putting Blutak on the contact points. It would make the CB move forward through the collision on stun shots. Is that a frictional effect or a restitutional effect or a bit of both?

btw: Where did you get the e=0.94 and mu=0.06 from and what kind of variability might we find between new balls and old pitted and greasy balls?

Thanks,
Colin

 

[dr_dave]

restitution and friction Q&A

what about the effect of the OB pulling the CB forward?

I'm not sure what you mean about the OB "pulling" the CB forward. The CB moves forward more (i.e., less momentum is transferred to the OB) for a smaller coefficient of restitution (i.e., less "elastic" balls, where e<1.0).

I used to play around with putting Blutak on the contact points. It would make the CB move forward through the collision on stun shots. Is that a frictional effect or a restitutional effect or a bit of both?

Ball friction has no effect (practically speaking anyway) with a full-ball stun shot. There is no relative sliding between the ball surfaces during impact. If the CB moves forward with a stun shot, either the CB is heavier or the coefficient of restitution (e) is less than 1.

btw: Where did you get the e=0.94 and mu=0.06 from and what kind of variability might we find between new balls and old pitted and greasy balls?

I have a summary of generally accepted values for most pool physical parameters here:

http://billiards.colostate.edu/threads/physics.html#properties​

Sorry, but I haven't seen published data, nor have I taken measurements, concerning how much the parameters change for different conditions.

Regards,
Dave

 

[Jal]

...btw: Where did you get the e=0.94 and mu=0.06 from and what kind of variability might we find between new balls and old pitted and greasy balls?

Colin, your compatriot Mac (Cushioncrawler) has measured e for a variety of ball compositions over a range of speeds. He hung the balls up by threads and had them collide, pendulum style. You may want to ask him for his graphs if you're interested. It can also be measured fairly easily by setting up the balls up in a certain way at the table, which I haven't done though (natch).

I believe Dr. Dave obtained his data for mu from tests done by Wayland Marlow and reported in his book "The Physics of Pocket Billiards". (Didn't you do some tests too for the contour plots?) Marlow's tests show a large variation in mu with surface speed, ranging perhaps as much as .01 to .1, with the largest values at slow surface speeds and the least at fast surface speeds. There may also be a somewhat significant variation with normal speed along the impact line (line of centers velocity component).

Jim
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