WHY DO THE HEATHEN CONTINUE TO RAGE....Part Two

[BC21]

I think you've reversed those.

pj
chgo

Lol....yes I did. Since the post-impact action of the balls was not my concern, I just skimmed through Dave's documents for the info. And I did have it backwards, which is weird because I already knew that a rolling cb hitting an ob at a halfball hit results in both balls going about equal distance. But I didn't know it was at exactly 33.5° that this happens. Anyway, I edited my reply to fix that error.

 

[Patrick Johnson]

If I had been struck at a glancing blow, where the other ball's center of mass is lined 1.125" away from my center mass, then I'd only feel half the impact that I'd feel with a head on collision.

Dr. Dave says the OB travels 3 times as far as the stunned CB with a half ball hit. Hard for me to reconcile that with half the impact.

pj
chgo

 

[bbb]

Well, for what it’s worth...

At 45 degrees the CB and OB deflect at equal angles from the CB’s original path.

45 degrees is midway between zero and 90 degrees.

Maybe some other things if I think of any.

pj
chgo

But the overlap is not 50/50
Therefore the collision isnt 50/50

 

[bbb]

Dr. Dave says the OB travels 3 times as far as the stunned CB with a half ball hit. Hard for me to reconcile that with half the impact.

pj
chgo

This is what i am saying

 

[BC21]

Dr. Dave says the OB travels 3 times as far as the stunned CB with a half ball hit. Hard for me to reconcile that with half the impact.

pj
chgo

Yeah, that's due to the friction between the balls. Still, it's half of each ball's mass interacting (overlapping) at impact. Factoring in the fact that one ball is moving and the other is static, and factoring in the friction at the contact patch on a stun shot, the cb naturally slows down more after impact.

But with a rolling cb, where the contact friction is reduced, the cb's post impact rolling distance is closer to the ob's. There is still friction, so that's why the equal roll occurs at 33.5° and not 30°. Put some silicone on the balls and it'll occur much closer to 30°, a halfball hit.

 

[chefjeff]

Nope, place two balls touching each other, look straight down at them from overhead.

This 1/2 ball, full ball concept only exist in a person s head, and not the real world. It is a visualization technique only.

So it's a technique?

That's how it works. It's not as if it doesn't work. It does work. It works so players use it. Wowser.

Visualize the plane, put the cue in that plane, shoot. Kerplunk. shoot again.

What's the problem with that? Are you upset that a half ball hit isn't really a half ball hit? Who cares if the ball goes in?


Jeff Livingston

 

[duckie]

Ever think that telling someone something is a 1/2 ball hit means nothing to them?

Therefore your attempt at communicating something aint worth shit?

The point so overlooked is how many different types of visualization techniques are used in pool. Visualization only can occur in a person minds eye.

It also shows how limited thinking is.

Im not upset, I just stand my ground......

 

[chefjeff]

Ever think that telling someone something is a 1/2 ball hit means nothing to them?

Therefore your attempt at communicating something aint worth shit?

The point so overlooked is how many different types of visualization techniques are used in pool. Visualization only can occur in a person minds eye.

It also shows how limited thinking is.

Im not upset, I just stand my ground......

I point the center of my tip at the edge of the ob (that's the part where the ball ceases to be a ball) and call that a half ball hit. When I show that to anyone, they immediately grasp it and see it and can try to do it.

Now you know what it means, so it is no longer worth just shit. It is now an expanded thinking technique that works to pot balls.

I like potting balls. Much better than missing.


Jeff Livingston

 

[Patrick Johnson]

Ever think that telling someone something is a 1/2 ball hit means nothing to them?

Everybody seems to understand it except you.

Therefore your attempt at communicating something aint worth shit?

Talking to yourself again...

pj
chgo

 

[dr_dave]

For those interested, the info, video, article, and analyses here answer all the questions in this thread:

Ball Speeds and Travel Distances

Ball-Hit Fraction​

Enjoy!

 

[dr_dave]

Dr. Dave says the OB travels 3 times as far as the stunned CB with a half ball hit. Hard for me to reconcile that with half the impact.

It might be easier to reconcile it considering the post-impact speeds rather than distances travelled after skidding. FYI, I just added a new page (page 2) to TP 3.2 – Ball speeds and distances after stun-shot impact that shows stun-shot speed ratios for various ball-hit fractions.

For a 1/2-ball-hit stun shot, compared to the incoming CB speed, the post-impact CB and OB speeds are 50% and 87%. So the OB has 1.7-times the CB speed immediately after impact.

Yeah, that's due to the friction between the balls.

Actually, the TP 3.2 analysis assumes perfect (frictionless) balls.

But with a rolling cb, where the contact friction is reduced, the cb's post impact rolling distance is closer to the ob's. There is still friction, so that's why the equal roll occurs at 33.5° and not 30°. Put some silicone on the balls and it'll occur much closer to 30°, a halfball hit.

Not true. TP A.16 – Final ball speeds, distances, and directions for natural roll shots also assumes perfect (frictionless) balls. With or without friction, the ball-hit fraction resulting in equal post-impact ball speeds is slightly thinner than a half-ball hit.

Regards,
Dave

 

[Imac007]

Everybody seems to understand it except you.

Talking to yourself again...

pj
chgo

Words are just that, they represent something.
The map is not the territory.
Language is basically description.
It’s a communication tool whose intent is to share understandings.
The fact that duckie could even disagree means he understood the description’s intended message.

 

[BC21]

Actually, the TP 3.2 analysis assumes perfect (frictionless) balls.

.....with a rolling cb, where the contact friction is reduced, the cb's post impact rolling distance is closer to the ob's. There is still friction, so that's why the equal roll occurs at 33.5° and not 30°. Put some silicone on the balls and it'll occur much closer to 30°, a halfball hit.

Not true. TP A.16 – Final ball speeds, distances, and directions for natural roll shots also assumes perfect (frictionless) balls. With or without friction, the ball-hit fraction resulting in equal post-impact ball speeds is slightly thinner than a half-ball hit.

Regards,
Dave

According to TP A.16, on page 7 (post speed results accounting for friction), you write, "...the CB picks up a little speed after impact for all ball-hit-fractions, except for a 0% ball-hit (90 degree cut angle)."

But the document shows that when shooting a 30° shot (halfball hit) with "perfect" balls (no friction), the post impact cb speed is 55.8% of its orginal speed (as described on page 2 of TP A.16). When accounting for friction the post impact cb speed is 53.6% (page 8). The document also shows that the ob does travel a little more than the cb when the impact friction is accounted for (except for thinner cut shots).

You show it being 33.5° for no friction and 33.2° with friction. Seems like there should be a larger difference between the two, but math is math.

According to the document, with zero friction on a 30° cut shot the difference between ob and cb post impact speeds is 6.1% (ob leaves the scene 6.1% faster than the cb leaves, both compared to the cb's orginal speed). At the same 30° shot angle, but this time accounting for friction, the ob leaves at a speed 6.5% faster than the cb. Both balls have lower post impact speeds when friction is factored in, but the ob speed is affected less. (With no friction you show ob speed as 61.9% and cb speed as 55.8%. Factoring in friction those numbers become 60.1% and 53.6%.)

Interesting stuff nonetheless.

 

[dr_dave]

According to TP A.16, on page 7 (post speed results accounting for friction),

Wow! I'm impressed you ventured all the way to page 7. Not many people do. You are a brave soul.

you write, "...the CB picks up a little speed after impact for all ball-hit-fractions, except for a 0% ball-hit (90 degree cut angle)."

That makes sense since the topspin remaining on the CB after impact accelerates the ball forward a little after the collision.

But the document shows that when shooting a 30° shot (halfball hit) with "perfect" balls (no friction), the post impact cb speed is 55.8% of its original speed (as described on page 2 of TP A.16). When accounting for friction the post impact cb speed is 53.6% (page 8). The document also shows that the ob does travel a little more than the cb when the impact friction is accounted for (except for thinner cut shots).

That all makes sense too.

You show it being 33.5° for no friction and 33.2° with friction. Seems like there should be a larger difference between the two, but math is math.

It is not just math. It is physics too. The friction between the balls has two competing effects:
1.) It slows the CB along the tangent line a little (due to the thrown OB pushing back on the CB). This narrows the carom angle.
2.) It reduces the topspin slightly. This reduces the curves forward slightly, which widens the angle.

These effects partly cancel each other.

Check out the 3:28 and 4:08 points in my CB Control video. It shows how greater friction due to slower speed widens the carom angle (showing that effect 2 is stronger than effect 1):

According to the document, with zero friction on a 30° cut shot the difference between ob and cb post impact speeds is 6.1% (ob leaves the scene 6.1% faster than the cb leaves, both compared to the cb's orginal speed). At the same 30° shot angle, but this time accounting for friction, the ob leaves at a speed 6.5% faster than the cb. Both balls have lower post impact speeds when friction is factored in, but the ob speed is affected less. (With no friction you show ob speed as 61.9% and cb speed as 55.8%. Factoring in friction those numbers become 60.1% and 53.6%.)

That all makes sense too. Friction from the rolling CB imparts slight bottom spin to the OB, which slows it slightly. But the sideways throwing force on the OB (also due to friction) tends to imparts a little extra speed to the OB. The math and physics analysis show that the backspin effect is stronger.

Interesting stuff nonetheless.

Agreed.

 

[Patrick Johnson]

Wow! I'm impressed you ventured all the way to page 7. Not many people do. You are a brave soul.

I have to admit I'm not so brave, so I'll ask a question that you might have already answered...

At what amount of CB/OB overlap is half the CB's momentum transferred to the OB (ignoring other effects like friction and CB rotation)?

Thanks for your thoughts, Dave.

pj
chgo

 

[dr_dave]

According to TP A.16, on page 7 (post speed results accounting for friction), you write, "...the CB picks up a little speed after impact for all ball-hit-fractions, except for a 0% ball-hit (90 degree cut angle)."

But the document shows that when shooting a 30° shot (halfball hit) with "perfect" balls (no friction), the post impact cb speed is 55.8% of its orginal speed (as described on page 2 of TP A.16). When accounting for friction the post impact cb speed is 53.6% (page 8). The document also shows that the ob does travel a little more than the cb when the impact friction is accounted for (except for thinner cut shots).

You show it being 33.5° for no friction and 33.2° with friction. Seems like there should be a larger difference between the two, but math is math.

According to the document, with zero friction on a 30° cut shot the difference between ob and cb post impact speeds is 6.1% (ob leaves the scene 6.1% faster than the cb leaves, both compared to the cb's orginal speed). At the same 30° shot angle, but this time accounting for friction, the ob leaves at a speed 6.5% faster than the cb. Both balls have lower post impact speeds when friction is factored in, but the ob speed is affected less. (With no friction you show ob speed as 61.9% and cb speed as 55.8%. Factoring in friction those numbers become 60.1% and 53.6%.)

Interesting stuff nonetheless.

It can be difficult to figure out some pool physics stuff with just intuition and logic. That's why I do the math and physics analyses. They often yield interesting and useful results. Of all the analyses I have done over the years, the throw (friction effects) analysis has taught me the most. A complete summary of everything I learned from the throw analysis is summarized here:

complete list of throw effects (with supporting resources)
​

You might also find some of the answers to common questions related to throw interesting. Check it out. Especially interesting are some of the conclusions and Q&A related to inside spin. Check that out also. I honestly found it hard to believe some of this stuff at first after doing the analyses, but experiments, lots of pool experience, and some intuitive understanding (helped by the analysis) validate all of it.

 

[dr_dave]

I have to admit I'm not so brave, so I'll ask a question that you might have already answered...

At what amount of CB/OB overlap is half the CB's momentum transferred to the OB (ignoring other effects like friction and CB rotation)?

From page 3 of TP A.16, I think the answer you are looking for is:

cut angle = 33.2˚
ball-hit-fraction = 0.45​

In other words, slightly thinner than a 1/2-ball hit.

Thanks for your thoughts, Dave.

You're welcome. I aim to swerve (but that is a totally different topic).

 

[Patrick Johnson]

From page 3 of TP A.16, I think the answer you are looking for is:

cut angle = 33.2˚
ball-hit-fraction = 0.45​

In other words, slightly thinner than a 1/2-ball hit.

So like Brian said. Guess that'll teach me to second guess him. Maybe I misread the .45 ball fraction...?

I still wonder how both balls travel the same distance after a 45-degree stun hit. And how their combined speed is more than the CB's original speed.

pj
chgo

 

[BC21]

So like Brian said. Guess that'll teach me to second guess him... Maybe I misread the .45 ball fraction...?

I still wonder how both balls travel the same distance after a 45-degree stun hit. And how their combined speed is more than the CB's original speed.

pj
chgo

I think, according to Dave's TP A.16 work, the equal distance happens at 33.5°. That's when the ratio is 1:1.

 

[Patrick Johnson]

I think, according to Dave's TP A.16 work, the equal distance happens at 33.5°. That's when the ratio is 1:1.

Yes, for a rolling CB. I'm confused by Dave's TP 3.2 (for stun hits), which seems to say the CB's and OB's after-collision speed is equal at a 45° cut - which I assume means they have equal momentum.

pj
chgo