Ball Phases and transitions and how they are defined

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JohnP
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Re: RG Values

Post by JohnP »

What you're ignoring is that very few (if any) bowlers can generate a rev rate that's higher than the "true roll" rate (~525 rpm). I've seen some pretty strong crankers, including Rudy Revs and Jason Belmonte, and I've never seen a ball that I thought sped up. -- JohnP
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Re: RG Values

Post by MWhite »

JohnP wrote:What you're ignoring is that very few (if any) bowlers can generate a rev rate that's higher than the "true roll" rate (~525 rpm). I've seen some pretty strong crankers, including Rudy Revs and Jason Belmonte, and I've never seen a ball that I thought sped up. -- JohnP
Both of those people also throw pretty high ball speed.

Back in the 80's, there was a new Casino in on the North West side of Vegas that had an early version of CATS. I think it was Fiesta, or something similar.

It would identify the ball speed at different points on the lane as well as which board the ball was on.

My readout indicated that early on the lane I was at 16 mph, but further down I was at 17 mph.

17 mph requires 665 rpm.

With Reactive Resin balls, and the THS, there isn't a reward for an "over revved" ball, and consequently very few people shooting on a THS are capable of "over revving" the ball.

However out on the tour, with much heavier oil, and tamed down balls, you see more of a reward for high revs.

The point is, even if it's rare, the definition of phases should be physics based so high rev / low speed needs to fit in the criteria.

The current criteria for 1st transition of "force from speed = force from revs" is wrong.
The current criteria for 2nd transition of "maximum revs" is also wrong.

You can set up a simple example in Blue Print and see that the ball begins changing direction well before the speed / rev graphs flatten out.

And if you create a bowler capable of over revving, the maximum rpm occurs at the foul line, and has decreased before the ball goes into the roll phase.

Depending on how much over rev you select, Blueprint can generate some unusual graphs.
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Re: RG Values

Post by Mo Pinel »

MWhite wrote: These descriptions about forces are very vague.
Forces have both a magnitude, and a direction.

The 2nd transition description is wrong, because if someone threw the ball 13 mph, with 550 rpm, the ball will be in the skid phase, but will also be at it's maximum rev rate.

Personally in the 80's, my ball speed was about 16 mph, and the rev rate was approximately 650.
I haven't attempted to accurately measure it since returning to the sport, but since I can still hook a plastic ball as much as I used to, either my rev rate is still up there, or my speed has reduced.

This is me in Aug 2012, about 6 weeks back into the sport after about 20 years away.

[youtube][/youtube]
Maximum rev rate is definitely achieved at the second transition when the skid factor is entirely gone. No slip any more after that. The rev rate will decrease very slightly after that because the ball speed continues to decrease slowly. There is no other accurate explanation as much as you want to massage it.
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Re: RG Values

Post by MWhite »

Mo Pinel wrote:Maximum rev rate is definitely achieved at the second transition when the skid factor is entirely gone. No slip any more after that. The rev rate will decrease very slightly after that because the ball speed continues to decrease slowly. There is no other accurate explanation as much as you want to massage it.

This is true if you assume the ball starts with more speed, than rev rate.

The problem is, there is no physical law that requires the rev rate to be less than the ball speed.

My main point of the thread was trying to define the 1st transition, and if a drill pattern has an effect on that location.

I think the definition of the 1st transition from the graphic, should actually be the definition of the 2nd transition. When the ball speed is equal to the rev rate. The wording needs to be clearer, but the concept is good.

The best I can see for the 1st transition is when the ball sees enough friction to potentially change direction.

A zero axis of rotation ball wouldn't change direction when encountering friction.
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Re: RG Values

Post by Mo Pinel »

MWhite wrote:
This is true if you assume the ball starts with more speed, than rev rate.

The problem is, there is no physical law that requires the rev rate to be less than the ball speed.

My main point of the thread was trying to define the 1st transition, and if a drill pattern has an effect on that location.

I think the definition of the 1st transition from the graphic, should actually be the definition of the 2nd transition. When the ball speed is equal to the rev rate. The wording needs to be clearer, but the concept is good.

The best I can see for the 1st transition is when the ball sees enough friction to potentially change direction.

A zero axis of rotation ball wouldn't change direction when encountering friction.

For a ball to have less rev rate than ball speed, it must be spinning backwards! Nobody trying to effectively deliver a bowling ball would throw it this way. Let's keep this discussion practical or it becomes useless! I'm not interested in wasting my time on that type discussion. Have at it solo!
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Re: RG Values

Post by JohnP »

I think when he says "more speed than rev rate" what he's talking about is the rotational speed of the ball track in mph compared to the linear speed of the ball, also in mph. -- JohnP
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Re: RG Values

Post by MWhite »

Mo Pinel wrote:
For a ball to have less rev rate than ball speed, it must be spinning backwards! Nobody trying to effectively deliver a bowling ball would throw it this way. Let's keep this discussion practical or it becomes useless! I'm not interested in wasting my time on that type discussion. Have at it solo!
WOW…

Has someone hacked Mo's account?

Maybe Mo just had a bad moment there.

Or, maybe it's a tactic to burn the messenger, and ignore the message.


Nobody said anything about the ball spinning backwards.
Stating that less revs than ball speed means the ball is spinning backwards, is a gross misinterpretation of the facts.


At ball thrown at 16 mph with 0 axis of rotation, and 0 tilt would require 625 RPM for the rev rate, and ball speed to be equal at release.

There is no physical law that states the rev rate can't be higher, or lower than 625 when thrown at 16 mph.

For most people, their rev rate is lower than their ball speed at release.

The transition definitions should be true for all people.
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Re: RG Values

Post by LabRat »

The transitions are about friction. If you want a meaningful definition, it needs to be within the context of the ball motion you are examining. The transitions for a straight ball while they exist, are unimportant in the context of the motion of a hook ball.
To me, the first transition happens when the ball encounters enough friction to start to hook. There will always be some friction even in the skid phase, but at some point there will be a discontinuity in the friction profile as the ball travels down the lane. That first 'corner' in the friction profile is the first transition. The second happens when the friction profile drops back to a minimum value, and the ball is rolling.
However, these are not 'useful' definitions. Where the ball starts to hook is useful, because we can see it and point it out to others.

To the original question - Drill angle doesn't matter in syms. The rg contours are circular (neglecting the effect of drilled holes, which will be the same for all drill angles), so the PAP migration path is dynamically the same for any given pin-PAP distance. This is not true for asyms - first, the rg of the PAP is different for different drilling angles (actually larger for smaller DA's) and the RG contours are elliptical. Higher drilling angles place the PAP on a lower differential (rg contours are further apart) but higher eccentricity (rg contours are more curved) pathway than low drill angles. I don't yet fundamentally understand why this translates to higher drill angles going longer than low angles. There is certainly a physical difference in the ball dynamics when the DA changes, which will affect the response of the ball to friction, and in the real world this shows up as larger DA = more skid.
Steve could probably explain this a lot better than me.
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Re: RG Values

Post by MWhite »

LabRat wrote:The transitions are about friction. If you want a meaningful definition, it needs to be within the context of the ball motion you are examining. The transitions for a straight ball while they exist, are unimportant in the context of the motion of a hook ball.
To me, the first transition happens when the ball encounters enough friction to start to hook.
I completely agree with that if you agree hook means deviate from it's original straight path (assuming it is going to deviate)
LabRat wrote:There will always be some friction even in the skid phase, but at some point there will be a discontinuity in the friction profile as the ball travels down the lane. That first 'corner' in the friction profile is the first transition. The second happens when the friction profile drops back to a minimum value, and the ball is rolling.
I don't think there is lane to ball friction in the skid phase, the only friction is could see potentially if from the viscosity of the oil. The skid phase implies the ball is hydroplaning on the oil.
LabRat wrote:However, these are not 'useful' definitions. Where the ball starts to hook is useful, because we can see it and point it out to others.
It's pretty hard to see where the ball starts to first deviate from the straight path. Estimating it however isn't very hard. It's usually where you cross from heavy oil, out to lighter oil.

LabRat wrote:To the original question - Drill angle doesn't matter in syms. The rg contours are circular (neglecting the effect of drilled holes, which will be the same for all drill angles), so the PAP migration path is dynamically the same for any given pin-PAP distance. This is not true for asyms - first, the rg of the PAP is different for different drilling angles (actually larger for smaller DA's) and the RG contours are elliptical. Higher drilling angles place the PAP on a lower differential (rg contours are further apart) but higher eccentricity (rg contours are more curved) pathway than low drill angles. I don't yet fundamentally understand why this translates to higher drill angles going longer than low angles. There is certainly a physical difference in the ball dynamics when the DA changes, which will affect the response of the ball to friction, and in the real world this shows up as larger DA = more skid.
Steve could probably explain this a lot better than me.
I think it's unfortunate you chose the word skid in "larger DA = more skid"
While I can see what you are talking about, I think it highlights an area where the choice of words creates confusion.

Your using the word skid to describe a condition similar to the tires on an airplane.

The 1st phase is while the tires are in the air (no friction)
The 2nd phase starts the instant the wheels touch down ( what we normally think of as skidding)
The 3rd phase begins when surface speed of the tires match the ground speed of the airplane. (rolling)


If the phases were called

1) Pure Slide (no friction)
2) Skid going to Roll (increasing friction as oil reduces)
3) Pure Roll (friction forces due to speed and RPM have reached equilibrium)

then it would probably match our normal use of the words.

The larger DA appears to lengthen the distance between the 1st and 2nd transition.

I don't think it increases the distance between the foul line, and the 1st transition.

The odd thing is, one would assume the smaller drill angle (translating to a larger RG for the PAP) would cause the ball to take longer to transition thru the 2nd phase due to more energy required to change the RPM rate.

So clearly there is something else in play.

Mo has mentioned something about when the migrating PAP reaches the line between the PSA, and the Pin, the ball changes it's rev rate faster.

I don't remember him stating why this happens.

It would explain more length in the 2nd phase for the larger DA since the original PAP is further from this PSA to Pin line as the DA increases.
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Re: RG Values

Post by Daryl »

This is a great thread. I love this stuff. I am not sure about some of the conclusions. I throw a bowling ball at around 15 mph off the hand at about 390 RPM's. For a bowling ball to travel 15 mph how many revolutions per minute would the ball be rotating to reach that speed without the help of a bowler? Is it more than 390 RPM's or less than 390 RPM's?

If a ball traveling down the lane at 15 MPH would naturally generate more RPM's (say 500 RPM's) than I throw it at (390 RPM's) than I would assume in my case that as soon as my ball (traveling at 15 MPH, but only with 390 RPM's) hit the end of the oil pattern it's speed would begin slowing down (not enough RPM's to maintain 15 MPH) and the RPM's would begin to pick up due to friction to match the speed of the ball. Eventually the two forces would equalize. The speed of the ball would equal the RPM's of the ball at some point down the lane.

On the other hand if a ball traveling down the lane at 15 MPH naturally generates less RPM's (say 300 RPM's) than I throw it at (390 RPM's) I would assume that as soon as my ball (traveling at 15 MPH with 390 RPM's) hit the end of the oil pattern it's speed (MPH) would increase (because 390 RPM's naturally produces a faster speed than 15 MPH in this example) and the RPM's would begin to decrease due to friction. Again at some point the speed and RPM rates would equalize down lane. Just like a spinning car tire does going from skidding (slipping) on the ice to a dry patch.

I am assuming that the oil pattern in these examples would generate no friction throughout the entire length of the pattern.

Am I on the same page here? If not please explain.

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Re: RG Values

Post by MWhite »

Daryl wrote:This is a great thread. I love this stuff. I am not sure about some of the conclusions. I throw a bowling ball at around 15 mph off the hand at about 390 RPM's. For a bowling ball to travel 15 mph how many revolutions per minute would the ball be rotating to reach that speed without the help of a bowler? Is it more than 390 RPM's or less than 390 RPM's?
A ball traveling at a rate of 15 mph is traveling at a rate of 15840 inches per minute.
The ball circumference of the ball is about 27 inches. 15480 / 27 = 587 RPM

Those numbers need to be increased when considering tilt and axis rotation.
Daryl wrote:If a ball traveling down the lane at 15 MPH would naturally generate more RPM's (say 500 RPM's) than I throw it at (390 RPM's) than I would assume in my case that as soon as my ball (traveling at 15 MPH, but only with 390 RPM's) hit the end of the oil pattern it's speed would begin slowing down (not enough RPM's to maintain 15 MPH) and the RPM's would begin to pick up due to friction to match the speed of the ball. Eventually the two forces would equalize. The speed of the ball would equal the RPM's of the ball at some point down the lane.

On the other hand if a ball traveling down the lane at 15 MPH naturally generates less RPM's (say 300 RPM's) than I throw it at (390 RPM's) I would assume that as soon as my ball (traveling at 15 MPH with 390 RPM's) hit the end of the oil pattern it's speed (MPH) would increase (because 390 RPM's naturally produces a faster speed than 15 MPH in this example) and the RPM's would begin to decrease due to friction. Again at some point the speed and RPM rates would equalize down lane. Just like a spinning car tire does going from skidding (slipping) on the ice to a dry patch.

I am assuming that the oil pattern in these examples would generate no friction throughout the entire length of the pattern.

Am I on the same page here? If not please explain.

Daryl

We are on the same page, however oil patterns tend to thin at it's borders (both sideways and lengthwise) so there would usually be friction before the end of the pattern.
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Re: RG Values

Post by RobMautner »

The only place where there is no friction is in a vaccuum: space. When the ball is skidding in the oil, it is still encountering friction, just not enough to allow the rotational energy to become equal to the linear energy created by the bowler, so that the ball can begin to hook.

Imagine a bowling lane that is one mile long, and oiled the entire length. Would the ball go the entire mile and not stop? Of course not. It would eventually stop by the limited friction that it is finding in the oil, and in the air. A space ship, on the other hand, once leaves the earth's atmosphere, just needs to fire its rockets once, and it will keep going in one direction until it hits something. A meteorite, travelling at thousands of miles per hour, will burn up when it enters the Earth's atmosphere as a result of the friction that it encounters from the molecules that make up the atmosphere.

Part of the challenge of evaluating ball motion is that we try to deal in absolutes, when there really aren't any.
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Re: RG Values

Post by Daryl »

So since my rev rate off the hand (390 RPM's) is less than a normally rolling 15 MPH ball (587 RPM's) than the statements in the Understanding Ball Motion graph are true for my delivery. But if I had a rev rate of 650 RPM's and threw the ball only at 15 MPH than some of the statements in the Understanding Ball Motion graph would need to be adjusted.

Are there really any who over rev the ball beyond the speed at which they through it normally? If so I guess there needs to be an asterisk in the ball motion graphs.

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Re: RG Values

Post by bowl1820 »

I got to say this is the most confusing thread.

First it starts off talking about RG, then changes into a debunking of the ball motion definitions.
(The Mod. should really split this into two different threads)

And the rewriting of the ball motion definitions is really convoluted, One minute it seems to be talking about the skid phase up to the first transition,

Then you see things like this:
A ball traveling at a rate of 16 mph ball having a rpm of 625
A ball traveling at a rate of 15 mph ball having a rpm of 587

Those figures appear to be based on balls in the roll phase which would be after the 2nd transition, so they can't be talking about the skid phase leading up to the 1st transition .

:?:so confusing :?:
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Re: RG Values

Post by MegaMav »

bowl1820 wrote:(The Mod. should really split this into two different threads)
Care to tell me which posts numbers should be in their own topic?
Im lost as well.
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Re: RG Values

Post by bowl1820 »

MegaMav wrote: Care to tell me which posts numbers should be in their own topic?
Im lost as well.

I'd say about everything after post #15 , from post #16 and on The RG stuff just vanishes , from there on is where all the talk about the transitions and the ball phases, rev rates rpms etc start.

Have thread called something like "Ball Phases and transitions and how they are defined" maybe.
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Re: RG Values

Post by MWhite »

bowl1820 wrote:I got to say this is the most confusing thread.

First it starts off talking about RG, then changes into a debunking of the ball motion definitions.
(The Mod. should really split this into two different threads)

And the rewriting of the ball motion definitions is really convoluted, One minute it seems to be talking about the skid phase up to the first transition,

Then you see things like this:
A ball traveling at a rate of 16 mph ball having a rpm of 625
A ball traveling at a rate of 15 mph ball having a rpm of 587

Those figures appear to be based on balls in the roll phase which would be after the 2nd transition, so they can't be talking about the skid phase leading up to the 1st transition .

:?:so confusing :?:

Those numbers are talking about the threshold between an under revved ball, and an over revved ball just off the hand.

An over revved ball, and an under revved ball will both hydroplane in the early part of the lane sensing effectively 0 friction.

The differences occur once the depth of the oil is low enough to end the hydroplaning.
At that moment (the 1st transition), an over revved ball will begin to increase speed, while an under revved ball will begin to decrease speed.

Unless the ball has an axis of rotation of 0 degrees, there is a sideways component to the forces created by friction. That sideways component causes the ball's path to become non-linear. (the hook phase)

During the hook phase, the balls speed and rev rate are coming into equilibrium.
For most people, the ball is slowing down, and the revs are increasing.

Once the ball speed, and rev rate have reached equilibrium (the 2nd transition) the ball has 0 degrees axis of rotation relative to the balls direction, and there is no more sideways component.

In the roll phase, the speed, and rev rate are in equilibrium, so there is no significant acceleration nor deceleration to either the rev rate, or ball speed.

However, there is some energy loss during the roll phase, because if the ball begins to roll too far from the pins, it can hit very weakly.

See, not confusing at all.
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Re: RG Values

Post by bowl1820 »

MWhite wrote:See, not confusing at all.
Your right it's not confusing.....you've taken it to a ridiculously confusing level!

you've tried to show how the definitions of ball motion are either wrong and/or poorly written and You've taken the points your trying to make and made them so technical their just a jumble.

IMO while I think you might have some valid points mixed in there, I think you are also not applying the math/physic's your trying to use to the correct things.

But I'm not going to be the one to sort it out, their are others here that know more about it than Me & You that can handle it.
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Re: RG Values

Post by EricHartwell »

(edit):My comments in color
MWhite wrote:
Those numbers are talking about the threshold between an under revved ball, and an over revved ball just off the hand.

An over revved ball, and an under revved ball will both hydroplane in the early part of the lane sensing effectively 0 friction.

The differences occur once the depth of the oil is low enough to end the hydroplaning.
At that moment (the 1st transition), an over revved ball will begin to increase speed, while an under revved ball will begin to decrease speed.
Unless the overrevved ball is revving at 0* rotation it will start to change direction, slow down and decrease in revs. The underrevved ball will also start to change direction, slow down and decrease in revs.

Unless the ball has an axis of rotation of 0 degrees, there is a sideways component to the forces created by friction. That sideways component causes the ball's path to become non-linear. (the hook phase)

During the hook phase, the balls speed and rev rate are coming into equilibrium.
For most people, the ball is slowing down, and the revs are increasing.

Once the ball speed, and rev rate have reached equilibrium (the 2nd transition) the ball has 0 degrees axis of rotation relative to the balls direction, and there is no more sideways component.
When tilt=rotation and not = to zero there is no more sideways component(nonlinear ball movement). When tilt=rotation the nonlinear effect on the ball from each component cancel each other out and thus the roll phase.

In the roll phase, the speed, and rev rate are in equilibrium, so there is no significant acceleration nor deceleration to either the rev rate, or ball speed.

the loss of energy in the roll phase shows up as the ball slowing down as well as rev rate.

However, there is some energy loss during the roll phase, because if the ball begins to roll too far from the pins, it can hit very weakly.

See, not confusing at all.
See, just as confusing as ever.
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Re: Ball Phases and transitions and how they are defined

Post by LabRat »

Not sure this thread is particularly useful any more - the number of bowlers that can overrev a ball, given tilt and rotation effects, is zero for all practical purposes. That said, I'm going into physics nit mode about a couple of points. :D

MWhite: I don't think there is lane to ball friction in the skid phase, the only friction is could see potentially if from the viscosity of the oil. The skid phase implies the ball is hydroplaning on the oil.
The depth of oil and ball speed is not enough for hydroplaning to occur. It's simply a lubricated friction skid. Friction is low, so the ball path is linear within the limits of experimental measurement. Obviously it won't be exactly linear, since there is always some friction and any friction will cause the ball to deviate from an exactly straight path, but the deviation is small enough to be ignored.
As an aside, meteors burn up through compressive heating of the atmosphere, not friction with air molecules.
If the phases were called

1) Pure Slide (no friction)
2) Skid going to Roll (increasing friction as oil reduces)
3) Pure Roll (friction forces due to speed and RPM have reached equilibrium)

then it would probably match our normal use of the words.
Our 'normal' use of the words is skid, hook, roll. They have accepted bowling specific definitions that work in all practical cases. There is no point changing them especially when one of the proposed changes (1) is simply incorrect.
MWhite - My readout indicated that early on the lane I was at 16 mph, but further down I was at 17 mph.

17 mph requires 665 rpm.
But to get from 16 to 17, given some axis rotation and tilt, requires 800+ revs at release. So, either you had more revs than anyone in the history of the sport, or the machine didn't work very well. Occams Razor applies here.
Eric - When tilt=rotation and not = to zero there is no more sideways component(nonlinear ball movement). When tilt=rotation the nonlinear effect on the ball from each component cancel each other out and thus the roll phase.
Eric - your statement that the ball rolls when axis tilt and rotation are equal is simply not true. A ball with 30* tilt and 30* axis rotation still has the surface of the ball in contact with the lane moving in a different direction from the linear direction of the ball. Blueprint has shown that this is the case.

So - skid phase - there is enough oil to reduce friction to the point where the ball path is practically straight;
Hook phase - friction increases to the point that the sideways rotation on the ball begins to noticably change the ball pathway;
Roll phase - axis rotation has decreased to zero, the ball is rolling and will no longer change direction due to lane friction.

Where skid and hook occur depends on the amount of oil, ball surface, and ball dynamics. Higher flare layouts, aggressive surfaces and layouts that burn rotation and tilt fast will reduce the length of the skid phase ON A PROPERLY CONSTRUCTED PATTERN because they increase the ball-to-lane friction and thus require more oil to keep the in the skid phase.

If we had 100 units flat to 40' then no oil, virtually all 'strike' balls would skid to 40' then hook. Nothing you could do would affect the first transition meaningfully. But, because we don't set lane conditions up this way, it's not helpful to talk about this example. Any meaningful discussion about ball roll and transitions relies on properly constructed lane patterns to allow changes in the location of the first and second transitions as the ball factors are manipulated to stay relevant. Just as discussion about ball roll should really not be concerned with one off cases that we will never see on a real life bowling lane.
Just MHO.
Cheers, Robbie.
Chemistry is like cooking - just don't lick the spoon.
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