RG of the PAP

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RG of the PAP

Post by deanchamp » May 1st, 2019, 11:06 pm

Simple question, does the RG of the PAP change as the ball flares going down the lane?
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Re: RG of the PAP

Post by MeNoRevs » May 2nd, 2019, 1:39 am

deanchamp wrote:Simple question, does the RG of the PAP change as the ball flares going down the lane?
Not for the length that the ball has to travel. If the lane was infinite, then it would eventually migrate to the low RG

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Re: RG of the PAP

Post by imagonman » May 2nd, 2019, 3:53 am

No. it stays on the same RG contour. See this video by MO roughly 20mins in :



[youtube][/youtube]

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Re: RG of the PAP

Post by PlainUgly » May 4th, 2019, 10:31 pm

example of migration path stays along RG contour....think itook this from blueprint ad one day..
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Re: RG of the PAP

Post by deanchamp » May 5th, 2019, 5:54 am

So we agree the RG value doesn't change because the PAP stays on the same RG contour.

My next question is: does the moment of interia in a ball get smaller as it goes down the lane, which leads to an increase in the angular or rotational velocity?

Answering this question from my understanding, if the moment of inertia (I) is defined as I = (mass)×(radius of gyration)2, and as the mass and RG are both constant, then the moment of inertia can't change.

This is all about understanding why it is stated that a ball revs up when the migrating axis crosses the pin-to-spin line. And yes revs up is the term Mo uses in the attached slide.

And also on the layouts for Radical balls:

"The “X” on the diagrams indicates the Preferred Spin Axis (PSA / Mass Bias) of the drilled ball, and the line that connects the PSA and PIN after drilling is referred to as the “Pin to Spin Line”. The important feature of the “Pin to Spin Line” is that the ball revs up when the migrating axis crosses this line so the sooner the migrating axis crosses the “Pin to Spin Line”, the sooner the ball rev up."

I know this has been discussed before, but the answers have never been satisfactory to me and I've been thinking about it again recently.

viewtopic.php?f=13&t=4227" onclick="window.open(this.href);return false;

Mo talks about linear velocity here but not angular velocity:

"The linear velocity will decrease as it approaches the minor axis (pin to spin line) and start to increase as it passes the minor axis. It will achieve maximum velocity as it approaches the major axis."

And this is from RG Contour Explanation in the Wiki:

"We believe that the ball will increase angular velocity as the spin axis passes the Pin to Spin Line."

How is this increase in angular velocity achieved if the moment of intertia doesn't change?

The formula for the conservation of angular momentum (L) is L = I w, where I is the moment of inertia and w is angular velocity. As L is conserved, a decrease in 'I' leads to an increase in 'w', but as shown earlier, 'I' doesn't decrease so 'w' can't increase as a result of this.

So is it just the ball encountering friction that makes a ball increase its RPM?
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Re: RG of the PAP

Post by DarkHorse » May 5th, 2019, 1:16 pm

In order to conserve angular momentum, the objects must be operating in a closed system with no external forces.
Since we have an external force (friction) acting upon the bowling ball, angular momentum is not conserved, and therefore not a constant.
Sounds official, right?
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Re: RG of the PAP

Post by deanchamp » May 6th, 2019, 2:38 am

It does.

So is friction adding energy to the system or taking it away?
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Re: RG of the PAP

Post by JohnP » May 6th, 2019, 3:51 pm

I don't understand the relationship of the migrating axis crossing the pin to spin line to revs. What I at least think I understand is that as the ball loses axis tilt and lane friction slows it down the revs increase because the track is rolling on almost its full circumference. Calculate the revs generated at a given speed as the track circumference increases and I think you'll see that it's higher than most (or all) players generate off the hand. -- JohnP

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Re: RG of the PAP

Post by snick » May 6th, 2019, 7:27 pm

It is my understanding that the asymmetry at the PSA flattens the ellipse of the RG plane (axis migration path) in the area of the pin-to-spin line, making the path less circular, and therefore shorter from point A to point B.
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Re: RG of the PAP

Post by deanchamp » May 7th, 2019, 4:52 am

snick wrote:It is my understanding that the asymmetry at the PSA flattens the ellipse of the RG plane (axis migration path) in the area of the pin-to-spin line, making the path less circular, and therefore shorter from point A to point B.
And how does this equate to a higher rev rate? And where is point A to point B?

Does this also mean that once the PAP passes this flat spot on the RG contour and the ellipse is more circular again, the rev rate drops as a consequence?
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Re: RG of the PAP

Post by snick » May 7th, 2019, 12:07 pm

A flat spot on an otherwise circular RG plane is essentially a shortcut from one point (A) on a circle to another (B), so the track would flare farther/faster as the PAP migrates through that area if the rate of precession remains constant. Which would be the case if the RG of the PAP remains constant.

The acceleration of rev-rate is likely a product of the increased friction/traction/track flare within the pin-to-spin zone rapidly converting ball speed (energy) to roll.

The rev-rate will decay according to the laws of physics after it maximizes.
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Re: RG of the PAP

Post by MegaMav » May 7th, 2019, 2:32 pm

snick wrote:A flat spot on an otherwise circular RG plane is essentially a shortcut from one point (A) on a circle to another (B), so the track would flare farther/faster as the PAP migrates through that area if the rate of precession remains constant. Which would be the case if the RG of the PAP remains constant.

The acceleration of rev-rate is likely a product of the increased friction/traction/track flare within the pin-to-spin zone rapidly converting ball speed (energy) to roll.

The rev-rate will decay according to the laws of physics after it maximizes.
Other than air resistance and very little lane friction, will it really lose that much if there is no skid left in the roll phase?
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Re: RG of the PAP

Post by snick » May 7th, 2019, 7:10 pm

Other than air resistance and very little lane friction, will it really lose that much if there is no skid left in the roll phase?
Probably not.
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Re: RG of the PAP

Post by deanchamp » May 7th, 2019, 11:21 pm

snick wrote:A flat spot on an otherwise circular RG plane is essentially a shortcut from one point (A) on a circle to another (B), so the track would flare farther/faster as the PAP migrates through that area if the rate of precession remains constant. Which would be the case if the RG of the PAP remains constant.

The acceleration of rev-rate is likely a product of the increased friction/traction/track flare within the pin-to-spin zone rapidly converting ball speed (energy) to roll.
Thanks for your clarification. So when Mo talks about linear velocity, is he referring to the linear velocity of the PAP moving along the RG contour?
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Re: RG of the PAP

Post by DarkHorse » May 8th, 2019, 12:48 am

deanchamp wrote:Thanks for your clarification. So when Mo talks about linear velocity, is he referring to the linear velocity of the PAP moving along the RG contour?
That was my understanding of his comment.
And I'd say friction is taking away energy, specifically linear energy (mph), not rotational energy (rpm).

Also, let's consider that during the Roll phase, as has been stated here before, 1" of ball surface equals 1" of lane surface traveled.

Bowling ball circumference = 27" = 2.25'
1 rpm - 2.25'/minute
1 mph - 5280'/60 minute = 88'/min

Consider a hypothetical "Matched" bowler of 18 mph and 350 rpms off the hand, and the oft-estimated loss of 2 mph at the pins (aka monitor speed)

Generated RPM => 350 rpm = 350 x 2.25 = 787.5'/min = 13.125'/sec
Generated MPH => 18 mph = (18 x 88)/1 minute = 1584'/min = 26.4'/sec
Final MPH (mon) => 16 mph = (16 x 88)/1 minute = 1408'/min = 23.47'/sec
Final RPM (calc) => 23.47'/sec = (23.47 x 60)/2.25 = 625.87 rpm

So, if my logic is correct, we can see a significant difference between the generated RPM and final RPM in the Roll phase.

I'm not sure how this helps understand the Pin-to-Spin Line portion, but here's my completely unsubstantiated guess:

The Low-RG axis (Pin) essentially represents the easiest point on the ball around which to rotate the ball. As the PAP crosses the Pin-Spin line, it is the closest to the Low-RG axis as it will get, and thus lane friction will have its greatest effect upon the ball at that time, providing the bowler with the strongest visual indication of "revving up."

That sounds really good, and I'd wager will convince most people. Maybe not everyone here, but it might get some intellectual juices flowing.
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Re: RG of the PAP

Post by deanchamp » May 8th, 2019, 2:40 am

DarkHorse wrote: Also, let's consider that during the Roll phase, as has been stated here before, 1" of ball surface equals 1" of lane surface traveled.

Bowling ball circumference = 27" = 2.25'
1 rpm - 2.25'/minute
1 mph - 5280'/60 minute = 88'/min

Consider a hypothetical "Matched" bowler of 18 mph and 350 rpms off the hand, and the oft-estimated loss of 2 mph at the pins (aka monitor speed)

Generated RPM => 350 rpm = 350 x 2.25 = 787.5'/min = 13.125'/sec
Generated MPH => 18 mph = (18 x 88)/1 minute = 1584'/min = 26.4'/sec
Final MPH (mon) => 16 mph = (16 x 88)/1 minute = 1408'/min = 23.47'/sec
Final RPM (calc) => 23.47'/sec = (23.47 x 60)/2.25 = 625.87 rpm

So, if my logic is correct, we can see a significant difference between the generated RPM and final RPM in the Roll phase.
Thanks for your input and calculations. When you calculated the RPM and MPH off the hand, is this assuming 0 degrees tilt, that the ball is tracking the full circumference of the ball? Or is it independent of initial tilt?
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Re: RG of the PAP

Post by snick » May 8th, 2019, 2:52 am

"linear velocity" is ball speed.
PAP migration along an RG plane is precession.
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Re: RG of the PAP

Post by snick » May 8th, 2019, 3:05 am

The Low-RG axis (Pin) essentially represents the easiest point on the ball around which to rotate the ball. As the PAP crosses the Pin-Spin line, it is the closest to the Low-RG axis as it will get, and thus lane friction will have its greatest effect upon the ball at that time, providing the bowler with the strongest visual indication of "revving up."

That sounds really good, and I'd wager will convince most people. Maybe not everyone here, but it might get some intellectual juices flowing.
That is an entirely logical conclusion.
But it contradicts the assertion that the RG plane that the PAP migrates along has a constant RG value. It would therefore follow that RG of the PAP at the pins is substantially the same as it is at the foul line, even though the PAP migrates closer to the PIN.
Maybe the true low RG axis shifts away from the PIN, post-drilling, and the LowRG-PAP distance is actually constant.

The ball "revving-up" at the break-point might just be a product of one type of energy (linear velocity) being rapidly converted to another (angular velocity), due to the introduction of friction (dry back end) and faster track flare in the pin-to-spin zone.
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Re: RG of the PAP

Post by deanchamp » May 8th, 2019, 10:56 am

snick wrote:"linear velocity" is ball speed.
PAP migration along an RG plane is precession.
If linear velocity is ball speed, what do you make of Mo's statement:

"The linear velocity will decrease as it approaches the minor axis (pin to spin line) and start to increase as it passes the minor axis. It will achieve maximum velocity as it approaches the major axis."
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Re: RG of the PAP

Post by JohnP » May 8th, 2019, 4:23 pm

Darkhorse's calculations are confirmation of the verbal explanation in my earlier post, except that he didn't take axis tilt (reduced initial diameter of the track) into account in the original RPM calculation. -- JohnP

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