"weak" ball and "strong" ball

Which layout is right for me?

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TonyPR
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Re: "weak" ball and "strong" ball

Post by TonyPR »

I believe the answer has to do with how the drilling angle affects the RG of the PAP and how this will make the core rev up sooner with a low drilling angle thus creating more friction and a shorter skid phase as a result. The opposite would happen with a larger drilling angle. I am no expert on this so could someone please add to my answer, Eric, Blueprint, Mo???
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Re: "weak" ball and "strong" ball

Post by EricHartwell »

MWhite wrote:My question better worded remains... how does the layout effect the length of the skid phase.
The coverstock is the coverstock it has a coefficient of friction. It is going to grab the lane and hook when the friction says to hook.

Now the core either works with or against the friction of the coverstock. Gyroscopic effects.

The Rg of the PAP is either round or elliptical in its path of migration. If it is round it will not affect the rate in which the core wants to "rev up" as much as compared to a path that is elliptical. This is the difference between a ball being more Symmetrical or more Asymmetrical after drilling. The more Asymmetrical the more elliptical the path of migration is. I think of it as the slingshot affect of the planets in their elliptical orbits. It is the manipulation of gravity and spinning objects.

The drilling angle controls when the migrating axis gets to the pin to spin line, the point at which the core starts to "rev up".
So a smaller drilling angle is to get the core to aid the coverstock early, while larger drilling angles delay the aid of the core until the ball is completely out of the pattern.
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Re: "weak" ball and "strong" ball

Post by MWhite »

TonyPR wrote:I believe the answer has to do with how the drilling angle affects the RG of the PAP and how this will make the core rev up sooner with a low drilling angle thus creating more friction and a shorter skid phase as a result. The opposite would happen with a larger drilling angle. I am no expert on this so could someone please add to my answer, Eric, Blueprint, Mo???
The amount of friction between the ball and the lane has to do with the mass of the ball, surface of the ball, the surface of the lane, and the amount of oil between the ball and the lane.

Since the drilling angle doesn't influence any of those items, it doesn't create more or less friction.

The drilling angle can effect the RG of the PAP, so it can influence the effect (increase rev rate) that friction has on the ball. But that happens during the hook phase, not the skid phase.
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Re: "weak" ball and "strong" ball

Post by MWhite »

EricHartwell wrote: The coverstock is the coverstock it has a coefficient of friction. It is going to grab the lane and hook when the friction says to hook.

Am I correct by interpreting the above as saying only the coverstock influences the length of the skid phase?

Now the core either works with or against the friction of the coverstock. Gyroscopic effects.

Please explain what you mean by "the core can work with the friction vs work against friction."

The Rg of the PAP is either round or elliptical in its path of migration.

This is going to need some clarification. The RG is a number, which is neither round nor elliptical.

The migration path can be round or elliptical, but one thing to note is at every point along the migration path has the same RG value as the initial PAP. This is an important piece of information.


If it is round it will not affect the rate in which the core wants to "rev up" as much as compared to a path that is elliptical. This is the difference between a ball being more Symmetrical or more Asymmetrical after drilling. The more Asymmetrical the more elliptical the path of migration is. I think of it as the slingshot affect of the planets in their elliptical orbits. It is the manipulation of gravity and spinning objects.

I don't see how the shape of the migration path, round or elliptical can cause a difference in how fast (both how soon, and at what rate) the ball will increase the rev rate.


The drilling angle controls when the migrating axis gets to the pin to spin line, the point at which the core starts to "rev up".
So a smaller drilling angle is to get the core to aid the coverstock early, while larger drilling angles delay the aid of the core until the ball is completely out of the pattern.

What is so different about the "pin to spin line" that on one side the ball doesn't "rev up" while on the other side, it does "rev up"
Friction is caused by factors external to the core of the ball.

Friction causes a force opposite to the direction the ball is moving.

In physics, we learn that Torque equals moment of inertia multiplied by angular acceleration.

Stated another way, angular acceleration equals Torque divided by moment of inertia.


Since all of the points along the migration path have the same RG, all of the points along the migration path would have the same moment of inertia.

Since the moment of inertia remains constant as the ball travels down the lane, the angular acceleration is proportional to Torque.

Please reconcile the idea of "pin to spin line" in a way that is consistent with accepted "laws of physics"
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Re: "weak" ball and "strong" ball

Post by EricHartwell »

https://en.wikipedia.org/wiki/Moment_of ... nertia.gif" onclick="window.open(this.href);return false;

This was interesting. I would have liked to see an Asymmetrical cored ball in this presentation. I think it could be rolled in different orientations ie. Different PAPs and would have different results. A ball with a faster spin time would probably win the race down the ramp.

You state some very rudimentary mathematics for moment of inertia. When I look it up I find some pretty high level Calculus that I am not going to even try to interpret.
MWhite wrote:Am I correct by interpreting the above as saying only the coverstock influences the length of the skid phase?

Incorrect interpretation. Coverstock is not the only influence.

The Gyroscopic effects of the core also affects the length of the skid.

The coverstock and surface grit is the main influence, now you use the core with layout to affect the skid phase. Smaller drilling angles shorten the skid and larger Drilling angles increase the length.

Another influence on the length of the skid zone is the release specs. Low rotation and low tilt will have a shorter skid zone than a ball released with high tilt and high rotation. Or more obviously a speed dominant delivery will skid longer than a rev dominant release.
MWhite wrote:Please explain what you mean by "the core can work with the friction vs work against friction."
Controlling the amount of flare affects friction. More flare is more friction, less flare less friction.

Gyroscopic effects seem to defy laws of physics/gravity. If you have ever played with gyroscopes/tops might help to understand this better. There was a product out called Shake Weights. You get the gyroscope spinning inside the Shake Weight and resistance to motion was increased. The faster you got them spinning the effort it took to move them, change direction is much more than the 2.5 lb static mass. Get it moving and it feels way heavier than that.

Even with a perfectly symmetrical ball ex. baseball/softball. Catching a ball that has a lot of spin feels heavier than one thrown the same speed but with less spin. Gyroscopic inertia of a spinning object.
MWhite wrote:This is going to need some clarification. The RG is a number, which is neither round nor elliptical.
The Rg is a number, a measurement to the center of mass. The path of the axis is what is round to elliptical around the given axis.
Wikipedia....
"Mathematically the radius of gyration is the root mean square distance of the object's parts from either its center of mass or a given axis, depending on the relevant application."

The distance from the center of mass is what we are looking at here because in an elliptical path the distance to the given axis obviously changes.
MWhite wrote:I don't see how the shape of the migration path, round or elliptical can cause a difference in how fast (both how soon, and at what rate) the ball will increase the rev rate.
In a round path the axis migrates at a constant speed and the flare rings are a consistent distance apart.
In an elliptical path the axis migration accelerates as it passes closer to that axis. I wish I had Blueprint to make showing examples easier. The flare rings will be closer together when the migrating axis is furthest from the major/minor axis of the ball. It is progressing at a slower rate, less flare.

The more elliptical the path the more Asymmetrical the ball is.

http://wiki.bowlingchat.net/wiki/index. ... xplanation" onclick="window.open(this.href);return false;

"The spin axis migrates along an elliptical path and the point which intersects the Pin to Spin Line is the endpoint of the minor axis of the ellipse. If we consider all of the RG contours then the Pin to Spin Line represents all of the endpoints of the minor axes and I view this like a “ridge” on the RG profile. We believe that the ball will increase angular velocity as the spin axis passes the Pin to Spin Line. Therefore, smaller drilling angles are used to shorten the length of the 1st transition (skid to hook) because the PAP is closer to the Pin to Spin Line. Larger drilling angles increase the length of the 1st transition because the PAP is farther away from the Pin to Spin Line."
MWhite wrote:Since the moment of inertia remains constant as the ball travels down the lane, the angular acceleration is proportional to Torque.
Please expound on this some more.
I do not believe the moment of inertia remains constant unless the migrating path is round and the same distance from the low Rg axis. In an elliptical path the distance from the axis varies and this affects how fast the migrating axis moves.
MWhite wrote:Please reconcile the idea of "pin to spin line" in a way that is consistent with accepted "laws of physics"
Please explain why you think the idea of the "pin to spin line" is not consistent with accepted "laws of physics"
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Re: "weak" ball and "strong" ball

Post by Qman »

EricHartwell wrote:https://en.wikipedia.org/wiki/Moment_of ... nertia.gif

This was interesting. I would have liked to see an Asymmetrical cored ball in this presentation. I think it could be rolled in different orientations ie. Different PAPs and would have different results. A ball with a faster spin time would probably win the race down the ramp.

You state some very rudimentary mathematics for moment of inertia. When I look it up I find some pretty high level Calculus that I am not going to even try to interpret.


Incorrect interpretation. Coverstock is not the only influence.

The Gyroscopic effects of the core also affects the length of the skid.

The coverstock and surface grit is the main influence, now you use the core with layout to affect the skid phase. Smaller drilling angles shorten the skid and larger Drilling angles increase the length.

Another influence on the length of the skid zone is the release specs. Low rotation and low tilt will have a shorter skid zone than a ball released with high tilt and high rotation. Or more obviously a speed dominant delivery will skid longer than a rev dominant release.


Controlling the amount of flare affects friction. More flare is more friction, less flare less friction.

Gyroscopic effects seem to defy laws of physics/gravity. If you have ever played with gyroscopes/tops might help to understand this better. There was a product out called Shake Weights. You get the gyroscope spinning inside the Shake Weight and resistance to motion was increased. The faster you got them spinning the effort it took to move them, change direction is much more than the 2.5 lb static mass. Get it moving and it feels way heavier than that.

Even with a perfectly symmetrical ball ex. baseball/softball. Catching a ball that has a lot of spin feels heavier than one thrown the same speed but with less spin. Gyroscopic inertia of a spinning object.


The Rg is a number, a measurement to the center of mass. The path of the axis is what is round to elliptical around the given axis.
Wikipedia....
"Mathematically the radius of gyration is the root mean square distance of the object's parts from either its center of mass or a given axis, depending on the relevant application."

The distance from the center of mass is what we are looking at here because in an elliptical path the distance to the given axis obviously changes.


In a round path the axis migrates at a constant speed and the flare rings are a consistent distance apart.
In an elliptical path the axis migration accelerates as it passes closer to that axis. I wish I had Blueprint to make showing examples easier. The flare rings will be closer together when the migrating axis is furthest from the major/minor axis of the ball. It is progressing at a slower rate, less flare.

The more elliptical the path the more Asymmetrical the ball is.

http://wiki.bowlingchat.net/wiki/index. ... xplanation" onclick="window.open(this.href);return false;

"The spin axis migrates along an elliptical path and the point which intersects the Pin to Spin Line is the endpoint of the minor axis of the ellipse. If we consider all of the RG contours then the Pin to Spin Line represents all of the endpoints of the minor axes and I view this like a “ridge” on the RG profile. We believe that the ball will increase angular velocity as the spin axis passes the Pin to Spin Line. Therefore, smaller drilling angles are used to shorten the length of the 1st transition (skid to hook) because the PAP is closer to the Pin to Spin Line. Larger drilling angles increase the length of the 1st transition because the PAP is farther away from the Pin to Spin Line."


Please expound on this some more.
I do not believe the moment of inertia remains constant unless the migrating path is round and the same distance from the low Rg axis. In an elliptical path the distance from the axis varies and this affects how fast the migrating axis moves.


Please explain why you think the idea of the "pin to spin line" is not consistent with accepted "laws of physics"
He can't because most of us know the relation of pin placement is equal to ball reaction.
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Re: "weak" ball and "strong" ball

Post by MWhite »

EricHartwell wrote: Please expound on this some more.
I do not believe the moment of inertia remains constant unless the migrating path is round and the same distance from the low Rg axis. In an elliptical path the distance from the axis varies and this affects how fast the migrating axis moves.
Lets work on this point first.

http://usbcongress.http.internapcdn.net ... namics.pdf
USBC wrote:The graphs for Test I show that despite having different distances or migratory paths, a trend is noticed. The most important overall factor that was consistent through out this test was noticed:
“While on the lane, RG values of the migratory path remained approximately constant at each migratory axis point for all core geometries and drillings.”
Based on the variety of ball tested, an accurate analysis concluded that the axis migration path was INDEPENDENT (not dependant) of the following:
1) Core shape
2) Core angle/orientation
3) Cover stock
4) Asymmetry
5) Symmetry
6) Mass bias strength/intermediate differential

7) Spin Time
8) Oil Pattern
9) Mass Distribution of the Core.
10) Ratio of the Intermediate to Total Differential
Regardless of the above factors, the same trend in axis migration occurred. While the approved ball is on the lane, the bowling ball flared and created an axis migration to yield approximately the same RG value that the ball was initially rotating on from the bowlers PAP.
http://usbcongress.http.internapcdn.net ... nstudy.pdf
USBC wrote:Based on the results of the recent axis migration study, USBC research determined that a ball migrates downlane on approximately the same RG profile. Therefore, the RG value on the positive axis point gave a clear representation of the RG value the bowling ball is rotating around during its entire trip toward the pins.
https://en.wikipedia.org/wiki/Radius_of_gyration

Image

If the RG, and the mass of the ball is constant as the ball travels down the lane, then it logically follows that the moment of inertia is constant as the ball travels down the lane.

Is that good enough for you to accept that the moment of inertia remains constant?
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Re: "weak" ball and "strong" ball

Post by rrb6699 »

funny I just started a thread symmetrical vs asymmetrical. basically covering this very thing. ball reaction.
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Re: "weak" ball and "strong" ball

Post by EricHartwell »

MWhite wrote:https://en.wikipedia.org/wiki/Radius_of_gyration



If the RG, and the mass of the ball is constant as the ball travels down the lane, then it logically follows that the moment of inertia is constant as the ball travels down the lane.

Is that good enough for you to accept that the moment of inertia remains constant?
Rg is not equal to the Moment of inertia. It is way more complicated than that.

https://en.wikipedia.org/wiki/Moment_of_inertia" onclick="window.open(this.href);return false;

"The moment of inertia, otherwise known as the angular mass or rotational inertia, of a rigid body is a tensor that determines the torque needed for a desired angular acceleration about a rotational axis. It depends on the body's mass distribution and the axis chosen, with larger moments requiring more torque to change the body's rotation. It is an extensive (additive) property: the moment of inertia of a composite system is the sum of the moments of inertia of its component subsystems (all taken about the same axis). One of its definitions is the second moment of mass with respect to distance from an axis r,......

For bodies constrained to rotate in a plane, it is sufficient to consider their moment of inertia about an axis perpendicular to the plane. For bodies free to rotate in three dimensions, their moments can be described by a symmetric 3 × 3 matrix; each body has a set of mutually perpendicular principal axes for which this matrix is diagonal and torques around the axes act independently of each other."
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Re: "weak" ball and "strong" ball

Post by elgavachon »

TonyPR wrote:I believe the answer has to do with how the drilling angle affects the RG of the PAP and how this will make the core rev up sooner with a low drilling angle thus creating more friction and a shorter skid phase as a result. The opposite would happen with a larger drilling angle. I am no expert on this so could someone please add to my answer, Eric, Blueprint, Mo???
Actually, It has very little to do with changing the RG of the PAP.
I have deanchamp to thank for talking me into ordering the digital Bowling This Month issues. I don't know how many of you know this, but Bill Sempsrott (Blueprint) has taken over the publication & is one of the major contributors. Here are some interesting quotes from the November article Pin-to-PAP Distance and Its Effect on Bowling Ball Motion


"The second thing we often hear about pin-to-PAP distance relates to its effect on the RG
of the PAP axis. Shorter pin-to-PAP distances place the low RG axis of the ball close to the PAP and cause the RG of the PAP to be lower: conversely, longer pin-to-PAP distances place the high or intermediate RG axis closer to the PAP and cause the RG of the PAP to the higher. So, the conventional thinking here goes that shorter pin distances will make the ball rev up earlier (due to the lower RG of the PAP axis) and that
longer pin distances will make the ball rev up later (due to the higher RG of the PAP axis).

So, is any of this RG-of-the-PAP stuff true? Well, it’s certainly true that the RG of the PAP changes as pin distance changes. It’s also true that shorter pin distance layouts can sometimes produce a ball motion that looks earlier and smoother and that longer pin distance layouts can sometimes produce later and more
angular ball motion. But, it turns out that this has very very little to do with the RG of the PAP. So,
this theory, too, while still being somewhat useful, is a little bit inaccurate. We’ll explain this in more detail a bit later with some examples."

Mo said the same thing on Facebook in his video presentation (at about 30 minutes). He had 3 charts about the numbers of the Fix. He said the last chart he did not use much (which was the RG of the PAP). He said they just included it for those who got that WARM AND FUZZY FEELING knowing their RG.
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Re: "weak" ball and "strong" ball

Post by MWhite »

EricHartwell wrote:
Rg is not equal to the Moment of inertia. It is way more complicated than that.
diagonal and torques around the axes act independently of each other."

I didn't say the RG was equal to the Moment of inertia.

I said since the Radius of Gyration has a constant value as the ball travels down the lane, and the mass has a constant value as the ball travels down the lane, then the moment of inertia has a constant value as it travels down the lane.

The relationship between Radius of Gyration, mass, and moment of inertia is as follows:

RG = Radius of Gyration around a specific axis
m = mass of the ball
I = moment of inertia of around a specific axis.

RG = square root of (I / m)

With a little algebra we can say I = (RG^2) / m

Let take two axis points on the axis migration path, as the ball travels down the lane.

RG_1 is the RG at the 1st axis
m_1 is the mass of the ball
I_1 is the moment of inertia at the 1st axis

RG_2 is the RG at the 2nd axis
m_2 is the mass of the ball
I_2 is the moment of inertia at the 2nd axis

According to USBC axis migration study, RG_1 is equal to RG_2
m_1 is equal to m_2 since it's the same ball, and no part has failen off the ball.

I_1 = (RG_1^2)/m_1
I_2 = (RG_2^2)/m_2

since RG_2 = RG_1 we can substitute RG_1 in place of RG_2

I_2 = (RG_1^2)/m_2

since m_2 = m_1 we can substitute m_1 in place of m_2

I_1 = (RG_1^2)/m_1 = I_2

I_1 = I_2

While we don't know what exactly the RG values, the mass values, or the moment of inertia values are, what we do know is they remain constant the whole was down the lane.

The calculation to determine the moment of inertia value is not complex, but it is tedious, and the information to calculate the value directly isn't provided by the ball manufacturers.

What the USBC does to discover the RG values is a form of reverse engineering.

[youtube][/youtube]

They apply a known torque on the ball and measure the acceleration around a specific axis.

From that they can deduce what the moment of inertia must be to achieve the acceleration measured.

Then using the moment of inertia, and mass of the ball, they can calculate the RG of that axis.

I tried to find a video that explains the way a CAD program would calculate the moment of inertia, but everyone I found that covered the topic was hard to understand the words spoken due to accents.

Understanding the process doesn't require calculus, but if you know calculus the symbols are much easier to understand.

The general idea is you take the ball, and chop it up into MANY equal size pieces, then one by one you find the mass of each little piece, and multiply that by the distance squared the piece originally was from the specific axis in question.

The smaller you chop up the pieces, the more accurate the answer, but the more work required, hence the computer.

The part the manufacture doesn't release is the differing densities of each of those tiny bits.
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Re: "weak" ball and "strong" ball

Post by EricHartwell »

MWhite wrote:The general idea is you take the ball, and chop it up into MANY equal size pieces, then one by one you find the mass of each little piece, and multiply that by the distance squared the piece originally was from the specific axis in question.

The smaller you chop up the pieces, the more accurate the answer, but the more work required, hence the computer.

The part the manufacture doesn't release is the differing densities of each of those tiny bits.
As you chop up the pieces, the core shape of those pieces in an Asymmetrical drilling are different.
Spinning those different shapes will produce different forces.
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Re: "weak" ball and "strong" ball

Post by MWhite »

EricHartwell wrote: As you chop up the pieces, the core shape of those pieces in an Asymmetrical drilling are different.
Spinning those different shapes will produce different forces.
That's not how moment of inertia works.

Where the pieces are located isn't as important as how far they are from the axis.

When your ball has for example an RG of 2.50, what that means is the force required to accelerate the rev rate a given amount is the same as if you positions ALL of those pieces as a hollow sphere with a radius of 2.50", you would have to mash those bits together so the thickness of the sphere was very thin.

Once the calculations are completed, the shape of the core is no longer a factor when it comes to the force required to increase in rev rate.
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Re: "weak" ball and "strong" ball

Post by EricHartwell »

Moment of inertia is in a declining state as the ball slows down.
What happens to the moment of inertia as revolutions increase as the ball approaches roll?
I don't see where any of your calculations take into consideration the rate of spin or speed of the ball. Both of which vary as a ball travels down the lane. Torque is being applied to the ball as it encounters friction.

You are trying to use the moment of inertia to argue that the core does not help the ball to rev up to roll.
You quote the usbc study like it is a bible. They needed to quantify things. They also used only one layout for Asymmetrical balls. We are getting caught up on the moment of inertia. There is more to it than that.

There are several points, the primary axis, the spin axis and the center of mass.
The spin axis is changing positions and not lined up with the center of mass or a primary axis. The distance to the primary axis is variable along the line of the Rg contour. These forces are yet to be quantified. I do not know how to do that but I believe they affect the revolutions.

The disconnect here is the Rg about the primary axis vs the spin axis.
The distance between those to axis varies when the migrating spin axis is elliptical.
The relationship of these axis to each other is what is not really understood.
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Re: "weak" ball and "strong" ball

Post by MWhite »

EricHartwell wrote:Moment of inertia is in a declining state as the ball slows down.
What happens to the moment of inertia as revolutions increase as the ball approaches roll?
I don't see where any of your calculations take into consideration the rate of spin or speed of the ball. Both of which vary as a ball travels down the lane. Torque is being applied to the ball as it encounters friction.


I think you're confusing moment of inertia with momentum.
Moment of inertia is a property of the ball, much like mass is a property of the ball.
The mass of the ball doesn't change because it is moving or not.
Likewise the moment of inertia doesn't change because the ball is rotating or not.
A force is required to accelerate the velocity of ball.
A torque is required to accelerate the rev rate of the ball.

If the ball were to somehow change to an axis with a different RG contour, then the moment of inertia would change, but that doesn't occur while the ball is traveling down the lane.



You are trying to use the moment of inertia to argue that the core does not help the ball to rev up to roll.

Right. I think the torque that makes the ball rev up is external to the core, it's the friction between the surface of the ball, and the lane.

The larger the moment of inertia, the slower the ball revs up, and the lower the moment of inertia, the faster the ball revs up.

As the ball travels down the lane, the PAP migrates, but does so along an "RG contour" meaning the RG is a constant value for all the points on the path of migration. And since RG is a function of mass and moment of inertia, the moment of inertia has to be a constant value for all points on the path of migration.


You quote the usbc study like it is a bible. They needed to quantify things. They also used only one layout for Asymmetrical balls. We are getting caught up on the moment of inertia. There is more to it than that.

The only thing I'm pulling out of the USBC study is the fact that the RG value is constant for all points on the migration path of the PAP

There are several points, the primary axis, the spin axis and the center of mass.
The spin axis is changing positions and not lined up with the center of mass or a primary axis. The distance to the primary axis is variable along the line of the Rg contour. These forces are yet to be quantified. I do not know how to do that but I believe they affect the revolutions.

The disconnect here is the Rg about the primary axis vs the spin axis.
The distance between those to axis varies when the migrating spin axis is elliptical.
The relationship of these axis to each other is what is not really understood.
I ran a simulation in blueprint, where I only modified the drill angle from 15 to 90 in steps of 15.

In all cases the ball revved up a little at 30 feet, and a lot more at 40 feet.

Both of those distance has a corresponding increase in friction.

30 feet is when it reached the 10 board area, and 40 feet was the end of the pattern.

That tells me that in Blueprint's opinion, the ball revs up due to friction, not due to drill angle.

I'm not saying Blueprint is right, just that Blueprint doesn't seem to be aware of whatever you claim is happening at the pin to spin line.
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Re: "weak" ball and "strong" ball

Post by ads »

I failed in both maths and physics so I skipped them and go into something a layman may imagine but not necessarily mature or comprehensive.

1. Think of driving on ice. No matter how hard one step on the acceleration paddle, the car will skid only if the tyres can't grasp the ice surface.

2. Change tyres (coverstock) which can get friction with the icy surface, the car can change direction then.

3. When tyres can grasp road surface, will accelerating the wheels (spin time/small drilling angle?), give a comparatively faster direction change?
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Re: "weak" ball and "strong" ball

Post by EricHartwell »

The Gyroscopic effect either resists or will aids the external forces, friction.
MWhite wrote:That tells me that in Blueprint's opinion, the ball revs up due to friction, not due to drill angle.
The Drilling angle is affecting how the ball reacts to friction. Will it succumb to Friction quickly or will it resist and push longer down the lane. Friction and its variable property is part of the equation as well. It is an external torque. There is no simple way of describing how much is going on.
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Re: "weak" ball and "strong" ball

Post by rrb6699 »

in an attempt to answer the first post, there are two way of looking at how a ball is strong or weak. the manufacturers overall rating and the bowlers use of strong or weak balls while bowling.

strictly from a bowlers point of view, a "strong" ball is the best ball you choose on a certain day that hits the hardest at the pins. there are conditions where a weaker cover will hit harder than a stronger cover ball.
on drier conditions the right cover/layout works better. (does this) by delaying the "release" of skid to roll at the right point for optimum hit on the pins with power. a stronger rated ball would enter the roll phase too early and lose energy, thus hit flat and weak. it may be rated a stronger ball but, when it rolls out too early it actually hits weaker.

on tighter conditions the right cover/layout does not try to delay the skid to roll but helps shorten the skid to roll as soon as possible. so, the ball can get to roll phase as quick as possible - meaning designed to react in a shorter distance in feet as it would on a dry lane, except instead of entering roll phase a few feet from the foul line or at the arrows it will much closer to the pins.

I look at it this way: the ball skids until the roll phase. what is defined as the hook phase is really still skid, only skid is decreasing with friction and roll starts to take over more than skid. take out bowler factors such as amount of speed, turn & revs imparted on the ball. the hook phase is a combination of skid and roll with skid becoming less dominant as friction is encountered or speed of momentum slows.

as a bowler I only care about my skid to roll when I'm bowling. I choose the ball that works for the condition and know what to look for or what I want to see. again, I'm speaking from a bowlers point of view on a given day as to what ball is hitting the stongest or hardest.

as for ball rating strong vs weak we all know coverstock dominates strength of the ball. with core contributing to dynamics also affecting ball strength vs weaker balls. the layout can maximize aggressiveness (roll) or weakness (prolong skid) but only as much as the coverstock and core will allow.

to me strong vs weak ball ratings for the most part are determined from the point of a long oil pattern with not much friction distance to the pins most likely 15' of friction average. nothing hooks on oil even heavily sanded balls to a point.

ball ratings help the bowler determine the right ball for the condition irregardless of the math or physics used to design balls. ball ratings are the final result of the ball design process. in this way balls are marketed by weak or strong overall performance.

from there the bowler has to choose from the point of his or her game what works best. a good instructor can help but not many instructors are independent of being ball reps so it depends on the brand they are pushing.

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Re: "weak" ball and "strong" ball

Post by MWhite »

ads wrote:I failed in both maths and physics so I skipped them and go into something a layman may imagine but not necessarily mature or comprehensive.

1. Think of driving on ice. No matter how hard one step on the acceleration paddle, the car will skid only if the tyres can't grasp the ice surface.

2. Change tyres (coverstock) which can get friction with the icy surface, the car can change direction then.

3. When tyres can grasp road surface, will accelerating the wheels (spin time/small drilling angle?), give a comparatively faster direction change?

Let me create a scenario where your intuition about a car's performance on ice might match a bowling balls performance on the lane.

I'll use your specs in the example.

One early winter morning you're driving to work.

You're driving at about 14 mph (14 mph off hand) because you're not a complete idiot, and realize there could be patches of ice on the road, and you would rather not make your debut on Youtube's idiot drivers of the month.

As you approach a patch of ice, you try to move to the right half of the road (stand 20, hit 10) but just as you reach the ice things get weird.

You crank the wheels 60 degrees to the left (axis rotation).
You tap on the brakes, enough to slow the rotation of the wheels from 14 mph down to 7 2/3 mph. (equivalent to 300 rpm)

In the process of cranking the wheel, the steering wheel breaks off, the brake line fails, the car pops out of gear, and the engine seizes.

Now I don't think it is actually possible to turn the wheels 60 degrees, but this is my story, so go with it.

Also remember where I said you weren't an idiot, turning the wheel that far makes me unsure about that opinion.

The reason all these things broke, is because on bowling, once you let go of the ball, you have no more input into what the ball will do.

You notice that the ice doesn't cover the road all the way to the curb so you're somewhat comforted knowing you're probably not going to go off the road.

At that moment you see standing on the side of the road, some buttmunch with his cell phone recording.
Yep that Youtube thing is happening.

However you notice a 2nd buttmunch on the other side of the street also recording.
You think since buttmunch #2 is nowhere near buttmunch #1, he must figure buttmunch #1 is standing in a bad place.

That means that while you may be destined for Youtube, you're likely to take down buttmunch #1 in the process, a comforting thought.

You also notice that the ice stops ahead about 2/3 of the way to an intersection filled with stopped cars except for a gap (pocket if you will) near the center of the road, just large enough for your car to pass thru without damage, if you get lucky.

Let me also state, I'm from Southern California, I have never driven on ice. So welcome to my perceived nightmare.


"3. When tyres can grasp road surface, will accelerating the wheels (spin time/small drilling angle?), give a comparatively faster direction change?"

Remember, you're engine seized so pressing on the accelerator will have no effect on the wheels.

Spin time is a measurement of how quickly track flare will occur.
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Re: "weak" ball and "strong" ball

Post by todvan »

MWhite wrote:
When I looked in the wiki I saw the page from Mo saying "from skid to hook to roll" and I interpreted that as the length of the hook phase, starting at "skid to hook", and ending at "hook to roll", and that doesn't include the length of the skid phase.

So I come back here, and what you said "from skid to hook to roll" was Mo's comment verbatim.

Somehow I interpreted the words different the first time, my bad.

However, Tony goes so far as to claim the layout does effect the length of the skid phase. "The length of the skid phase is controlled by coverstock (chemical composition and surface grit) and the drilling angle."

You appear to agree, and add in "To add to what Tony said, the angle to the VAL affects the transition from hook to roll. The smaller the angle the faster the transition (the shorter the hook zone)."

But you don't comment about the length of the skid phase.

My question better worded remains... how does the layout effect the length of the skid phase.

Tony, your answer was along the lines of what to adjust to get different lengths, as opposed to why those adjustments cause the different lengths.

I'm interested in the why.
So you are asking why a smaller drill angle causes the ball to hook sooner?
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