Author Topic: Do you agree? (Edit: some agree) (Read 5435 times)

Re-Evolution · « **on:** October 27, 2004, 02:32:30 AM »

Many say that to find your PAP you should measure over 6 3/4" from your track.
Well by my thinking this is incorrect unless you have a 13 1/2" track since PAP is defined as the equidistant point from all points of your track. With that in mind lets say you have a 10 1/2" track wouldn't you then measure over 7 3/4" since it is 15 1/2" from track to track at a point 180Â° around your track measuring around the positive side. Other ways to find your PAP more accurately would be to use the Ball Spinner technique or find it based on your NAP.

Here is a pic that I made that explains my thoughts.

http://get.filehosting.org/?ks6895

All the tracks that are represented in the pic have the same NAP so they would have the same PAP, but if you used the 6 3/4" method they would have very different PAPs.

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STORMIN1

Edited on 10/28/2004 1:09 AM

Re-Evolution · « **Reply #1 on:** October 27, 2004, 04:21:23 PM »

Doesn't anyone have any comments.
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STORMIN1

charlest · « **Reply #2 on:** October 27, 2004, 06:03:29 PM »

FWIW, as far as the pictures are concerned, and your text, that seems to me to be the proper way to specify a bowler's PAP. Just be careful not to make unwarranted assumptions. Directions to find a PAP are described in the website from most of the ball manufacturers.

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"When we choose an action,
we also choose the consequences of that action.

khamûl · « **Reply #3 on:** October 27, 2004, 06:04:35 PM »

I like pie...

It is a sound theory.
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two of nine

sote

Burak Natal · « **Reply #4 on:** October 27, 2004, 07:01:18 PM »

Stormin1, that theory ignores the axis tilt completely. If we consider just the vertical axis which is axis rotation, you are 100% correct. However, as you know there is another axis created by the release, which is axis tilt. The horizontal axis.. PAP is the result of the both.

Burak

Edited on 10/27/2004 7:28 PM

charlest · « **Reply #5 on:** October 27, 2004, 08:29:41 PM »

quote:
Stormin1, that theory ignores the axis tilt completely. If we consider just the vertical axis which is axis rotation, you are 100% correct. However, as you know there is another axis created by the release, which is axis tilt. The horizontal axis.. PAP is the result of the both.

Burak

Burak,

I am not sure I understand your point. Yes, the tilt, rotational axis and a couple of other factors (like difference between the points where the thumb exits and where the fingers exit) determine the bowler's track, but there is only one initial track for each thrown ball. That track (essentially) determines the bowler's axis points, both negative and positive, regardless of the individual constituents of the thrown ball's properties.

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"When we choose an action,
we also choose the consequences of that action.

Re-Evolution · « **Reply #6 on:** October 28, 2004, 12:07:02 AM »

quote:
Stormin1, that theory ignores the axis tilt completely. If we consider just the vertical axis which is axis rotation, you are 100% correct. However, as you know there is another axis created by the release, which is axis tilt. The horizontal axis.. PAP is the result of the both.

Burak

Edited on 10/27/2004 7:28 PM

Tilt and rotation don't factor in to it, because I am only concerned with the fact that the 6 3/4" from track to find PAP is inaccurate.

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STORMIN1

Re-Evolution · « **Reply #7 on:** October 28, 2004, 12:47:33 AM »

Since there are some that agree with my thought process I will get to my hidden agenda.

With the predescribed thoughts 3 3/8" (leverage) from PAP is only accurate if you have a 13.5" track, since leverage is the half way point between your track and PAP. Therefore if you have a track of 10.5" your leverage point would be 4 3/8" from PAP.

Normal leverage:
27 - 13.5 = 13.5
13.5 / 2 = 6.75
6.75 / 2 = 3.375 (3 3/8)

Modified "leverage"
10.5" track leverage:
27 - 10.5 = 15.5
15.5 / 2 = 7.75
7.75 / 2 = 4.375 (4 3/8)

It makes sence to me.
What about you?
I know this is different from what has been said for years, but isn't it possible that the thoughts have been wrong for years.
I am only using LOGIC to come to these thoughts.
If you THINK about what "leverage" means you will agree.
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STORMIN1

Edited on 10/28/2004 0:50 AM

Burak Natal · « **Reply #8 on:** October 28, 2004, 03:21:17 AM »

Stormin1,

I wish I could tell you all the calculations and stuff. However I'm not the right person to do it.

On the other hand, my point was;
Initial track's axis based on two planes. Horizontal and vertical. Initial track rotates on not only one axis. Therefore, the PAP, which maybe I should call "the absolute center of the very first track/ring" is in some way the combination of two axis. Thus it can not be calculated with simply adding and substracting.

This can be explained with the sphere dynamics which I'm certainly not knowledgeable about to make scientific explanations.

The leverage position you were referring is locating the core in 45 degrees position in relation to your initial track. I'm sure you know it very well that this is called leverage because it is the most unstable position. The leverage position that takes longest time to reach the natural axis of the ball, longest migration time..
360 degrees = 27inches
45 degrees = 3-3/8
That is why it is always 3-3/8..

Hope I could make my point clearly.

That is a great post in anyway. And very clever question..

Burak

Re-Evolution · « **Reply #9 on:** October 28, 2004, 08:09:15 AM »

quote:
Stormin1,

The leverage position you were referring is locating the core in 45 degrees position in relation to your initial track. I'm sure you know it very well that this is called leverage because it is the most unstable position. The leverage position that takes longest time to reach the natural axis of the ball, longest migration time..
360 degrees = 27inches
45 degrees = 3-3/8
That is why it is always 3-3/8..

Hope I could make my point clearly.

That is a great post in anyway. And very clever question..

Burak

Yes 45Â° is the leverage point but that is because it is based on PAP being
6 3/4"/90Â° from your track. With that in mind if your PAP is lets say 100Â°
from your track then the leverage point would be 50Â°/the mid way point.
I think it is easier to agree with if you go to extremes with it.
Lets say someone is a PURE spinner with a track dot their PAP would then be
180Â°/13.5" from track the most unstable point would then be 90Â°/6.75" from PAP.
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STORMIN1

Ishmael · « **Reply #10 on:** October 28, 2004, 08:21:36 AM »

quote:
Yes 45Â° is the leverage point but that is because it is based on PAP being 6 3/4"/90Â° from your track.

Not so. The leverage point is 3 3/8 because of the geometry of the bowling ball. It has no relation to the track. 3 3/8 places the core at a 45 degree angle to the PAP. The actual location of the track has no bearing.

Burak Natal · « **Reply #11 on:** October 28, 2004, 09:02:01 AM »

Nope
It is NOT based on PAP being 6 3/4"/90Â° from your track.
Actually everything comes from degrees. 0-45-90 these are constant points. Distance calculation is made based on these.
That's why I wrote:
360degrees = 27 inches
45degrees = 3-3/8

6-3/4 is based on 3-3/8 being 45degrees from track, the leverage position. If you go 3-3/8 to reach 45 degrees, you go two times of it to reach 90 degrees

Think about a football. If you throw it with lying horizontally, 0 degrees angle from your palm directly (which will be 90 degrees angle to the direction that can be named as your track) it will go smooth and stable, spinning faster around its axis. If you throw it 90 degrees angle in relation to your palm (it rotates parallel to its axis, which can be measured 0) it will still go smooth but with less rotations. In both cases, since its axis is stable, it will not try to migrate anywhere. Now think about throwing it in 45 degrees..

45 degrees is the breakeven point. The more you get closer to each side more stable position you get. This is a physical fact.
PAP calculation is based on this fact. As I've written previously, it is sphere dynamics and I do not know enough physics to prove it to you with formulas..

Please check the diagrams in http://www.bowling-tr.com/yazilar/bn-yazilar06.html
Unfortunately it's not in English but pictures may help..

Why don't you find a college professor or mail some of the manufacturers?
I'd really like to hear their answers.

My best regards,

Burak

Re-Evolution · « **Reply #12 on:** October 28, 2004, 10:12:47 AM »

Burak
I agree with what you are saying if you are dealing with a sphere that is in balance and has no preferred spin axis (PSA) but as we all know bowling balls are not in balance and do have a PSA.
So as you put it earlier "The leverage position that takes longest time to reach the natural axis of the ball, longest migration time" the longest migration time would always be the half way point between the initial axis (release axis) and the PSA (ball dynamics).
Not sure if you are familiar with the new MoRich ball Weapon of Mass Bias that is coming out soon it has a Mass Bias position that is 7 1/4" from the pin so it would have a leverage point of 3 5/8" considering that a balls mass bias is the axis when the ball is rotating around it's PSA. I think that adds some validity to my thoughts.
This just makes sense to me but I could be wrong it wouldn't be the first time. If I only had the resources/$ to research this theory. I think I will see if I can talk with a Physics professor at the University of Illinois they should be able to prove/disprove my thoughts. If I am able to get an answer I will let you know what they say.
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STORMIN1

cgilyeat · « **Reply #13 on:** October 28, 2004, 10:58:44 AM »

Burak, Since I see you name at the top of the page, would it be possible for you to translate it to English for us?

Thanks

Re-Evolution · « **Reply #14 on:** October 28, 2004, 12:53:32 PM »

quote:
Stormin1, the PAP is NOT "defined as the equidistant point from all points of your track".

That is funny because I was able to find multiple places that say it is.

quote:
"POSITIVE AXIS POINT (P.A.P.)
This is the axis of the ball during the first few revolutions that is created totally by the bowler's release. The point on the ball that is equidistant from all points of the release ball track."

Therefore your PAP is only over 6.75" from your track if you have a 13.5" track, because that is the equidistant point between your track at points 180Â° from each other.
See for yourself:
www.youthbowling.com/glosary2.htm
http://www.gayaba.org/terms.htm
http://host155.ipowerweb.com/~sdb300co/proshop_glossary.htm
www.bowling2u.com/trivia/glossary/glossary.asp?OpMode=List&Key=P
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STORMIN1

Author Topic: Do you agree? (Edit: some agree) (Read 5435 times)

Re-Evolution

Do you agree? (Edit: some agree)

Re-Evolution

Re: Do you agree? (Edit: some agree)

charlest

Re: Do you agree? (Edit: some agree)

khamûl

Re: Do you agree? (Edit: some agree)

Burak Natal

Re: Do you agree? (Edit: some agree)

charlest

Re: Do you agree? (Edit: some agree)

Re-Evolution

Re: Do you agree? (Edit: some agree)

Re-Evolution

Re: Do you agree? (Edit: some agree)

Burak Natal

Re: Do you agree? (Edit: some agree)

Re-Evolution

Re: Do you agree? (Edit: some agree)

Ishmael

Re: Do you agree? (Edit: some agree)

Burak Natal

Re: Do you agree? (Edit: some agree)

Re-Evolution

Re: Do you agree? (Edit: some agree)

cgilyeat

Re: Do you agree? (Edit: some agree)

Re-Evolution

Re: Do you agree? (Edit: some agree)