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Author Topic: AXIS TILT  (Read 3489 times)

BXTECH

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AXIS TILT
« on: September 03, 2007, 11:34:21 AM »
I am trying to calculate the amount of rpm & the amount of axis tilt i produce. If anyone can assist i do appreciate it. Also how do you locate one's PAP?

 

J_Mac

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Re: AXIS TILT
« Reply #1 on: September 03, 2007, 07:50:33 PM »
Locating your PAP is the first step in calculating axis tilt and track diameter...

PAP video - http://www.brunsnick.com/positive_axis_point.html

Track Diameter/axis tilt chart - https://www.buddiesproshop.com/36/Asymetrical_Layout_Guide.htm
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kleptic

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Re: AXIS TILT
« Reply #2 on: September 12, 2007, 09:22:51 AM »
could somebody explain how to measure track diameter better than that buddies link.

dR3w

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Re: AXIS TILT
« Reply #3 on: September 12, 2007, 12:20:42 PM »
How about this, measure your first oil ring with a tape measure, and take the number and divide by pie (3.14159). [Circumference = pi x Diameter ] If you don't have a tape measure use a piece of string and measure the length of the string with a ruler.  Using a spare ball with no flare would make seeing and measuring the oil ring easier.
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JohnP

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Re: AXIS TILT
« Reply #4 on: September 12, 2007, 08:47:40 PM »
Using a white or yellow grease pencil, draw the first oil ring all the way around its circumference (it will be a circle).  Place the ball so that the ring is up.  Now use a tape measure (or a ProSect if you have one) to measure the diameter of the circle along the surface of the ball.  --  JohnP

kleptic

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Re: AXIS TILT
« Reply #5 on: September 12, 2007, 10:17:38 PM »
oh okay, so you are actually measuring the partial circumference of the ball starting and stopping at your track lines and calling it the track diameter.

I was confused by the actual diameter going through the ball and was wondering how you could get a 13.5 inch diameter of a 8 inch diameter bowling ball.

thanks for explaining it for me. all this bowling science is very interesting, if only I could aim better...

JohnP

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Re: AXIS TILT
« Reply #6 on: September 13, 2007, 02:41:17 PM »
That's right.  I think circumferential diameter would be a better term to use.  --  JohnP