## RG of the PAP

Which layout is right for me?

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snick Posts: 753
Joined: August 31st, 2014, 8:00 pm
THS Average: 190
Positive Axis Point: 5.625" x 0
Speed: 17 off hand
Rev Rate: 360
Axis Tilt: 7
Axis Rotation: 55
Medium Oil Ball: Storm JoyRide
Light Oil Ball: Storm Tropical Breeze
Preferred Company: Storm
Location: Tucson, AZ

### Re: RG of the PAP

deanchamp wrote:
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."
Clear as mud. Not sure what to make of it.
Benchmark Bowling Pro Shop
Byron

RH
PAP: 5.625" x 0
REVRATE: 360
SPEED: 17mph at release
AR: 55º
AT: 7º

DarkHorse
Member Posts: 221
Joined: June 15th, 2013, 12:06 am
Location: USA

### Re: RG of the PAP

deanchamp wrote: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?
JohnP wrote: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
Doh!
I forget about that. Nice catch, guys!
The RPM values will be affected by Axis Tilt, but MPH will not.

From viewtopic.php?p=23272#p23272" onclick="window.open(this.href);return false;
Post #15 by Mo:
In practical terms, if a ball is released with 15* of tilt, an asymmetrical ball with hit the pins with about 11* of tilt, while a symmetrical ball will reach the pins with about 8* of tilt

Track Length = Pi x Ball Dia. x cos(Tilt)
[from http://wiki.bowlingchat.net/wiki/index. ... tguide.xls" onclick="window.open(this.href);return false;]
Which simplifies to:
Track Length = 27" x cos(Tilt)

Initial Track Length == 27" x cos(15*) = 26.1" = 2.17'
Generated RPM => 350 rpm = 350 x 2.17 = 759.5'/min = 12.7'/sec
Generated MPH => 18 mph = (18 x 88)/1 minute = 1584'/min = 26.4'/sec

Final Track Length (Asym) == 27 x cos(11*) = 26.5" = 2.21'
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)/2.21 = 637.2 rpm

Final Track Length (Sym) == 27 x cos(8*) = 26.7" = 2.23'
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)/2.23 = 631.5 rpm

Look better?
deanchamp wrote: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."
Since he made that as a general physics statement, I'd say he's talking about the rate of travel for the PAP (the object) along that particular RG contour (elliptical orbit), aka PAP migration or Precession, if you like.
Right Handed
Speed: 18 mph (foul line)
Rev Rate: ~350 rpm
Tilt: 10*
Rotation: 55*
PAP: 5 1/8" right, 1/2" up

JohnP
Trusted Source Posts: 3432
Joined: January 31st, 2010, 1:04 am
Positive Axis Point: 15 15/16 x 3/16
Speed: 13.5 (Qubica)
Axis Tilt: 13
Axis Rotation: 45
Location: Hawesville KY/Tell City IN

### Re: RG of the PAP

Final RPM (calc) => 23.47'/sec = (23.47 x 60)/2.23 = 631.5 rpm
From the data on the TV broadcasts Belmo gets around 550 RPMs, so even it revs up when it rolls. The less revs a bowler has the more dramatic the increase will be. -- JohnP