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 Post subject: Re: RG of the PAP
 Posted: Wed May 08, 2019 6:43 pm Post Number: #21 Post
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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.

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 Post subject: Re: RG of the PAP
 Posted: Wed May 08, 2019 10:43 pm Post Number: #22 Post
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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
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]
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

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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.

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Rev Rate: ~350 rpm
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Rotation: 55*
PAP: 5 1/8" right, 1/2" up


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 Post subject: Re: RG of the PAP
 Posted: Thu May 09, 2019 4:05 pm Post Number: #23 Post
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Quote:
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


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