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Re: 351C

To: James Barrett <jamesbrt@mindspring.com>
Subject: Re: 351C
From: Bob Palmer <rpalmer@ames.ucsd.edu>
Date: Fri, 26 Mar 1999 13:16:34 -0800
At 02:37 PM 3/26/99 -0500, James Barrett wrote:
>        Bob,
>        The rear end is 2.88.  I also came up with fractional
>number of teeth.  This student is brain dead, that is why I asked for
>help.  Do not trust the speedometer 1020 calibration.  The speedometer
>has been apart many times.
>
>James Barrett Tiger II 351C and others
>
>
Jim, et Listers,

If I use the number for the mi/rev Laifman quotes for the stock Tiger, and
invert this to give rev/mi, you get 876.8853. Multiply by your 2.88s to get
2525.43 revolutions of the drive shaft per mile. Now, to get your 22 tooth
gear to turn 1020 revolutions per mile, we need to drive it with
1020/2525.43 as many teeth (or the equivalent). This comes out to be 8.88
teeth; i.e., pretty close to nine, but not quite. I tried to look up the
Reference on page 23 of Taylor's book; not there, but he I did find the
speed at 1000 rpm spec (same for both MkI and MkII), which is 23.92 mph @
1000 rpm. This means 2508 revolutions of the motor and, in fourth gear
(1:1), of the drive shaft per mile. Then, (1020/2508)x22 = 8.946, even
closer to an even number. Steve tells me the helical drive gear has "lots
of teeth" and is about 2" in diameter, so somehow it must be geared down a
lot from the rotational speed of the drive shaft. I think it's time to call
a moratorium on the math exercises and research how the drive gear works.

TTFN,

Bob
Robert L. Palmer
Dept. of AMES, Univ. of Calif., San Diego
rpalmer@ames.ucsd.edu
rpalmer@cts.com

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