At 04:58 PM 3/25/99 -0800, you wrote:
>Jim,
>
>You may have given this information in an earlier post, but you need to
>also multiply the number of wheel revolutions by the rear end gear ratio to
>get the number of drive shaft revolutions. Let's say you have 3.54 gears,
>then 111x3.54/2 = 196.47. Now, if the driving gear makes one revolution per
>revolution of the drive shaft, then I think its teeth would have to be
>55/196.47 = 0.2799 as many as the driven gear, or 6.16. Now, it doesn't
>make sense to have any answer other than integers, the right angle drive
>notwithstanding. However, there may be some leeway for measurement error. I
>won't go through all the possible rear end ratios to see which one gives
>closest to an integer value; hopefully your actual ratio will make more
>sense than the 6.16 answer. Also, I have no idea whether the gear in the
>tranny turns at one turn per drive shaft rotation; do you or anyone else
>know the answer?
>
>I could go through the math, but it seems that if you took the known
>original setup, wheel diameter (actually rolling radius), odometer calib
>(1020 rev/mi), and if you knew the number of teeth originally on the driven
>gear at the end of the cable, then you could figure out the number of
>teeth, or at least the equivalent number of teeth on the driving gear in
>the tranny. Maybe if I'm feeling ambitious later I'll see how this works
>out in my case. But, as we often say in academia, "This problem is left as
>an exercise for the student."
>
>TTFN,
>
>Bob
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
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