Posted by Ken in CO on Thursday, September 27, 2001 at 4:30PM :
In Reply to: 2 nominations for 'Best of the Forum' posted by Tim Way on Thursday, September 27, 2001 at 3:15PM :
Thanks for the compliment Tim. It is a tricky one to explain(atleast that's what I'll claim in response to my screw-up), and I would personally edit my own post, especially that last paragraph. I was just plain wrong with that last paragraph. It doesn't matter whether you turn the driveshaft or the wheel, the other will turn at half the rate(with the one tire on the ground). I simply got caught reversing a rotation direction in my visualization of turning the driveshaft. I had the rotation of the spider adding to the rotation of the side gear(it rotates against it), which of course violates the average of the side gears rotation being the translation of the spider.
In my example I used equal tooth count for the spider and side gears as this is most common(perhaps always?) I then later elude to the degrees of rotation and translation of the spider as 10 and 10, which would be correct for my example. However, this may initiate a false conclusion in that the rotation/translation of the spider is always a 50/50 deal. Well, it is if the side and spiders are the same tooth count, and the translation of the spider will always be 50% of the side gear rotation(in the condition so described), but the rotation of the spider is dependent on the relationship of tooth count with the side gear. That being said, the spider's rotation factor to the side gear has no effect on the factor between the driveshaft and wheel.
Whew! I'm long-winded today.
Ken
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