Re: Re: Gear Ratio Math?

Posted by Alberta Mike on January 19, 2002 at 19:53:05 from (209.47.89.119):

In Reply to: Re: Gear Ratio Math? posted by Kelvin on January 19, 2002 at 18:39:26:

Kelvin, your values are the same as K-Mo's except your division is reversed. He divided the input teeth by the output teeth to get his numbers, you divided the output teeth by the input teeth to get yours. Both of you are correct since you're looking at the same ratio from different ends of the problem, but the numbers are correct if you just take those two gears into account. But looking through an old automotive theory text I've got here, I have now clarified that the 3 gears in question are (in order as the power moves through), ONE: the countershaft reverse gear, TWO: the idler gear, and THREE: the low/reverse gear (on the mainshaft). So the 3 gears do work together, # one drives # two which drives # three right? If the idler gears on his two gear sets were the same then the idler could be ignored in the math, but since the two idlers have different numbers of teeth, then they must be calculated into the overall gearing ratio. Although my wording on my original post was confusing, I think the math was correct since you've got two ratios involved in each set, the ratio between Gear #1 and Gear #2, then the next ratio between Gear #2 and Gear #3 (where #2 is the idler). What do you think?

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