**Gold stars to ConorP & John Coffey...**

Black stinky rotten egg to Al 260Z, your favorite foot-in-mouth Z professor.

John gets a gold star for the Gordian Knot solution: The **** with mental gymnastics - cut out all the bullshit and check the nos. on the gearset. (easier even than counting teeth).

ConorP is right on on all counts, especially in pointing out that you want to make multiple revolutions of the wheel for a single revolution of the driveshaft. (Shame on me!)

As I posed the question, I could not for the life of me remember whether to multiply the ration by two or by one-half with one wheel blocked. (One-half is correct.) Nor could I visualize the diffy action that causes this. I've been playing with this mental visualization all weekend, and just barely grasp it. See what you can do when you're not distracted by some lousy radio station blaring at you! ;-). I certainly can't explain it any better than ConorP did.

Recall that the pinion carrier shaft revolves with the ring gear. If one side gear (output shat) is blocked, then the free side gear shaft must advance twice as much. The ratio of the side to pinion gear is immaterial because the ratio of the side gears to each other is always one-to-one.

Gear ratios are interesting as they illustrate some basic characteristics of numbers. The number of teeth on a gear must always be an integer number. The ratio of two integers is a rational number. Different gear sets can give the same number ratios such as 40/10 or 20/5. (4/1 is the same, but a gear with one tooth would not have very smooth action!) Remember simplifying fractions to their lowest possible denominator? (Sorry to remind you!) Well, in the 260Z case 37/11= 3.363636363636.... (isn't that fun - the 36363636 repeats infinitely!). You could get the same ratio by doubling the numbers to 74/22, but there is no reducing this pair. Furthermore, 37 and 11 are both prime numbers - no factors at all for either number!

Later all!

:-D

Al