http://www.bikeforums.net/showthread.php?t=242123
This theorem has it that only the numerator of the chainring teeth/ cog teeth fraction affects whether ambidextrous skidding doubles the number of skid patches. This is contrary to how Rabbit calculates this (it assumes that both the numerator and the deniminator must be odd), and to what Sheldon says (that only the deniminator must be odd). The proof looks right to me but I suck at numbers and am lazy. Can anyone tell me if it is correct?
Here it is:
Originally Posted by fraction
Let a / b be the reduced gear ratio (that is, a and b are integers with no common divisors other than 1). Then,
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(2) Ambidexterous skidding doubles the number of skid patches if and only if a is odd.
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Proof of (2):
As above, turning the pedals through one revolution turns the rear wheel through a / b revolutions. Turning the pedals through one half-revolution turns the rear wheel through half as many revolutions. So the number of skid patches with ambidexterous skidding should be the same as that with single-sided skidding on a gear ratio half as large. Now to apply (1) to this situation, we need to know how 1/2 * a/b reduces as an integer ratio. This depends on whether a is even or odd. If a is even, (a/2) / b is the reduced ratio, so there are b skid patches, as in the single-sided case. If a is odd, a / (2b) is the reduced ratio, so there are 2b skid patches.