Theorem:

Let *a* / *b* be the reduced gear ratio (that is, *a* and *b* are integers with no common divisors other than 1). Then,

(1) With single-sided skidding, there are *b* skid patches, and

(2) Ambidexterous skidding doubles the number of skid patches if and only if *a* is odd.

(1) is already well-known. It is proved here for completeness.

(2) is generally not known and disproves several conjectures seen recently.

Proof of (1):

Turning the pedals through one revolution turns the wheel through *a* / *b* revolutions. Turning the pedals through *b* revolutions turns the wheel through *b* * (*a* / *b*) = *a* revolutions. That is, after *b* pedal revolutions, the wheel is returned to the same position it was originally (since *a* is an integer). So there must be no more than *b* skid patches, since the same cycle of *b* wheel positions will be repeated through every *b* pedal revolutions.

Now suppose that two of the intermediate wheel positions were the same, say, after *i* and *j* pedal revolutions (0 <= *i* < *j* < *b*). Then *j* - *i* pedal revolultions also returns the wheel to its original position, so (*j* - *i*) * *a* / *b* is an integer. Thus *b* must evenly divide (*j* - *i*) * *a* / b. However, *a* and *b* have no common divisors, so *b* must evenly divide *j* - *i*. But *j* - *i* is less than b, so this cannot happen. Therefore, all *b* of the intermediate wheel positions (after 0, 1, 2, ..., and *b-1* pedal revolutions) are different. So there must be no fewer than *b* skid patches.

There are no more than *b* skid patches and there are no fewer than *b* skid patches, so there must be exactly *b* skid patches.

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* / (2*b*) is the reduced ratio, so there are 2*b* skid patches.