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 / (2b) is the reduced ratio, so there are 2b skid patches.
yup. we were just talking about this earlier in the week. i was thinking that a AND b need to be odd, but i agree with you that it's just a (not that it matters what i agree with, nice proof )
Originally Posted by Sheldon Brown
Because when fashion conflicts with function, I vote for function.
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This new learning amazes me, Sir Bedevere. Explain again how sheep's bladders may be employed to prevent earthquakes.
This is Africa, 1943. War spits out its violence overhead and the sandy graveyard swallows it up. Her name is King Nine, B-25, medium bomber, Twelfth Air Force. On a hot, still morning she took off from Tunisia to bomb the southern tip of Italy. An errant piece of flak tore a hole in a wing tank and, like a wounded bird, this is where she landed, not to return on this day, or any other day.
I read that book and the only part I found really interesting was the story about the one mathematician who had a bad tooth on his cog and a stiff link in his chain which, when they hit in a particular spot, would drop the chain. He would keep track in his head how many pedal revolutions it would take between instances where the bad tooth and stiff link would meet. Whenever it came around he would stop, get off and rotate the cranks one turn and get back on.
This was when I was a math major but yet to get into bikes.
Originally Posted by Josh Frank
I will derive power from their cries of despair. My crank a speedy dervish, spinning and spinning through the darkest night that anyone with the audacity to try and suck my wheel will ever see...
ok. I am having a little trouble understanding exactly what a skid patch is. I'll take a guess and say every rotation of the pedals you lock up the wheels once or twice (single-sided or ambidextrous)?
edit: if the theorem above is correct, and it looks like it is, the sentence should read:
If you are an ambidextrous skidder, and the simplified ratio has an even numerator, your number of skid patches will be the same.
If you are an ambidextrous skidder, and the numerator and denominator is odd, the number of possible skid patches will be doubled.
to be fair, i was the one who called sheldon out on this originally, and being the reasonable person he is, he changed his site to reflect it. turns out we were both wrong
Originally Posted by Sheldon Brown
Because when fashion conflicts with function, I vote for function.
ok. I am having a little trouble understanding exactly what a skid patch is.
A skid patch is the spot on your tire where the rubber wears down from skidding. Since skidding generally only happens at two different points on the chainring rotation, the location of skid patches on the tire (and their quantity) can be calculated by looking at the number of teeth on the cog and chainring (the gear ratio).
Originally Posted by Dr Irwin Goldstein
Men should never ride bicycles. Riding should be banned and outlawed. It is
the most irrational form of exercise I could ever bring to discussion.
edit: if the theorem above is correct, and it looks like it is, the sentence should read:
If you are an ambidextrous skidder, and the simplified ratio has an even numerator, your number of skid patches will be the same.
If you are an ambidextrous skidder, and the numerator and denominator is odd, the number of possible skid patches will be doubled.
to be fair, i was the one who called sheldon out on this originally, and being the reasonable person he is, he changed his site to reflect it. turns out we were both wrong
edit: if the theorem above is correct, and it looks like it is, the sentence should read:
If you are an ambidextrous skidder, and the simplified ratio has an even numerator, your number of skid patches will be the same.
If you are an ambidextrous skidder, and the numerator and denominator is odd, the number of possible skid patches will be doubled.
to be fair, i was the one who called sheldon out on this originally, and being the reasonable person he is, he changed his site to reflect it. turns out we were both wrong
Look here, according to Fraction the denominator doesn't have to be odd. Only the numerator matters. I just want to get clear on this because everyone's website says a different thing and Fraction's result is backed up by his proof (which looks OK to me, but ought to be checked by someone better at math) as well as Rabbit's brute force algorithm.
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,
...
(2) Ambidexterous skidding doubles the number of skid patches if and only if a is odd.
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Originally Posted by Sheldon Brown
