Fraction

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

dirtyphotons

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 )

maxknee

nerdy

dylandom

omg what did u get in differential and calc 3. don't tell me, an A. i'm going to go over ur theorem with the math club tonight. hopefully ur right.

dobber

This new learning amazes me, Sir Bedevere. Explain again how sheep's bladders may be employed to prevent earthquakes.

lasertotheface

i didnt even read it.

endform

Jesus, I was trying to blow off my number theory homework not do extra curricular work.

Anyways, nice proof.

dommer

no ****ing way am i reading that

Analog

I nominate this thread for "Best Title of the Week"

Moximitre

just rotate your tire a few degress every so often. prob. = solved

stevo

or your cog (relative to the same links)

noisebeam

Maybe its time to update my SP chart and put the ones to double for ambi in italics or somesuch. Last update was may-05

WakeUpOnFire

testtube

I've added a "skid patch analysis" function to rabbit showing the conclusion of Fraction's SPT is valid.

Check it out at http://software.bareknucklebrigade.com if interested.

jamey

i like bikes...

nexus6

You just run a 17t cog, problem solved!

thatcher

theres a book called http://en.wikipedia.org/wiki/Cryptonomicon it has a cool discription of how code machines work n it uses a bike to describe it.

TimArchy

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.

shogun17

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)?

dirtyphotons

ask and ye shall be linked:
http://sheldonbrown.com/fixed.html#skid

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

Aeroplane

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).

shogun17

ah thankyou. That explains a lot.

get a brake.

oldsprinter

The photos on that page are shocking!

FridgeRobot

truth

mander

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.