Or just really, really dull.

Nice.

yes!!! i was looking for this this morning.

I like the math thing but in reality all you have to do is put your bike in a stand and go through the revolutions and count the different skid patches you get (probably not as much fun though)

To be honest I hadn't really ever paid any attention to skid patches.
Interesting though

spoken like someone that can't skid. it's way fun, son.

Executive summary: if the number of teeth on the cog is a prime number, the skid patches will be equal to the number of teeth on the cog. If the # of teeth on the chainring is a prime, the # of skid patches is equal to the # of teeth on the cog. In either case, if one is a prime, the other can be anything without reducing the # of patches. So to max your patches, use a 13, 17, 19, or 23 on the rear OR a 41, 43, 47, 51 on the front.

Right?

Wrong. Try again.

i kinda just spin my pedals by hand n count the revolutions till i get to my original spot.

Out of curiosity, what does the highlighted yellow mean?

i think it's about the gear ratios, showing what most fixed riders use, 2.9:1 to 3.0:1....so on the bottom one, it'll show where a lot of people will be looking, and then on the top, it'll show where their skid patches stand.

that's my guess. i don't remember just where, but the "most fixed riders use...." part is from an old poll on here...

I should add that most of my skid patch theory is based on prime numbers. So far it works...

This is above me.

For the sake of setting a possibly obvious conclusion.
The point of this theory is so that if you understand it, you can make sure you don't put on a chainring and sprocket combination that gives you only 1 or 2 or 3 skid spots and that wears out your tire too quickly.

When you look at the chart, look for a combination where there is a good number of skid spots, to spread the wear over more of the tire. Remember the chart is only for single-sided skidding.

Follow this rule for ambidextrous skidders.
ODD CHAINRING NUMBER = DOUBLE THE SKID-PATCHES

Hope this helps, but I'm just rehashing the original poster in different words so credit to him for the workings out.

these number are only effective in a perfect world or for those who only skid with their feet at
3'o clock and 10 o'clock

I should only have three skid patches due to the fact i run 40x13

but on a good day i skld with my rear foot anywhere from 1 o'clock to 5''o clock
so if you look at my tire there are effectively 6 skid patches. it is really odd




but dude
the number skills i have seen thrown aroudn this thread are amazing. good work guys.

fanciest way to say, "you have b skid patches" i've ever seen. =P

Wrong.

You have 13 skid patches with a 40x13. However because you skid anywhere from 1 to 5 o'clock a lot of them overlap and it looks like 6.

Also, I think you installed your cranks wrong if you can get your feet at 3 o'clock and 10 o'clock.

nah i just cant tell time

not running 700c tires so I have fewer skid patches due to smaller rim size and fewer rotations in relation to the cranks.

like i stated 650c rims front and rear, it throws the math off for me. thats why i have 74 gear inches at that ratio

There's nothing sexy about applying really basic abstract algebra.

Great proof. Elegant and well-written (I didn't check every line, but I believe it is correct.) And contrary to the last poster, there is something sexy about basic algebra.

wheel size has nothing to do with skid patch count. calculations dependent on wheel diameter will change (gear-inches, development, &c...) but skid patch count is not one of them. it's solely related to chain ring tooth count related to sprocket tooth count...

basic algebra rocks.

A smaller wheelsize is going to result in greater overlap between skid patches but won't affect the actual number of skid patches.

An Equivalence
 

My aim in this post is to couch the original theorem in language of cyclic groups (Fraction probably had this in mind). The only real benefit of this is that it reduces Part (2) of the theorem to an immediate corollary.

Corollary to 1: Let a be the number of teeth on the chainring and b the number of teeth on the rear sprocket, and s=# skid patches. Then s=ord(a) in Z/bZ, where Z is the group of integers. In particular, s=b/gcd(a,b).

Proof: The equivalence is immediate when you recognize that b/gcd(a,b) is the b from the reduced fraction a/b as per the theorem, Part (1).

Corollary to 2: Suppose the rider is an ambidextrous skidder. If a is even, then in general ord(a) in Z/2bZ is the least k such that ka equiv 0 (mod 2b), and since b is even we have ka/2 equiv 0 (mod b). The converse is similarly shown. Suppose that a was odd. Recall that the order of an element has to divide the order of the group, which in the ambidextrous skidder case is 2b. So in particular, k is even. Thus, letting k be as above, we have ka equiv 0 (mod 2b) which implies k/2 a equiv 0 (mod b).

Or, for a more direct number theoretic description, simply recognize that 2b/gcd((a,2b)=b/gcd(a,b) if and only if a is even.

by golly, i think you're on to something!

Right, but any non-trivial case of gear ratios will involve at least one odd number. Hm. Shouldn't there be some kind of probabilistic analysis? Meaning, even if one skids in the same position every time, and I'll accept that, and even if each skid patch is a discrete point, which I'm also willing to buy for the sake of argument, does each of the b skid patches occur with equal frequency? One would think so, but it might be worth looking into.

Best thread ever, IMHO.

What's The Derivative Of My Bike Nigguh

SS = 1 x 1 = 1^2 so SS' = 2 x 1^1 = 2 :)