# Skid Patch Theorem

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**51**moving target

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

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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.

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basic algebra rocks.

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

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**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

Proof: The equivalence is immediate when you recognize that

Corollary to 2: Suppose the rider is an ambidextrous skidder. If

Or, for a more direct number theoretic description, simply recognize that 2

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/*b*Z, 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 2*b*), 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 2*b*. So in particular,*k*is even. Thus, letting*k*be as above, we have*ka*equiv 0 (mod 2*b*) which implies*k/2 a*equiv 0 (mod*b*).Or, for a more direct number theoretic description, simply recognize that 2

*b*/gcd(*(a,2b)*=*b/gcd(a,b)*if and only if*a*is even.
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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

Best thread ever, IMHO.

*b*skid patches occur with equal frequency? One would think so, but it might be worth looking into.Best thread ever, IMHO.

*Last edited by BatNastard; 04-26-08 at 07:35 PM.*

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**59**SWAAAAAAAAAAAT

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What's The Derivative Of My Bike Nigguh

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**61**so it goes

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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/

*b*Z, 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 2

*b*), 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 2

*b*. So in particular,

*k*is even. Thus, letting

*k*be as above, we have

*ka*equiv 0 (mod 2

*b*) which implies

*k/2 a*equiv 0 (mod

*b*).

Or, for a more direct number theoretic description, simply recognize that 2

*b*/gcd(

*(a,2b)*=

*b/gcd(a,b)*if and only if

*a*is even.

**reduced**ratio, but you define

*a*to be the actual number of teeth on the chain ring. The reduced numerator is

*a/gcd(a,b)*in this case, which may be odd even though

*a*is even.

E.g., suppose you have a chain ring with

*a*= 42 teeth and a sprocket with

*b*= 14 teeth, then

*a*is even but the reduced ratio 42/14 = 3/1 has an

**odd**numerator. Note that this also gives a counter example to your final claim:

2*14/gcd(42,2*14) = 2

**does not equal**1 = 14/gcd(42,14)