Skid Patch Theorem
11-13-07, 09:07 PM
#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
11-13-07, 11:18 PM
#52
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11-13-07, 11:21 PM
#53
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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.
11-14-07, 11:45 AM
#54
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basic algebra rocks.
11-14-07, 12:41 PM
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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.
04-26-08, 04:50 PM
#56
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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 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.
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.
04-26-08, 07:29 PM
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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 b skid patches occur with equal frequency? One would think so, but it might be worth looking into.
Best thread ever, IMHO.
Best thread ever, IMHO.
Last edited by BatNastard; 04-26-08 at 07:35 PM.
04-27-08, 12:23 AM
#59
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What's The Derivative Of My Bike Nigguh
09-10-08, 05:00 AM
#61
so it goes
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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.
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.
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)
