Singlespeed & Fixed Gear "I still feel that variable gears are only for people over forty-five. Isn't it better to triumph by the strength of your muscles than by the artifice of a derailer? We are getting soft...As for me, give me a fixed gear!"-- Henri Desgrange (31 January 1865 - 16 August 1940)

Skid Patch Theorem

Old 11-13-07, 09:07 PM
  #51  
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Originally Posted by Yoshi View Post
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.
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
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Old 11-13-07, 11:18 PM
  #52  
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Originally Posted by mander View Post
A lot of people just associate math with the boring number crunching they did in high school, but it *is* sexy when you get into the good stuff. the power and beauty is really summinelse.
There's nothing sexy about applying really basic abstract algebra.
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Old 11-13-07, 11:21 PM
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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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Old 11-14-07, 11:45 AM
  #54  
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Originally Posted by c0urt View Post
not running 700c tires so I have fewer skid patches due to smaller rim size and fewer rotations in relation to the cranks.
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.
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Old 11-14-07, 12:41 PM
  #55  
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Originally Posted by c0urt View Post
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
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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Old 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.
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Old 04-26-08, 05:20 PM
  #57  
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by golly, i think you're on to something!
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Old 04-26-08, 07:29 PM
  #58  
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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.

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Old 04-27-08, 12:23 AM
  #59  
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What's The Derivative Of My Bike Nigguh
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Old 04-27-08, 05:37 AM
  #60  
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SS = 1 x 1 = 1^2 so SS' = 2 x 1^1 = 2
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Old 09-10-08, 05:00 AM
  #61  
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Originally Posted by Satyr View Post
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.
I like the pun, but I don't think your proof of part 2 is correct. I don't understand your proof, but I can explain why it can't make sense: part two of the theorem concerns the numerator in the 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)
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