Skid Patch Theorem
11-02-06, 03:08 PM
#1
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Skid Patch Theorem
Theorem:
(2) is generally not known and disproves several conjectures seen recently.
Proof of (1):
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.(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.
(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):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.
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.
11-02-06, 03:11 PM
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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 
)
)
11-02-06, 05:01 PM
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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.
11-02-06, 05:09 PM
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This new learning amazes me, Sir Bedevere. Explain again how sheep's bladders may be employed to prevent earthquakes.
__________________
This is Africa, 1943. War spits out its violence overhead and the sandy graveyard swallows it up. Her name is King Nine, B-25, medium bomber, Twelfth Air Force. On a hot, still morning she took off from Tunisia to bomb the southern tip of Italy. An errant piece of flak tore a hole in a wing tank and, like a wounded bird, this is where she landed, not to return on this day, or any other day.
This is Africa, 1943. War spits out its violence overhead and the sandy graveyard swallows it up. Her name is King Nine, B-25, medium bomber, Twelfth Air Force. On a hot, still morning she took off from Tunisia to bomb the southern tip of Italy. An errant piece of flak tore a hole in a wing tank and, like a wounded bird, this is where she landed, not to return on this day, or any other day.
11-02-06, 08:57 PM
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Jesus, I was trying to blow off my number theory homework not do extra curricular work.
Anyways, nice proof.
Anyways, nice proof.
11-03-06, 12:48 AM
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I nominate this thread for "Best Title of the Week"
11-03-06, 01:23 AM
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just rotate your tire a few degress every so often. prob. = solved
11-03-06, 07:13 AM
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Originally Posted by Moximitre
just rotate your tire a few degress every so often. prob. = solved
11-03-06, 08:31 AM
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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
03-27-07, 12:17 PM
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I've added a "skid patch analysis" function to rabbit showing the conclusion of Fraction's SPT is valid.
Check it out at https://software.bareknucklebrigade.com if interested.
Check it out at https://software.bareknucklebrigade.com if interested.
03-27-07, 07:33 PM
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theres a book called https://en.wikipedia.org/wiki/Cryptonomicon it has a cool discription of how code machines work n it uses a bike to describe it.
03-27-07, 09:26 PM
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Originally Posted by thatcher
theres a book called https://en.wikipedia.org/wiki/Cryptonomicon it has a cool discription of how code machines work n it uses a bike to describe it.
This was when I was a math major but yet to get into bikes.
03-28-07, 01:39 AM
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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)?
03-28-07, 05:49 AM
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ask and ye shall be linked:
https://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
https://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
03-28-07, 05:51 AM
#21
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Originally Posted by shogun17
ok. I am having a little trouble understanding exactly what a skid patch is.
03-28-07, 05:51 AM
#22
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ah thankyou. That explains a lot.
get a brake.
get a brake.
03-28-07, 06:45 AM
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Originally Posted by dirtyphotons
ask and ye shall be linked:
https://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
https://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
03-28-07, 08:06 AM
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Originally Posted by nexus6
You just run a 17t cog, problem solved!
03-28-07, 10:20 AM
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Originally Posted by dirtyphotons
ask and ye shall be linked:
https://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
https://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
Originally Posted by fraction
Let a / b be the reduced gear ratio (that is, a and b are integers with no common divisors other than 1). Then,
...
(2) Ambidexterous skidding doubles the number of skid patches if and only if a is odd.
...
(2) Ambidexterous skidding doubles the number of skid patches if and only if a is odd.
