Skid Patch Theorem
03-28-07, 10:24 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, 11:01 AM
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Originally Posted by mander
Look here, according to Fraction the denominator doesn't have to be odd. Only the numerator matters. I just want to get clear on this because everyone's website says a different thing and Fraction's result is backed up by his proof (which looks OK to me, but ought to be checked by someone better at math) as well as Rabbit's brute force algorithm.
this is for the most part academic, though. anything more than about 8 skid patches is probably overkill. if you're worried but don't want to do the math just run a 17t cog (and not a 51t chainring) and rest easy.
03-28-07, 11:21 AM
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Seriously, this is the sexiest post I've read on this forum. Excuse me while I sneak off to the little girls room.
Originally Posted by Fraction
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.
05-16-07, 12:31 PM
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Originally Posted by Maracski
Seriously, this is the sexiest post I've read on this forum. Excuse me while I sneak off to the little girls room.
05-16-07, 01:40 PM
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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.
05-16-07, 02:46 PM
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Originally Posted by mander
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.
05-16-07, 02:48 PM
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Nice.
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05-16-07, 03:00 PM
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Originally Posted by noisebeam
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
yes!!! i was looking for this this morning.
05-16-07, 03:05 PM
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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
To be honest I hadn't really ever paid any attention to skid patches.
Interesting though
05-16-07, 04:19 PM
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Originally Posted by shogun17
ah thankyou. That explains a lot.
get a brake.
get a brake.
05-16-07, 04:29 PM
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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?
Right?
05-16-07, 04:31 PM
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Originally Posted by dwainedibbly
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?
Right?
05-16-07, 04:52 PM
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Originally Posted by noisebeam
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
05-16-07, 07:33 PM
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Originally Posted by missej
Out of curiosity, what does the highlighted yellow mean?
that's my guess. i don't remember just where, but the "most fixed riders use...." part is from an old poll on here...
05-16-07, 11:35 PM
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I should add that most of my skid patch theory is based on prime numbers. So far it works...
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11-13-07, 02:12 PM
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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.
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.
Last edited by lisitsa; 11-13-07 at 02:22 PM. Reason: Add Information
11-13-07, 03:27 PM
#45
moving target
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.
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.
11-13-07, 03:42 PM
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Originally Posted by Fraction
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.
11-13-07, 05:46 PM
#47
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Originally Posted by c0urt
I should only have three skid patches due to the fact i run 40x13
11-13-07, 05:57 PM
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Originally Posted by c0urt
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.
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.
11-13-07, 07:47 PM
#49
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Also, I think you installed your cranks wrong if you can get your feet at 3 o'clock and 10 o'clock.
11-13-07, 09:02 PM
#50
moving target
Originally Posted by mathletics
Also, I think you installed your cranks wrong if you can get your feet at 3 o'clock and 10 o'clock.
