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 03-28-07, 10:24 AM
  #26  
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Originally Posted by dirtyphotons
ask and ye shall be linked:
http://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
Yeah, Sheldon has two places he describes skid patch calculations on his site and they aren't consistent. Fraction's SPT added a 3rd inconsistent calculation which is why I added the 'skid patch analysis' to figure it all out for myself. Low and behold, Fraction is spot on.
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Old 03-28-07, 11:01 AM
  #27  
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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.
yep, a bit of careless cut and pasting. agreed that fraction is correct, and that only the numerator matters when deciding whether or not to double skid patches.

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.
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Old 03-28-07, 11:21 AM
  #28  
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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:
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.
(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):
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.
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Old 05-16-07, 12:31 PM
  #29  
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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.
plus one.
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Old 05-16-07, 01:40 PM
  #30  
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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.
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Old 05-16-07, 02:46 PM
  #31  
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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.
Or just really, really dull.
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Old 05-16-07, 02:48 PM
  #32  
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Nice.
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Old 05-16-07, 03:00 PM
  #33  
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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.
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Old 05-16-07, 03:05 PM
  #34  
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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
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Old 05-16-07, 04:19 PM
  #35  
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Originally Posted by shogun17
ah thankyou. That explains a lot.

get a brake.
spoken like someone that can't skid. it's way fun, son.
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Old 05-16-07, 04:29 PM
  #36  
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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?
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Old 05-16-07, 04:31 PM
  #37  
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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?
Wrong. Try again.
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Old 05-16-07, 04:52 PM
  #38  
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i kinda just spin my pedals by hand n count the revolutions till i get to my original spot.
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Old 05-16-07, 04:52 PM
  #39  
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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

Out of curiosity, what does the highlighted yellow mean?
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Old 05-16-07, 07:14 PM
  #40  
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good thing i didn't forget my high school quantam physics
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Old 05-16-07, 07:33 PM
  #41  
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Originally Posted by missej
Out of curiosity, what does the highlighted yellow mean?
i think it's about the gear ratios, showing what most fixed riders use, 2.9:1 to 3.0:1....so on the bottom one, it'll show where a lot of people will be looking, and then on the top, it'll show where their skid patches stand.

that's my guess. i don't remember just where, but the "most fixed riders use...." part is from an old poll on here...
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Old 05-16-07, 11:35 PM
  #42  
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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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Old 05-17-07, 12:13 AM
  #43  
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This is above me.
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Old 11-13-07, 02:12 PM
  #44  
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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.

Last edited by lisitsa; 11-13-07 at 02:22 PM. Reason: Add Information
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Old 11-13-07, 03:27 PM
  #45  
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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.
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Old 11-13-07, 03:42 PM
  #46  
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Originally Posted by Fraction View Post
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.
fanciest way to say, "you have b skid patches" i've ever seen. =P
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Old 11-13-07, 05:46 PM
  #47  
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Originally Posted by c0urt View Post
I should only have three skid patches due to the fact i run 40x13
Wrong.
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Old 11-13-07, 05:57 PM
  #48  
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Originally Posted by c0urt View Post
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.
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.
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Old 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.
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Old 11-13-07, 09:02 PM
  #50  
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Originally Posted by mathletics View Post
Also, I think you installed your cranks wrong if you can get your feet at 3 o'clock and 10 o'clock.
nah i just cant tell time
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