Singlespeed & Fixed Gear - More information about # of skidding spots with gearing...

Headache inducing...

http://www.dartmouth.edu/~ccatalan/skid.html

that rocked! hooray for math-geek cyclists!

43 teeth here i come!

that page seems to make it a lot more complex than it needs to be.

48/16 = 3/1 = 3
so the back wheel goes around three times with each crank rotation and ends up in the same place. so if you always skid with the same foot forward, you always hit the same spot. this is true for all ratios that are whole numbers.

45/18 = 5/2 = 2.5
now the crank has to go around two times to get the rear wheel back in the same position so you have two places on the wheel.

48/15 = 16/5 = 3.2
here, the crank has to go around 5 times to get back to the same spot on the rear wheel so you've got 5 spots to hit.

48/17 = something with a lot of decimals.
it won't reduce so the crank has to go around 17 times to hit the same place on the rear wheel.

easy formula:
reduce your ratio to the smallest denominator (chainring/cog). the denominator will tell you how many places you will skid on the rear wheel. (if you skid with either foot forward, double the number)

much easier right?

a month ago I taught a lesson to a 6th grade math class about ratios using my fix.

tim

nice use of applied math with the kiddies

yeah, my mentor teacher liked it a lot. I'm teaching high school geometry and algebra next spring. I'm trying to think of lessons using cycling that I can use on them.

tim

44/14 = 3.1428571428571428571428571428571 - dammit - I almost had the fabled pi gearing!!!!

According to the last paragraph, taking the highest denominator (2) I have 7 patches to skid on...

I don't see how he connects Sheldon with brakeless though...

HereNT, he doesn't connect scheldon with brakeless;

<snip>
Though Sheldon Brown, the laws of several states and plain old common sense urge otherwise, many fixed gear riders choose to pilot their machines without brakes.
</snip>

Dang, losing that contact lens earlier must be affecting my eyesight or something. Found it again, but everythings kind of blurry out of that eye...

I'm glad I found that elusive 43 tooth 118bcd chainring on Ebay.

damn...
this thread's been Pi-jacked

what's 52-17, for a math ******?

you've got 17 patches, since 17 is a prime number.

hooray!

yup i think trevor worked all this out for us a while back. totally awesome.

and now as of last week my new chainring is 43t. i also changed to a 3/32" drivetrain. hopefully ill take some pics in a week or so..

Headache inducing...

http://www.dartmouth.edu/~ccatalan/skid.html

Fan-diddely-tastic. Now i'm gonna translate this into my native language and put it in our magazine and pretend that i figured it out myself

*drool*

you've got 17 patches, since 17 is a prime number.
it wouldn't matter if 17 was a prime number. if you ran a 51/17, you'd only have one skid patch since 51/17 = 3/1 = 3. whole number = one skid patch.

52/17 indeed gives 17 skid patches, but it's because 52 and 17 are mutually prime.

tim

relatively prime. :)

edit: hey, lookit that, indeed "mutually prime" is also an acceptable term. I don't think I've ever heard it called thaht.

heh heh, i was gonna point that out too, nice to know that both are correct.

btw, TimArchy, thanks for correcting my comment. nice catch

mutually prime, relatively prime...
I can't remember which is correct.
I took abstract algebra twice, but it was two years ago
I figured that since they were prime relative to each other...

tim

A prime chainring is what you want. Otherwise you'll need to do the calculation at the bottom.

you don't need a prime chainring OR cog as long as they're prime relative to each other

Yes, but what is most flexible is a prime chainring. Then you can run pretty much any size cog and not have to worry.

Yes, but what is most flexible is a prime chainring. Then you can run pretty much any size cog and not have to worry.

For anyone wondering, the prime chainrings of reasonable size would be 41, 43, 47, 53.
Justification of reasonable size: 37 seems a bit small, 61 a bit big. Certainly a good spread though.

For anyone wondering, the prime chainrings of reasonable size would be 41, 43, 47, 53.
Justification of reasonable size: 37 seems a bit small, 61 a bit big. Certainly a good spread though.

for a while i was riding a 47t chainring but then i wanted a 94t cog and this issue of 'skid spots' came up! im am SO PERPLEXED! should i uprgrade to a 51t chainring, or maybe a 93t cog wouldnt be tooo hard on the knees... who knew riding a shiftless bike would require so much math! please advice.

im am SO PERPLEXED!

me too

easy formula:
reduce your ratio to the smallest denominator (chainring/cog). the denominator will tell you how many places you will skid on the rear wheel. (if you skid with either foot forward, double the number)


I like your easy formula, but skidding with either foot forward only doubles the number of "skid patches" on your rear wheel if the numerator of your reduced ratio is odd.

So, if you skid with either foot forward,
46/16 = 23/8 gives you 16 skid patches,
but 44/14 = 22/7 only gives you 7 skid patches.

I like your easy formula, but skidding with either foot forward only doubles the number of "skid patches" on your rear wheel if the numerator of your reduced ratio is odd.

So, if you skid with either foot forward,
46/16 = 23/8 gives you 16 skid patches,
but 44/14 = 22/7 only gives you 7 skid patches.

Hmm, If you use the formula in the article 46/16 has a highest factor of 2. 16/2 gives 8 skid patches using both feet, 4 using one one foot forward. Is this not correct?

Hmm, If you use the formula in the article 46/16 has a highest factor of 2. 16/2 gives 8 skid patches using both feet, 4 using one one foot forward. Is this not correct?

The article was only talking about skidding with the same foot forward. Using the formula from the article, 46/16 has a highest common factor of 2, so 16/2 = 8 skid patches (if you always put the same foot forward to skid).

But because 46/2 = 23, which is odd, you get 8 different skid patches with the other foot forward, for 16 total.

woah. cool. now it is perfect. lets release it on the world!

[edit: oh wait. but how long are these 'skid patches'? how many possible patches are there on a 700c tire? wouldnt they overlap?]

It depends on how large your contact patch is which is a function of tire size and pressure as well how precisely you position your feet for each skid. Whether they overlap or not, well I'll leave that for you to calculate. ;)

47 is a prime number. Are 47 and 15 relatively prime? i guess what i'm asking is, how many skid patches do i have?

47 is a prime number. Are 47 and 15 relatively prime? i guess what i'm asking is, how many skid patches do i have?

you have 15. thinking about the primeness of either number of teeth, or their relative primeness, is not necessary for solving this problem. you only have to list the factors of each number, and divide the rear teeth by the largest common factor. in the process of doing that, you may discover something about the primeness of the numbers.

if you already know that the chainring number is prime (as you do, crusters) then a shortcut is available - you always have as many skid patches as you have teeth on the rear cog.

Unless the rear cog is a multiple of the front ring... You know, on your stump-pulling bike.

much obliged, cicadashell. oh, and in passing, is there much of a fixed gear, eh, "community" in ann arbor?

Unless the rear cog is a multiple of the front ring... You know, on your stump-pulling bike.

you east-coasters and your progressive ways! we still use oxen for stump-pulling out here.



...oh, and in passing, is there much of a fixed gear, eh, "community" in ann arbor?

probably - i see a variety of homemades around town. i'm unaware of any professional courier service here, so i figure it's just the usual self-styled hipsters. myself, i'm a bit of a loner (read: adult) so i don't really "make the scene" as they say.

Even adults like to ride together sometimes.

Another thought about skid patches. Sometimes I push hard enough backwards that the rear wheel actually rotates backwards while I skid. I'd imagine the swiping action helps even out some of the patches... maybe.

Even adults like to ride together sometimes.



you are correct, sir. my remark was not very well though out. i guess i just seem to remember having more time to just ride and hang out 20 or 30 years ago. but i still appreciate having riding partners.

easy formula:
reduce your ratio to the smallest denominator (chainring/cog). the denominator will tell you how many places you will skid on the rear wheel. (if you skid with either foot forward, double the number)


Skidding with either foot forward will not necessarily double the number of skid patches.

It's ~4:30am, so there are probably a lot of errors in the following...

For those who have taken a course in Abstract Algebra, consider the points on the rear wheel as a group isomorphic to a dihedral group (but we are really only interested in the group operation of rotation). The order of the dihedral group is 2n, where n = the number of teeth in the cog. Our ultimate goal is to find the cardinality of the orbits of the points on the wheel. In other words, we want to find the number of points fixed by all possible rotations of the crank. There are 2^n possible permutations of this group through rotation. Since we are now considering *two* possible crank positions, we only have to consider the identity, R_0, and the group generated by a half-rotation of the crank, <R_{m/n**>, where m = the number of teeth in the chainring. From Burnside's theorem, we see that there should be at least 1/(2^n)*(1 + 2^n) patches, which does not equal a strict doubling.

My math is probably all horribly wrong, but I'm currently too tired to check it over.

However, for the less Math-inclined, here is a simple counter-example:

Consider a chainring with twice as many teeth as the cog. When you skid with one foot forward, that creates one patch. Then the crank rotates 180 degrees and you skid again with the other foot forward. However, since the chainring is twice as big as the cog, the wheel has made one full rotation since the previous skid and the same patch will be skidded upon once again. Therefore, even though you are now skidding with both foot positions, there is still only one skid patch.

There are 2^n possible permutations of this group through rotation.
Why is this?


Since we are now considering *two* possible crank positions, we only have to consider the identity, R_0, and the group generated by a half-rotation of the crank, <R_{m/n**>, where m = the number of teeth in the chainring. From Burnside's theorem, we see that there should be at least 1/(2^n)*(1 + 2^n) patches, which does not equal a strict doubling.
I'm not familiar with Burnside's Theorem and didn't find much through Google, but how does that equation really tell us anything?

[1/(2^n)]*(1+2^n) ~= 1

I like your counter-example though.