BostonFixed

Headache inducing...

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

dolface

that rocked! hooray for math-geek cyclists!

ink1373

43 teeth here i come!

TimArchy

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

adamkell

nice use of applied math with the kiddies

TimArchy

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

HereNT

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...

dolface

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>

HereNT

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...

BlastRadius

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

TimArchy

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

gh-ap

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

dolface

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

gh-ap

hooray!

ephemeralskin

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..

Jumbo

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

Gravityhatesme

*drool*

TimArchy

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

bostontrevor

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.

dolface

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

TimArchy

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

BlastRadius

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

dolface

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

BlastRadius

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

ryan_c

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.