Originally Posted by 80sMutRider
I figured, but that does mean that I have to ride the evil bike tomorrow to my LBS.
With the skid spot thing:
48/18 = 24/9 = 8/3 = 2.6666666666
Just wondering though, if I divide 48/18 I get a number with a decimal, doesn't that mean that the wheel goes around a non-even amount, making the spots where you stop pedaling slightly different each time (in my case it goes an extra two-thirds around each time). I don't know much about it, but I may just be drawing completely incorrect conclusions.
Yeah, your math is right, but its that "slightly different each time" part that gets you. In your case, there are only 3 spots where that decimal amount can take you.
First, the math for calculating skid patches:
Take your two numbers and find the largest common factor (in this case its 6. 6 is the largest number that both 48 and 18 are divisible by). Then take your cog size and divide by 6. 18 divided by 6 is 3. That is how many skid patches you have. (this is assuming you skid with the same leg forward each time. double that if you skid with either leg forward).
The idea is is that every time you turn the front crank one full rotation, it spins the back cog. when you go 48 teeth around in front, it spins the back one around 48 teeth as well, in this case that is 2 2/3 times (as your math showed). 18 + 18 + 12 (12 being 2/3 of 18). its that 12 you should be concerned about.
If you look at your cog and just pick a tooth and start counting out those 12, you'll hit the spot where your cog will be after one rotation of the chainwheel. Rotate your chain wheel again, and you will go around 2 more times exactly + another 12, so count out another 12. Repeat and repeat. You will see that there are only 3 spots that the number 12 will fall on. That means there are only three spots where your cog is always stopping, hence 3 skid patches.
My awesome drawing may help, or may confuse the hell out of you.
