A few weeks ago, on a regular Saturday ride we were talking crap as always and I said something along the lines of:

” blah blah f@#$ed blah coffee blah blah 17 tooth cog, which is great for skid patches.”

Des immediately stopped me with: “Why is a 17t good for skid patches?” My response was pretty much equivalent to “er – cos, um… 17 patches”. Naturally, he was not convinced, so I showed him. We flipped the bike, chose a reference and then started turning the cranks. Luckily, there were indeed 17 distinct places the wheel would rotate to, which for someone skidding with the same foot forward would result in 17 skid patches. Des then said something about having a 20t cog and therefore it being just as good, if not better, to which I replied something like “no, because like… gear ratio, something you know?”. But when he asked again for an explanation, the best I could do was a change of topic. We were out riding, and I felt pretty ashamed…

So why do you get skid patches? Why is it that some gears are better than others for skid patches? And why is it that you get a certain number for a given chainring/cog combo? Why is it that sometimes the number of patches is not simply the number of teeth on the rear cog? And how do you know ahead of time (ie. without flipping and turning) which combos are good and bad?

If you don’t really care about all the whys – you can just skip to the tables (note some people call them ‘charts’) published on Bike Forums and friends:

And choose a combo with a higher number of patches. But as good as those tables are, and as thoroughly distracting as the logical proofs are.. they really don’t explain why. Even Sheldon Brown’s glossary entry is less than satisfying on the why:

Simplify the gear ratio to the smallest equivalent whole number ratio. The denominator of the resulting fraction is the number of skid patches you will have on your rear tire.

Fraction? Denominator? Simplify? That’s all annoying maths isn’t it? Yes, and a somewhat distracting smokescreen. But why?

The following text is an attempt to answer in terms we all understand why it is that you get skid patches, and why the number is not always the same as the number of teeth on your cog. Where possible I’ve tried to eliminate mathematics, or rely on hidden assumptions. If it’s too simple please just ignore it and simplify your fractions. If I’m wrong, and that’s completely possible (just ask Blakey), please tell me and I’ll retract in a sec.

Disclaimer:

We all use cogs with more than 10 teeth. If we skid, it’s never precisely in the same crank position. Most of us will move the rear wheel every now and then to clean the chain, repair a flat or change gear. So: All this skid patch crap is irrelevant. However, it is good to know that it is possible to have a particularly bad combination, that over the course of a day might lead to a single worn out patch on a tyre. So please forget everything you know and read on…

Basics

On a fixed gear bicycle, and for the purpose of this (long-winded) explanation, the drivetrain consists of three major elements:

1. Chainring
2. Chain
3. Cog

The cranks are fixed to the chainring, and for normal forward pedaling, they drive the rear wheel by rotating the chainring, which pulls on the chain. The chain advances, and pulls on a cog affixed to the rear wheel, hence turning the rear wheel.

Teeth on the chainring and cog slot into matching holes in the chain. Since it is not desirable (or practical for most purposes) for a chain to engage with a partial tooth, cogs and chainrings have a whole number of teeth. For example, 20, 21, 22, etc. Never half or partial counts like 34.6 or ten and a half. While this is a blindingly obvious statement to most people, it is important to remember, because it means that every full rotation of the driving cranks will advance the chain by exactly the number of teeth on the chainring.

If you were to mark the chainring at the top, cut the chain at that link and then rotate the crank until the mark was at the original position, the number of ‘teeth’ of chain that would have been advanced past the mark would be the same as the number of teeth on the chainring.

Let’s assume that when skidding you always have the same foot forward. This means that the chainring will always have done a complete revolution when the skid happens. So some whole multiple of that many ‘teeth’ of chain will have been pulled over the rear sprocket. For the purpose of this explanation, let’s assume you are (say) going down a very long hill and do a little skid on each pedal revolution, what happens at the rear sprocket?

The cog, like the chainring has a whole number of teeth. Again, this is blindingly obvious, BUT important because since the chainring is in the same place for each skid – it means that no matter what your gear ratio is, the rear wheel can only possibly ever be in as many positions as there are teeth on the cog. The chainring is in the same position, the chain is the same length, the cog has teeth that slot into the chain, so only one of those teeth can be the ‘topmost’.

Imagine a 5 tooth cog, for the same chain position, it can only ever be in 5 rotational positions, so there are only 5 possible wheel positions. This means only 5 possible places to skid, and for certain chainring sizes, 5 skid spots will end up on the tyre. In the diagram below, these are marked on the imaginary tyre 1 to 5.

That is why you get skid spots: The chain is, for the same crank position, in the same spot, so there are only as many possible positions for the rear wheel as there are teeth on the cog.

But now imagine you have a 10 tooth chainring, that means for each pedal revolution, 10 teeth of chain will be advanced. If the rear wheel started in position 1, then it would rotate once for the first 5 teeth of chain, and then once again for the next five, stopping back at position one. As you skid down the hill, each skid would be in the same place on the tyre! Even though there were 5 possible positions, the chain only advanced in whole lots of five (ten fits two fives in perfectly , only one of the five possible positions is ever reached.

So that’s it! The reason the number of skid patches is not necessarily equal to the number of teeth on your cog is that, depending on the number of teeth on the chainring, some of the possible positions for the rear wheel are never reached. They are never reached because the number of teeth on the cog divides ‘neatly’ into the number of teeth on the chainring. Just how ‘neatly’ it divides up will give you an indication of the number of skid patches.

Practical Examples:

Before looking at how to determine the number of skid spots for a chosen chainring/cog combo, let’s look at some practical combinations and how ‘neatly’ they work out.

A fairly typical street gear could be 42:16, that’s 42 teeth on the chainring and 16 on the cog. One full revolution of the cranks advances the chain 42 teeth and there are 16 possible positions for the rear wheel. Number all 16 possible wheel positions 1 – 16, now what happens on that long descent skidding each pedal revolution? For simplicity let’s assume the wheel skids first in position 1. On the next skid, 42 teeth of chain will have been advanced. So that means the wheel will do two full revolutions and then a further 10 teeth worth of revolution counting around the cog (16 + 16 + 10 = 42), taking it to position 11. On the next skid, it will do two full revolutions back to position 11, then ten more teeth, stopping at position 5. On the next skid it will rotate twice to position 5, then ten teeth to position 15…

• pos 15 + two rotations + ten teeth = pos 9
• pos 9 + two rotations + ten teeth = pos 3
• pos 3 + two rotations + ten teeth = pos 13
• pos 13 + two rotations + ten teeth = pos 7
• pos 7 + two rotations + ten teeth = pos 1

This sequence of positions repeats after each 8 steps:

1, 11, 5, 15, 9, 3, 13, 7, 1, 11, 5, 15, 9, 3, 13, 7, 1, …

This gear combination has 8 skid spots. What if a 17t cog was used instead? In this case each skid rotates the wheel twice and then 8 teeth (17 + 17 + 8 = 42). In this case the wheel position sequence goes:

1, 9, 17, 8, 16, 7, 15, 6, 14, 5, 13, 4, 12, 3, 11, 2, 10, 1, 9, 17, …

The sequence repeats after 17 steps – that’s 17 skid patches.

Interesting? Maybe. Tedious? Most definitely! About now I’d recommend flipping your bike and watching the sequence for yourself. Oh, and have a coffee, LSD, or ‘chai-tea’ while you’re at it

Working It Out:

Turn your gear ratio (eg 48:20) into a fraction (eg 48/20), and (take it away Sheldon) …

Simplify the gear ratio to the smallest equivalent whole number ratio. The denominator of the resulting fraction is the number of skid patches you will have on your rear tire.

For example: 48:20, that’s the fraction 48/20, which simplifies to 12/5 – five skid patches. Or, if you use a fancy calculator, matlab, excel, 123 or etc, divide the number of teeth on the cog by the greatest common divisor of the teeth on the chainring and cog:

ss = R / gcd (F, R)

Where ss is the number of skid spots, F is the teeth on the chainring and R is the teeth on the cog.

Skidding With Both Legs

“So, if I skid with both legs, do I get twice as many skid spots?”

Short answer: maybe. In this case you are now advancing the cranks half turns when skidding. Assuming the long downhill short skids again, that means you advance the chain half as many links per skid than before. If the chainring has an even number of teeth, the number of skid spots can be worked out as above, just half the teeth on the chainring first.

For example, the 42:16 becomes 21:16. gcd(21, 16) is 1, so there are 16 skid spots, double the amount for one-legged skid. For the ratio 48:20, it becomes 24:20, gcd(24, 20) is 4 so there are still 5 spots when skidding two-legged.

If the chainring has an odd number of teeth, just double the number of spots worked out for one-legged skids.

Note: this does not agree with Sheldon’s method. See the skid patch theorem post for more.

Not Working It Out:

Ride your bike and clean your chain. Change your tyre when it gets too worn anywhere. That should do the trick. If you are worried, look it up in a table.

ndf 26/11/2006