Shants almost got it right, but here's how skidpatches (SP) are determined. If the number of cog teeth (RT) divide evenly into the number of chainwheel teeth (FT), say 51 x 17, we all agree that there is 1 and only 1 SP. If the remainder (R) of FT / RT is 1, say 52 x 17, then you go an extra 1 / RT revolution of the back wheel with each revolution of the chainwheel. So, after RT revolutions of the chainwheel you are back where you started and that means you have RT SPs, in our case 17 SPs. Things get more complicated when R is greater than 1, however. Consider 52 x 14, in which case R = 10. Every revolution of the chainwheel gives 3 of the cog plus an extra 10 cog teeth. Now 7 revs up front produces 21 in back (7 x 3) plus an extra 70 (7 x 10) cog teeth. But 70 teeth on a 14 toothed cog is 5 complete revolutions, so we are full cycle (no pun intended) with an SP of 7, only half as many as you might expect from a 14 toothed cog. The rule for remaindered ratios is:

SP = RT / gcd(RT, R) (and this is where gorn's chart came from)

where gcd is the greatest common divisor (think 8th or 9th grade here). For the 52 x 17 case we get SP = 17 / gcd(17, 1) = 17 / 1 = 17, so that works. For 52 x 14 we get SP = 14 / gcd(14, 10) = 14 / 2 = 7, and that's good, too.

So how do you maximize SPs. It's simple. Use a cog with a lot of teeth and one that yields a remainder of 1 (which always makes gcd = 1). Another way is to use a cog with a prime number of teeth, like 17, in which case gcd always = 1 for any remainder.