At what cable lengths are termination resistors required for RS-485 networks?
In general, for short cables (< 20-30m) and low baudrates (< 115200) you can leave them out without much trouble. But:
It is useful to put some kind of load on the signal lines to improve noise immunity (the RS485 driver will supply enough current to switch the voltage on the differential line, many noise sources will not). But you do not need this load to be equal to any "characteristic impedances", \$200-500\,\Omega\$ will be ok.
When you go for high speed or long cabling you will need proper termination that depends on the cable you use. So this should be \$100\,\Omega\$ for Cat 5 cables (not \$120\,\Omega\$).
Don't forget about the pull-up and pull-down resistors. They are required unless all receivers used in the system give a well defined (high-level) output for \$0\,\mathrm{V}\$ input. Their values should be chosen so that (when connected together with the "terminating" resistors) the un-driven line is properly polarized (\$> 0.3\,\mathrm{V}\$ for most receivers)
All RS-485 cables require termination. Some may just happen to work without them, but all should have them.
Since terminating resistors load down the network, they should not be used unless they are required. Since reflected waves will dampen in 3-4 cycles, if the time for this to occur is less than one data bit width (or one half the bit width if sampling in the middle), the reflected waves will not interfere and terminating resistors are not required.
It is a simple enough calculation, figuring on the propagation velocity averaging around 65% of the speed of light: For a 9600 bps communication rate, on a 1000 foot cable, you have a round trip time of 3 usec, a dampening time between 9-12 usec, and a bit width of 10 msec. Therefore, each reflected wave will dampen out prior to you sampling each bit, so termination resistors are not required.