As Paul pointed out, you really want to compare the attenuation of the two channels in dB. I assume you measured resistance at the center lug. The formula for attenuation of a channel would be 20*log(R/R100%) - e.g. 20*log(12.9/96.5).
If the resistance at 100% is the same, the difference in attenuation between the two channels can be calculated directly using the formula 20*log(RL/RR) – e.g. 20*log(12.9/12.6). The louder channel will have the higher resistance.
Paul, please correct me if I got it wrong.
Doc has pointed out that most people will have little trouble hearing a 1 dB difference in volume. at 0.5 dB it gets harder. Here is a website where you can here different dB shifts in volume for a test tone:
http://www.audiocheck.net/blindtests_index.phpThe above will work for most pots, I've read that the TKD 2511 is an exception. A more accurate equation would be 20*(log (Rout to grnd/(Rin to Rout + Rout togrnd)). But I am not sure if this works for the TKD pot. The sure fire way is to play a 150hz or thereabouts test tone and measure input and output AC voltage; using the equation 20*log(Vout/Vin) to calculate attenuation for each channel.