I would look at the methanol as a percentage of the total fuel. The total fuel would be 196 from meth + 378 from gas = 574 lb/hr. So the meth % would be 196/574 = 34%. That would be the meth% to use for a/f calcs if all the meth injected were burned in the combustion process. But as you alluded to this still seems like too high a %. The corresponding safe a/f ratios would be much lower than what we know works.
I did some quick calcs for the amount of fuel I'm injecting and what I know some of my friends are injecting for low 10sec cars. Using some calculators to guess what fuel the engine needs for those Hp levels I come out with about 22% more fuel needed than what we are injecting. So if we use 22% meth for a dual nozzle setup the equivalent a/f ratios correspond pretty well to what a/f ratios we know works at those power levels (10.3-10.4 or so).
So we are injecting 34% meth but only burning 22% in this case it seems. So what happens to the unaccounted for meth? Maybe even though that extra meth doesn't get burned its still needed to run the higher boost/power levels on pump gas as it cools down the cylinders and prevents detonation.
(Here's where I got my numbers from:
For my low 10sec car, I run around 274lb/hr of total fuel injector when on alky (~55% duty cycle on 83lb injectors). Now, if I use the handy TRX Performance calculator, it says I should be using a total of 350 lb/hr of injector to support my horsepower. I'm missing 76lb/hr, about 22%. If that 76 lb/hr is coming from meth, that number seems more along the lines of the % of meth we are burning to use in the a/f calculations. The corresponding a/f ratios for the 22% meth seem to follow what people have run/recommended for low 10 sec alky cars (say 10.3 for a 11.76 gas a/f ratio). For a high 9 sec car you might want to run closer to an 11.5 gas a/f ratio which would give you an corrected 10.1 a/f ratio for 22% meth.)