Thanks for the info.- so doing the math:
2 cups=473 ml = 473 gm water and
70 F to 212 F is the same as 21.1 C to 100 C = 78.89 C degree change.
So 473 gms of water X 78.89 degrees C = 37.314 Kcals required
(Since I don't have a value for "less than 3/4 ounce" I will use 3/4 oz)
3/4 fl oz = 22.1 ml fuel used
methanol's density is 0.7918 gm/ml
So 22.1 ml X 0.7918 gm/ml = 17.5 gm of fuel used
17.5 gm x 4.7 Kcal/gm of methanol = 82.25K Kcal available in the alcohol used.
37.314 Kcals required / 82.25 Kcals used = 0.45 or about 45% efficient.
Which is not going to win any prizes but is fairly respectable.
(For comparison a Caldera Keg system gets closer to 60%. and the most efficient beer can alcohol stove I am aware is about 80% efficient)[NOTE: THIS IS AN ERROR, SEE MY 3RD POST DOWN BELOW}]
Concerning the wind- As you well know, wind has a huge effect on the performance of alcohol stoves. With 45% efficiency and the hard to heat beer can pot, I would think you will need to have a good wind screen before you market this stove. To my knowledge there is no standard for wind in stove testing circles (Is there? You would know better than me), but a light breeze (defined as 3-7 mph or 6-11 km/h on the Beaufort Scale) would be a reasonable design goal IMO.
P.S. - Out of curiosity, with the beer can mod and top jets, what kind of bloom time do you get now? You should have better heat transfer, but then again with the insert you have more thermal mass.