I realized that when I posted about calculating extract yield, that I forgot to explain how I calculated brewhouse effieiency. I use the same formula

S.G. = ((((DBCG – Mousture Content – 0.002) * Brewhouse Efficiency * 46.214) * 0.001) +1)

Except that I remove the variable for brewhouse efficiency. The forumla will then assume that the mashing efficiency is at 100%

S.G. = ((((DBCG – MC – 0.002) * 46.214) * 0.001) +1)

This will give us the theoretical maximum ppg of 1 lb of grain in 1 gallon of water. For example, Maris Otter Pale Malt.

S.G. = ((((0.805 – 0.03 – 0.002) * 46.214) * 0.001) +1)

S.G. = 1.035723422 or 35.7ppg

In my bitter 2 recipe I mashed 3 lbs. of Maris Otter Pale Malt, and 1 lb of Briess 10 L crystal malt collecting a final 2.5 gallons of wort at a S.G of 1.043 .

The theoretical maximum ppg of 10 L crystal malt is

S.G. = ((((0.75 – 0.07 – 0.002) * 46.214) * 0.001) +1)

S.G = 1.031333092 or 31.3 ppg

Now, using the formula from John Palmer’s How To Brew, we calculate the maximum theoretical points for our brew using the following formula

((ppg Malt1 x lbs Malt1) / gallons of wort) + ((ppg Malt2 x lbs Malt2) / gallons of wort) = Total Points

((35.7 x 3)/2.5) + ((31.3 x 1)/2.5) = 55.36

This is our theoretical maximum number of points per gallon.

In the wort for the Bitter 2 recipe the S.G was 1.043 or 43 ppg. If we divide our real ppg by the theoretical ppg this will yeild our percentage yeild.

43/55 x 100 = 77.75%