I'm trying to get a handle on ibu calculations. The Palmer book has given me these solid estimation equations which consider many variables so I like them.
AAU= aa% * oz
IBU = (AAU * u * 75) / 5 (final volume in gallons)
u = f(g) * f(t) (where g is gravity and t is time)
f(g) = 1.65 * 0.000125^(g - 1)
f(t) = (1 - e^(-.04) * t) / 4.15
Now to do this right I need to know g (the gravity of the boil... I assume a the time of hop addition) and t. The t is easy... it's just the time the hop spends boiling.
The g is tough because it is a function of volume (which changes through out the boil for each hop addition). My boil volume can be estimated by f(t) = (7/240)t + 5. Assuming evaporation is linear. And I'd bet it Isn't.
Now if I used 12 lbs of grain and achieved about a 1.052 SG reading in 5 gallons...
What is my gravity at any point in the boil?
I came up with .438 points per lb/gal. But that can't be right because my sg at 6.75 gallons would be about 0.779 which is less dense than water. Let's assume that heat doesn't factor into it.
Thanks guys!
AAU= aa% * oz
IBU = (AAU * u * 75) / 5 (final volume in gallons)
u = f(g) * f(t) (where g is gravity and t is time)
f(g) = 1.65 * 0.000125^(g - 1)
f(t) = (1 - e^(-.04) * t) / 4.15
Now to do this right I need to know g (the gravity of the boil... I assume a the time of hop addition) and t. The t is easy... it's just the time the hop spends boiling.
The g is tough because it is a function of volume (which changes through out the boil for each hop addition). My boil volume can be estimated by f(t) = (7/240)t + 5. Assuming evaporation is linear. And I'd bet it Isn't.
Now if I used 12 lbs of grain and achieved about a 1.052 SG reading in 5 gallons...
What is my gravity at any point in the boil?
I came up with .438 points per lb/gal. But that can't be right because my sg at 6.75 gallons would be about 0.779 which is less dense than water. Let's assume that heat doesn't factor into it.
Thanks guys!