Hey,
I calculated the continous hopping formula if anybody wants it. It just turned out to be simple Riemann integrals. This is based on Tinseth's approximation.
k := 1.65*0.000125^(SG-1) * AA * 1000 / (4.15*V)
IBU_c = k*h*[ 1 - 1/(0.04*T) * (1-exp(-0.04*T)) ]
Where SG is in specific gravity, AA is the decimal alpha acid rating (i.e. 0.07 for 7%), V is the final volume of the wort in liters, h is the amount of hops in grams, and T is the length of time over which you are continuously hopping. Also, exp() is the exponential function with base e (Euler's constant). This formula assumes you hop from time T to time 0.
As an example, suppose h = 30g, SG = 1.050, AA = 7%, V = 19L, and T = 60 min
If I added them all at once, I would get 25.49 IBUs according to Tinseth. If I add them continuously over 60 minutes, using the above formula I would get 17.41 IBUs.
Someone else can do the calculus, but I'm pretty sure this is it.
I calculated the continous hopping formula if anybody wants it. It just turned out to be simple Riemann integrals. This is based on Tinseth's approximation.
k := 1.65*0.000125^(SG-1) * AA * 1000 / (4.15*V)
IBU_c = k*h*[ 1 - 1/(0.04*T) * (1-exp(-0.04*T)) ]
Where SG is in specific gravity, AA is the decimal alpha acid rating (i.e. 0.07 for 7%), V is the final volume of the wort in liters, h is the amount of hops in grams, and T is the length of time over which you are continuously hopping. Also, exp() is the exponential function with base e (Euler's constant). This formula assumes you hop from time T to time 0.
As an example, suppose h = 30g, SG = 1.050, AA = 7%, V = 19L, and T = 60 min
If I added them all at once, I would get 25.49 IBUs according to Tinseth. If I add them continuously over 60 minutes, using the above formula I would get 17.41 IBUs.
Someone else can do the calculus, but I'm pretty sure this is it.