MMW
Well-Known Member
It's a bit of a slow day at work and I got to thinking about the IBU contribution of continual hopping. I've seen brewsmith style printouts where the hops are divvied up in 10 or so increments over an interval, but thought there had to be a cleaner method.
****WARNING: MATH FOLLOWS*******
Let's think this through:
Tinseth's hop utilization formula looks like this:
i(t) = K*W*u(t)
where:
K is a constant in respect to time (it is a function of sg and alpha acid rating of the hops)
W is the weight of hops.
u(t) = 1-exp(-0.04*t)
Let's define the equation for continual hopping:
i(t) = K*w(t)*u(t)
The hop addition now comes over an interval from time T to flameout, so w(t) is W/T during that interval and zero elsewhere.
This function gives the bittering contribution at the specific time t. We'll need to add the contributions across all of the time together to get the total contribution- we will simply integrate the function over the time w(t) is non-zero:
I(T) = int( K*w/T*u(t)*dt) over 0..T <-- this is the total bittering contribution of the continual hopped processes
most of that comes out, leaving us only needing to integrate u(t)
that is: t + 25*(exp(-0.04t)) eval over 0..T--> (25 = 1/0.04)
so...
T + 25*(exp(-0.04T)) - 25 = T + 25*(exp(-0.04T) - 1)
that exp(-0.04T) - 1 term is interesting...it is -u(T)!
I(T) = K*T*W/T - 25*K*W/T*u(T)
the second term is easy...it is the 25 times the time T addition of W/T of hops in the standard Tinseth formula.
The first term is the 'perfect' utilization of W oz. (or grams if you're metric) of hops.
We can 'cheat' brewing software by picking the time when we extract half the alpha acids from the hops. Solve this equation for t: 0.5 = exp(-0.04*t). (The answer is 17.33) Alternatively, you could calculate a very early hop addition like 1000 minutes or so.
This leaves us with the IBU contribution of continual hopping as:
IBU = 2*i(17.33) - 25*i(T) using W as the weight of hops for the first term and W/T as the weight of the hops in the second.
Like I said...bored at work :fro:
****WARNING: MATH FOLLOWS*******
Let's think this through:
Tinseth's hop utilization formula looks like this:
i(t) = K*W*u(t)
where:
K is a constant in respect to time (it is a function of sg and alpha acid rating of the hops)
W is the weight of hops.
u(t) = 1-exp(-0.04*t)
Let's define the equation for continual hopping:
i(t) = K*w(t)*u(t)
The hop addition now comes over an interval from time T to flameout, so w(t) is W/T during that interval and zero elsewhere.
This function gives the bittering contribution at the specific time t. We'll need to add the contributions across all of the time together to get the total contribution- we will simply integrate the function over the time w(t) is non-zero:
I(T) = int( K*w/T*u(t)*dt) over 0..T <-- this is the total bittering contribution of the continual hopped processes
most of that comes out, leaving us only needing to integrate u(t)
that is: t + 25*(exp(-0.04t)) eval over 0..T--> (25 = 1/0.04)
so...
T + 25*(exp(-0.04T)) - 25 = T + 25*(exp(-0.04T) - 1)
that exp(-0.04T) - 1 term is interesting...it is -u(T)!
I(T) = K*T*W/T - 25*K*W/T*u(T)
the second term is easy...it is the 25 times the time T addition of W/T of hops in the standard Tinseth formula.
The first term is the 'perfect' utilization of W oz. (or grams if you're metric) of hops.
We can 'cheat' brewing software by picking the time when we extract half the alpha acids from the hops. Solve this equation for t: 0.5 = exp(-0.04*t). (The answer is 17.33) Alternatively, you could calculate a very early hop addition like 1000 minutes or so.
This leaves us with the IBU contribution of continual hopping as:
IBU = 2*i(17.33) - 25*i(T) using W as the weight of hops for the first term and W/T as the weight of the hops in the second.
Like I said...bored at work :fro:
