Continuous hopping exact formula

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.

Wow, no wonder we homebrewers get the geek label... Seems like a useful formula, but I would have been awfully tempted to just add hops every 5 minutes and use the Tinseth formula programmed into my calculator...

Wow thank you. I tried this once, and failed. For whatever reason, Chem majors don't have to take much calculus. I tried to do this exact thing once, and failed miserably. I really should have just tossed it into mathcad or something and let it do the work. I am positive I made some stupid mistake.

Living up to your name!
Well done.

I used your formula and the constants you used for your example and I got a value of 14.87176 IBUs.

I'm sure I got my order of operations messed up somewhere.

I got 0.9346 for k......

any suggestions?

cruelkix said:

I used your formula and the constants you used for your example and I got a value of 14.87176 IBUs.

I'm sure I got my order of operations messed up somewhere.

I got 0.9346 for k......

any suggestions?

Click to expand...

Your k is right. You performed this calculation:
0.934600023*30*[1 - 1/(0.04*60)]*(1-exp(-0.04*60))

instead of

0.934600023*30*[1 - 1/(0.04*60)*(1-exp(-0.04*60)) ]

rocketman768 said:

Your k is right. You performed this calculation:
0.934600023*30*[1 - 1/(0.04*60)]*(1-exp(-0.04*60))

instead of

0.934600023*30*[1 - 1/(0.04*60)*(1-exp(-0.04*60)) ]

Click to expand...

Gotcha! Thanks. I knew I had the order of operations messed up somewhere!