
10282009, 05:34 PM

#1

Feb 2008
Evanston, IL
Posts: 1,083
Liked 25 Times on 22 Posts

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^(SG1) * AA * 1000 / (4.15*V)
IBU_c = k*h*[ 1  1/(0.04*T) * (1exp(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.



10282009, 05:43 PM

#2

Mar 2009
Allentown, PA
Posts: 178
Liked 2 Times on 2 Posts

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...



10292009, 06:07 AM

#3

Feb 2008
Chicago, Illinois
Posts: 103
Liked 8 Times on 4 Posts

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.



10292009, 07:33 AM

#4

Sponsor
Jul 2006
Nor*Cal
Posts: 7,976
Liked 1113 Times on 778 Posts

Living up to your name!
Well done.



07252010, 08:00 PM

#5

May 2009
Aurora, CO
Posts: 524
Liked 16 Times on 15 Posts

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?



08012010, 03:54 PM

#6

Feb 2008
Evanston, IL
Posts: 1,083
Liked 25 Times on 22 Posts

Quote:
Originally Posted by cruelkix
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?

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



08012010, 04:15 PM

#7

May 2009
Aurora, CO
Posts: 524
Liked 16 Times on 15 Posts

Quote:
Originally Posted by rocketman768
Your k is right. You performed this calculation:
0.934600023*30*[1  1/(0.04*60)]*(1exp(0.04*60))
instead of
0.934600023*30*[1  1/(0.04*60)*(1exp(0.04*60)) ]

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





