**From:**whg@sunFb.tellabs.com (Walter H. Gude)

**Subject:**Hop Utilization

**Date:**1992-03-16 17:14:31 GMT

`What follows is a summary of responses I received to my original hop
utilization questions. There seems to be enough interest so I will post
to the digest. I have edited responses down to a minimum. Apologies in
advance if I quote anyone out of context.
P.S. I hope no one who responded to me minds my posting of their response.
Let me know if this is not proper net etiquette.
>From: joshua.grosse@amail.amdahl.com
>
> IBU = HBU * (%utilization / (gallons * 1.34))
>archive site) gives 30% for pellet and 28% for leaf for a 60 minute boil in a
>
############################################################################
And from:
>From srussell@snoopy.msc.cornell.edu Thu Mar 12 11:25:17 1992
>
>Jackie Rager's article in the Hops special issue says to divide the
>factor you would get w/o considering gravity by a correction factor of:
>
>1 + 5(G-1.050)
>
>for G > 1.050 (and leave it at 1 for G < 1.050)
>
>
>So, if you added 10 AAUs, w/ 30% utilization that's 46 IBUs, except that
>you have a 1.090 gravity wort, so you really have 46/[1+5(1.090-1.050)] = 23
>IBUs, a considerable difference!!
Actually ---> = 38
>
>I don't trust his figures on utilization (I use Burch's), but the time range
>where the two diverge is 30 to 50 minutes, and hops need not be added in
>there anyhow. Burch says utilization increases linearly from 0 min to 30
>min (from 5% to 12%) then jumps, attaining 29% at 60 min. I suspect that
>differences are the result of boil vigor, wort pH, moon phase, etc, etc.
>and either boil for < 30 min or a full 60.
##########################################################################
Finally for those who want to be REALLY accurate:
>From: Frank Tutzauer <uunet!ubvms.cc.buffalo.edu!COMFRANK>
>
>First, there is a table in the new edition of Papazian that gives percent
>utilization as a function of gravity (of the boil) and time. If I can find it
>on disk, I'll send it to you.
>
>Then, some guy named Tim (I would have to look up his last name--anyway, he's
>a statistician) sent me a formula that explains 99 percent of the variance in
>the table. It's still a linear model, but he transformed some of the
>variables. The formula is:
>
>U = exp{-23.63 + .12896*t + 37.76*s - .00068496*t^2 - 18.01*s^2 -.04187*t*s}
>
>where U, t, and s are of course utilization, time, and s.g, and where exp
>means raise e to the bracketted power (e = 2.7182...). Now, many other things
>besides time and s.g. affect the utilization--things like kettle geometry and
>vigor of the boil--but if you keep these things constant, then the formula
>should give you a good means of altering your utilization.
`

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