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

**Subject:**Re: HBUs and IBUs???

**Date:**1992-04-23 16:38:43 GMT

`Funny you should ask. I dredged the collective knowledge of the HBD on this
very subject a while ago. Following are my responses.
So with 35 IBUs = 7.8 - 8.9 AA pellets
(8.2 - 9.5 AA whole leaf) or about 1 oz.
If your not doing a full boil (or if the recipe was formulated that way)
you'll have a higher gravity in the boil and utilization will go down.
(If you boil 2.5 gallon and the 5gal have a 1.045 G then the boil has
a gravity of 1.090. This will cause hop utilization to go down by 20%,
and you'll need around 10 AAUs to get 35-40 IBUs) Confused? I know I am? :-)
________________________________________________________________________
To: whg%tellabf.tellabs.com@juts.ccc.amdahl.com, *@amail.amdahl.com
Status: RO
IBU = HBU * (%utilization / (gallons * 1.34))
One number I remember for utilization is 30%, for 60 minute boils of standard
(1.040) worts. In that case, then:
IBU = HBU * (30 / (gallons * 1.34))
archive site) gives 30% for pellet and 28% for leaf for a 60 minute boil in a
>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)
_______________________________________________________
From: Frank Tutzauer <uunet!ubvms.cc.buffalo.edu!COMFRANK>
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
`

The posts that comprise the Homebrew Digest Searchable Archive remain the
property of their authors.

This search system is copyright © 2008
Scott Alfter; all rights reserved.