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Old 09-07-2007, 02:59 AM   #1
ohiodad
 
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What is the calculation to caculate this?

 
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Old 09-07-2007, 04:16 AM   #2
Iordz
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(OG-FG)/(OG)=%A
I have a beer that started at 1.040, ended at 1.010, that is the attenuation?
(40-10)/(40)=75%A

 
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Old 09-07-2007, 03:37 PM   #3
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Quote:
Originally Posted by Iordz
(OG-FG)/(OG)=%A
I have a beer that started at 1.040, ended at 1.010, that is the attenuation?
(40-10)/(40)=75%A
To clarify:

(OG-FG)/(OG) * 100 = %A

 
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Old 09-07-2007, 04:22 PM   #4
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Quote:
Originally Posted by Got Trub?
To clarify:

(OG-FG)/(OG) * 100 = %A
Thanks a bunch.. I thought that was it.. Doesn't work out the same though if you put the 1. in front of the og and fg.. Doh! Thanks again!

 
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Old 09-07-2007, 05:18 PM   #5
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isn't that apparent attenuation?
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Old 09-08-2007, 02:03 AM   #6
Iordz
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Sorry, I was too vague. You multiply by 100, I just thought it was pretty much a given. I believe this is actual attenuation, it makes sense, you are calculating how much the beer, actually, attenuated.

 
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Old 09-08-2007, 04:10 PM   #7
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DB is correct - the formula is for apparent attenuation. That is what we measure with our hydrometer. As alcohol is less dense then water and our fermenting/ed beer has alcohol in it it will throw off your hydrometer reading. While of theoretical importance actual attenuation has little practical use in homebrewing as all attenuation numbers we use are of the apparent variety.

 
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