Subject: The Gravity of the Situation
Date: 1989-08-16 13:09:00 GMT
Doug Roberts writes about his sweet stout recipe -
> The start S.G. was 1.057, which translates to a potential alcohol of 7.8
> percent. The end S.G. was 1.022 prior to kegging, six weeks after the boil.
> The 1.022 S.G. meant a residual of 3.0%, for an alcohol content of 4.8%.
Can someone help me out with the math here? Working backwards from these
numbers, I assume that potential alcohol is the starting (original) gravity
divided by 7.31, or more precisely, potential alcohol = (OG-1)*1000/7.31.
Residual is final gravity divided by 7.31, or (FG-1)*1000/7.31. Alcohol
content is the potential alcohol minus the residual, giving
(OG-FG)*1000/7.31. Is this by weight, or by volume?
Alcohol by volume gives a higher figure than alcohol by weight, because
alcohol is lighter than water. %-by volume is around 20% higher than
Doug's technique seems reasonable, but the resulting numbers seem high. As
an example, Budweiser has a starting gravity around 1.045, and a final
gravity around 1.005. Using Doug's numbers, Budweiser has an alcohol
content of 5.5%. I've always thought that Bud was around 4.0 by weight,
4.8 by volume.
Something is wrong here. Either my numbers are wrong on Budweiser, or my
calculator is broken. I use a similar equation to Doug's, but divide by
10.0, instead of 7.31. I would estimate Doug's recipe as yielding
(57-22)/10 = 3.5 percent by weight. Adding an additional 20%
gives 4.2 percent by volume.
I probably should just use the potential alcohol numbers on my hydrometer
and not worry!
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