**From:**roberts%studguppy@LANL.GOV (Doug Roberts @ Los Alamos National Laboratory)

**Subject:**The Gravity of the Situation

**Date:**1989-08-18 04:14:18 GMT

`> 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
> %-by-weight.
>
> Doug's technique seems reasonable, but the resulting numbers seem high.
You know, those numbers _do_ seem high. After I read your message, I
went home & looked at my hydrometer a little more closely. Using your
Budweiser numbers, It says a starting SG of 1.045 represents a
potential alcohol of 6.1 percent; SG of 1.005 is 0.6 percent. This
would mean that Budweiser is 5.5 percent. My (cheap little) hydrometer
doesn't say if the alcohol percentages are by weight or volume.
However, since the specific gravity of a liquid is defined as the
ratio of it's density with respect to pure water, I would think that
the alcohol percentages would be weight percents.
BTW: where does your magic 7.31 number come from?
>
> I probably should just use the potential alcohol numbers on my hydrometer
> and not worry!
>
That's what I was doing, but now I'm a little suspicious of them :-).
--Doug
`

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