Tuesday, December 14, 2010

Computing ABV

A recent question on http://homebrew.stackexchange.com/ asked for a simple formula for calculating the percentage alcohol by volume (ABV) of a beer given the original gravity (OG) and final gravity (FG). Here was the complicated formula which was to be simplified:

$ABV=\frac{76.08\cdot(OG-FG)}{1.775-OG}\cdot\frac{FG}{0.794}$

OK, so it's not that complicated, but a linearization would be good. The original post suggested a range of 1.035-1.095 for OG and 1.002-1.028 for FG. I chose the midpoints of these intervals to get the point (1.065,1.015) and linearized the given formula at this point. Here's what I got:

$ABV \approx -17.1225210+146.6266588\cdot OG-130.2323766\cdot FG$

It's simpler, but not easy to remember. If we round the coefficients, we get an easier to handle expression.

$ABV \approx 147\cdot OG - 130\cdot FG -17$

Analysis using Maple shows that this simplified linear approximation has an error of at most 0.78 from the original equation. This is not too hard to remember, since 147-130=17. Also, if both OG and FG are equal to one (the specific gravity of water), then the formula predicts 0% ABV.

1 comment:

1. Way to discover a simpler equation. But! Why no parenthesis?