From the HBD Archive
Subject: formula for calories
Date: 1992-06-01 20:04:02 GMT

for the caloric content of a 12 oz. bottle of beer in a form which
is _linear_ with respect to starting and ending gravity.

Of course, George's fromula can be solved in terms of OG and FG only,
or in terms of OE and AE only. The answer isn't linear, and no amount
of wishing will make it so, especially as there is a term involving OE
which is the denominator of a rational function.

On the other hand "every differentiable function is locally linear".
This was Newton's great insight (although Archimedes, Gallileo and
Pascal, among others, probably had a gut feeling to this effect).
If it were not for this important principle, there would be no such
thing as an economist. Or at least one who is earning a salary :-)

So....assuming that your original gravity is in the range 1.036 - 1.060,
the following is a good approximnation to Goerge's Law:

calories/12 oz. = FG[12.876*OE + 1.324*AE - 1.42]

This still doesn't satisfy Dan's wish for a linear function.
However, within reasonable limits (FG in the range 1.005 - 1.015),
we can drop the multiplier and constant term to get

calories/12 oz. = 12.876*OE + 1.324*AE
= 3219*OG + 331*FG - 3550 (*)

where OE = hydrometer reading before fermentation, degrees Plato
AE = hydrometer reading after fermentation, degrees Plato
OG = hydrometer reading before fermentation, specific gravity
FG = hydrometer reading after fermentation, specific gravity

With the given example OG=1.045, OE=11.25
FG=1.010, AE= 2.5,

We hav calories = 148.165 in either case. Of course, we have no reason in
the world to trust those final digits. 148 calories is probably even
more accuracy than we're entitled to (this is not to casr aspersions
on the accuracy of George's coefficients, rather a reflection of the
fact that we're approximating a rational function by a polynomial).

For barley wines or ultra-light brews, different fudge factors would
be needed, although the numbers won't change too much.

It's interesting to note that original gravity tells almost the whole
story when it comes to calories.

Starting Gravity 1.050 Final Gravity Calories in 12 oz.
1.030 (heavy!) 171
1.020 168
1.010 164
1.005 163
1.000 (yuck!) 161

Moral, when a yeast eats a sugar, it doesn't use much of the stored
energy.

Now a question:
- --------------
a number of sources (incl. TCJOHB and George Fix's posting) give
or use the formula

degrees Plato = (fractional part of gravity * 1000) / 4

(I used it too, to derive formula (*) from the one above it.)
Nevertheless, every hydrometer I've ever seen contradicts this.
1.047 appears to be 12 degrees Plato or even 12 1/4 instead
of the predicted 11.75.

SO.....are the hydrometers off, or is the formula above just a
rough and ready approximation? Just curious.

Rob