**Specific Gravity and Attenuation**

*Specific gravity* (SG) is a measure of the density of a liquid, with respect to
water. Pure water has a specific gravity of 1.000; a specific gravity of 1.050 would mean
that the liquid being tested is 1.050 times as dense as pure water. In brewing, specific
gravity is used to tell us how much dissolved sugar is in our wort.

The *original gravity* (OG) of a wort is the specific gravity reading prior to
fermentation. The *final gravity* (FG) is the specific gravity of the beer after
fermentation. You can calculate the percentage of alcohol in your finished beer by
subtracting the final gravity from the original gravity, and multiplying by 105 to get
percentage alcohol by weight (ABW). Multiply ABW by 1.25 to get alcohol by volume (ABV).

The easiest way to work with specific gravity is by using *points*. Simply drop
the leading "1." from the specific gravity reading, and use the 3 digits after
the decimal - so a specific gravity of 1.050 is equivalent to 50 points.

In home brewing, the amount of sugar in brewing ingredients is usually expressed as *points
per pound per gallon* (PPG). A malt extract which gives you 45 PPG will result in a
wort with a specific gravity of 1.045, when mixed with 1 gallon of water.

**Estimating OG from Ingredients**

To estimate the OG of a recipe, multiply the PPG rating of each ingredient by the number of pounds of that ingredient, and sum the results for each ingredient. Divide this total by the number of gallons of wort you are making (including top-up water, if any). The result will be the estimated OG, in points.

Example:

A 5 gallon recipe calls for 6 lbs of dried malt extract and 1 lb of steeped crystal malt; what will the OG be?

( ( 6 lbs DME x 45 points ) + ( 1 lb of crystal x 32 points x 0.70 efficiency ) ) / 5 gallons = 58

So we would expect the original gravity to be approximately 1.058.

**Estimating Ingredients from OG**

You can also work things backwards, to come up with the amount of extract (or grain) required to hit your desired OG.

Example:

How many pounds of liquid malt extract are required to achieve an OG of 1.050 in a 5 gallon batch of beer, assuming we will also be adding 1/2 lb of roasted malt for flavor?

We need a total of ( 50 points x 5 gallons ) = 250 points, to achieve our desired OG. The roasted malt will give us approximately ( 0.5 lbs x 30 points x 0.70 efficiency ) = 10 points. The remaining 240 points must come from the malt extract.

240 points / 37 PPG = 6.5

So we would need 6.5 lbs of LME.

**Converting Recipes**

You can use the points system to convert recipes between all-grain and malt extract.

Example:

How much dry malt extract should be substituted for 7 lbs of pale 2-row malt?

From the table, DME gives us 45 PPG; assuming 70% efficiency, pale malt gives us approximately ( 37 PPG x 0.70) = 26 PPG. So we need approximately ( 26 / 45 ) = 0.58 lbs of DME for each pound of grain, for a total of ( 0.58 x 7 ) = 4 lbs of DME.

**Attenuation**

*Attenuation* is a measure of how much of the sugar in the wort has been fermented by
the yeast.

In home brewing, we usually deal with *apparent attenuation*:

apparent attenuation = ( ( OG in points - FG in points ) / OG in points ) x 100

Apparent attenuation is different (higher) than the *actual attenuation*, because
the alcohol produced during fermentation is lighter than water, and throws the reading
off.

The apparent attenuation you will get on a given batch depends on a number of factors, including the types of grains/extracts used, mash temperature (if brewing all-grain), and the strain of yeast. Typically, apparent attenuation will range from 65 to 80%. Beers with a lower apparent attenuation will be sweeter, fuller-bodied, and lower in alcohol. Higher apparent attenuation will result in a drier, thinner, higher alcohol brew.

Example:

What is the apparent attenuation of a beer with an OG of 1.055 and a FG of 1.016?

( 55 - 16 ) / 55 x 100 = 71%

*Copyright © 1999-2000 by Michael Uchima, All Rights Reserved*

*(Last updated January 11, 2000)*