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%

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Copyright 1999-2000 by Michael Uchima, All Rights Reserved

(Last updated January 11, 2000)