Beer Forum

John Palmer gives an equation for determining the gravity of a beer based on the amount of water and the amount of yield from the extract. What is the formula for determining gravity?

Every fermentable has a yield value expressed in ppg (points per pound per gallon). So, for example most DME is 44 ppg. One pound of DME in one gallon of water will yield 44 points, or 1.044. A five gallon recipie with 6 lbs of DME will be 1.053 (ie. 6 * 44 = 264 / 5 = 52.8 or 1.053).

It gets a little trickier with all grain - because that theoretical yield is effected by your actual efficiency. In other words, if you are only 75% efficient at extracting the sugars from 2 Row malt (36 ppg), then you would only get 27 ppg, and you would need 1.96 pounds for every gallon of wort, or 9.8 pounds for the 5 gallon previous example (1.053).


Mylo

okay so I'm trying to grasp this.

Hypothetically (or maybe not), if I want a beer at a gravity of 1.06 and I'm brewing 3 gallons. and lets say my DME yields 1.04 ppg or is 40 pts how many pounds of DME do I need to use?

???????

1.06 @ 3 gallons from 40 pts = ??? lbs?

???????

Ugh, I fell so retarded I can't figure this equation out.

You want 3 gallons of 1.060 beer. 3 * 60 = 180 points. 180 / 40 (ppg of extract) = 4.5 pounds of DME. You can start with 4 gallons to account for boil off. As long as you boil down to 3 gallons, you should hit 1.060 with that amount of DME.


Mylo

4.5 pounds I believe that's the answer.

Size of batch (gallons) x OG (estimated original gravity) = Points (the amount of points it will take to make) / PPG (the yield of the yeast) = The amount of extract needed.

Wahoo!

This is correct right?

Woha, good job on the post. You just validated my answer. I finally get it. Thanks Mylo!