Extract Potential 1 pound per gallon - Math help

General Homebrew Discussion
1

Hi, guys!

I am a Spanish homebrewer and I read a lot of American publications like this forum and some books.
When I am learning to calculate the malt bill, all the publications use tables with examples of Extract Potential of Fermentable Materials.

For example:
Extract Potential Pale Malt: 1.035 – 1.037

And I know that figure is because if all the extract derived from mashing 1 pound of grain were contained in 1 gallon of water, it would have an OG in the range given (1.035 – 1.037).

Ok…
But the problem is in Spain we use kilograms (or grams) instead of pounds.
And liters instead of gallons.

So…
How can I convert the 1.035 – 1.037 range got from ppg in grams per liter???

Any clever guy can help me???
I hope…

Best regards,

2

Palmer gives a quick way on this link: http://www.howtobrew.com/section1/chapter3-4.html .

Paul

3

1 lb/gallon(1 kg/2.2lbs)(1 ~~gallon/~~3.8 liters)=1/8.36 (kg/liters)

I used rounded conversion values, so if you are the type that wants more digits you can use John Palmers number.

4

Thank you very much!!!

5

So…

1,037 Potential Extract is…

6

multiply 8.36 by whatever the ppg # you are converting. In your case 37*8.36=309.32

7

It’s easier for me to remember to multiply by 2.2 and 3.8.  It’s the same thing, but that way I don’t have to remember 8.36.  The conversion of 2.2 lb = 1 kg and 1 gal = 3.8 liters I know already though.

So 372.23.8 = 309.32

Like I said, same thing, but easier (for me) to remember.

8

Yes, it is true.
But if really we want to do properly, we have to think in 100 grams per liter (not kilograms per litre, that is crazy).
Then, the new calculation will be:

37 * 2.2 * 3.8
-------------- = 30.932
      10

Therefore:

1,037 potential extract with the experiment 1 pound per 1 gallon will be 1,031 if we would use 100 grams of malt in 1 liter of water!!!

Thank you everybody for your help!!

anKalagon