Author Topic: Extract Potential 1 pound per gallon - Math help  (Read 1745 times)

Offline ankalagon

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Extract Potential 1 pound per gallon - Math help
« on: April 11, 2012, 11:59:27 AM »
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,

Offline Slowbrew

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Re: Extract Potential 1 pound per gallon - Math help
« Reply #1 on: April 11, 2012, 12:43:28 PM »
Palmer gives a quick way on this link: http://www.howtobrew.com/section1/chapter3-4.html .

Paul
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Online hopfenundmalz

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Re: Extract Potential 1 pound per gallon - Math help
« Reply #2 on: April 11, 2012, 12:53:28 PM »
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.
Jeff Rankert
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Home-brewing, not just a hobby, it is a lifestyle!

Offline ankalagon

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Re: Extract Potential 1 pound per gallon - Math help
« Reply #3 on: April 11, 2012, 12:55:25 PM »
Palmer gives a quick way on this link: http://www.howtobrew.com/section1/chapter3-4.html .

Paul

Thank you very much!!!!!

Offline ankalagon

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Re: Extract Potential 1 pound per gallon - Math help
« Reply #4 on: April 11, 2012, 01:02:07 PM »
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.

So...

1,037 Potential Extract is...

Offline repo

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Re: Extract Potential 1 pound per gallon - Math help
« Reply #5 on: April 11, 2012, 01:35:38 PM »
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.

So...

1,037 Potential Extract is...

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

Offline tschmidlin

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Re: Extract Potential 1 pound per gallon - Math help
« Reply #6 on: April 11, 2012, 06:25:04 PM »
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 37*2.2*3.8 = 309.32

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

Offline ankalagon

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Re: Extract Potential 1 pound per gallon - Math help
« Reply #7 on: April 12, 2012, 08:36:52 AM »
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 37*2.2*3.8 = 309.32

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

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