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Old 06-21-2010, 08:52 PM   #1
kmacva
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Default LME to All grain equivalents

I searched for an answer to this but did not find one. I've done quite a few partial mash and extract brewing kits. I have now gone all grain. I want to duplicate some of the partial/extract brews recipes I have used . If I had a recipe that uses 4 pounds of LME how many pounds of base grain would I need to use to equal the 4 pounds of LME to get the same OG.
Thanks for the advice in advance.


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Old 06-21-2010, 08:56 PM   #2
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Generally, a good rule of thumb is 1 pound grain = .75 pound LME = .6 pound DME. That's a good way to start.

Once you have experience, you can adjust the amount of grain to get to your efficiency. It takes a few batches to "know" your system, though.

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Old 06-21-2010, 08:58 PM   #3
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I am trying to remember this off the top of my head, and I'm sure someone will correct me if I'm wrong. First, you really need to have light or x-light malt extract to convert because any of the medium or dark extracts have mixed grains and will depend on brand. But, if you do have light or x-light, I think it is .75lb of LME to 1lb 2-row base malt, and .60lb of DME to 1lb 2-row base malt.

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Old 06-21-2010, 08:59 PM   #4
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Quote:
Originally Posted by YooperBrew View Post
Generally, a good rule of thumb is 1 pound grain = .75 pound LME = .6 pound DME. That's a good way to start.

Once you have experience, you can adjust the amount of grain to get to your efficiency. It takes a few batches to "know" your system, though.
Ahhhhhh.......beat me to it.
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Old 06-21-2010, 09:03 PM   #5
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Here's a reference I used to start converting recipes myself. I think you can skip through most of it. Reading an example helps. I then used it to create a spreadsheet to auto calculate for the 3-4 malts I'd be converting most often (munich, wheat, two row, amber). Hope it helps.

http://members.cox.net/steve.krieske/Extract Brewing Guide.pdf

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Old 06-21-2010, 09:06 PM   #6
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According to my chart: 4.13 lbs of LME equal 5.5 lbs of grain.

4.13 is really closer to 4 lbs, 2.4 oz (@.15).

Email me if you want a copy of my chart: homebrewer_99@yahoo.com

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Old 06-21-2010, 10:32 PM   #7
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LME usually gives 38 points/lb (YMMV)

lbs grain to use = (lbs LME)*(38 pts/lb) / [(the malt's etract potential)*(your system efficiency)]

The first part, (lbs LME)*(38 pts/lb), is the total points from the LME. The second part, [(the malt's etract potential)*(your system efficiency)], is the extract yield per lb of grain on your system. The first part divided by the second part is the answer.

For example, your efficiency is 70% and you're using 6.6 lbs LME, and your malt has a potential of 30 points per lb:

6.6 lbs of LME x 38 points/lb = 250.8 points

lbs grain = 250.8/(.7*30) = 11.95 lbs grain.

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Old 06-22-2010, 09:47 PM   #8
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Thanks for your replies. The information you've given looks like it will get me where I need to be. I going to use it as a starting point and modify it to my tastes as needed.


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