exract potential calculation (by hand)

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garcara

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Hi,

If I have the amount of each grain in my recipe, my water volumes and the potential in SG of each grain, how can I calculate the extract potential % for the entire recipe?
 
(grain potential x system efficiency x # of pounds) / gallons of water

Sorry if I'm misunderstanding, it's not fully clear in your question. You're asking how to calculate the SG of your wort after mashing - right?
 
(grain potential x system efficiency x # of pounds) / gallons of water

Sorry if I'm misunderstanding, it's not fully clear in your question. You're asking how to calculate the SG of your wort after mashing - right?

No, sorry, I am asking for the extract potential of the fermentables, not the "brewhouse efficiency".
 
Your extract potential (or how much sugar can you yield) is depended on efficiency of your mash, so brycelarson gave you right formula in 2nd post.
Unit is point per pound per gallon (ppm), multiply it by efficiency and you"ll get your actual potential.
 
Your extract potential (or how much sugar can you yield) is depended on efficiency of your mash, so brycelarson gave you right formula in 2nd post.
Unit is point per pound per gallon (ppm), multiply it by efficiency and you"ll get your actual potential.

I think I am just not being complete clear as to what I means. Try checking out this chart: https://www.homebrewtalk.com/wiki/index.php/Malts_Chart and look at the yield column. This isn't taking taking my brewhouse into consideration. I would like to know what the yield is for my entire recipe.
 
I think I am just not being complete clear as to what I means. Try checking out this chart: https://www.homebrewtalk.com/wiki/index.php/Malts_Chart and look at the yield column. This isn't taking taking my brewhouse into consideration. I would like to know what the yield is for my entire recipe.

I think I know what you're asking for... but I'm not sure.

Let's say your recipe is 80% 2-row and 20% Special B. (It doesn't matter if this is gross or not... just follow the math.)

According to that chart, 2-row has a potential of 1.036, and Special B a potential of 1.030.

So you take 80% of 1.036 and 20% of 1.030 and add the result.

1.036 * .8 = 0.8288
1.030 * .2 = 0.206

.8288 + .206 = 1.0348

That's the theoretical potential of the recipe. If there are more ingredients, just follow the formula.

If anyone sees a problem with this, feel free to tell me I'm smokin' hops. :mug:
 
I think I know what you're asking for... but I'm not sure.

Let's say your recipe is 80% 2-row and 20% Special B. (It doesn't matter if this is gross or not... just follow the math.)

According to that chart, 2-row has a potential of 1.036, and Special B a potential of 1.030.

So you take 80% of 1.036 and 20% of 1.030 and add the result.

1.036 * .8 = 0.8288
1.030 * .2 = 0.206

.8288 + .206 = 1.0348

That's the theoretical potential of the recipe. If there are more ingredients, just follow the formula.

If anyone sees a problem with this, feel free to tell me I'm smokin' hops. :mug:

I think I am idiot. This sounds right. I'd like to see if someone else agrees. So simple.. then I can use a formula I found on another page to get the % from that.
 
I think I am idiot. This sounds right. I'd like to see if someone else agrees. So simple.. then I can use a formula I found on another page to get the % from that.

yup, he's giving you the formula to find the weighted average. that's correct. The other way to do it is simply multiply the points by pounds of each grain then divide by gallons.

for example:

8 pounds of XXX malt @ potential of 1.035 (35 points)
2 pounds of YYY malt @ potential of 1.028 (28 points)
5 gallons

formula is (8 x 35) + (2 x 28) / 5 = 67.2 or a SG of 1.067

cool?
 
I think I am just not being complete clear as to what I means. Try checking out this chart: https://www.homebrewtalk.com/wiki/index.php/Malts_Chart and look at the yield column. This isn't taking taking my brewhouse into consideration. I would like to know what the yield is for my entire recipe.

This is theoretical maximum potential of grain, or what can be extracted in perfect circumstances. Since we don't have perfect conditions that percentage is decreased by our efficiency.

Above example gives you maximum potential of grain (67.2 pppg in 5 gallons), by multiplying that with efficiency (eg. 75%) you will get your actual extraction (67.2 x 0.75 = 50.4 or 1.050 OG)
 
I think the hangup is that the OP was looking how to find the theoretical extract potential for a given grain bill prior to considering losses at a personal brewery level. I just multiply the individual grains' points by how many pounds. Then sum to a total points value and divide by how many gallons the batch is.

1.036 grains... 36 x 8 pounds = 288
1.034 grains... 34 x 2 pounds = 68
total grain bill is 356 points / 5 gallons = 1.071 at 100% efficiency
 
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