(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".
http://www.beersmith.com/Grains/Grains/GrainList.htm
does this help? the potential SG column is pretty useful.
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 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.
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 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.
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