Utilization Equation question - Home Brew Forums

 Home Brew Forums > Utilization Equation question

12-29-2012, 05:58 PM   #1
richwyso
Recipes

Dec 2012
Vernon Hills, IL
Posts: 27

I have been reading this book from how to brew from john palmer and i wonder if anyone out there is familiar with the equation below. More specifically, its been driving me crazy all morning trying to figure out where the 7/10 came from. Does anyone know? I have posted below the quote from the book:

For example, to calculate the utilization for a boil gravity of 1.057 at 30 minutes, look at the utilization values for 1.050 and 1.060. These are .177 and .162, respectively. There is a difference of 15 between the two, and 7/10ths of the difference is about 11, so the adjusted utilization for 1.057 would be .177 - .011 = 0.166.

dont know where the 7/10 came from. thank

12-29-2012, 07:35 PM   #2
ajdelange
Recipes

Aug 2010
McLean/Ogden, Virginia/Quebec
Posts: 9,442
Liked 1566 Times on 1192 Posts

He is interpolating. The densities of worts are quite close to being linearly dependent on the 'points' obtained by subtracting 1 from the specific gravity and multiplying by 1000. Thus a wort of SG = 1.050 has a density proportional to 50 points, one at 1.060 a density proportional to 60 points and one at 1.057 a density proportional to 57 points. Fifty seven (57) lies 7/10 of the way between 50 and 60 so we suppose that the density of 1.057 wort is equal to the density of 1.050 wort plus 0.7 times the difference between the density of 1.060 and 1.050 worts. The formula assumes the utilization is also linear with density so that the utilization for a 1.057 wort will be the utilization for a 1.050 wort plus 0.7 times the difference between the utilization at 1.060 and that at 1.050.

12-31-2012, 03:07 PM   #3
richwyso
Recipes

Dec 2012
Vernon Hills, IL
Posts: 27

Thanks so much for that clarification.. that makes a lot more sense now.. You are da bomb!