Determining malt quantities in John Palmer's book

Hi guys,

John explained in his book if we want to get a 5 gallon of 1.050 wort, we just use the formula:

50 x 5 = 250points
250 points / 30ppg (anticipated yiled) = 8.33 lbs

Is the 30ppg ("anticipated yield") the default value in determining the malt quantities?

How to Brew - By John Palmer - Planning Malt Quantities for a Recipe

Thanks!

That's a somewhat outdated way of thinking about things, tbh. In any case, it still works, but people tend to talk about the numbers in a different way now.

The "anticipated yield" here is just the base yield of 2-row malt (37ppg) multiplied by some baseline efficiency (80%ish, by the looks of it).

In any case, different malts will give you significantly different numbers.

It's a starting point. YMMV. You can make necessary adjustments to the equation based on the conversion efficiency of your brewery.

The anticipated yield is basically a stand-in for your brewhouse efficiency. What you REALLY want to do is figure out the maximum potential extract of your grain bill, and then multiply by your efficiency to determine the actual starting gravity points. Then you can play with those numbers until you get close to 250, or 1.050 in 5 gallons.

As a rule of thumb, with 70% efficiency you are going to need a bit under 10 lbs of grain to get 1.050.

2-row has a potential extract of 37 so 10 * 37 = 370
370 * .7 = 259

EDIT: sorry it's 37 I suck at remembering things

Generally if you have something like 9 lbs of 2-row and a pound of specialty grain (lower potential extract; more like 29-30 for roasted malt and 34-35 for crystal), you'll be very close to 1.050 with a 10 pound grain bill.

Thank you all very much for the replies. Second, please bear with me as I am quite new to AG. My first 3 gallon AG batch of Pale Ale was very close. The grain bill was as follow:

5lbs of Briess Pale Malt 2 Row
(the FGDB is 80.5% of 37 points or 29.7ppg)
.63lbs of Caramel/Crystal 60L
(the FGDB is 74% of 34points or 25.16ppg)

So, I calculated (29.78x5lbs)/3gal = 49.63ppg and (25.16x.63lbs)/3gal = 5.28ppg. Therefore, the OG should be around 1.055 based on the calculation. My actual OG reading was 1.056. My preboil gravity based on 75% efficiency should be 1.042 (1.056x75%) which was exactly the same with the actual reading.

The problem arose when I brewed an ESB yesterday (my 2nd AG batch). I wanted to brew a 3.50gal batch. With my equipment setup, beersmith suggested a preboil volume of 4.35 gal. According to the software, I should get a wort with OG of 1.059 and preboil gravity of 1.048 based on the following grain bill.

5.33lbs of Maris Otter
(FGDB is 82.5% of 38 or 31.35ppg)
1.06lbs of Biscuit Malt
(FGDB is 79% of 36 or 28.44ppg)
1.06lbs of Crystal 70L
(FGDB is 74% of 34 or 26.66ppg)

so if we were to use the above formula, I should get an OG of about 1.063.97 or 1.064 and preboil gravity of 1.048 (assuming 75%).

I only collected 4gal preboil using the suggested sparge water amount instead of the suggested 4.35gal. As a result, I only got a 2.5gal wort instead of 3.50gal. I lost 1gal to the 70mins boil and .50gal to cooling and trub.

My actual preboil gravity was 1.051 and OG was 1.070. Am I on the right track of formulating a recipe? Thanks again and sorry for the long post.

Hmm...there's some voodoo math here.

You want to multiply your efficiency by the theoretical potential, not the FGDB. FGDB is useful for determining the extract potential of a given grain. Sugar, for example, has a ppg of 45 or so because it dissolves completely. A 2-Row with 80.5% FGDB would, under ideal circumstances, produce 36 ppg or so. This is the number you want to multiply by your efficiency.

You are right. FGDB is the maximum potential extract so I should only be targeting 75% out of the 2 Row 80.5% in this case.