The Home Brew Page Brewing Guide

PG/P - Contribution and Efficiency

I'm sure that, unless you don't get around much, you've heard someone somewhere refer to PGP, conversion efficiency or mash efficiency. And you probably wonder what the heck they're referring to.

Fret no longer - it's all about to be revealed!

First, though, a nit: PGP is incorrect. PGP suggests a factor of points-gallon-pound when, in fact, the factor is points-gallon-per-pound. Whenever you write about it, for the greater good of mankind, please write it as PG/P and call it points-gallon per pound. (Thanks!)

Now! On to the details.

As PG/P might suggest, it is a determination of the POINTS of specific gravity contributed to the GALLONS of wort by the POUNDS of grain or extract used. This is powerful. Once you know of this, you can convert recipes readily from all grain to extract; from DME to LME; from brewer to brewer. Let's start with extract brewing. We'll learn how to determine the expected SG of a recipe here.

DME

One pound of DME dissolved in one gallon of water is purported to yield a wort of SG 1.045. On average, this is true with some variation, brand to brand, style to style. Think of what this is telling us: If I add one pound of DME to one gallon of water, I get a 1.045 wort.

But wait! There's more!

Relying on simple algebra, you can determine, for instance, how much DME to add to four gallons of water to yield a 1.060 wort. How. you ask? Here's how:

Subtract one from the gravity you wish to achieve, then multiply the result by 1000 to get a whole number. Ours is 60.
We need to get 4 gallons at 1.060, so we need to have 60*4 points to divide into the batch. 60*4=240 (ie, if added to one gallon, the necessary amount of DME would create a wort of SG 1.350).
Each pound of DME will give us 45 points, so we need 240/45=5.333 pound of DME to make our four gallons of 1.060 wort.

LME

As you might expect, the case is similar for liquid malt extract. Typically, LME is said to have a contribution of 30 pg/p. What's that you say? That means that one pound of LME dissolved in one gallon of water will yield a wort of 1.030. Again, it varies maker to maker, style to style, but this is a pretty good rule of thumb average.

Let's say your best buddy just whipped up a ten gallon batch of the most bodacious pale ale you've ever allowed across your tongue. You want to duplicate his efforts, but he can't remember how much LME he used! The nerve! He can tell you that the SG of the wort was 1.035. Being no dummy, you immediately calculate how much he must have used:

Ten gallons at 1.035 means he had 350 points from the extract.
LME is credited with providing 30 pg/p, so he must have used 350/30=11.666 pounds of LME!

Converting from LME to DME and back again

OK, so now your friend wants to know how much DME he would have needed to do the same batch. "Easy," you say! DME gives 45 pg/p and you need a total of 350 points. So you'd need 350/45=7.777 pounds!

The concept here should be clear, and it is the concept for converting recipes between methods and ingredients:

Determine the one gallon equivalent points for your target by multiplying points of the batch by its volume in gallons: (((1.045-1)*1000)*size of batch in gallons)
Divide by the contribution of the ingredient or method (oe, 45 for LME, 30 for DME - we'll get to the "method" part under "All Grain" below...)

The concept above (plus conversion information for the units of measure used) can also be used to develop starters of various gravities from nothing but the weights of the ingredients. This is precisely what was done to develop the starter recipe I use in the "Canning Starters" chapter of this Brewing Guide.

All Grain Brewing - EFFICIENCY

In all grain brewing, the brewer need to understand the efficiency of the breweries involved - theirs and that of the brewer whose recipe you may be trying to duplicate. Let me explain: If you are trying to duplicate Johanne's recipe made from all grain, you cannot possibly achieve the same results without knowing Johanne's yield form the mashed grains - her efficiency. If Johanne say you need 6.5 pounds grist to achieve a five gallon batch of her 1.035 beer, you need to look at her efficiency versus yours to determine how much of that pale malt YOU will need.

There are two points in the brewing process in which efficiency can be meaningfully measured: before and after the boil. Prior to the boil, the number reports your mash efficiency. This is the number most home brewers report as their efficiency. The post-boil efficiency can be called your system efficiency. It includes the boil and theoretically should be the same number, except there are usually losses incurred: wort trapped in the hops, boilovers. etc. System efficiency can be highly variable for the same recipe brewed by the same brewer using the same techniques and ingredients. For this reason, I personally prefer the mash efficiency number when attempting to replicate a recipe. (I also prefer grains to be reported as "percent of grist" rather than in ounces or pounds. But I'm a picky kind of guy.)

When Johanne told you the particulars of her recipe - pounds of grain, final volume and final gravity - she told you all you need to know in order to calculate her system efficiency. This is actually "close enough for horse-shoes and hand-grenades" and is likely what you'll use. Some thoughtful brewers will also tell you their mash efficiency - much closer! (And some will give you the recipe in "percent of grist" format. Sigh...)

Let's say we know that our mash efficiency is is on the order of 34 pg/p. Johanne's system efficiency is 35*5/6.5 or 26.9 pg/p. Even though the two efficiencies are different numbers, we'll assume that everything in Johanne's boil was perfect, and that her system efficiency and mash efficiency are the same. We'll need 35*5/34=5.15 pounds of grain to achieve the same results. If Johanne's recipe contains more than simply pale malt, you can directly convert her grist to a percent grist format by adding up the weights of each grain, then dividing the weight of each grain by the total to yield the percent makeup. To translate the ratios to OUR recipe, we then multiply the percentage of each grain type by our total grist to determine how much of each to include. For example:

Johanne's Recipe (6.5 lb grist) Our Conversion (5.15 lb grist)

4 # Pale Malt = 4/6.5*100 = 61.5% 5.15*61.5% = 3.17 # Pale Malt

2 # 80L Caramel = 2/6.5*100 = 30.8% 5.15*30.8% = 1.59 # 80L Caramel

0.5 # Roast Malt = 0.5/6.5*100 = 7.7% 5.15*7.7% = 0.40 # Roast Malt

Total: 6.5 Total: 5.16 (difference due to rounding)

Now, if you brew it, your batch should turn out pretty durned close to the original.

Efficiency as a percent

Some brewers and most software report efficiency as a percentage. This is tricky business as the true percentage is determined by dividing your yield by the lab yield of the grains involved. This would require that those reporting their efficiencies as a percentage find the lab yield for each of the grains their using. Guess what. No-one (or, at least very few) does this. They use a theoreticla average - much like we use for the contributions of DME and LME. It's 40 pg/p. So, if you divide your yield by 40 pg/p, you can represent yours as a percentage, too. But why go through the extra work?

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