We use the efficiency concept in reverse when designing a recipe to achieve a targeted OG. Let's go back to our Short Stout example.

To produce a 1.050 wort, how much malt will we need?

1. First, we need to assume an anticipated yield (e.g. 30 ppg), for the recipe volume (e.g. 5 gallons [18.9271 litres]).
2. Then we multiply the target gravity (50) by the recipe volume (5) to get the total amount of sugar. 5 x 50 = 250 pts.
3. Dividing the total points by our anticipated yield (30 ppg) gives the pounds of malt required. 250 / 30 = 8.3 lbs. (I generally round up to the nearest half pound, i.e. 8.5)
4. So, 8.5 lbs. of malt will give us our target OG in 5 gallons. Using the malt values for 85% Efficiency in Table 9, we can figure out how much of each malt to use to make up our recipe.

Remember though that this is the post-boil gravity. When you are collecting your wort and are wondering if you have enough, you need to ratio the measured gravity by the amount of wort you have collected to see if you will hit your target after the boil. For instance, to have 5 gallons of 1.050 wort after boiling, you would need (at least):
6 gallons of 1.042 (250 pts/6g) or
7 gallons of 1.036 (250 pts/7g)

So, when planning to brew with grain, you need to be able to figure how much malt to use if you are going to collect 6-7 gallons of wort that will boil down to 5 gallons at a target OG. (Actually you need 5.5 gallons if you plan for fermentation losses from the hops and trub.) These considerations are taken into account in Chapter 19 - Designing Recipes.