
x = volume of water you will boil, eg: 4Lt
y = batch volume, eg: 23LT
z = total amount of fermentable sugars to be used in the batch, eg 3 kilos
f = amount of fermentable sugars to be used in your boil, eg: 522g
f = (x/y)*z
So for a 4L boil and a 23L batch with 3 kilos of fermentables you divide 4 by 23 and get 0.174 multiplied by the total fermentables in the batch which is 3 kilos meaning you need to use 522g in the boil. Of course this will vary if you are using amounts of things that are not fully fermentable as in the following recipe:
Coopers Draught
1.5 kilo LLME
500g LDME
500g Dextrose
Different malts and sugars will yield different amounts of final solids so we can not assume 100% across the board.
Kit = 1.7 kilos of LLME which yields 78% to the batch or 1.326 kilos
LLME is another 1.5 kilos at 78% yielding 1.17 kilos
LDME at 95% yields 475g
Dex is 500g because it yields 100%
This gives us a total of 4.2 kilos of addition but only 3.471 kilos are yielded to the batch. So to do the maths properly we need to use the equation as follows:
f = (4/32)*3.471
f = 604g and if we use LDME in the boil (which I would recommend) and it yields 78% then we need to use 775g.
Hope this helps, and if anyone needs anything explained more clearly feel free to ask!
Kev