222 SCIENCE AND PRACTICE OF CHEESE-MAKING 



made from loo pounds of such milk? Using the 

 formula, we have [0.93X4 (fat)-j-2.5 (casein) — 

 o. 10] X 1 .09= (3.72+240) X 1 .09=6.67 pounds. 



It remains now simply to calculate the cheese- 

 solids into cheese with a given percentage of water. 

 This can be done by subtracting from i.oo the per- 

 centage of water desired in the cheese, expressed as 

 hundredths, and then dividing by the result the solids 

 in the cheese, as obtained above. The formula, thus 

 amended, becomes : 



(0.93 Fat + Casein — 0.10) X 1.09 



100 — W (water- in cheese) 



Continuing the illustration in which we have found 

 6.67 pounds of cheese-solids, we will suppose that we 

 wish to know how much cheese, containing 37 per 

 cent of water, can be made from this amount of 

 cheese-solids. We simply divide 6.67 by 0.63 (i.oo — 

 0.37), which gives 10.6 pounds. To find the equiv- 

 alent amount of cheese containing 35 per cent of 

 water, divide by 65 (i.oo — 0.35) ; for cheese con- 

 taining 40 per cent of water, divide cheese-solids by 

 0.60 (i.oo — 0.40). 



If, then, we wish to have a method for calculating 

 yield of cheese when the cheese contains a definite 

 amount of water, say 37 per cent, which is the 

 average amount in green cheddar cheese, we can use 

 the formula : 



(0.93 Fat + Casein — 0.10) X 1.09 

 0.63 



This can be further simplified by dividing 1.09 by 

 0.63, when the formula becomes 



(0.93 Fat+Casein — o.io)Xi-73. 

 In other words, find, in the manner indicated, the 



