CALCULATING YIELD OF CHEESE 221 



the results of the work done at the New York ex- 

 periment station, to work out a method of deter- 

 mining cheese yield which should be simple and 

 at the same time more accurate than the methods 

 previously used. This method is based upon (i) 

 the per cent of fat and of casein in milk; (2) a loss 

 of fat proportional to the amount of fat in milk, 

 based upon average results; (3) a uniform loss of 

 casein; (4) an amount of salts and albumin in cheese 

 proportional to the available fat and casein in 

 the milk; and (5) a uniform percentage of water in 

 cheese. 



We will now briefly consider the details upon 

 which the method is based, under the two following 

 divisions: (i) Calculation of cheese-solids, and (2) 

 calculation of water in cheese. The amount of 

 solids in cheese is calculated by the formula, 

 (0.93 Fat -j- Casein 0.10X1.09. This is based upon 

 the following details : ( i ) Of the fat in milk, 7 per 

 cent (0.07 pound for each pound of milk-fat) is lost 

 in whey and 93 per cent (0.93 pound for each 

 pound of milk- fat) remains in cheese (p. 190). 

 (2) Of the milk-casein, about o.io pound for 100 

 pounds of milk is lost, the rest going into the 

 cheese (p. 195). (3) The other constituents of 

 cheese-solids, consisting mostly of salts (p. 187), 

 form about 9 per cent (0.09) of the fat and casein 

 present in cheese. Therefore, if we multiply the 

 amount of fat and casein in cheese by 1.09 we ob- 

 tain the total amount of cheese-solids (fat, casein, 

 salts, etc.) in cheese. For example, suppose we* have 

 milk containing 4 per cent of fat and 2.5 per cent of 

 casein, how many pounds of cheese-solids can be 



