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-f- 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 fo.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 



