2 4 o THE CHEMISTRY OF CATTLE FEEDING 



ratio number (in this case 6) of the dietary ; put the quotients 

 to the percentages of protein, each to each, and subtract the 

 less from the greater in each case. 



The following example will make this clear : 



Linseed cake. Rice meal. 



Percentage of " starch " . . . 56*00 65^00 



Divided by 6 9-33 IO'85 



Percentage of protein .... 25*00 6 - oo 



By subtraction 1 5'&7 4'&3 



By division 3-24 i - oo 



As before, the quantities are to be taken inversely, i.e. 

 3*24 parts of meal to i of cake. 



Substitution of Foods. It was shown at the beginning 

 of this chapter that one food cannot replace another without 

 altering the amount of one of the nutrients in the ration, unless 

 they have the same N-ratio. Such a condition is found in 

 the case of malt dust and cocoanut cake, and i\ Ibs. of the 

 former is equivalent to i Ib. of the latter. 



One food, however, can always be replaced by two others, 

 provided the N-ratio of one of them is closer, and that of the 

 other is wider than the N-ratio of the food for which they are 

 to be substituted. 



Suppose, for example, that the rations of an animal include 

 a quantity of pollards, and that it is desired to substitute rape 

 cake and locust beans for the same, it is easy to calculate the 

 quantities required. Let the data be as follows : 



Protein. " Starch." N-ratio. 



Pollards 12-0 63 5-25 to i 



Rape cake 22-0 44 2'OO ,, i 



Locust beans 3-5 73 20-85 l 



The problem is to calculate what quantities of rape cake 

 and locust beans together contain the same amounts of protein 

 and " starch," as, say, 100 Ibs. of pollards, i.e. 12 Ibs. of protein, 

 and 63 Ibs. of starch. It can obviously be worked out by the 

 methods described above. 



0-2 2* -f 0-035^=12; 0-44^ -1-073^=63 



*== 45-14 and jy = 59'i 

 or 22* + z'Sy = I ', 44*+ 73^ = 5' 2 5 



