128 THE CHEMISTRY OF CATTLE FEEDING 



about five times the radiating surface, and other facts of a 

 similar kind, it seems probable that their food requirements 

 will stand in a like relationship. In other words, when the 

 mass is increased ten times, the food should be increased only 

 about five times, and vice versd. In that case ten animals, each 

 of 100 Ibs. live weight, would require twice as much food as 

 one similar animal of 1000 Ibs. live weight. 



On this basis it is easy to construct a formula by which to 

 calculate the food requirements of oxen of any given size 

 (weight). The following 1 is, perhaps, the simplest mode of 

 expression ; and it corresponds with the available data regard- 

 ing the maintenance requirements of oxen of different sizes. 

 log E = 07 log M 0-556 



E is the amount of available energy (kt.) of the food 

 required, and M is the live weight (Ibs.) of the animal. 



The graph A (Fig. 8) was plotted from the values of E, 

 calculated by the formula, for each 100 Ibs., up to 2000 Ibs. 

 live weight. It shows the amount of food (kt. per head) re- 

 quired. B gives the same results expressed as kt. per 1000 Ibs. 

 live weight. The dotted line corresponds to the formula 

 35 M -7- 1000, and shows how the curve A deviates from the 

 line of strict proportion. 



The formula based upon the requirements of oxen is not 

 applicable to sheep, because in the case of the latter the 

 radiation of heat is greatly diminished by their thick coats of 



1 Assuming that the relation of M and E is that described in the text, 

 we have the following corresponding values : 



M = i 10 100 1000 10,000 



E = 0-28 1-4 7 35 175 



0-28M o-28M o-28M o'28M o'28M 



~T~ J~ ~4~ ~8~ ~16^ 

 Or if M = 10, this may be put as follows : 



_ o-28M 0-28M 2 0-28M 8 



~2~~ ~tf~ ~2~ 



.. log E = log M + log 0*28 log M X log 2 

 = log M + T'4472 log M x 0-301 

 = log M (i - 0-301) + (0-4472 - i) 

 = 0-699 log M 0*5528 



