STATICS r 39 



the formula in terms of available energy, may be translated 

 into the corresponding amounts of nutrients as follows l : 



log E = 07 log M 0*556 



P_ JL"c-.JL 



-49-3' - 4 -i8 



P is the amount (Ibs.) of protein and C is the amount 

 (Ibs.) of non-nitrogenous nutrients reckoned as starch. A 

 certain amount of the latter may, of course, be given in the 

 form of fat. (See also note at end of chapter.) 



Example. In the case of oxen of 1000 Ibs. live weight, 

 E = 35 kt. Therefore 



In the case of sheep the formulae would be as follows 2 : 

 log E = 07 log M 0*8 



p= _E_. C= A 

 39*44' 4'3 



Rations. Rations for maintenance may be determined 

 from the feeding standards, or from the results calculated by 

 means of the formulae, provided the composition of the feeding 

 stuffs are known. If the available energy of the food is 

 known, it is not necessary to translate it into terms of non- 

 nitrogenous nutrients. 



On reference to the tables (p. 167) it will be seen that 

 average meadow hay yields 1-83 kt. of available thermic 

 energy per lb., so [35 -f- 1*83 ] 19*1 Ibs. will be required 

 for the maintenance of oxen of 1000 Ibs. live weight; and as 

 such hay contains 21 '8 Ibs. of protein per 1000 kt. of energy, 



1 For reasons given in the text, not less than 10 per cent, of the total 

 available energy must be derived from protein, which yields 4*93 kt. per lb. 

 The remaining T 9 n ths, reckoned as starch, yields 376 kt. per lb. Therefore 



P=*x-!- = ^ C= 7 ' 5 Ex-^ = ^ 



10 4-93 49-3 376 4I 



2 P = IE x - = C = IE x --, = ~ 



4'93 39 '44 376 4*3 



