NORMAL PROCESSES OF ENERGY METABOLISM 565 



The heat production then would be: 



1)0.56 X 3.29 4 =- 297.9 Gal. 

 467.12 X 3.56 = 1662.9 " 

 157.0 X 3.19 = 481.7 " 



Total 2442.5 " 



Kauffmann confirmed the correctness of this method of calculation by 

 means of a calorimeter (p. 571) suitable for dogs. His results may be 

 summarized thus: 



TABLE 6 



The discrepancy between the two methods is only one per cent. 



b. Method of Zuntz and Schumberg (b). In their study of the meta- 

 bolism of a marching soldier Zuntz and Schumberg developed a somewhat 

 different method of calculation based, however, upon essentially the same 

 principles as the method of Kauffmann. All calculations are on the basis of 

 one hour. 



The !N" in the Urine (per hour) (a) X 2.56 = C from protein in the res- 

 piration. 



The CO 2 output in grams per hour X 3/11 = C output in grams per hour. 

 The C of respiration C of protein in respiration C of carbohydrate 



and fat in respiration (b). g|w'*J 



N" in urine X 8.45 = O 2 from protein in respiration. 

 Total O 2 absorbed O 2 from protein == 2 absorbed for carbohydrate and 



fat (c). V. 



The O 2 for oxidation of one gram of fat = 3.751 4 (average). 

 The O 2 for oxidation of one gram of CH 2.651 (average). 

 Let x = number of grams G-from fat (1 gm. C from fat = 32.3 Cal.). 

 Let y = number of grams C from CH (1 gm. C. from CH = 9.5 Cal.). 



x + y = b. (1 gm. N. from Prot. = 26.0 Cal.) 

 3.751 x + 2.651 y = c 



Solving for x and y, a X 26 = Cal. from Prot. 



x X 12.3 = Cal. from fat. 

 y X 9.5 = Cal. from CH 

 Total = Cal. per hour. 

 4 Compare the thermal quotients (see page 556). 



