INJURY AND RECOVERY 103 



On replacing the tissue in sea water, therefore, we 

 start with M = 70.69 and A =2,027.96, but this value of 

 A is at once augmented by the conversion of 8 into A. 

 In order to find the amount of this augmentation we 

 must know the value of S. 



During exposure to NaCl the reaction R >8 >T 

 occurs. The value of S may be easily calculated by 

 employing formula (1) and substituting the appropriate 

 constants. We thus obtain 



\ 



(3) 



The value of R at the start in sea water is taken as 1,041.77 

 and that of S as 2.7. In the solution of NaCl the values of 

 K R (the velocity constant of the reaction R >S) and K s 

 (the velocity constant of the reaction S > T) are taken 

 as 0.04998 and 0.02856 respectively (see Table V). Hence 

 the value 27 of 8 at the end of 15.9 minutes is 447.26. 

 When the tissue is replaced in sea water 8 is rapidly 

 converted into A so that the total value of the latter 

 becomes 447.26 + 2,027.96 == 2,475.22. 



On replacing the tissue in sea water A = 2,475.22 and 

 M = 70.69. The resistance due to A and M after any 

 given time T in sea water is obtained by modifying 

 formula (1) which becomes 



Resistance = 2,475.22! - A - } ( C ~ KA TR _ -K*fTR \ 



\^KM KA/ \ / 



+70.69 e- +I o (4) 



in which T denotes the time which has elapsed after 



R 



27 In general the greater the rise in recovery the greater the value of 8, 

 while the greater the fall the less the value of 8. 



