114 INJURY, RECOVERY, AND DEATH 



(2), (3) and (5) should now correspond to the total 

 exposure to NaCl, and is 20.8 + 21.2 = 42. 



These data were employed in calculating the second 

 recovery curve and the results are shown in Fig. 44. The 

 third recovery curve was calculated in the same fashion. 



Instead of waiting for the establishment of equilib- 

 rium we may replace the tissue in NaCl after it has 

 been for a short time in sea water. During the fourth 

 recovery, after the tissue had been 10.2 minutes in sea 

 water and the resistance had risen to 54.92%, it 

 was replaced in sea water : the subsequent fall in resist- 

 ance was calculated by means of formula (9). For the 

 value 77.1 in this formula we must substitute the observed 

 resistance less 10, or 55.89 10 = 45.89 ; and in place of 

 2313 we must substitute the present value of A. We 

 assume that at the beginning of the fourth exposure to 

 NaCl equilibrium had been reached in sea water: hence 

 as the resistance was 68.10 the value of A (which we call 

 A, ) is, A 1 = 30 (68.10 10) . During the fourth exposure 

 to NaCl (lasting 20.4 minutes) the value of A l diminished 

 to A 2 according to the formula 



- (0.018)20.4 



On replacing the tissue in sea water A 2 was augmented 

 by the conversion of 8 into A. The value of S is found 

 according to formula (3) in which T E is equal to the total 

 time of exposure (20.8 + 21.2 + 20.8 + 20.4=83.2). We 

 may call this 8^ Hence the value of A immediately after 

 replacement in sea water is A s = A 2 -{- 8 lt During the 

 subsequent 10.2 minutes in sea water A 3 diminished to 

 A 4 according to the formula 



-(0.0036)10.2 

 A z e = A t 



But at the same time it received an addition from the 



