W. J. V. OSTERHOUT 615 



we find that if the time is multiplied by 1.06 (making it 21,2 minutes) 

 the observed resistance (64.18) agrees with the standard curve. 

 This figure is therefore adopted. The value of Te in formulas (2), 

 (3) and (4) 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, 1. The third recovery curve 

 was calculated in the same fashion. 



Instead of waiting for the establishment of equilibrium 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 per cent 

 it was replaced in sea water: the subsequent fall in resistance was 

 calculated by means of formula (7). 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 resist- 

 ance was 68.10 the value of A (which we may call A^ is, Ai = 

 30(68.10-10). During the fourth exposure to NaCl (lasting 20.4 

 minutes) the value of Ax diminished to ^2 according to the formula 



, -(0.018)20.4 , 

 Aie — Ai. 



On replacing the tissue in sea water A^ was augmented by the con- 

 version of S into A. The value of S is found according to formula 

 (3) in which J^ is equal to the total time of exposure (20.8 + 21.2 

 + 20.8 + 20.4 = 83.2). We may call this 5*1. Hence the value of A 

 immediately after replacement in sea water vst Az = Ai -\- S\. Dur- 

 ing the subsequent 10.2 minutes in sea water Az diminished to .4 4 

 according to the formula 



-(0.0036)10.2 

 A%e = A\ 



But at the same time it received an addition from the decomposition 

 of 0; the amount of this may be found as follows: The loss of ^ in 

 sea water under normal conditions^ in 10.2 minutes is 



^ The principle upon which this formula is based is explained in a previous 

 paper in discussing the loss of M and its replacement by 0. In the present case 

 the effect of S is negligible since the amount of 5 in sea water is only 2.7. 



