468 SMITHSONIAN MISCELLANEOUS COLLECTIONS VOL. 5 1 



The computation of the example graphically worked out by Hertz 

 in his table will now be given, in order, on the one hand, to discuss 

 the accuracy of this table and, on the other hand, to show the new- 

 method of computation and the use of the tables. 



We consider a mass of air that is expanding adiabatically as it 

 rises and for whose initial condition we have p = 75o mm , / =27°. 

 and the relative humidity 50 per cent. According to the table 1 

 the temperature 27 corresponds to a vapor pressure e m = 26.5 mm 

 and to a total quantity of vapor at saturation x = 22.8 grams. 

 Hence for 50 per cent relative humidity we have the existing vapor 

 pressure e = 13.5 and the mixing ratio x = 11.4 grams. Corre- 

 sponding to these figures there results from table 4 



m 1 = 3.46 in the dry stage, 

 m n = 3.64 in the rain stage, 

 m i\ = 3-5 2 * n ^ ie snow stage, 



in which we assume that the condensed water remains suspended 

 in the air. 



According to equation (6) 5 is found to be 7.4470, which value 

 according to table 3 corresponds to the temperature of saturation 

 t t = 13. 2 . The corresponding pressure as given by equation (4) is 

 P = 637.5 and the partial pressure for e — 11. 3 is p' = 62 6.2 mm . 

 From the barometric formula there follows the difference of alti- 

 tude H = 1403 meters. 



We further compute the final pressure for three values t = io° C, 

 t = o° C, and t — — 20 C, by the general adiabatic equation (11) 

 since the temperature of saturation represents the beginning or 

 initial condition of the condensation stage. If we consider the 

 hail stage, then in place of 



T 

 m log — - 



we have the value — 1.82 f; the number corresponding to this is 

 made conspicuous by bold-face type in column 3 of the following 

 table F. 



