MOVEMENT OF AIR IN ANTICYCLONES POCKELS 609 



velocity, and no further difficulties arise in computing* from it the 



distribution of pressure in the inner region of the anticyclone 



and in so determining the constants of our previous solution for 



the exterior region whose form remains the same that V n , 



dp 

 V t and p and now also y-pass continuously over at the boundary 



circle. 



I will not now go more precisely into this computation but will 

 only communicate some results for the special case of n =2. In 

 this case the velocity of the descending air is 



w = y l — — 

 ' V R 2 



Since now we put y = k therefore, by retaining the values of k and 

 R adopted in the previous example we obtain for the center, r = o, 

 a descending velocity twice as large as then ; on the other hand the 

 total mass of the descending air for the whole interior region 



J'R 

 wrdr. We have also 

 



y / , \ r 2 



V n = ~ - r 1 



2 \ 2 R 2 



V t = A ( 2 ^ - 1 ) { r 2 + 2 R 2 log nat ( 1 - — - ) \ 

 1 2r\ r 2 / 1 \ 2R 2 / ( 



Therefore in the interior region also the angle of deflection is no 

 longer constant: let the "normal" angle of deflection be 



a = tg-'X/k 



then for our two special values of r we have 



f or r = tg fa =-tg a 



iotr = R tg<{> R = 0.7726 tg a 



whence by introducing the numerical values above given (i. e., k 

 = 0.00012 and k = 0.00008) we find the angles <p = 36 32' and 

 (f , R = 49 13' respectively. 



