G. I. TAYLOR ON EDDY MOTION IN THE ATMOSPHERE. 15 



If we assume the motion to be horizontal equations (9) become 



0= -2<usinX + ^^, (10) 



p dz 2 



G u. d 2 V / 1 , \ 



= 2a>u sm X --- h- -r-5 ........ (11) 



p ^ rtz j 



Eliminating u the equation for v becomes 



,--04 i '> wo sin X 



^ -- h4B w = where 



^ 

 az* 



Taking into consideration the fact that v does not become infinite for infinite values 

 of z, the solution of this is 



v = A 2 e- B/ sin Bz + A 4 e- B * cos Bz ........ (12) 



Differentiating this twice with respect to z we find 



f| =2B 2 (-A,e- B ~- cos Bz + A 4 e- B -- sin Bz). 



Cv% 



Substituting this value in (ll) we find 



G 

 u = A 2 e~ B " cos Bz A 4 e~ Bz sin Bz + ....... (13) 



f ' 



The quantity or -- : -- is the gradient velocity, so that at great heights, 



" sin X 



v = and u is equal to the gradient velocity. 



The values of A 3 and A 4 will be found by imposing suitable boundary conditions. If 

 there is slipping at the earth's surface it seems natural to assume that it is in the 

 direction of the stress in the fluid. In this case one boundary condition will be 



L u Jz=o L v J-=o. 



Where the square brackets are intended to show that the values of the quantities 

 contained in them are to be taken at the surface of the ground, z = 0. 



Substituting for u, r, dujdz and dv/dz, and putting 2 = 0, equation (14) becomes . 



A G 



A 2 + 



A 



-^1-4 



. . 

 \ ' 



where Q G represents the gradient velocity. 



In order to determine the motion completely one more relation between A 2 and A 4 is 

 necessary. Let the wind at the surface be deviated through an angle a from the 



