42 Lord Rayleigh on the Incidence of Aerial 



whence 



c 2 = a 2 — Z> 2 , tanb £ = b/a. 

 Thus 



2^R=f-fo=log^; .... (82) 



and then (79) gives as applicable at a great distance 



e~ ikr / ir \i 



^ = 7 + log{^(a + 6)} ( 2l£r) ' * * l ' 83 ^ 



The result for an infinitely thin blade is obtained by merely 

 putting b = in (83). 



For some purposes the imaginary part of the logarithmic 

 term may be omitted. The realized solution is then 



_/ 7T Y cosk(Yt — r — l\) 

 Y \2kr) y + log\tk{a + b)\' ' ' ' [b) 



corresponding, as usual, to 



<f> = cosk(Yt + a:) (85) 



Electrical Applications. 



The problems in two dimensions for aerial waves incident 

 upon an obstructing cylinder of small transverse dimensions 

 are analytically identical with certain electric problems which 

 will now be specified. The general equation (V 2 -4-A 2 )=0 is 

 satisfied in all cases. In the ordinary electrical notation 

 V 2 = l/K/i, V /2 =1/K / /a / ; while in the acoustical problem 

 V' 2 = m/a ) Y'" 2 = m / /cr'. The boundary conditions are also of 

 the same general form. Thus if the primary waves be denoted 

 by y = e ikx , y being the magnetic force parallel to z, the con- 

 ditions to be satisfied at the surface of the cylinder are the 

 continuity of 7 and of K _1 dy/dn. Comparing with the 

 acoustical conditions we see that K replaces <x, and conse- 

 quently (by the value of V 2 ) //, replaces 1/m. These sub- 

 stitutions with that of 7, or c (the magnetic induction), for yfr 

 and <j> suffice to make (66), (70) applicable to the electrical 

 problem. For example, in the case of the circular cylinder, 

 we have for the dispersed wave 



corresponding to the primary waves 



c = e ikx (87) 



