Ogilvie 



is the normal velocity component on the y = plane caused by the 

 distribution of dipoles, |jl|(x,z), over the same plane. Now we 

 would presumably restrict the dipole distribution to the region H , 

 and so (2-29) is valid if the range of integration is reduced to just 

 H, since the integrand is identically zero outside H. But the same 

 is not true in (2-28). There is a generally non-zero nornial com- 

 ponent of velocity, 'Y|(x,z), over the entire plane, and the range of 

 integration in (2-28) cannot be reduced to just H. Unfortunately, 

 we know Y|(x,z) only on H, from (2-33), and so we have solved 

 nothing. 



This difficulty is hardly surprising, since we are really 

 formulating here the classical lifting -surface problem, and its 

 solution requires either the solution of a two-dimensional singular 

 integral equation or the introduction of further simplifications -- 

 which will be discussed presently. 



In the lifting- surface problem, we really should distribute 

 dipoles over two regions , the centerplane H and the part of the 

 plane y = which is directly downstream of H. Let the latter be 

 called W, Pressure must be continuous across W» since there is 

 no body there to support a pressure jump. In the usual aerodynamics 

 manner, one can then show that 9|j.i/9x nnust be zero on W » In 

 this way, the integration range in (2-29) can be reduced to an integral 

 over just H . 



Of course, lifting surface theory is usually worked out in 

 terms of vorticity distributions. I happen to prefer using dipole 

 distributions, mainly because then I do not have to worry about 

 whether a vortex line might be ending in the fluid region. The con- 

 nection is fairly simple between the two versions, of course, A 

 single discrete horseshoe vortex extending spanwise between z = s 

 and z = - s and downstream to x = oo corresponds to a sheet of 

 dipoles of uniform density, spread over the plane region bounded 

 by the vortex line. The potential function can be written, for unit 

 vortex strength, 



00 



♦(x,y.z)=ir d; r . , "^ 



4wJ., Jo j,^_^)2^y2^.,^.^,2]y2 



r»s 



= JL ^ 



V ^ r.'+,/ + (,.M2i"2J 



*^^-^ y^U^-%f L [^2+/ + ,,.^f] 



4ir L 



tan' -2L. - tan' -^ . tan"' vVL x' | y' M^-sfl 

 z-s z+s x(z-s) 



. tan' yV[x^yMz.s)^]] . 



X(ZTS) J 



694 



