14 



formula C(b/a) = 1 + 3b 2 /a 2 may fit the facts reasonably well. For fuller 

 bodies lower values seem to be suitable. 

 The constant 



is therefore obtained as 



u = ^c£, p*)b 2 U [7a] 



The flux [7] or the strength of the doublet distribution at the 

 midship section must be somewhat higher than the product of the cross section 

 times the speed of advance. 



For the source , we have 



1 n/b ..», b" 



% = i c ^"*)r u ^ 



3. EVALUATION OF HAVELOCK'S INTEGRAL 



3.1. GENERAL CONSIDERATIONS 



The wave resistance experienced by a continuous doublet sheet //, 

 distributed over a vertical plane and moving uniformly on a straight hori- 

 zontal path, has been calculated by Havelock. 3 Concentrating the distribu- 

 tion u(x) along a horizontal straight line we obtain immediately 



R = i6ttpK* I** 72 < P 2 + Q 2 L see 5 0d0; K = -£- 



°io I 1 M ° u 2 



[9] 



with 



r+a 

 P = exp(-K f sec 2 0) /u(x) cos (K x sec 0)dx = exp(-K f sec 2 0)p [9a] 



r + a 



Q = exp(-K f sec 2 0) fi(x) sin (K x sec 0)dx = exp(-K f sec 2 0)q [9b] 



1 u J _ a vj 1 



hence 



R = l67rpK£ f n/2 (p 2 + q 2 ) exp(-2K Q f sec 2 0) sec 5 0d0 



Using a source-sink distribution we obtain similarly 



R = l67rpK 2 \ n/2 (p 2 + q 2 ) exp(-2K f sec 2 0) sec 3 0d0 [10] 



J 



r + a 

 p = <j(x) sin (K Q X sec 0)dx [10a] 



/+a 

 a(x) cos (K x sec 0)d0 [10b] 



- a ^ 



[9c 



