Murthy 



The line integral is obviously zero, since 



h (x', z') = for z' ^ 



and we may combine the second surface integral with the first giving 

 the result 



//«■ 



'zO-l^-.dx'dz' 



m 



ax 1 



(z-x^)f(x' 



■■■>]-&<='-•>&] 



(z-x'0)-^— , | dx'dz' 

 a z 



1o 



■f=^>-|w^-6^]-^ *••■•>] 



dx'dz 1 



In the case of 



f(x', z") 



ah 



dx'dz 1 



the integral may be written 

 z'ro 



L 



ah 



f(x'.z') -"— dz 1 

 dz" 



which can be approximated as before to 



ah 



i. e. 





f(x', z 1 ) 



az- j 



dx' 



f (x'.o) -^ 



az 1 



dx 1 



It does not seem easily possible to convert this into a surfa- 

 ce integral for merging with the main integral over S. as has just 

 been done in the previous case. 



Substituting for p+ and p~ in terms of the potentials and 

 reducing the integration to the known surface & after correcting 

 for the strips S'< and S'.j we have the following results for the 

 longitudinal force and the vertical force. 



It is assumed here that the potentials <&.jqqq and 4> , mQ 

 will be continuous across the longitudinal planes for these are the 

 potentials of the side hulls in the absence of the air cushion. The 



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