J dx j it^-J" f » d * 



M y = " P ] X dX J f£ I^T dS = " j f « X dX 



-S. C(x) -£ 



where C(x) is the contour of each transverse section. The function f(x) 

 gives the force per unit length at the section. If we divide the velocity 

 potential in two-dimensional and three-dimensional portions as 



= e ($ v y +$ v ') (178) 



we can write 



,(2D) 3z , , . .(3D) 9z , 



f (x) = 1 p w | $ -5— r ds + i p (o $ -r— r ds 



C(x) J C(x) 



= f x (x) + f 2 (x) (179) 



Now take the added mass m and the damping coefficient N for a heaving 

 cylinder of infinite length. Then the two-dimensional portion of the 

 sectional force is expressed by 



f , (x) = - (iOJm +N ) [V(x)-W(x)] (180) 



i z z 



while the three-dimensional portion has the expression 



f 2 (x) = i p co S(x) - i £& B(x) W(x) (181) 



where S(x) is the area and B(x) is the waterline width of each section. 



64 



