Forces on an A.C.V. Executing an Unsteady Motion 



= — )Q(w, u, t) P(w,u, -0) - P(w, u, t) Q(w, u, -0)}' 



rcosj(7+wc)t} cos{(7+wc)(t+T)j cos)(7-wc)t( cos{(7-wc)(t+T)}l 

 7 +wc 7 + wc 7-wc 7-wc 



. . . + — )P(w, u, t) P(w, u, -0) + Q(w, u, t) Q(w, u, -0)[. 



sin|(7+wc)t[ sin) (7+wc)(t+T)} sin{(7-wc)t} sin{ (7-wc)(t+T)| 

 7+wc 7+wc 7-wc 7-wc 



We consider first the case when t = and T — ► oo (that is, 

 a steady state). The four terms containing the cosine factors, and the 

 first and third sine factors are zero. The fourth sine term is the only- 

 one that gives a non-zero result in the wu integral of Eq. (46) as 

 T — > oo . The steady-state forces may be obtained in the same manner 

 as the limit of Eq. (25) for large time : 



R 



S Zir p 



"*1 */2' 



_-7T/2 



'1 -I 



n 3 , cos 6 > 



k • ( n ) 



V sin 6' 



2 2, , 



1 - k d -see 6 * sech (kd) 

 o 



•{P (k cos 6, k sin 6) + Q (k cos 6, k sin 6) } d9 , (50) 



where k , k and 6y are given by Eqs (30), (31) and (32). 



If we now assume that the ACV starts yawing at t = , then 

 as T— > oo , the second and fourth cosine terms, and the second sine 

 term contribute nothing to the wu integral in Eq. (46). The expres- 

 sion for the forces after t = becomes : 



1 



R 



S 27rpg 



r -«i *fi 



hi 



-"/2 e. 



*/? 1 



,cos 6 



2 2. , 



1 - k d *sec 6 • sech (kd) 



o ' 



•{P(w,u, t) P(w,u, -0) + Q(w,u, t) Q(w,u, -0)} dd +... 



(cont'd over) 



55 



