he = ^ Lgc 2^/4ax f°^ Z < Z^_^^/2 



and for the lower half; 



2 2 



for Z > Z„^^/2 

 max 



By substituting I , £ , and 2.gj, (neglecting the weight of lift line w) , 

 the coordinates or the payload and the guideline are: 



(A-3) 



payload 

 1 



^ (Dp Z + Dj, Z/2) sin ^ 



Yp = So +- (Dp Z + Dj, Z/2) cos i|j 



(A-4) 



Lgo Z sin 



D„ Z sin ^ 

 2~T 



L Z cos i D Z cos i) 



— + — 



Z„o^ 2 T 



^hen Z < Z^3^/2 



(A-5) 



(A-6) 



LgQ Z sin 

 ^ax 



2 



Dg (^max "2) sin i|) 



2~T 



Lgo Z cos ({) Dg (Z^Qy. - Z) cos i[) 



2 T 



when Z > Z^^^/2 



(A-7) 



(A-8) 



Now the angular displacement 6 can be obtained by Equation A-1 and the 

 distance between the payload and the guideline is 



S = \|(Xp - Xg)2 + (Yp - Yg)2 



(A-9) 



If S < Sq, frame orientation is not stable because of compressive forces 

 acting on the frame. 



This method should give the general orientations of the guide frame 

 at various depths during a lowering and recovery operation. More exact 

 prediction can be made by employing more tedious catenary computations for 

 both lift line and guideline under distributed current loads. 



29 



