From equation (13) the horizontal length for the imaginary segment 



is; 







X, = X - X 



a 









The 



vertical height is; 













d, = d - d 

 a 



1 







The 



tension 



at A, the anchor, 



, is 







t' = y'w 











The 



tension occurs 



at 



an 







e = tan"-^ 



W 

 H 







X' = c In lf- + t/(^) - 1 I (30) 



Where- y' = c + d' 



Therefore, the horizontal distance for the actual catenary segment 



AB is: 



(31) 



(32) 



(33) 



e 9 given by equation (6) as: 



(34) 



The tension at point A may be divided into horizontal and vertical 

 components. The vertical component tends to lift the depressor weight 

 decreasing the horizontal load that it can hold. The tension components 

 are given by: 



P,^ = t'cosQ (35) 



Py = t'sinQ (36) 



Equations (26) through (36) may be used to analyze the sweepline 

 tensions and geometry. Equations (7) through (25) of Case 1 may be used to 

 calculate the rest of the cable parameters (substituting x for x and d for 

 d where required) . 



In summary, the following values are assumed: 



s = length of lower catenary segment including imaginary 



portion and sweepline 

 = angle of sweepline at sentinel 



The known values are: 



s = length of sweepline 



w = linear density of cable 



P = weight of sentinel 



D = depth from surface to bottom 



