Consider a uniform current of velocity V and width X (see Figure 2). 

 The horizontal force, F, which is equal to the total drag of the current on the 

 cable, is given by: 



F = a a p V^ X (2) 



where: 



a = Drag coefficient. It is a function of the Reynolds 



2aV 

 number, , where v is the kinematic viscos- 



V 



ity of water. 

 a = Radius of the cable . 

 p = Density of water. 



The stress at the top of the cable (which is the maximum stress) will 

 be altered only slightly by the current. The offset, D, is given approximately 

 by: 



_ aapV^X^ 



L (w-b) + W - B ^ ^ 



a a 



which is an overestimate of the actual D. 



As a numerical example, let us consider a current with V = 1 ft/sec 

 and X = 5, 000 feet. * The cable described in Section II will be used. The 

 drag of this current on the cable is 0.225 lb/ft, which shows that the above ap- 

 proximation is valid. The horizontal force, F, is equal to 1, 100 pounds. For 

 the 50-ton array without any buoyancy in the cable and array and for L = 20, 000, 

 10, 000 and 5,000 feet, D is equal to 18, 28, and 37 feet, respectively. These 

 offsets are, indeed, very small. Even for lighter cables and arrays the offsets 

 will be small as compared to the length of the cable. 



^A current with this velocity and width is about the largest that will be found in 

 the Atlantic. 



Arthur B.liiuU3nc 



S-7001-0307 



