^f 



= 



4- A S 

 2 s u 



for clay 



^f 



= 



y N 



for sand 



(1) 



(2) 



where F = ''friction'' force (lb) 



2 

 A = skid area (in. ) 

 s ^ 



S = surface shear strength (0.5 to 1.0 psi) 

 N = weight of machine (lb) 

 y = coefficient of friction (%0.5) 



The problem with using these formulas directly is that the effect 

 of water and bearing load on the coefficient of friction and on the shear 

 strength is not known. To get an estimate of these effects, data from 

 towing Sea Plow III with the plowshare disengaged show the following: 



N = 28,600 lb 



A = 12,500 in.^ 

 s 



F = 5,000 lb (no plowing, water drag, or cable drag 

 avg forces included) 



This force may be used to obtain effective friction and shear strength 

 values from Equations 1 and 2. 



2F 

 S = -; — - = 0.8 psi 

 "eff ^s 



^eff =ir-= °-2 



The target cable burial system weighs 17,300 pounds wet; therefore, 



N = 17,300 lb 



Selecting an allowable ground bearing pressure of 2 psi, the skid area 

 required is 



17,300 lb „ ,„„ . 2 

 A = X- = 8,700 xn. 



^ 2 lb/in. 



Therefore, 



F^ = 4- A S = 3,500 lb 

 f T 2 s u ^^ 

 clay eff 



F^ = y ^^ N = 3,500 lb 

 f , eff 

 sand 



30 



