F = V p sin (j) 



lift/bucket b s 



3 

 where p = soil density =100 lb/ft 



^lift " ^°"^^ ft'^/bucketXIOO Ib/ft^) sin 60° = 16 lb/bucket 



The maximum force which will be applied is the sum of the lift force and 

 total force times the number of buckets in contact with the soil. 



N, = (bucket distribution) = (1 bucket/ft) 



sin (j) sin 60° 



= 3.5 buckets 



F = (F + F ) N 

 max ^ TOT liff^ b 



= (864 lb/bucket + 16 lb /bucket) (3.5 buckets) = 3,080 lb 

 The torque required 



T = F (moment arm) = F (r + d) 



max max max ' 



where r is the chain drive radius = 0.75 ft (Reference 47) 



T = (3,080 lb) (0.75 ft + 0.82 ft) = 4,840 ft-lb 

 max ' ' ^ ' ' 



The angular velocity is 



^ ^ chain speed / 2 tt rad \ ^ / 9.2 ft/sec> ^ ^ tt = 12.3 rad/s 

 2 TT r \ rev / \t^0 .5 ft) 



Then, 



P = T W 



lax max 



= (4,840 ft-lb) (12.3 rad/sec) ' ^^ 



550 ft-lb/sec 



P = 108 hp 



max 



The forces acting on the trencher are as follows: 



64 



