Answer Posted / manumitra singh

case-1 say komatsu 450 is used
1)bucket capacity-2.4 cum
2)ideal output=load per cycle*no. of cycle/hour
=2.4*100 (considering 35 second cycle say)
=240 cum
3) apply correction factors
a)excavator correction=0.8(if area is not a trench exactly bucket size in which case it is1
b)load factor=0.8(if load has to be dumped in tippers and not kept aside)
c)soil factor=0.67 in case of hard digging and0 .85 in medium digging
d)swing factor=1.07 for 75degree and 0.71 in case of 180 degree.
e)performance efficiency=0.83 if 50 minutes work is done in one hour.
4)corrected output=ideal ouput*a*b*c*d*e
ex: corrected output=240*0.8*0.8*.67*1.07*0.83=91.7 cum

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