Problem: Determine the theoretical capacity for a 20.0-m wide cultivator traveling at 9.6 km/hr.

Solution:

CT — = S — x w(m) x --x hr / V hr J km 10,000 m-

10,000 hr

Problem: What is the effective capacity for the cultivator in the previous problems if the operator wasted 0.75 hr per 10.0 hr working day?

Solution: The first step is to solve for the time efficiency. output

The effective capacity is:

hr km 10,000 m2

177,600 ha

10,000 hr

Problem: How many hours will it take to cultivate 20.0 ha with an 8.0-m cultivator traveling at 10 km/hr? Assume a field efficiency of 85%.

Solution: The first step is to determine the effective field capacity.

hr km 10,000 m2

68,000 ha

10,000 hr

The second step is to calculate the amount of time it will take to cultivate 20.0 ha with a field capacity of 6.8 ha/hr.

1 hr

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