ABE TUTORIALS

AREA 1 PROBLEM 55: HARROWING PERFORMANCE

A TRACTOR PULLING A 2-METER WIDE HARROW COVERS ONE TRIP (END-TO-END LENGTH) OF A FIELD IN 5 MINUTES. IF THE TRIP HAS A LENGTH OF 300 METERS, WHAT IS THE AVERAGE SPEED OF THE TRACTOR?

a. 3.6 kph

b. 4 kph

c. 4.5 kph

d. 5 kph

CORRECT ANSWER: A. 3.6 KPH

Solution:
• Distance = 300 m = 0.3 km
• Time = 5 min = 5/60 hr = 0.0833 hr
• Speed = Distance / Time = 0.3 / 0.0833 = 3.6 kph

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DETERMINE THE FIELD EFFICIENCY IN THE PRECEDING ITEM IF THE HARROWING OF 1-HECTARE AREA CAN BE FINISHED IN 1 HR AND 51 MINUTES.

a. 65%

b. 70%

c. 75%

d. 85%

CORRECT ANSWER: C. 75%

Solution:
• Effective Capacity (EFC) = 1 ha / 1.85 hr = 0.54 ha/hr
• Theoretical Capacity (TFC) = (W × S) / 10 = (2 m × 3.6 kph) / 10 = 0.72 ha/hr
• Efficiency = (EFC / TFC) × 100 = (0.54 / 0.72) × 100 = 75%

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IF THERE IS NO LOST TIME, WHAT IS THE THEORETICAL FIELD CAPACITY?

a. 2 ha/hr

b. 3.6 ha/hr

c. 0.72 ha/hr

d. 0.54 ha/hr

CORRECT ANSWER: C. 0.72 HA/HR

Solution:
• TFC = (Width × Speed) / 10
• TFC = (2.0 m × 3.6 kph) / 10 = 0.72 ha/hr

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WHAT IS THE THEORETICAL TIME (NO TIME LOSSES) TO COVER 1 HA?

a. 1 hr

b. 1 hr and 10 min

c. 1 hr and 15 min

d. 1 hr and 23 min

CORRECT ANSWER: D. 1 HR AND 23 MIN

Solution:
• Time = Area / TFC
• Time = 1 ha / 0.72 ha/hr = 1.3888 hr
• 0.3888 hr × 60 min = 23.33 min
• Total = 1 hr and 23 min

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WHAT IS THE EFFECTIVE OPERATING TIME IF THE OVERLAP IS 10% OF THE RATED WIDTH?

a. 1 hr and 40 min

b. 1 hr and 33 min

c. 1 hr and 15 min

d. 1 hr

CORRECT ANSWER: B. 1 HR AND 33 MIN

Solution:
• Effective Width = Rated Width × (1 - overlap)
• We = 2 m × 0.90 = 1.8 m
• New Capacity = (1.8 × 3.6) / 10 = 0.648 ha/hr
• Time = 1 ha / 0.648 ha/hr = 1.543 hr
• 0.543 hr × 60 min = 32.58 min ≈ 33 min
• Total = 1 hr and 33 min