ABE TUTORIALS

AREA 1 PROBLEM 54: GRAIN DRILL MECHANICS

A 10-25 CM GRAIN DRILL HAS A WHEEL RADIUS OF 25 CM. WHAT LENGTH OF STRIP IS COVERED BY THE MACHINE IN 670 REVOLUTIONS OF THE GROUND WHEEL WHEN THE WHEEL SLIPPAGE IS 5%?

a. 500 m

b. 700 m

c. 800 m

d. 1000 m

CORRECT ANSWER: D. 1000 M

Solution:
• Wheel Radius (r) = 25 cm = 0.25 m
• Circumference (C) = 2 × π × r = 2 × 3.1416 × 0.25 = 1.57 m
• Theoretical Distance = 670 revs × 1.57 m/rev = 1,051.9 m
• Actual Distance (Da) = Theoretical × (1 - slippage)
• Da = 1,051.9 m × (1 - 0.05) = 999.3 m ≈ 1000 m

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WHAT PART OF A HECTARE WAS COVERED BY THE GRAIN DRILL IN THE PRECEDING ITEM AFTER 670 REVOLUTIONS OF THE GROUND WHEEL AT A WHEEL SLIPPAGE OF 5%?

a. 0.2 ha

b. 0.25 ha

c. 0.36 ha

d. 0.5 ha

CORRECT ANSWER: B. 0.25 HA

Solution:
• Number of rows = 10
• Row spacing = 25 cm = 0.25 m
• Rated Width (W) = 10 × 0.25 m = 2.5 m
• Distance (D) = 1000 m
• Area (A) = W × D = 2.5 m × 1000 m = 2,500 m²
• Area in ha = 2,500 / 10,000 = 0.25 ha

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IF THE AMOUNT OF SEEDS DISPENSED BY THE GRAIN DRILL IN THE PRECEDING TWO ITEMS WAS 20 KG, WHAT IS THE SEEDING RATE IN KG/HA?

a. 20

b. 40

c. 60

d. 80

CORRECT ANSWER: D. 80

Solution:
• Weight of seeds = 20 kg
• Area Covered = 0.25 ha

Calculation:
Seeding Rate = Total Weight / Total Area
Seeding Rate = 20 kg / 0.25 ha = 80 kg/ha