Question

An 8 x 20 cm seed drill has a 0.60 m diameter ground wheel. On a hard-surface calibration stand, the drill delivers 814 g of seeds in 30 wheel revolutions.
When operated in a field having soft soil, the wheel's effective rolling circumference reduces by 4%. Neglecting other losses, the actual-field seed application rate (in kg/ha) is ________. (Rounded off to two decimal places)
(Take \(\pi = 3.14\))

Hint:

Find the calibration seed rate from the hard-surface distance and drill width, then adjust for the reduced field circumference.

Answer 1

Correct Answer: 93.76

Step 1: Find the ground wheel's nominal circumference.
\[ C = \pi D = 3.14 \times 0.60 = 1.884\ m \]

Step 2: Find the distance and area covered during calibration.
In 30 revolutions on the hard stand, distance covered \(= 30 \times 1.884 = 56.52\) m.
The drill has 8 rows at 20 cm spacing, so its working width is \(8 \times 0.20 = 1.60\) m.
Area covered \(= 56.52 \times 1.60 = 90.432\ m^2 = 0.0090432\) ha.

Step 3: Find the calibration (hard-surface) seed rate.
\[ Rate_{hard} = \frac{814\ g}{0.0090432\ ha} = 90012\ g/ha = 90.01\ kg/ha \]

Step 4: Adjust for the reduced rolling circumference in the field.
The metering mechanism is geared to the wheel, so the same 814 g is delivered per 30 revolutions regardless of soil.
But in soft soil the wheel's effective circumference is only 96% of nominal, so the same 30 revolutions cover 4% less ground distance and 4% less area.
Delivering the same seed mass over a smaller area raises the rate:
\[ Rate_{field} = \frac{Rate_{hard}}{0.96} = \frac{90.01}{0.96} = 93.76\ kg/ha \]

Final Answer:
The actual-field seed application rate is \[ \boxed{93.76\ kg/ha} \]