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: Work out the wheel's field circumference directly.
Nominal circumference: $C = \pi D = 3.14 \times 0.60 = 1.884$ m.
In soft field soil this drops by 4%: $C_{field} = 1.884 \times 0.96 = 1.809$ m.

Step 2: Find the actual ground distance covered for 30 wheel turns in the field.
$d_{field} = 30 \times 1.809 = 54.26$ m.

Step 3: Find the actual field area covered for those 30 turns.
Working width $= 8$ rows $\times 0.20$ m spacing $= 1.60$ m.
$A_{field} = 54.26 \times 1.60 = 86.82\ m^2 = 0.008682$ ha.

Step 4: The metering delivers a fixed seed mass per 30 turns, set by the calibration.
Since the drill mechanism is gear-driven off the wheel, 30 turns always release 814 g of seed, in the field just as on the stand.
\[ Rate_{field} = \frac{814\ g}{0.008682\ ha} = 93760\ g/ha = 93.76\ kg/ha \]

Final Answer:
The seed drill's actual field application rate works out to \[ \boxed{93.76\ kg/ha} \]