An 8×20 cm seed drill has a .60 m diameter ground wheel. A hard-surface calibration delivers 814 g of seed in 30 wheel revolutions. In soft soil, effective rolling circumference decreases by 4%. Ignore other losses and use π=3.14. Find actual field application rate in kg/ha, to two decimals.
Working width is $8(.20)=1.60$ m. Effective travel per revolution is $.96\pi D=.96(3.14)(.60)=1.80864$ m. Area for 30 revolutions is $30(1.80864)(1.60)=86.81472$ m². Seed output remains .814 kg, so rate is $.814(10000)/86.81472=\mathbf{93.76}$ kg/ha. The smaller circumference reduces covered area and increases application rate.
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