Farm Power and Machinery
Farm Power and Machinery tests how power is generated, transmitted and matched to implements, and how field performance and costs are calculated. Expect numerical items on engine cycles, tractor systems, draft, field capacity and implement selection, plus practical judgments on safety and maintenance.
Core rules and concepts
- A four-stroke diesel engine completes suction, compression, power and exhaust strokes in two crankshaft revolutions; ignition is by compression heat, not a spark plug.
- Tractor power reaches the ground as drawbar power and the implement through the PTO. The standard independent PTO speed is 540 rpm; 1000 rpm systems exist but are not the default.
- Drawbar power (kW) = draft (kN) × forward speed (m/s). Convert speed from km/h by dividing by 3.6.
- Theoretical field capacity (ha/h) = working width (m) × speed (km/h) ÷ 10. Effective field capacity = theoretical × field efficiency.
- Draft rises with width, depth, speed and soil strength; a wider or deeper implement needs more power and may exceed traction.
Worked example
A 3 m wide harrow works at 6 km/h with 80% field efficiency for 8 hours. Theoretical capacity = 3 × 6 ÷ 10 = 1.8 ha/h. Effective capacity = 1.8 × 0.80 = 1.44 ha/h. Daily work = 1.44 × 8 = 11.52 ha. If draft were 9 kN at 6 km/h (1.667 m/s), drawbar power = 9 × 1.667 = 15 kW. Reasoning: calculate capacity before daily area; keep width in metres and speed in km/h for the shortcut.
Traps that cost marks
Do not use engine horsepower as drawbar power; transmission and traction losses intervene. Do not insert speed in m/s into the field-capacity shortcut. Field efficiency is below 100% because of turning, loading, repairs and overlap. Mixing hp and kW without converting, or forgetting that 1 hp ≈ 0.746 kW, is a common numerical error.