Three-Bottom Plough Behind a 45 hp Tractor: Power, Ballast or a Bigger Machine?
A 45 hp (33.6 kW) two-wheel-drive tractor weighing 2,000 kg carries 70% of its weight on the rear axle and pulls a three-bottom mouldboard plough, each bottom 35 cm wide. The owner wants 22 cm depth at 5 km/h on 6 ha/day in a sandy clay loam where soil resistance is 0.55 kg/cm². Field efficiency is 78% over a 10-hour day, drawbar power is 72% of engine power, rolling resistance is 8% of tractor weight, the coefficient of traction on the firm moist seedbed is 0.65, and the target wheel slip is 15%. There is no ballast. The owner complains that the outfit 'spins and stops' at the headland and wants to buy either a fourth bottom or a 60 hp tractor. A second field on the same soil series, planned for the same week, measured 0.90 kg/cm² resistance at 28% gravimetric moisture.
1. Compute the draft and drawbar power needed. Can the tractor pull three bottoms at the stated depth and speed, and will it cover 6 ha/day?
Read the worked answer
Soil area cut = 3 × 35 cm × 22 cm = 2,310 cm².
Draft = 0.55 kg/cm² × 2,310 cm² = 1,270 kgf ≈ 12.46 kN.
At 5 km/h = 1.389 m/s, drawbar power = 12.46 × 1.389 ≈ 17.3 kW, against 33.6 × 0.72 = 24.2 kW available. Power is adequate with a 40% margin.
But width 1.05 m and 5 km/h over 10 h gives 5.25 ha gross; at 78% efficiency the real output is about 4.1 ha/day, not 6 ha. So the 6 ha/day target fails even if traction were perfect, and adding a fourth bottom widens the cut to 1.40 m - 5.5 ha/day - while raising draft to 16.6 kN, making the traction problem worse.
The complaint is thus about traction, not engine power, so the 60 hp tractor is the wrong fix.
2. Quantify the traction deficit and recommend a ballast figure. Would ballasting alone solve the problem?
Read the worked answer
Available gross tractive force = 0.65 × (0.70 × 2,000 × 9.81) = 0.65 × 13.73 kN = 8.93 kN. Deduct rolling resistance (0.08 × 19.62 kN = 1.57 kN) and net usable pull is only about 7.4 kN against 12.46 kN required - hence very high slip.
For 15% slip, the gross force must be roughly draft/(1 − slip) = 12.46/0.85 = 14.66 kN, so the rear axle needs 14.66/0.65 = 22.6 kN, i.e. about 2,300 kg rear weight. Current rear load is 1,400 kg, so roughly 900 kg of rear wheel ballast (water-filled tyres plus wheel weights) is needed, bringing the tractor to about 2,900 kg.
Ballasting works on this field, but it raises rolling resistance and soil compaction, and the tractor is then near its tyre load limit. The honest answer is that the machine is under-weighted for a 1.05 m plough; ballast is the short-term fix, a heavier tractor only if the 6 ha/day target is truly binding.
3. In the wet field (0.90 kg/cm²) what happens, and what is the correct operational decision?
Read the worked answer
Draft = 0.90 × 2,310 = 2,079 kgf ≈ 20.4 kN, needing 20.4 × 1.389 = 28.3 kW drawbar, above the 24.2 kW available. Required rear weight for 15% slip would be about 20.4/(0.85 × 0.65) = 36.9 kN ≈ 3,760 kg - impossible for a 2,000 kg tractor, and wet soil would drop the traction coefficient well below 0.65 anyway.
Reducing to two bottoms (draft 13.6 kN) or to 15 cm depth at 4 km/h (draft 13.9 kN, 15.4 kW) brings power inside the envelope but still exceeds traction, and ploughing at 28% moisture smears the furrow base and creates a compacted plough pan that costs the next crop far more than a week's delay.
Recommendation: take the tractor off this field and wait for moisture to fall; schedule the drier field first, use the ballasted three-bottom set there, and only enter the wet field with reduced depth once it is workable. Neither a fourth bottom nor a 60 hp tractor fixes a soil-moisture problem.