Tractors and Power Tillers: Transmission, Traction, Three-Point Hitch, Hydraulics and Steering, Testing and Safety
2. Clutches, brakes and the power-transmission system
The clutch connects the engine to the gearbox smoothly and disconnects it for gear changes. A single-plate dry friction clutch transmits T = μ·W·R_m·n, where W is the axial spring force, n the number of friction surfaces (two for a single plate) and R_m the mean radius — (R₁ + R₂)/2 under the uniform-wear assumption used for a worn-in clutch. For μ = 0.3, W = 3000 N, R₁ = 150 mm and R₂ = 100 mm, T = 0.3 × 3000 × 0.125 × 2 = 225 N·m. A dual clutch has two plates operated by one pedal in stages, so the gearbox can be disengaged while the PTO keeps running — the live PTO a mower or baler needs. Brakes (drum, or wet disc in oil) act on each rear wheel independently, so braking one wheel sharpens a turn; the two pedals are locked together for road travel.
The gearbox (sliding-mesh, constant-mesh or synchromesh, often with high and low ranges) multiplies torque and sets the travel speed. The differential lets the two driving wheels turn at different speeds on a turn while sharing the torque equally; the sum of the two wheel speeds equals twice the crown-wheel speed, so if one wheel stands still the other turns at twice the crown-wheel speed. That same property lets a wheel on a slippery patch spin while the other stops, which the differential lock prevents by locking both half-axles together — for straight-line pulling only, never on a turn. Final drives give a last large reduction just before the wheels. The PTO shaft runs at a standard speed of 540 rev/min (or 1000 rev/min for larger machines), at a specified engine speed, so any PTO machine fits any tractor; a belt pulley serves stationary machines.
Travel speed follows from the chain of reductions. An engine at 2100 rev/min through an overall reduction of 70:1 turns the wheels at 30 rev/min; with a wheel diameter of 1.4 m the theoretical speed is π × 1.4 × 30 × 60/1000 = 7.92 km/h, and the actual speed is lower by the wheel slip.
3. Mechanics of the tractor chassis and traction theory
A driving wheel on soil always travels less than its rolling circumference per turn, because it shears the soil to develop pull. The wheel slip is s = (V_t − V_a)/V_t, where V_t is the theoretical (no-slip) speed and V_a the actual speed, equivalently (distance per revolution unloaded − loaded)/unloaded. The soil reaction on the wheel gives a gross traction; part is spent on motion resistance (rolling resistance, largely the work of making the rut), and the remainder is the net traction or drawbar pull. The coefficient of net traction is pull divided by the dynamic weight on the driving wheels, and the tractive efficiency is drawbar power over axle power: TE = (P·V_a)/(T·ω) = (P/GT)(1 − s). Tractive efficiency peaks at a moderate slip — roughly 8 to 15 per cent, lower on firm soil and higher on loose soil — because at low slip little pull is developed and at high slip speed is thrown away; this is the basis for ballasting a tractor.
Weight transfer. Taking moments about the rear-wheel contact of a two-wheel-drive tractor with wheelbase L, a horizontal drawbar pull P applied at height h lifts the front axle by P·h/L and adds the same to the rear. With static reactions of 10 kN (front) and 20 kN (rear), L = 2 m, h = 0.5 m and P = 10 kN, the transfer is 2.5 kN, so the rear carries 22.5 kN and the front 7.5 kN. The transfer improves traction, but if P·h approaches the front static reaction times L the front wheels lift and the tractor can overturn backwards — so the hitch must stay low, and a tractor pulling a stuck load must never be hitched above the drawbar.
Fuel consumption follows from the engine’s specific fuel consumption: litres per hour = bsfc (kg/kWh) × power (kW)/density (kg/L). At 30 kW and 0.25 kg/kWh, with diesel of density 0.84 kg/L, a tractor burns 7.5 kg/h = 8.93 L/h. Dividing by the effective field capacity turns this into litres per hectare, the figure that decides the cost of an operation.
4. The three-point hitch, hydraulic control and steering
The three-point linkage carries a mounted implement on two lower links and one top link. In side view the lower and top links, extended, meet at the virtual hitch point (instantaneous centre), about which the implement pivots; it decides how the implement responds to changes in soil force. The linkage is used in two ways. In free-link (floating) operation the hydraulic system is released, the links can move freely, and the implement finds its own depth on its gauge wheel or its own soil forces, following the ground. In restrained-link operation the hydraulic system holds the links, so the tractor controls the implement’s depth.
| Mode | What it holds constant | Suits |
|---|---|---|
| Position control | the height of the implement relative to the tractor, whatever the draft | level ground; implements carried clear of the soil, sprayers, graders |
| Draft control | the draft: rising draft raises the implement slightly, falling draft lowers it | ploughs and cultivators in soil of varying resistance; protects against overload |
| Mixed (intermix) control | a blend of the two | uneven fields where pure draft control varies depth too much |
Draft is sensed through the linkage itself — through the top link on smaller tractors, whose compression varies with draft, and through the lower links on larger ones. The hydraulic system is a gear pump, a control valve, a lift cylinder and a reservoir; its power is p × Q, so 15 MPa at 30 L/min (5 × 10⁻⁴ m³/s) is 7.5 kW. Steering is mechanical (worm-and-sector or recirculating-ball boxes), power-assisted, or fully hydrostatic. Correct steering geometry follows the Ackermann principle: all wheels roll about one centre on the rear-axle line, which requires cot φ_o − cot φ_i = w/L for the outer and inner steered angles, with w the kingpin spacing and L the wheelbase. For L = 2 m, w = 1.2 m and φ_i = 30°, cot φ_o = 1.732 + 0.6 = 2.332, so φ_o = 23.2°.
5. Tractor tests and performance; human engineering and safety
Tractors are tested to standard codes — internationally the OECD codes, in India by the national testing institutes to the Bureau of Indian Standards procedures, the central one being the Central Farm Machinery Training and Testing Institute at Budni. The main tests are: PTO performance (maximum power, and power and fuel consumption at varying load and speed, giving the torque and specific-fuel-consumption curves); drawbar performance on a test track in each gear (pull, speed, slip, drawbar power and fuel consumption); hydraulic lift capacity and maintenance of lift; brake performance; turning radius and turning space; noise at the operator’s ear and in the surroundings; vibration at the seat; and field trials with implements. The ratio of drawbar to PTO power in these tests is the best single number for comparing tractors’ ability to pull.
- Human engineering (ergonomics): seats, pedals, levers and steering placed within the reach and strength of the operator population, sized from anthropometric data (for example the 5th to 95th percentile), with forces on controls kept within comfortable limits.
- Noise and vibration: sound levels are measured in dB(A); whole-body vibration at the seat is reduced by suspension seats and cab mounts.
- Safety: roll-over protective structures (ROPS) with a seat belt against overturning; shields on the PTO shaft and moving parts; a low hitch and front ballast against backward overturning; slow-moving-vehicle marking and lights on roads; safe starting only in neutral.
- Implement design: guards on cutter bars, chaff-cutter feed rollers and thresher feeding chutes — the thresher feeding chute is a documented cause of hand injuries, and a long covered chute keeps the hand away from the cylinder.
Key takeaways
- Wheel slip = (V_t − V_a)/V_t; tractive efficiency = drawbar power/axle power, and it peaks at moderate slip.
- A drawbar pull P at height h transfers P·h/L from the front axle to the rear; too high a hitch overturns the tractor backwards.
- Clutch torque T = μWR_m n with R_m = (R₁ + R₂)/2 for uniform wear; the PTO standard is 540 rev/min (or 1000).
- Free-link lets the implement float on its own forces; restrained-link lets the hydraulics hold it — by position or by draft.
- Fuel use in L/h = bsfc × kW/density; Ackermann steering needs cot φ_o − cot φ_i = w/L.
Practice questions (13)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Unloaded, a tractor’s driving wheels carry it 100 m in 20 revolutions; pulling an implement, 20 revolutions carry it only 85 m. What is the wheel slip, in per cent?
Numerical answer — type the value.
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Answer: 15
s = (100 − 85)/100 = 0.15 = 15 per cent, measured against the no-slip (unloaded) travel. Dividing by the loaded distance gives 17.6 per cent, the common wrong option.A tractor exerts a drawbar pull of 12 kN at 4.5 km/h. What is its drawbar power, in kW?
Numerical answer — type the value.
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Answer: 15
Drawbar power = pull × speed = 12 kN × 4.5/3.6 m/s = 12 × 1.25 = 15 kW. Multiplying 12 by 4.5 directly gives 54, which forgets that km/h must be turned into m/s.A tractor delivers 15 kW at the drawbar when the power at its driving axle is 22.5 kW. What is its tractive efficiency, in per cent (to one decimal place)?
Numerical answer — type the value.
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Answer: 66.7
TE = drawbar power/axle power = 15/22.5 = 0.667 = 66.7 per cent; the remaining 7.5 kW goes into slip and motion resistance. Comparing drawbar power with engine or PTO power gives a lower figure that includes transmission losses — a different efficiency.A two-wheel-drive tractor of wheelbase 2.0 m has static axle loads of 10 kN on the front and 20 kN on the rear. It exerts a horizontal drawbar pull of 10 kN at a height of 0.5 m above the ground. What is the load on the rear axle while pulling, in kN?
Numerical answer — type the value.
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Answer: 22.5
Moments about the rear contact: the pull’s moment P·h = 10 × 0.5 = 5 kN·m is balanced by a reduction of 5/2 = 2.5 kN in the front reaction, and vertical equilibrium adds the same 2.5 kN to the rear: 20 + 2.5 = 22.5 kN. Subtracting from the rear instead reverses the direction of weight transfer.A tractor engine delivers 30 kW with a specific fuel consumption of 0.25 kg/kWh. Taking the density of diesel as 0.84 kg/L, what is the fuel consumption, in L/h (to two decimal places)?
Numerical answer — type the value.
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Answer: 8.93
Mass rate = 0.25 × 30 = 7.5 kg/h; volume rate = 7.5/0.84 = 8.93 L/h. Multiplying by the density instead of dividing gives 6.30 L/h, which would make diesel denser than water.A tractor engine runs at 2100 rev/min and the overall reduction from engine to driving wheels is 70:1. The wheels are 1.4 m in diameter. What is the theoretical travel speed, in km/h (to two decimal places)?
Numerical answer — type the value.
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Answer: 7.92
Wheel speed = 2100/70 = 30 rev/min; distance per minute = π × 1.4 × 30 = 131.9 m; speed = 131.9 × 60/1000 = 7.92 km/h. Using the diameter as a radius doubles it to 15.83 km/h; the actual speed is lower by the wheel slip.A single-plate clutch with both faces effective has friction linings of outer radius 150 mm and inner radius 100 mm. The axial spring force is 3000 N and the coefficient of friction is 0.3. Assuming uniform wear, what torque can it transmit, in N·m?
Numerical answer — type the value.
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Answer: 225
Uniform wear gives R_m = (0.15 + 0.10)/2 = 0.125 m; with two friction surfaces T = μWR_m n = 0.3 × 3000 × 0.125 × 2 = 225 N·m. Forgetting that both faces work gives 112.5 N·m; the uniform-pressure mean radius, (2/3)(R₁³ − R₂³)/(R₁² − R₂²) = 0.1267 m, gives a slightly higher 228 N·m.In free-link operation of a tractor’s three-point hitch:
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Answer: A — the hydraulic system is released and the implement finds its working depth from its own gauge wheel and soil forces
Free-link means the links float: the implement follows the ground on its own gauge wheel and the balance of its soil forces, with no hydraulic input. Holding a fixed height is position control and sensing draft is draft control — both restrained-link modes, which is the distinction the question tests.Which of the following statements about the hydraulic control of a tractor’s three-point linkage are correct?
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Answer: A — Under draft control, a rise in draft above the set value raises the implement slightly; B — Position control keeps the implement at a set height relative to the tractor whatever the draft; D — On smaller tractors the draft is commonly sensed through the top link
Draft control lifts the implement a little when draft rises, keeping the load on the tractor near constant; position control holds height regardless of draft; smaller tractors sense draft by the compression of the top link, larger ones through the lower links. In free-link operation the hydraulics are released, so the third statement is false.The differential lock on a tractor is engaged to:
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Answer: A — make both driving wheels turn together when one of them is slipping
The differential normally sends equal torque to both wheels, so a wheel on a slippery patch spins while the other stands; the lock joins the half-axles so both turn together and the gripping wheel pulls. Allowing different speeds on a turn is what the unlocked differential does, which is why the lock must be released before turning.A tractor has a wheelbase of 2.0 m and a kingpin spacing of 1.2 m. For correct Ackermann steering, if the inner front wheel is turned through 30°, through what angle must the outer wheel turn, in degrees (to one decimal place)?
Numerical answer — type the value.
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Answer: 23.2
cot φ_o = cot φ_i + w/L = cot 30° + 1.2/2.0 = 1.732 + 0.6 = 2.332, so φ_o = tan⁻¹(1/2.332) = 23.2°. The outer wheel turns less than the inner; subtracting w/L instead gives cot φ_o = 1.132 and 41.5°, which would make the outer wheel turn more.A tractor’s hydraulic pump delivers 30 L/min at a pressure of 15 MPa. What hydraulic power does it deliver, in kW?
Numerical answer — type the value.
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Answer: 7.5
Q = 30 L/min = 0.030/60 = 5 × 10⁻⁴ m³/s, and P = pQ = 15 × 10⁶ × 5 × 10⁻⁴ = 7500 W = 7.5 kW. Converting litres to m³ but forgetting to divide by 60 gives 450 kW, and leaving the flow in litres per second without the 10⁻³ gives 7500 kW.A roll-over protective structure (ROPS) on a tractor is intended mainly to:
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Answer: A — protect the operator, held by a seat belt, if the tractor overturns
A ROPS does not stop an overturn; it creates a survival zone around a seat-belted operator when one happens. Preventing a backward overturn is done by a low hitch and front ballast, and the PTO has its own shield.