Tractive effort, resistances and gear ratio selection

Why it matters

Every gearbox and final-drive ratio in a vehicle is chosen by balancing the force the engine can deliver at the wheels against the forces resisting motion. The same balance tells you a vehicle's top speed, the steepest gradient it can climb, how fast it accelerates and whether the tyres will spin before the engine runs out of torque. It is the starting calculation for any transmission design and a regular source of numerical questions.

Key ideas

• Tractive effort is the forward force at the tyre–road contact patch produced by the driving wheels. It is the engine torque multiplied through the gearbox and final drive, reduced by transmission losses, and divided by the wheel's dynamic rolling radius.
• Total resistance to motion is the sum of:
  • Rolling resistance – mainly hysteresis loss in the deforming tyre (plus road deformation and bearing friction). It is roughly proportional to the normal load and nearly independent of speed at moderate speeds. Typical coefficients: about 0.010–0.015 for car tyres on good asphalt, higher on poor or soft surfaces (take values from your data book).
  • Aerodynamic (air) resistance – grows with the square of speed relative to the air, so it dominates at highway speeds; the power to overcome it grows with the cube of speed.
  • Gradient resistance – the component of weight along the slope, m·g·sin θ. Road gradients are usually quoted as a percentage, 100·tan θ, or as "1 in n" (meaning sin θ = 1/n in most automotive texts; for small slopes tan θ and sin θ are almost equal).
  • Acceleration resistance – the force needed to accelerate the vehicle's mass plus the equivalent mass of rotating parts (wheels, shafts, gears, flywheel). The rotating-mass effect is largest in low gears because the engine's inertia is multiplied by the square of the overall ratio.
• Drawbar pull (surplus tractive effort) is the tractive effort minus the road resistances at a given speed. It is what is available for acceleration, climbing or towing.
• Adhesion limit. Tractive effort cannot exceed the friction available at the driving wheels: F ≤ μ·W_d, where W_d is the normal load on the driven axle. If the engine-limited tractive effort is higher than this, the wheels spin and the extra gear reduction is useless.
• Performance curves. Plotting tractive effort against vehicle speed for each gear (a family of hyperbola-like curves) and overlaying the resistance curve for different gradients shows top speed (where the top-gear curve meets the level-road resistance curve), gradeability in each gear and the acceleration reserve.
• Gear ratio selection:
  • Top gear (with the final drive) is set so the vehicle reaches its intended maximum speed at or near the engine speed of maximum power on a level road. A slightly taller ratio ("overdrive") gives better fuel economy at the cost of acceleration.
  • First (lowest) gear is set by the maximum gradient the vehicle must climb or start on, and by the adhesion limit.
  • Intermediate gears are commonly spaced in geometric progression, so the engine drops back to the same speed after every upshift and stays in its useful speed band. Real gearboxes often close up the upper steps slightly.
• Transmission efficiency (typically about 0.85–0.95 overall) reduces wheel torque; it does not change speed ratio.

Formulas

F_t = T_e · i_g · i_f · η_t / r

• F_t: tractive effort at the driving wheels (N); T_e: engine torque (N·m); i_g: gearbox ratio; i_f: final-drive ratio (both dimensionless, input speed ÷ output speed); η_t: transmission efficiency; r: dynamic rolling radius of the driving wheel (m).

v = 2π · r · N_e / (60 · i_g · i_f)

• v: vehicle speed (m/s); N_e: engine speed (rev/min). Assumes no tyre slip and a locked clutch or torque converter. Multiply by 3.6 for km/h.

R_r = f · m · g · cos θ

• R_r: rolling resistance (N); f: coefficient of rolling resistance (–); m: vehicle mass (kg); g = 9.81 m/s²; θ: road inclination. On level road cos θ = 1.

R_a = ½ · ρ · C_d · A · v²

• R_a: aerodynamic resistance (N); ρ: air density (kg/m³, about 1.2 kg/m³ at sea level, 20 °C); C_d: drag coefficient (–); A: frontal area (m²); v: speed relative to the air (m/s). With a head wind, use v + v_wind.

R_g = m · g · sin θ, with tan θ = gradient (%) / 100

• R_g: gradient resistance (N), positive uphill, negative downhill.

F_t − (R_r + R_a + R_g) = m_e · a, where m_e = m · (1 + δ)

• a: acceleration (m/s²); m_e: effective mass including rotating parts (kg); δ: rotational-mass factor (–), larger in low gears.

P = F_t · v / 1000 and P_e = R · v / (1000 · η_t)

• P: power at the wheels (kW); P_e: engine power needed to overcome resistance R at speed v (kW).

F_t,max ≤ μ · W_d

• μ: tyre–road adhesion coefficient (–); W_d: normal load on the driving axle (N).

k = (i_1 / i_n)^(1/(n − 1)), i_j = i_1 / k^(j − 1)

• Geometric step ratio k for an n-speed gearbox with lowest ratio i_1 and highest ratio i_n.

Worked examples

Example 1 (standard). A 1500 kg car climbs a 5% gradient at a steady 72 km/h in still air. Given f = 0.015, C_d = 0.30, A = 2.2 m², ρ = 1.225 kg/m³, η_t = 0.90. Find the tractive effort and engine power required.

1. Speed: v = 72 / 3.6 = 20 m/s.
2. Angle: tan θ = 0.05, so θ = 2.862°, sin θ = 0.04994, cos θ = 0.99875.
3. R_r = f · m · g · cos θ = 0.015 × 1500 × 9.81 × 0.99875 = 220.4 N.
4. R_a = ½ · ρ · C_d · A · v² = 0.5 × 1.225 × 0.30 × 2.2 × 20² = 161.7 N.
5. R_g = m · g · sin θ = 1500 × 9.81 × 0.04994 = 734.8 N.
6. Steady speed, so F_t = R_r + R_a + R_g = 220.4 + 161.7 + 734.8 = 1116.9 N.
7. Wheel power = F_t · v = 1116.9 × 20 = 22 338 W = 22.34 kW; engine power = 22.34 / 0.90 = 24.8 kW.

Answer: F_t ≈ 1117 N; engine power ≈ 24.8 kW.

Note that a 5% grade adds far more resistance than the air does at 72 km/h.

Example 2 (GATE level). A front-wheel-drive car of mass 1200 kg carries 60% of its weight on the front axle. Maximum engine torque is 140 N·m, first-gear ratio 3.5, final drive 4.1, η_t = 0.90, dynamic wheel radius 0.30 m, f = 0.015, adhesion coefficient μ = 0.70. Neglect air resistance (low speed) and load transfer on the slope. Find the steepest gradient (in %) the car can climb steadily in first gear.

1. Engine-limited tractive effort: F_t = T_e · i_g · i_f · η_t / r = 140 × 3.5 × 4.1 × 0.90 / 0.30 = 6027 N.
2. Engine limit on grade: F_t = m·g·(sin θ + f·cos θ) gives sin θ + 0.015 cos θ = 6027 / (1200 × 9.81) = 0.5120. Solving, θ = 29.9°, i.e. tan θ = 0.576, a gradient of 57.6%.
3. Adhesion limit: available traction = μ · 0.6 · m·g·cos θ. Setting this equal to the resistance m·g·(sin θ + f·cos θ) and dividing by m·g·cos θ: tan θ = 0.6μ − f = 0.6 × 0.70 − 0.015 = 0.405.
4. The adhesion limit (40.5%) is lower than the engine limit (57.6%), so the front wheels would spin first.

Answer: maximum gradient ≈ 40.5% (θ ≈ 22.0°), limited by adhesion, not by the engine.

Example 3 (ratio spacing). A four-speed gearbox has i_1 = 3.5 and i_4 = 1.0. Find geometric-progression intermediate ratios.

1. k = (3.5 / 1.0)^(1/3) = 1.518.
2. i_2 = 3.5 / 1.518 = 2.305; i_3 = 2.305 / 1.518 = 1.518.

Answer: i_2 ≈ 2.31, i_3 ≈ 1.52.

Common mistakes

• Treating a percentage gradient as an angle in degrees, or using sin θ = gradient/100 without saying it is a small-angle approximation (5% gives sin θ = 0.04994, nearly 0.05; 40% gives 0.371, not 0.40).
• Forgetting cos θ in rolling resistance on steep grades, or forgetting that air resistance uses speed in m/s, not km/h.
• Multiplying engine torque by the gear ratio but forgetting the final drive or the efficiency, or dividing by wheel diameter instead of radius.
• Using "higher gear ratio" loosely: first gear has the numerically largest ratio and the largest tractive effort; top gear has the numerically smallest ratio.
• Ignoring the adhesion limit, so a very low gear appears to climb impossible slopes.
• Using engine power at the wheels without the transmission efficiency, or mixing kW and W.

For GATE ME

Expect numericals on: tractive effort at the wheels from torque and ratios; vehicle speed from engine speed and ratios; total resistance on a grade at a given speed; power required at the wheels or the engine; maximum gradeability or acceleration in a gear; adhesion-limited traction for front-, rear- or four-wheel drive; and geometric spacing of intermediate gears. Practise solving the sin θ + f·cos θ equation and converting between percentage, "1 in n" and degrees quickly.

Quick check

1. Which resistance grows with the square of vehicle speed?
2. A car has T_e = 200 N·m, overall ratio 12, η_t = 0.9 and r = 0.3 m. What is the tractive effort?
3. Why does the top-gear ratio usually match maximum vehicle speed to the engine speed of maximum power?
4. For a 3-speed gearbox with i_1 = 4 and i_3 = 1, what is the geometric middle ratio?
5. What limits tractive effort other than engine torque?

Answers: 1. Aerodynamic resistance. 2. 200 × 12 × 0.9 / 0.3 = 7200 N. 3. So the full engine power is available exactly where the level-road resistance curve meets the top-gear tractive-effort curve. 4. 2.0. 5. Tyre–road adhesion, μ times the load on the driving wheels.