Interactive Lecture
Traction, contact-patch forces, and drivetrain layouts
by Dr Arnaldo Delli Carri
Every wheel on a vehicle belongs to one of two categories: it is either driven by a torque from the engine, or it is dragged along by the vehicle structure with no driving torque at the hub. The distinction matters because friction at the contact patch — the small area where tyre meets road — behaves very differently in each case, and that difference determines how a vehicle accelerates, how it loses traction, and why different drivetrain layouts (FWD, RWD, 4WD, AWD) suit different conditions.
Friction at the contact patch is what moves the car. The engine never touches the road — it is friction between tyre and road surface that converts hub torque into forward force. Without sufficient friction, no traction is possible regardless of how much power the engine produces.
When the engine delivers a torque \(T\) to a wheel hub, the wheel tends to rotate. The bottom of the tyre — the contact patch — tries to slip rearward relative to the road. Static friction resists this slip and acts forward on the wheel, propelling the vehicle. For a wheel rolling at constant speed, the torque from the engine and the friction moment about the wheel centre are in equilibrium:
\[F_\text{traction} = \frac{T}{r}\]
Where:
This traction force cannot exceed the maximum static friction the road can supply. The friction limit depends on the tyre–road static friction coefficient \(\mu_s\) and the normal force \(N\) pressing the wheel into the road:
\[F_\text{traction} \leq \mu_s N\]
Combining with the traction equation, the maximum hub torque before wheelspin is:
\[T_\text{max} = \mu_s \, N \, r\]
Where:
When \(T/r\) exceeds \(\mu_s N\), static friction is overcome and the wheel spins. Kinetic (sliding) friction takes over, which is lower than static friction — so the traction force drops, the car accelerates less than expected, the tyre wears rapidly, and steering authority is reduced. Set \(\mu_s = 0\) in the simulation below and you will see the limiting case: the wheel spins indefinitely in place because no forward force can be generated at all.
A non-driven wheel — say, the rear wheels of a front-wheel-drive car — has no engine torque at the hub. The chassis pushes the hub forward, and friction at the contact patch provides the moment needed to spin the wheel up to the correct rotational speed for the vehicle's velocity. In steady rolling the only force at the contact patch is the small rolling resistance:
\[F_\text{rr} = C_{rr} \, N\]
Where:
If the surface has no friction (\(\mu_s = 0\)), the contact patch cannot spin the wheel — it slides along without rotating, like a locked wheel skidding. No traction force is generated by a dragged wheel; it simply follows the vehicle.
Which wheels receive engine torque defines the drivetrain layout. Each arrangement distributes traction differently, and the total traction a vehicle can generate is limited by the sum of friction limits across all driven wheels:
\[F_\text{total,\,max} = \sum_{i=1}^{n_\text{driven}} \mu_s \, N_i\]
Where \(n_\text{driven}\) is the number of driven wheels and \(N_i\) is the normal force on wheel \(i\).
When a vehicle accelerates forward, inertia resists the change in velocity. Because the centre of gravity (CoG) sits above the ground, this inertial reaction creates a pitching moment about the contact patches, shifting load from the front axle to the rear. This is longitudinal weight transfer:
\[\Delta N = \frac{m \, a \, h_\text{cg}}{L}\]
Where:
The updated axle loads during acceleration are:
\[N_\text{front} = N_\text{front,static} - \Delta N \qquad N_\text{rear} = N_\text{rear,static} + \Delta N\]
Since the traction limit of each axle is \(\mu_s N_\text{axle}\), a loss of normal force is a direct loss of traction capacity. FWD is penalised under acceleration because the driven front axle loses load; RWD benefits because the driven rear axle gains it. A tall CoG or short wheelbase amplifies the transfer — the simulation below lets you see this directly.
Select the wheel type, then adjust the friction coefficient \(\mu_s\) and the hub torque. Watch the contact-patch force arrows and observe what happens when friction is too low to sustain rolling equilibrium. Try setting \(\mu_s = 0\) with a drive wheel to see wheelspin without forward motion.
The road is uniform — \(\mu_s\) is the same at every wheel. As throttle increases, weight transfers rearward: the rear axle gains load (and traction capacity), the front axle loses it. The axle load bars below the car show this shift in real time. Push the throttle until one axle spins, then switch drivetrain and compare. Try the SUV preset to see how a tall CoG amplifies the effect.
Two markers make the distinction between geometry and load visible. The yellow CoG dot sits at the geometric centre of mass and is fixed in the body — it pitches with the chassis but does not slide forward or back when you accelerate. The orange Load-Centre dot on the road shows where the resultant ground reaction is centred; as load shifts rearward, this point migrates towards the rear axle even though the CoG itself does not move.
| Concept | Equation | Description |
|---|---|---|
| Drive-wheel traction force | \(F_x = T / r\) | Forward friction force at contact patch produced by hub torque |
| Traction limit (static) | \(F_x \leq \mu_s N\) | Maximum traction before wheelspin; \(N\) is the wheel's normal force |
| Maximum hub torque | \(T_\text{max} = \mu_s N r\) | Highest torque a driven wheel can transmit without spinning |
| Weight transfer under acceleration | \(\Delta N = m a h_\text{cg} / L\) | Load shifted from front to rear axle; \(h_\text{cg}\) = CoG height, \(L\) = wheelbase |
| Rolling resistance | \(F_\text{rr} = C_{rr} N\) | Small opposing force at a dragged-wheel contact patch; \(C_{rr} \approx 0.010\text{–}0.020\) |
| Total vehicle traction | \(F_\text{total,\,max} = \displaystyle\sum_{i} \mu_s N_i\) | Sum of friction limits across all driven wheels |
A front-wheel-drive car has a total mass of 1 400 kg. The front axle carries 58% of the vehicle weight. The engine delivers a torque of 240 N·m to each front hub via the driveshafts, and each wheel has a rolling radius of 0.33 m. The road is dry tarmac with \(\mu_s = 0.85\). Take \(g = 9.81\) m/s².
(a) Calculate the traction force generated at each front wheel.
(b) Determine the maximum traction force available at each front wheel before wheelspin.
(c) State whether wheelspin occurs.
Show Solution
Solution:
Given:
(a) Traction force per front wheel:
\[F_x = \frac{T}{r} = \frac{240}{0.33} = 727 \text{ N}\]
(b) Maximum traction force before wheelspin:
Total weight on the front axle:
\[W_\text{front} = 0.58 \times 1400 \times 9.81 = 7966 \text{ N}\]
Normal force per front wheel (two wheels share the axle load):
\[N = \frac{7966}{2} = 3983 \text{ N}\]
Maximum traction per wheel:
\[F_\text{max} = \mu_s N = 0.85 \times 3983 = 3386 \text{ N}\]
(c) Wheelspin check:
Since \(F_x = 727 \text{ N} < F_\text{max} = 3386 \text{ N}\), the traction limit is not exceeded.
Answer: (a) \(F_x = 727\) N. (b) \(F_\text{max} = 3386\) N. (c) No wheelspin occurs.
A rear-wheel-drive sports car has a total mass of 1 600 kg, with 50% of the weight on the rear axle. The gearbox provides a ratio of 3.2:1 and the engine produces a peak torque of 450 N·m. The final drive (differential) splits torque equally between the two rear wheels. The wheel radius is 0.35 m. The road is wet: \(\mu_s = 0.55\). Take \(g = 9.81\) m/s².
(a) Find the torque delivered to each rear hub.
(b) Find the traction force at each rear wheel.
(c) Determine whether wheelspin occurs.
Show Solution
Solution:
Given:
(a) Hub torque per rear wheel:
\[T_\text{hub} = \frac{T_\text{eng} \times \text{gear ratio}}{2} = \frac{450 \times 3.2}{2} = 720 \text{ N·m}\]
(b) Traction force per rear wheel:
\[F_x = \frac{T_\text{hub}}{r} = \frac{720}{0.35} = 2057 \text{ N}\]
(c) Wheelspin check:
Normal force per rear wheel:
\[N = \frac{0.50 \times 1600 \times 9.81}{2} = \frac{7848}{2} = 3924 \text{ N}\]
Maximum traction per rear wheel:
\[F_\text{max} = \mu_s N = 0.55 \times 3924 = 2158 \text{ N}\]
Since \(F_x = 2057 \text{ N} < F_\text{max} = 2158 \text{ N}\), the rear wheels are just within the traction limit — but operating very close to it.
Note: if \(\mu_s\) dropped to 0.52 (slightly wetter surface), \(F_\text{max} = 0.52 \times 3924 = 2040 \text{ N}\), which is less than 2057 N, and the rear wheels would spin.
Answer: (a) \(T_\text{hub} = 720\) N·m. (b) \(F_x = 2057\) N. (c) No wheelspin — margin is 101 N (4.7%).
An SUV has a total mass of 2 200 kg, with 48% of the weight on the front axle and 52% on the rear axle. The wheel radius is 0.40 m. The surface is icy: \(\mu_s = 0.10\). Take \(g = 9.81\) m/s².
(a) Calculate the maximum total traction force available in RWD mode (rear wheels only).
(b) Calculate the maximum total traction force available in 4WD mode (all four wheels driven).
(c) Calculate the maximum hub torque that can be delivered to each rear wheel in RWD without wheelspin.
Show Solution
Solution:
Given:
Normal forces per wheel:
\[N_\text{front} = \frac{0.48 \times 2200 \times 9.81}{2} = \frac{10\,359.36}{2} = 5180 \text{ N}\]
\[N_\text{rear} = \frac{0.52 \times 2200 \times 9.81}{2} = \frac{11\,222.64}{2} = 5611 \text{ N}\]
(a) Maximum traction in RWD (two rear wheels driven):
\[F_\text{RWD} = 2 \times \mu_s N_\text{rear} = 2 \times 0.10 \times 5611 = 1122 \text{ N}\]
(b) Maximum traction in 4WD (all four wheels driven):
\[F_\text{4WD} = 2\,\mu_s N_\text{front} + 2\,\mu_s N_\text{rear} = 2(0.10)(5180) + 2(0.10)(5611)\]
\[= 1036 + 1122 = 2158 \text{ N}\]
4WD provides nearly twice the traction of RWD on this surface.
(c) Maximum hub torque per rear wheel before wheelspin in RWD:
\[T_\text{max} = \mu_s N_\text{rear} \times r = 0.10 \times 5611 \times 0.40 = 224 \text{ N·m}\]
Answer: (a) \(F_\text{RWD} = 1122\) N. (b) \(F_\text{4WD} = 2158\) N. (c) \(T_\text{max} = 224\) N·m.