Interactive Lecture
Understanding Longitudinal and Lateral Vehicle Motion
by Dr Arnaldo Delli Carri
Vehicle dynamics studies the motion of vehicles during acceleration, braking, and cornering. It combines rigid body dynamics, tyre mechanics, and control theory to predict and optimise vehicle behaviour for safety, performance, and comfort.
Vehicle behaviour is governed by the forces at the tyre–road contact patch. Everything else — cornering, braking, stability — flows from what happens at those four small footprints.
Vehicle motion is described in multiple coordinate systems:
Vehicle-Fixed Coordinates:
Primary DOFs:
The bicycle model is a simplified 2D representation that captures essential lateral dynamics:
Lateral Force Balance:
\[m(\dot{v} + ur) = F_{yf} + F_{yr}\]
Yaw Moment Balance:
\[I_z \dot{r} = l_f F_{yf} - l_r F_{yr}\]
where:
Tyre forces depend on the slip angle, which is the difference between tyre orientation and direction of travel:
Front Slip Angle:
\[\alpha_f = \delta - \frac{v + l_f r}{u}\]
Rear Slip Angle:
\[\alpha_r = -\frac{v - l_r r}{u}\]
where \(\delta\) is the front wheel steering angle.
Linear Tyre Model:
\[F_y = C_\alpha \alpha\]
where \(C_\alpha\) is the cornering stiffness (N/rad), typically 50,000-100,000 N/rad for passenger cars. With the slip-angle sign convention above (\(\alpha_f = \delta - (v + l_f r)/u\)), a positive slip angle produces a positive lateral force, so the coefficient is positive.
The steady-state cornering behaviour is characterised by the understeer gradient:
Understeer Gradient:
\[K = \frac{m}{L}\left(\frac{l_r}{C_{\alpha f}} - \frac{l_f}{C_{\alpha r}}\right)\]
where \(L = l_f + l_r\) is the wheelbase.
Classification:
Acceleration and braking performance depend on force balance:
Longitudinal Equation of Motion:
\[m\dot{u} = F_x - F_{drag} - F_{roll} - F_{grade}\]
where:
Weight Transfer During Braking:
\[\Delta F_z = \frac{m\dot{u} h}{L}\]
where \(h\) is the CG height above ground.
For an oversteering vehicle, there exists a critical speed:
\[u_{crit} = \sqrt{\frac{L}{-K}}\]
Above this speed, the vehicle becomes directionally unstable and cannot maintain steady cornering. This is why most production vehicles are designed with understeer.
This simulation shows a top-down view of a vehicle performing a set of test manoeuvres. Adjust parameters to see how vehicle characteristics affect handling behaviour.
Engineers reach for vehicle dynamics wherever handling, stability, or safety has to be designed in rather than discovered on the road:
| Concept | Equation | Description |
|---|---|---|
| Lateral Force Balance | \(m(\dot{v} + ur) = F_{yf} + F_{yr}\) | Bicycle model lateral dynamics |
| Yaw Moment Balance | \(I_z \dot{r} = l_f F_{yf} - l_r F_{yr}\) | Bicycle model yaw dynamics |
| Front Slip Angle | \(\alpha_f = \delta - \frac{v + l_f r}{u}\) | Tyre slip at front axle |
| Linear Tyre Model | \(F_y = C_\alpha \alpha\) | Lateral force from slip angle |
| Understeer Gradient | \(K = \frac{m}{L}\left(\frac{l_r}{C_{\alpha f}} - \frac{l_f}{C_{\alpha r}}\right)\) | Steady-state handling characteristic |
| Longitudinal Dynamics | \(m\dot{u} = F_x - F_{drag} - F_{roll} - F_{grade}\) | Acceleration/braking equation |
| Weight Transfer | \(\Delta F_z = \frac{m\dot{u} h}{L}\) | Load transfer during accel/braking |
A vehicle has mass m = 1500 kg, wheelbase L = 2.7 m, lf = 1.2 m, lr = 1.5 m, and equal front/rear cornering stiffnesses Cαf = Cαr = 60,000 N/rad. Calculate the understeer gradient K and classify the vehicle's handling behaviour.
Show Solution
Calculate understeer gradient:
\[K = \frac{m}{L}\left(\frac{l_r}{C_{\alpha f}} - \frac{l_f}{C_{\alpha r}}\right)\]
\[K = \frac{1500}{2.7}\left(\frac{1.5}{60000} - \frac{1.2}{60000}\right)\]
\[K = 555.6 \times (0.000025 - 0.00002)\]
\[K = 555.6 \times (0.000005) = +0.00278\ \text{rad·s}^2/\text{m}\]
Classification: K > 0, therefore the vehicle is understeering.
With \(l_r > l_f\) the CG sits closer to the front axle, so the static front load is the larger share: \(W_f = mg\,l_r/L = 1.5/2.7 = 55.6\%\) (≈ 8175 N front vs 6540 N rear) — this is a front-weight-biased car. Under equal cornering stiffnesses, the more heavily loaded front axle runs at the larger slip angle, so the front reaches its grip limit first: that is understeer, which is stable and predictable. Production vehicles are set up this way (through weight distribution or tuned cornering stiffnesses) for safety.
Answer: \(K = +0.00278\) rad·s²/m; the vehicle understeers (K > 0, stable).
For the understeering vehicle from the lecture (K = 0.002), calculate the required steering angle to maintain a 100 m radius turn at: (a) 50 km/h, and (b) 100 km/h. How does the steering angle change with speed?
Show Solution
Steady-state steering angle:
\[\delta = \frac{L}{R} + K\,a_y = \frac{L}{R} + K\frac{u^2}{R}\]
where R = turn radius, u = speed, \(a_y = u^2/R\) is the lateral acceleration, L = 2.7 m, K = 0.002 rad·s²/m. The understeer term scales with lateral acceleration, not with speed alone.
(a) At 50 km/h = 13.89 m/s:
\[\delta = \frac{2.7}{100} + 0.002 \times \frac{(13.89)^2}{100}\]
\[\delta = 0.027 + 0.002 \times 1.929 = 0.027 + 0.00386 = 0.0309 \text{ rad} = 1.77°\]
(b) At 100 km/h = 27.78 m/s:
\[\delta = \frac{2.7}{100} + 0.002 \times \frac{(27.78)^2}{100}\]
\[\delta = 0.027 + 0.002 \times 7.715 = 0.027 + 0.01543 = 0.0424 \text{ rad} = 2.43°\]
Observation: Doubling the speed quadruples the lateral acceleration (\(a_y = u^2/R\)), so the understeer term grows fourfold: from 0.00386 rad at 50 km/h to 0.01543 rad at 100 km/h. The steering angle climbs above the geometric Ackermann value (\(L/R\) = 0.027 rad = 1.55°) and keeps rising with speed — the signature of understeer.
Answer: (a) \(\delta \approx 0.0309\) rad = 1.77°; (b) \(\delta \approx 0.0424\) rad = 2.43°.
A 1600 kg vehicle with wheelbase L = 2.8 m and CG height h = 0.6 m brakes at 0.8g deceleration. Calculate: (a) the weight transfer from rear to front, and (b) the percentage of vehicle weight on the front axle during braking if the static weight distribution is 55% front / 45% rear.
Show Solution
(a) Weight transfer:
\[\Delta F_z = \frac{m\dot{u} h}{L} = \frac{m \times 0.8g \times h}{L}\]
\[\Delta F_z = \frac{1600 \times 0.8 \times 9.81 \times 0.6}{2.8}\]
\[\Delta F_z = \frac{7534.08}{2.8} = 2690.7 \text{ N}\]
(b) Front axle load during braking:
Static front load: \(F_{zf,static} = 0.55 \times 1600 \times 9.81 = 8632.8 \text{ N}\)
Dynamic front load: \(F_{zf,dynamic} = F_{zf,static} + \Delta F_z = 8632.8 + 2690.7 = 11{,}323.5 \text{ N}\)
Total weight: \(W = 1600 \times 9.81 = 15{,}696 \text{ N}\)
Front percentage: \(\frac{11323.5}{15696} = 0.721 = 72.1\%\)
Interpretation: During hard braking, the front axle load increases from 55% to 72%, while the rear decreases from 45% to 28%. This is why front brakes are larger and why anti-lock braking systems (ABS) are essential - the rear wheels are more prone to lockup due to reduced normal force.
Answer: (a) \(\Delta F_z \approx 2690.7\) N; (b) front axle carries ≈ 72.1% of the weight during braking.