Tyre forces, slip angle and cornering stiffness
How a tyre generates longitudinal and lateral force through slip: slip angle, the F_y–α curve and cornering stiffness, pneumatic trail and self-aligning moment, slip ratio and the friction circle.
Drafted with Aria, reviewed by the AiCanCode.org team. Spotted an error? Use Give Feedback at the bottom of the page.
Why it matters
Every force that accelerates, brakes or turns a road vehicle passes through four contact patches, each roughly the size of a palm. How a tyre turns slip into force decides cornering grip, braking distance and whether a car understeers or oversteers. Tyre test data and cornering stiffness are the inputs to every handling model you will meet later (steady-state cornering, the bicycle model, ESC).
Key ideas
Tyre axis system and forces. At the centre of the contact patch the road exerts on the tyre:
- a longitudinal force F_x along the wheel plane (tractive or braking force);
- a lateral force F_y perpendicular to the wheel plane, in the road plane (the cornering force);
- a vertical (normal) load F_z;
- moments, of which the important one is the self-aligning moment M_z about the vertical axis, plus rolling-resistance and overturning moments.
Slip angle. When a rolling tyre carries a side force, its tread elements deflect sideways in the contact patch, so the wheel travels in a direction different from the one it points in. The slip angle α is the angle between the wheel plane (heading of the wheel) and the velocity vector of the contact-patch centre. It is not the steering angle: a rear wheel that is not steered still runs at a slip angle in a corner. No slip angle, no cornering force — the slip angle is how the tyre "asks" the road for side force.
Lateral force versus slip angle. A typical F_y–α curve has three regions:
- Linear region (up to about 3–5° for car tyres): F_y rises in proportion to α, because most of the contact patch is still adhering and the tread deflection grows linearly.
- Transitional region: the rear part of the patch starts to slide, the curve bends over.
- Saturation (frictional) region: the whole patch slides; F_y peaks at about μ·F_z (typically at 6–10° on dry roads) and then falls slightly.
Cornering stiffness C_α is the slope of the F_y–α curve at zero slip angle, in N/rad (or N/deg). Typical passenger-car values are about 40 000–100 000 N/rad per tyre. Dividing by load gives the cornering coefficient C_α/F_z (per rad), which lets tyres of different loads be compared.
C_α increases with vertical load, but less than proportionally (the curve flattens at high load) — this is why load transfer across an axle reduces the axle's total cornering stiffness. It also depends on inflation pressure, tyre construction (radials and low-profile tyres are stiffer), tread depth (worn tread is stiffer laterally), rim width and, slightly, speed and temperature.
Self-aligning moment and pneumatic trail. The side force is not centred in the contact patch: tread deflection builds up towards the rear, so the resultant F_y acts a distance t_p (pneumatic trail) behind the patch centre. This gives a self-aligning moment M_z = F_y·t_p that tries to steer the wheel back to straight. As the rear of the patch begins to slide, t_p shrinks, so M_z peaks and then falls before F_y peaks — the steering goes "light", which is the driver's warning of front-tyre saturation.
Longitudinal slip. A driven or braked tyre also needs slip. The slip ratio compares wheel circumferential speed ω·R_e with vehicle speed V. F_x rises almost linearly with slip at small values, peaks at about 10–20% slip, then drops towards the sliding value at 100% (locked wheel). ABS and traction control hold the slip near the peak.
Combined slip and the friction circle. A tyre has only so much grip, roughly μ·F_z, to share between directions. To a first approximation the resultant of F_x and F_y must lie inside a circle (in practice an ellipse) of radius μ·F_z. Braking hard in a corner therefore reduces the available side force — the basis of "brake before the bend".
Camber thrust. A wheel tilted from the vertical produces a side force towards the tilted side even at zero slip angle (camber stiffness is much smaller than cornering stiffness for car tyres, but large for motorcycle tyres).
Formulas
α = δ − tan⁻¹(v_y / v_x) — slip angle of a wheel steered by δ, whose contact patch has velocity components v_x (along vehicle) and v_y (lateral). α, δ in rad; v in m/s. Sign conventions vary between textbooks; be consistent.
F_y = C_α · α — lateral force in the linear region. F_y in N, C_α in N/rad, α in rad. Valid only for small α (about < 4°).
CC = C_α / F_z — cornering coefficient, 1/rad.
M_z = F_y · t_p — self-aligning moment, N·m; t_p is pneumatic trail in m (typically 20–50 mm in the linear range).
s = (ω·R_e − V) / V (driving) and s = (V − ω·R_e) / V (braking) — longitudinal slip ratio (dimensionless, often quoted in %); ω in rad/s, R_e effective rolling radius in m, V vehicle speed in m/s.
√(F_x² + F_y²) ≤ μ·F_z — friction circle; μ is the tyre–road friction coefficient (dimensionless).
C_α (N/rad) = C_α (N/deg) × 180/π — unit conversion; 1 rad = 57.3°.
Worked examples
Example 1 (standard). A tyre has a cornering stiffness of 1000 N/deg and a pneumatic trail of 30 mm. Find the lateral force and self-aligning moment at a slip angle of 3°, and C_α in N/rad.
- Convert:
C_α = 1000 × 180/π= 57 296 N/rad. - Lateral force:
F_y = C_α · α= 1000 N/deg × 3° = 3000 N (same as 57 296 × 0.05236 rad). - Self-aligning moment:
M_z = F_y · t_p= 3000 N × 0.030 m = 90 N·m.
Answer: C_α ≈ 57 300 N/rad, F_y = 3000 N, M_z = 90 N·m.
Example 2 (GATE level). A 1200 kg car carries 55% of its weight on the front axle and rounds a flat curve of radius 100 m at 20 m/s. Each front tyre has C_α = 60 000 N/rad and the tyre–road μ = 0.9. Ignore lateral load transfer. Find (a) the front slip angle and (b) the largest braking force each front tyre can add without exceeding the friction circle. Take g = 9.81 m/s².
- Lateral acceleration:
a_y = V²/R= 20²/100 = 4.0 m/s². - In steady cornering each axle carries side force in proportion to its static load: front axle
F_yf = 0.55·m·a_y= 0.55 × 1200 × 4.0 = 2640 N, i.e. 1320 N per tyre. - Slip angle:
α = F_y / C_α= 1320 / 60 000 = 0.0220 rad = 1.26° (well inside the linear range). - Vertical load per front tyre:
F_z = 0.55 × 1200 × 9.81 / 2= 3237 N; grip limitμ·F_z= 0.9 × 3237 = 2914 N. - Friction circle:
F_x,max = √((μF_z)² − F_y²)= √(2914² − 1320²) = 2597 N.
Answer: (a) α ≈ 0.022 rad (1.26°); (b) about 2.6 kN braking force per front tyre.
Common mistakes
- Treating slip angle as the steering angle. A non-steered rear wheel still has a slip angle in a corner.
- Mixing N/deg and N/rad. A cornering stiffness in N/deg multiplied by an angle in rad gives an answer 57.3 times too small.
- Using F_y = C_α·α at large slip angles. Beyond about 4–5° the curve saturates and the force is limited by μ·F_z.
- Assuming cornering stiffness is proportional to load. It grows less than proportionally, which is why load transfer costs grip.
- Adding F_x and F_y arithmetically instead of vectorially when checking the friction limit.
- Quoting unrealistic magnitudes: a car tyre's C_α is tens of thousands of N/rad, not a few thousand.
For GATE ME
Expect short numericals on F_y = C_α·α with unit conversions, slip-angle calculations from axle load and lateral acceleration, friction-circle checks under combined braking and cornering, and conceptual questions on slip angle, pneumatic trail and why steering feel lightens near the limit. Practise converting between N/deg and N/rad and splitting cornering force between axles by static weight distribution.
Quick check
- A tyre with C_α = 50 000 N/rad runs at 0.04 rad slip angle. What is F_y?
- Why does the self-aligning moment fall before the lateral force peaks?
- Can a non-steered rear tyre have a slip angle?
- A tyre with F_z = 4000 N and μ = 1.0 carries 2400 N side force. What braking force is still available?
Answers: 1. 2000 N. 2. The rear of the contact patch starts sliding, so the pneumatic trail shrinks. 3. Yes — every tyre that produces side force runs at a slip angle. 4. √(4000² − 2400²) = 3200 N.
Interview questions
All Vehicle Dynamics, Body and Safety interview questionsTry answering each one aloud before you open it.
1.What is a slip angle in the context of vehicle dynamics?Concept
The slip angle is the angle between the wheel plane (the direction the wheel points) and the direction the contact patch actually travels. It exists because the tread elements deflect sideways in the contact patch whenever the tyre carries a lateral force, so the wheel crabs slightly relative to where it is pointed. Every tyre that produces cornering force, including an unsteered rear tyre, runs at a slip angle; in the linear range the lateral force is roughly proportional to it, F_y = C_α·α.
2.Explain the concept of cornering stiffness.Concept
Cornering stiffness C_α is the slope of the lateral force versus slip angle curve at zero slip angle, in N/rad or N/deg, so F_y ≈ C_α·α for small slip angles (up to about 3–4°). A stiffer tyre needs less slip angle for the same side force, which gives a quicker, more precise steering response. It rises with vertical load but less than proportionally, and depends on inflation pressure, construction, aspect ratio and tread depth. Typical car tyres are in the range of roughly 40 000–100 000 N/rad.
3.How do tire forces affect vehicle handling?Concept
All control forces act at the contact patches: longitudinal forces accelerate and brake the car, lateral forces bend its path and create the yaw moment that turns it. The ratio of front to rear lateral force capability, set by cornering stiffness and load on each axle, decides whether the car understeers or oversteers. Because each tyre's grip is limited to about μ·F_z shared between directions (the friction circle), braking or accelerating in a corner reduces the available side force and can tip the balance at the limit.
4.Why is it important to understand slip angles when designing a vehicle's suspension system?Application
Suspension geometry decides the slip angles each axle ends up running at: toe settings, roll steer, compliance steer and camber change all add to or subtract from the slip angle the driver commands. The roll-stiffness distribution sets how load is transferred across each axle, and since cornering stiffness grows less than proportionally with load, more load transfer at one axle reduces that axle's effective cornering stiffness. Designers use these effects to tune understeer and keep the car predictable at the limit.
5.What happens if a vehicle's tires have low cornering stiffness?Application
A low-cornering-stiffness tyre needs a larger slip angle to produce the same side force, so the car responds more slowly and less precisely to steering input and feels vague. If the low-stiffness tyres are on the front axle the car tends towards more understeer; if they are on the rear, it moves towards oversteer, which is dangerous. This is why mixing worn and new tyres or different tyre types between axles is discouraged, and why new tyres are usually fitted to the rear.
6.How does tire pressure influence slip angle and cornering stiffness?Application
Lower inflation pressure makes the carcass and sidewall more compliant, so cornering stiffness falls and the tyre needs a larger slip angle for a given side force; it also raises rolling resistance, heat and shoulder wear. Raising pressure generally increases cornering stiffness up to a point, but very high pressure shrinks the contact patch and can reduce peak grip, especially on rough or wet roads. Pressure differences between front and rear axles are therefore a simple way to shift a car's understeer balance, and the placard pressure is chosen for that balance as well as for load.
7.Calculate the lateral force generated by a tire with a cornering stiffness of 80,000 N/rad and a slip angle of 0.05 rad.Numerical
The lateral force (F_lat) can be calculated using the formula: F_lat = cornering stiffness × slip angle. Substituting the given values: F_lat = 80,000 N/rad × 0.05 rad = 4,000 N. Therefore, the lateral force generated by the tire is 4,000 N.
8.Explain how vehicle speed affects slip angle and cornering stiffness.Application
On a given radius the lateral force needed is m·V²/R, so it grows with the square of speed, and the slip angles at both axles grow with it. Cornering stiffness itself is mainly a tyre property and changes only slightly with speed in the linear range. As speed rises the tyres move towards the non-linear and saturation region, where extra slip angle gives little extra force; the axle that saturates first decides whether the car runs wide or spins.
9.What is pneumatic trail and why does steering feel light near the cornering limit?Application
Pneumatic trail is the distance by which the resultant lateral force acts behind the centre of the contact patch, because tread deflection builds up towards the rear of the patch. Multiplying it by the lateral force gives the self-aligning moment that returns the steering to centre. Near the limit the rear of the contact patch starts sliding, the pneumatic trail shrinks, and the aligning moment falls even though side force is still rising, so the steering goes light. That drop is a useful warning to the driver that the front tyres are close to saturation.
Finished this topic? Mark it so your progress, study plan and readiness keep up.
Stuck on something here?