Interactive Lecture
Pressure distribution, slip kinematics, and contact-patch distortion
by Dr Arnaldo Delli Carri
In Drive and Dragged Wheels the contact patch was treated as a point and friction as a single resultant capped by \(\mu N\). That description is enough for grip-limited acceleration problems, but it cannot tell you why a tyre still grips while it is "slipping", why a free-rolling wheel on a perfect road still loses energy, or why a steering wheel self-centres after a corner. To explain those, the patch has to be opened up.
Following Genta's Motor Vehicle Dynamics (Chapter 2), this lecture treats the contact patch as a finite strip that carries pressure and shear at every point along its length. A tyre rolling under tractive or braking torque distorts that distribution. A wheel that is driving or braking has a different velocity field around it from a free-rolling one, and the centre of instantaneous rotation moves accordingly. Sideslip stretches the patch laterally and gives rise to the self-aligning moment that a driver feels through the steering wheel.
The contact patch is a strip, not a point. Pressure varies along it; shear varies along it; and in general part of the patch sticks to the road while another part slides. Global rolling and local slipping coexist.
A stationary loaded tyre flattens against the road over a roughly rectangular footprint. Throughout this lecture we work with a one-dimensional strip model, so \(p(x)\) is the vertical load per unit length along the rolling direction (units N/m — the contact pressure already integrated across the patch width), not a true pressure in Pa. It is approximately parabolic — zero at the leading and trailing edges, peaking near the centre. The resultant vertical force \(N\) acts directly below the wheel axis, and no moment about the axis is produced by the road.
Set the same tyre rolling on a flat road and that symmetry breaks. The rubber takes a finite time to recover its shape after being squashed by the leading edge — this is hysteresis. Pressure under the leading half of the patch rises faster than it falls under the trailing half, and the resultant of the vertical reaction shifts forward of the geometric centre by a small distance \(e\). The forward-shifted resultant develops a moment about the wheel axis that opposes rotation. This is rolling resistance:
\[F_\text{rr} = \frac{N \cdot e}{r}, \qquad f_\text{rr} = \frac{e}{r}\]
Where:
Apply a tractive torque at the hub and the pressure distribution distorts further: longitudinal shear in the patch is no longer balanced, and the pressure peak shifts. Under a driving torque the peak biases forward; under a braking torque it biases rearward. Simulation 1 lets you watch these shape changes directly.
A rigid wheel rolling without slip on a flat surface has its centre of instantaneous rotation (CIR) exactly at the contact point. At that instant the contact point is stationary in the ground frame; every other point on the wheel orbits about it. The wheel's forward speed \(V\) and angular speed \(\Omega\) are locked: \(V = \Omega r\).
Real tyres do not roll without slip whenever they generate longitudinal force. Define the slipping speed at the contact as:
\[V_s = \Omega r - V\]
and the longitudinal slip ratio (SAE convention) as:
\[\sigma = \frac{\Omega r - V}{V} = \frac{V_s}{V}\]
Sign conventions:
The CIR is the point about which the wheel instantaneously rotates as a rigid body. With the road at \(y = 0\), the wheel centre at \(y = r\), and \(y\) measured upward, the CIR sits at:
\[y_\text{CIR} = r - \frac{V}{\Omega} = \frac{V_s}{\Omega}\]
Once the CIR is located, the velocity of any other point on the wheel follows from a single rigid-body rotation about it: \(\vec{v} = \vec{\Omega} \times \vec{r}_\text{from CIR}\). Simulation 2 makes this construction visible — velocity arrows sampled along the vertical line through the wheel centre, a marker at the CIR, and a reference point on the rim that turns at the wheel's rate. The arrow tips fall on a single straight line through the CIR, where the velocity is zero.
A real tyre is not rigid — the tread can shear sideways and lengthwise relative to the carcass. Picture the tread as a row of compliant "bristles" attached to a stiff belt: as a bristle enters the leading edge of the contact patch it is dragged through under the carcass, accumulating a shear deflection that grows linearly with the time it has spent in contact.
While the local shear stays below \(\mu\,p(\xi)\) — both quantities being per unit length of patch (N/m), as above — the bristle sticks to the road. This region — typically the front portion of the patch — is the adhesion zone. Where the linearly growing required shear would exceed \(\mu\,p(\xi)\), the bristle starts to slide back across the road surface. This trailing region is the slip zone.
Global slipping happens when the slip zone fills the whole contact patch. Up to that point the wheel still rolls and the adhesion zone still anchors part of the patch to the road, even though the slip ratio is non-zero.
Integrating the local shear over the patch gives the longitudinal force the tyre can transmit. In the small-slip regime the relation is linear:
\[F_x \approx C_\sigma \, \sigma\]
where \(C_\sigma\) is the longitudinal slip stiffness (units N). As \(\sigma\) grows the slip zone advances forward from the trailing edge; the curve bends over and reaches a peak \(F_x^\text{max} = \mu N\) when the whole patch slides. Beyond the peak the available force falls again — kinetic friction is lower than static.
When the wheel travels in a direction that differs from where it is pointing, the velocity at the contact patch has both a longitudinal component \(V_x\) and a lateral component \(V_y\). The angle between them is the slip angle:
\[\alpha = \arctan\!\left(\frac{V_y}{V_x}\right)\]
The same brush picture applies, rotated 90°. Bristles entering the patch are dragged sideways as well as backwards; lateral deflection grows linearly with distance from the leading edge in the adhesion zone, then saturates at \(\mu\,p(\xi)/k\) in the slip zone (where \(k\) is the lateral stiffness per unit length of bristle). The integrated lateral force, in the small-angle regime, is again linear in slip:
\[F_y \approx C_\alpha \, \alpha\]
where \(C_\alpha\) is the cornering stiffness (N/rad). At larger \(\alpha\) the slip zone takes over, the lateral force saturates near \(\mu N\), and beyond the peak it drops.
The lateral force resultant does not act at the centre of the patch. Because lateral deflection — and therefore lateral shear — is small at the leading edge and large further back, the centroid of the lateral shear sits behind the geometric centre by the pneumatic trail \(t\). The lateral force \(F_y\) acting at offset \(t\) produces a moment about the vertical axis:
\[M_z = F_y \cdot t\]
This is the self-aligning moment. It rotates the wheel back towards straight-ahead — the moment a driver feels resisting the steering input. In the linear range \(t \approx L/6\), where \(L\) is the patch length. As \(\alpha\) grows the slip zone advances and the trail collapses: the steering goes "light" just before the tyre saturates, which is the warning a skilled driver reads at the limit.
Longitudinal and lateral force are not independent — they share the same friction limit. To a first approximation the combined operating point is constrained by the friction circle:
\[\sqrt{F_x^{\,2} + F_y^{\,2}} \le \mu \, N\]
You can accelerate hard, or you can corner hard, but not both at once.
The circle is the isotropic idealisation — one \(\mu\) for every direction of sliding, which is exactly what the brush model in Simulation 3 assumes. Measured tyres are slightly anisotropic: the peak longitudinal grip is typically 10–20% higher than the peak lateral grip on dry asphalt, because tread blocks and the carcass distort differently under fore-aft and sideways shear, and the lateral peak arrives through carcass twist as much as through pure sliding. The envelope is then a friction ellipse:
\[\left(\frac{F_x}{\mu_x N}\right)^{2} + \left(\frac{F_y}{\mu_y N}\right)^{2} \le 1\]
With \(\mu_x = \mu_y\) this collapses back to the circle; with \(\mu_x > \mu_y\) it is a little longer along the drive-brake axis than across it. Vehicle-dynamics models such as Pacejka's combined-slip formulation fit this ellipse empirically.
Simulation 3 gives you \(\mu_x\) and \(\mu_y\) as separate sliders, both starting at the same value so the limit opens as a circle. Pull them apart and watch the inset stretch. Notice what happens at the same slip angle before and after you raise \(\mu_x\): the lateral force does not change, because a purely lateral operating point only ever meets the \(\mu_y\) end of the ellipse. Add drive slip as well and the extra longitudinal grip lets the same patch carry more combined force before the whole of it slides.
The strip below the wheel is the contact patch viewed from the side. The orange curve above the strip is the vertical pressure \(p(x)\) along the patch, from leading edge (right, in the direction of travel) to trailing edge (left). The red dot marks the centroid of the vertical reaction — the point at which the resultant \(N\) effectively acts. Set the hysteresis offset to zero and the centroid sits at the geometric centre; raise it and the centroid migrates forward, generating a backward moment about the wheel axis — rolling resistance. Switch between drive and brake to see the pressure peak bias forward or rearward.
The wheel rotates at angular speed \(\Omega\) while the chassis moves forward at \(V\). The yellow dot is the centre of instantaneous rotation (CIR): on the road for free-rolling, inside the wheel when driving, below the road when braking. Three velocity arrows sample the rigid-body field along the vertical line through the wheel centre — at the top of the wheel (blue), at the hub (green, the chassis velocity \(V\)) and at the contact patch (red, the ground-frame velocity of the contact point, \(V - \Omega r = -V_s\)). Their tips lie on a straight line through the CIR, where the velocity is zero. The view is fixed to the chassis: the road marks scroll backwards at \(V\) while the tread blocks turn at \(\Omega r\), so under free-rolling a tread block and a road mark stay locked together at the contact point, and under driving or braking you can watch them drift apart at the slipping speed \(V_s\) — switch on Slow-mo to follow a single block past a single mark. Try the Pure spin preset (\(V = 0\), the CIR rides up to the hub) and the Wheel-lock preset (\(\Omega = 0\), the CIR disappears to \(-\infty\) and every point on the wheel slides forward at the same speed).
Top-down view of the contact patch under combined longitudinal and lateral slip. The wheel travels to the right: bristles enter at the leading edge (right) and are dragged back through the patch to the trailing edge (left) — laterally if there is a slip angle \(\alpha\), longitudinally if there is a slip ratio \(\sigma\). The arrow ahead of the leading edge is the wheel-centre velocity \(V\), tilted by \(\alpha\) from the wheel plane; the bristle tips are dragged the opposite way, so the cornering force \(F_y\) acts against the lateral component of \(V\). The green region is the adhesion zone (bristles still sticking to the road); the red region is the slip zone (bristles sliding at the local friction limit). Drive \((\sigma > 0)\) pulls bristle tips towards the leading edge; brake \((\sigma < 0)\) pushes them toward the trailing edge. The thick blue arrow is the lateral force \(F_y\), drawn at the resultant point — behind the geometric centre by the pneumatic trail \(t\). The inset on the upper right shows the current operating point \((F_x, F_y)\) against the friction ellipse with semi-axes \(\mu_x N\) and \(\mu_y N\) — a circle while the two friction sliders agree. The operating point touches the boundary exactly when the slip zone has swallowed the whole patch; the boundary turns red when it does.
| Concept | Equation | Description |
|---|---|---|
| Longitudinal slip ratio | \(\sigma = \dfrac{\Omega r - V}{V}\) | Drive \(\sigma > 0\), brake \(\sigma < 0\), free-rolling \(\sigma = 0\) |
| Slipping speed | \(V_s = \Omega r - V\) | Speed of the contact point relative to the ground |
| CIR vertical offset | \(y_\text{CIR} = r - \dfrac{V}{\Omega}\) | On the road for free-rolling; inside the wheel for drive; below the road for brake |
| Rolling resistance | \(F_\text{rr} = \dfrac{N e}{r}\) | From the hysteresis-driven forward offset of the vertical resultant |
| Longitudinal force (linear) | \(F_x \approx C_\sigma \, \sigma\) | Small-slip regime; saturates at \(\mu N\) |
| Slip angle | \(\alpha = \arctan(V_y / V_x)\) | Angle between the travel direction and the wheel plane |
| Cornering force (linear) | \(F_y \approx C_\alpha \, \alpha\) | Small-angle regime; saturates at \(\mu N\) |
| Self-aligning moment | \(M_z = F_y \cdot t\) | \(t\) = pneumatic trail; tends to straighten the steering |
| Friction circle | \(\sqrt{F_x^{\,2} + F_y^{\,2}} \le \mu N\) | Combined longitudinal + lateral grip limit, isotropic friction |
| Friction ellipse | \(\left(\dfrac{F_x}{\mu_x N}\right)^{2} + \left(\dfrac{F_y}{\mu_y N}\right)^{2} \le 1\) | Same limit with \(\mu_x \ne \mu_y\); reduces to the circle when they are equal |
A driven wheel rolls at forward speed \(V = 20\) m/s. The rolling radius is \(r = 0.30\) m and the wheel spins at \(\Omega = 80\) rad/s.
Compute (a) the peripheral speed \(\Omega r\), (b) the slip ratio \(\sigma\), (c) the slipping speed \(V_s\) at the contact, and (d) the vertical position of the centre of instantaneous rotation relative to the road surface.
Show Solution
Given: \(V = 20\) m/s, \(r = 0.30\) m, \(\Omega = 80\) rad/s.
(a) Peripheral speed:
\[\Omega r = 80 \times 0.30 = 24 \text{ m/s}\]
(b) Slip ratio:
\[\sigma = \frac{\Omega r - V}{V} = \frac{24 - 20}{20} = 0.20 \quad (+20\%)\]
The wheel is driving (\(\sigma > 0\)).
(c) Slipping speed at the contact:
\[V_s = \Omega r - V = 24 - 20 = 4 \text{ m/s}\]
The contact patch moves rearward at 4 m/s relative to the road.
(d) CIR offset above the road:
\[y_\text{CIR} = r - \frac{V}{\Omega} = 0.30 - \frac{20}{80} = 0.30 - 0.25 = +0.05 \text{ m}\]
Since \(\sigma > 0\), the wheel is driving and the CIR sits inside the wheel, 0.05 m above the road surface (and 0.25 m below the wheel hub).
Answers: \(\Omega r = 24\) m/s, \(\sigma = 0.20\), \(V_s = 4\) m/s, CIR is 0.05 m above the road, inside the wheel.
A free-rolling tyre has a contact patch of length \(L = 150\) mm. Pressure-mat measurements show the centroid of the vertical reaction sits 6 mm forward of the geometric centre. The wheel carries a vertical load of \(N = 4000\) N and has a rolling radius of \(r = 0.32\) m.
Compute (a) the rolling resistance force \(F_\text{rr}\), and (b) the rolling resistance coefficient \(f_\text{rr}\).
Show Solution
Given: \(N = 4000\) N, \(e = 6\) mm \(= 0.006\) m, \(r = 0.32\) m.
(a) Rolling resistance force:
\[F_\text{rr} = \frac{N \, e}{r} = \frac{4000 \times 0.006}{0.32} = \frac{24}{0.32} = 75 \text{ N}\]
(b) Rolling resistance coefficient:
\[f_\text{rr} = \frac{e}{r} = \frac{0.006}{0.32} = 0.0188\]
Answers: \(F_\text{rr} = 75\) N, \(f_\text{rr} = 0.0188\). This is on the high end for a passenger-car tyre — typical values sit between 0.010 and 0.020.
A tyre operating in the linear range has a cornering stiffness of \(C_\alpha = 60\,000\) N/rad. The driver applies a steering input producing a slip angle of \(\alpha = 3°\).
Compute (a) the lateral force \(F_y\), and (b) the self-aligning moment \(M_z\) given a pneumatic trail of \(t = 30\) mm.
Show Solution
Given: \(C_\alpha = 60\,000\) N/rad, \(\alpha = 3°\), \(t = 0.030\) m.
Convert slip angle to radians:
\[\alpha = 3° \times \frac{\pi}{180°} = 0.0524 \text{ rad}\]
(a) Lateral force in the linear range:
\[F_y = C_\alpha \, \alpha = 60\,000 \times 0.0524 = 3142 \text{ N} \approx 3140 \text{ N}\]
(b) Self-aligning moment:
\[M_z = F_y \cdot t = 3142 \times 0.030 = 94.2 \text{ N·m}\]
Answers: \(F_y \approx 3140\) N, \(M_z \approx 94.2\) N·m. This is the moment the steering column transmits back to the driver's hands — the source of "steering feel".
A driven wheel carries \(N = 4500\) N and operates on a road with friction coefficient \(\mu = 0.9\). The driver is currently applying a longitudinal traction force of \(F_x = 2000\) N.
How much lateral force \(F_y\) does the tyre still have available before the combined operating point hits the friction circle?
Show Solution
Given: \(N = 4500\) N, \(\mu = 0.9\), \(F_x = 2000\) N.
Maximum total friction force:
\[F_\text{max} = \mu \, N = 0.9 \times 4500 = 4050 \text{ N}\]
The friction circle bounds the combined force:
\[F_x^{\,2} + F_y^{\,2} \le F_\text{max}^{\,2}\]
Available lateral force:
\[F_y^\text{avail} = \sqrt{F_\text{max}^{\,2} - F_x^{\,2}} = \sqrt{4050^2 - 2000^2}\]
\[= \sqrt{16\,402\,500 - 4\,000\,000} = \sqrt{12\,402\,500} \approx 3522 \text{ N}\]
Answer: \(F_y^\text{avail} \approx 3522\) N. If the driver tries to corner with more than this while still applying 2000 N of traction, the tyre saturates and slip ratio or slip angle (or both) will run away.