Why a cornering tire must slip
In Module 1 we treated the tire as a block of rubber with a friction coefficient \(\mu\). That tells us the maximum force, but not how a tire actually produces force below the maximum. For that we need the single most important idea in tire mechanics: the slip angle.
If a rolling wheel travels exactly in the direction it is pointing, it produces (ideally) no sideways force. To produce a lateral force, the wheel must be pointed at a small angle to its direction of travel. That angle is the slip angle, \(\alpha\).
Why does this produce a force? Think about a single piece of tread. It touches down at the front of the contact patch and, as long as it grips, stays stuck to that spot on the road. But the wheel is carrying it along a slightly different direction from where it is pointing, so as the tread element moves toward the back of the patch it gets dragged further and further sideways. The rubber deflects like a stiff spring, and the sum of all those little spring forces is the lateral force.
- At small slip angles every element grips, the deflections are proportional to \(\alpha\), and so is the force.
- As \(\alpha\) grows, elements near the back of the patch need more force than friction can supply, so they start to slide. The force keeps rising, but more slowly.
- Eventually the whole patch slides. The force peaks and then levels off or falls.
“Slip angle means the tire is skidding.” Not at small angles — most of the contact patch is gripping the road. The slip comes from rubber deforming, not sliding. Sliding only takes over near and beyond the peak.
The lateral force curve
Plotting lateral force against slip angle gives the most important graph in vehicle dynamics. It has three regions:
| Region | What the contact patch is doing | What the driver feels |
|---|---|---|
| Linear | Almost all elements gripping; \(F_y \approx C_\alpha\,\alpha\) | Precise, predictable response to steering |
| Transitional | Sliding spreads forward from the back of the patch | Steering goes lighter; the car starts to “move around” |
| Frictional | Most or all of the patch is sliding | More steering gives no more grip — the car slides |
The slope at the origin is the cornering stiffness:
Cornering stiffness is usually quoted in N/deg. Like a spring stiffness, it tells you how much “deflection” (slip angle) the tire needs to make a given force. The peak force is \(F_{y,\max} = \mu_{\text{peak}} F_z\), and it occurs at a slip angle that depends on the tire's construction — typically a few degrees for a race tire.
A tire produces force by operating at a slip angle. The cornering stiffness sets how much slip angle each axle needs at a given lateral force — and in Module 5 the difference between the front and rear slip angles determines whether a car understeers or oversteers.
Sign convention: in the SAE axis system (Module 0) a positive slip angle produces a negative lateral force. In this lesson we plot magnitudes to keep the pictures simple.
Load sensitivity: more load, less \(\mu\)
Push a tire harder into the road and it produces more lateral force — but not proportionally more. Double the vertical load and the peak force rises by less than double. Equivalently, the friction coefficient falls as load rises. This is called tire load sensitivity, and it is one of the most consequential facts in race car engineering.
A simple model that captures it is a friction coefficient that falls linearly with load:
Here \(F_{z0}\) is a reference load, \(\mu_0\) the friction coefficient at that load, and \(s\) the load sensitivity (0 means no sensitivity). Now consider two tires sharing a total load \(2F_{z0}\). If one gains \(\Delta F\) and the other loses \(\Delta F\), a few lines of algebra give their combined peak force:
The second term is always a loss, and it grows with the square of the load difference. Uneven loading costs grip. Module 3 shows that this is exactly what happens to the inside and outside tires in a corner — and how engineers use it to tune a car's balance.
A tire model you can play with
Real tire curves come from test machines. Engineers fit them with empirical formulas; the most widely used is the Magic Formula developed by Hans Pacejka and colleagues. A simplified version is:
\(D\) is the peak force, \(C\) controls the shape, \(E\) the curvature near the peak, and \(B\) is chosen so that the slope at the origin equals the cornering stiffness (\(BCD = C_\alpha\)). You do not need to memorize it — it is just a convenient curve with the right features. The explorer below uses it together with the load-sensitivity model above.
- At the default settings, where does the curve leave the dashed linear line? Read off the slip angle and the force there.
- Increase the load from 3000 N to 6000 N. Does the peak force double? What happens to \(\mu\)?
- Set the load sensitivity \(s\) to zero. What happens to the pair-of-tires loss? Why?
- With \(s = 0.12\), note the pair loss at \(\Delta F_z = 1000\) N, then at 2000 N. Does the ratio match the \(\Delta F^2\) prediction?
A front tire has a cornering stiffness of 1500 N/deg. In a gentle corner it must provide 2400 N of lateral force. Assuming it is in the linear region, what slip angle does it run?
\[ \alpha = \frac{F_y}{C_\alpha} = \frac{2400\ \text{N}}{1500\ \text{N/deg}} = 1.6^\circ \]
This is well inside the linear region of a typical race tire, so the linear model is a good approximation here.
Check your understanding
A tire has a cornering stiffness of 1200 N/deg. Assuming linear behavior, what slip angle is needed for 3000 N of lateral force?
In the linear region, \(F_y = C_\alpha \alpha\).
\(\alpha = F_y / C_\alpha = 3000 / 1200 = 2.5^\circ\).
Using the load-sensitivity model with \(\mu_0 = 1.6\), \(s = 0.15\) and \(F_{z0} = 3000\) N, find the combined peak lateral force of two tires loaded to 4500 N and 1500 N.
Find \(\mu\) for each tire separately, multiply by its own load, then add.
\(\mu(4500) = 1.6\,[1 - 0.15 \times 0.5] = 1.48\), so \(F_{y,\max} = 1.48 \times 4500 = 6660\) N.
\(\mu(1500) = 1.6\,[1 + 0.15 \times 0.5] = 1.72\), so \(F_{y,\max} = 1.72 \times 1500 = 2580\) N.
Total \(= 9240\) N, compared with \(2 \times 1.6 \times 3000 = 9600\) N if the loads were equal — a 3.75% loss. Check: \(2\mu_0 s\,\Delta F^2/F_{z0} = 2(1.6)(0.15)(1500^2)/3000 = 360\) N. ✓
Same tires as above, but now the load difference is halved: 3750 N and 2250 N. How much grip is lost compared with equal loads?
Use the pair formula: loss \(= 2\mu_0 s\,\Delta F^2 / F_{z0}\). What is \(\Delta F\) now?
\(\Delta F = 750\) N, so loss \(= 2(1.6)(0.15)(750^2)/3000 = 90\) N — one quarter of the 360 N loss, because the loss scales with \(\Delta F^2\).
A tire is operating at 1° slip angle, well inside its linear region. The driver adds steering so it runs at 2°. Approximately what happens to its lateral force?
- It roughly doubles.In the linear region \(F_y \approx C_\alpha \alpha\), so doubling \(\alpha\) doubles \(F_y\).
- It increases by about \(\sqrt{2}\).That square-root relationship was corner speed vs radius in Module 1, not force vs slip angle.
- It stays the same, because friction force does not depend on sliding speed.That is true in the fully sliding (frictional) region, not at 1–2°.
A load-sensitive tire's vertical load is doubled. What happens to its peak lateral force?
- It doubles, because \(F = \mu N\).That would be true only if \(\mu\) were constant. For a load-sensitive tire, \(\mu\) falls as load rises.
- It increases, but by less than double.More load still means more force, but the friction coefficient drops, so the increase is less than proportional.
- It decreases, because \(\mu\) falls.\(\mu\) falls, but the load rises more — for realistic tires the product \(\mu F_z\) still increases.
A driver keeps adding steering past the slip angle at peak lateral force on the front tires. What happens?
- The car turns more sharply, since more slip angle always means more force.Only up to the peak. Beyond it, extra slip angle produces the same or less force.
- The front tires produce no more (or slightly less) force, and the car runs wide — it understeers.The front tires are in the frictional region. Unwinding the steering back toward the peak slip angle is often faster.
- The rear of the car steps out.That would require the rear tires to exceed their peak. Here it is the fronts that are saturated.
Going further with Milliken
Read: RCVD Chapter 2, Tire Behavior — the material on slip angle, lateral force and the effect of load. For more on how raw test data is turned into usable curves, skim Chapter 14, Tire Data Treatment.
As you read, look for answers to these questions:
- How do the authors explain the build-up of force across the contact patch?
- How is lateral force data presented against load, and how is load sensitivity visible on those plots?
- Why is it useful to “normalize” tire data (divide by load, or by the peak), and what can go wrong if you do?
Summary
| Idea | Key relation |
|---|---|
| Linear tire | \(F_y \approx C_\alpha\,\alpha\) |
| Peak force | \(F_{y,\max} = \mu_{\text{peak}}\,F_z\) |
| Load sensitivity | \(\mu(F_z) = \mu_0[1 - s\,(F_z - F_{z0})/F_{z0}]\) |
| Pair of tires with load transfer \(\Delta F\) | Loss \(= 2\mu_0 s\,\Delta F^2 / F_{z0}\) |
Next in this module: the same ideas in the longitudinal direction (slip ratio), and what happens when a tire must brake and corner at once.