The other half of the friction circle
Lesson 2.1 explained how a tire makes sideways force. Every braking zone and every corner exit asks for force in the fore–aft direction instead. On the g-g diagram, that is the top and bottom of the envelope. The good news is that the mechanism is the same. Where lateral force needs a slip angle, longitudinal force needs a slip ratio.
Wheel speed vs vehicle speed
A wheel that is simply rolling, neither driven nor braked, turns at exactly the rate that matches the ground speed: \(\Omega r_e = V\), where \(r_e\) is the effective rolling radius. To make a longitudinal force, the wheel must turn slightly faster (driving) or slower (braking) than that. The slip ratio measures the mismatch:
| Condition | Slip ratio |
|---|---|
| Free rolling | \(\kappa = 0\) |
| Driving (wheel turning faster than the road) | \(\kappa > 0\) |
| Braking (wheel turning slower than the road) | \(\kappa < 0\) |
| Locked wheel | \(\kappa = -1\) (−100%) |
Why must a tire slip to make force? The same reason as in Lesson 2.1. Under braking, tread elements entering the contact patch stick to the road, but the wheel is turning a little slower than the road is passing. Each element is stretched more and more as it moves back through the patch, like a spring being pulled. The sum of those spring forces is \(F_x\). At small slip ratios almost the whole patch grips; as slip grows, the rear of the patch starts to slide.
The longitudinal force curve
Plot \(F_x\) against \(\kappa\) and you get a curve that looks very like the lateral force curve:
- Linear region: \(F_x \approx C_\kappa\,\kappa\), where \(C_\kappa\) is the longitudinal stiffness in N per unit slip. It is usually much stiffer than the lateral equivalent, so the slip is small: a few percent.
- Peak: \(F_{x,\max} = \mu_{x}\,F_z\), typically somewhere around 5–15% slip depending on the tire.
- Sliding: past the peak the force falls. A locked wheel (−100%) slides at a noticeably lower friction coefficient than the peak, and it flat-spots the tire.
Maximum braking and maximum traction both happen at a small slip ratio, just before the peak. A locked wheel or a spinning wheel is always slower than one held at the peak.
Why wheels lock so suddenly
Treat the wheel as a rigid body spinning about its axle. Brake torque \(T_b\) slows it down; the tire force \(F_x\) acting at radius \(r_e\) speeds it back up:
In steady braking the two torques balance and the wheel settles at whatever slip ratio makes \(F_x = T_b / r_e\). On the rising part of the curve that balance is stable. If the wheel slows slightly, slip grows, \(F_x\) rises, and the wheel is pushed back.
Now ask for more brake torque than the peak can balance. There is no operating point. The wheel slows, slip grows, \(F_x\) falls, so the wheel slows even faster. The wheel goes from turning to locked in a fraction of a second. The same instability in the driving direction gives sudden wheelspin.
ABS and traction control break this runaway: they sense the slip and trim the torque to hold the tire in a band just below its peak. Many racing series ban both, which is why threshold braking (holding the pedal right at the edge of lock-up) is a core racing skill.
- Slowly increase the brake torque (more negative). Note the slip ratio and stopping distance just before the wheel locks, then just after.
- With the wheel locked, switch on ABS. How much stopping distance does it save from 100 km/h?
- Increase the vertical load to 5000 N. Does the torque needed to lock the wheel go up or down? Connect this to longitudinal load transfer under braking (Module 3).
- Reduce the longitudinal stiffness coefficient. What happens to the slip ratio at a fixed braking force, and to the slip ratio at the peak?
- Try the driving side: find the torque at which the wheel starts to spin.
A car is braking at 30 m/s. Its wheel speed sensor reads 95 rad/s, and the effective rolling radius is 0.30 m. What is the slip ratio?
\[ \kappa = \frac{\Omega r_e - V}{V} = \frac{95 \times 0.30 - 30}{30} = \frac{28.5 - 30}{30} = -0.05 \]
The tire is running at −5% slip: braking, comfortably on the rising side of a typical curve.
Check your understanding
A car travels at 25 m/s. A braked wheel with effective rolling radius 0.32 m turns at 71.875 rad/s. What is its slip ratio, in percent?
Compute the wheel's surface speed \(\Omega r_e\) first, then use \(\kappa = (\Omega r_e - V)/V\). Watch the sign.
\(\Omega r_e = 71.875 \times 0.32 = 23.0\ \text{m/s}\), so \(\kappa = (23.0 - 25)/25 = -0.08 = -8\%\). Negative, because the wheel is braking.
A tire has a longitudinal stiffness of 90 000 N per unit slip. Assuming linear behavior, what slip ratio (in percent) does it need to deliver 1800 N of driving force?
In the linear region, \(F_x = C_\kappa \kappa\). Convert the result to percent.
\(\kappa = 1800 / 90\,000 = 0.02 = 2\%\). That is a very small slip, which is why a tire making plenty of force still looks as if it is just rolling.
A car stops from 100 km/h. With threshold braking its tires work at a peak \(\mu = 1.5\); with the wheels locked they slide at \(\mu = 1.2\). How much longer is the locked-wheel stop? Assume constant deceleration \(\mu g\).
Stopping distance \(d = v^2 / (2 \mu g)\). Remember to convert km/h to m/s.
\(v = 27.78\ \text{m/s}\), \(v^2 = 771.6\ \text{m}^2/\text{s}^2\).
Peak: \(771.6 / (2 \times 1.5 \times 9.81) = 26.2\ \text{m}\). Locked: \(771.6 / (2 \times 1.2 \times 9.81) = 32.8\ \text{m}\).
Difference \(\approx 6.6\ \text{m}\) — more than a car length, and the locked car also cannot steer (Lesson 2.3).
A driver brakes harder and harder in a straight line. Just past the peak of the \(F_x\)–\(\kappa\) curve, what happens to the wheel?
- It settles at a new, slightly higher slip ratio.Past the peak, extra slip reduces \(F_x\), so the tire can no longer balance the brake torque. There is no steady point to settle at.
- It locks almost immediately, because extra slip reduces the tire force, which lets the brake slow the wheel even more.This positive feedback is what makes lock-up so sudden, and why ABS must react within milliseconds.
- Nothing changes until the slip ratio reaches −100%.The curve starts falling right after the peak. That is where the trouble starts.
What does an anti-lock braking system actually control?
- The vertical load on each tire.ABS cannot change wheel loads; it modulates brake pressure.
- The brake pressure on each wheel, to keep that wheel's slip ratio in a band just below the peak.It watches wheel speed (and so slip) and releases or reapplies pressure to stay near maximum \(F_x\).
- The friction coefficient of the tire.\(\mu\) is a property of the tire and road. ABS chooses where on the curve the tire operates.
Why is the slip ratio for a given longitudinal force usually much smaller than the slip angle needed for the same lateral force (in radians)?
- The tread and carcass are much stiffer in the fore–aft direction, so the longitudinal stiffness is high.A high \(C_\kappa\) means a small \(\kappa\) gives a large force. That is why slip ratios are quoted in single-digit percent.
- Because longitudinal friction is higher than lateral friction.Peak friction sets the maximum force, not how much slip is needed below the peak.
- Because the wheel is rotating.A cornering tire is rotating too. The difference is in the stiffness of the structure in each direction.
Going further with Milliken
Read: RCVD Chapter 2, Tire Behavior, on longitudinal slip and traction/braking force. Then look at Chapter 20, Driving and Braking, which uses the same ideas at the scale of the whole car.
As you read, look for answers to these questions:
- How is slip ratio defined in the book? Compare it with the definition used here, and check that the signs agree.
- How do the shapes of the traction and braking sides of the curve compare?
- How do the authors connect the tire's longitudinal behavior to a car's brake balance and driveline layout?
Summary
| Idea | Key relation |
|---|---|
| Slip ratio | \(\kappa = (\Omega r_e - V)/V\); locked \(\kappa = -1\) |
| Linear tire | \(F_x \approx C_\kappa\,\kappa\) |
| Peak force | \(F_{x,\max} = \mu_x F_z\) at a small slip ratio |
| Wheel spin dynamics | \(I_w \dot\Omega = F_x r_e - T_b\); past the peak, lock-up runs away |
| Stopping distance | \(d = v^2 / (2\mu g)\), so peak beats sliding |
Next: what happens when a tire must brake and corner at the same time?