One contact patch, one budget
In Lesson 1.1 the friction circle said that a car cannot brake and corner at full strength at the same time. In Lessons 2.1 and 2.2 we met the two ways a tire makes force: slip angle for lateral force, slip ratio for longitudinal force. Real driving mixes them constantly: trail braking into a corner, feeding in power on the way out. This lesson puts the two together at the level of a single tire. The result explains locked-wheel understeer, power oversteer, and why smooth drivers are fast.
The friction ellipse
A tire has one contact patch. Whether the force it is asked for points forward, backward, sideways or somewhere in between, the rubber in that patch has only so much grip to offer. If the peak longitudinal and lateral forces are \(F_{x,\max}\) and \(F_{y,\max}\), a good first model of the limit is an ellipse:
When \(F_{x,\max} = F_{y,\max}\) this is the friction circle from Lesson 1.1. In practice tires usually have slightly more longitudinal than lateral grip, so the boundary is an ellipse. The car's g-g envelope is, roughly, the four tires' ellipses added together, with load transfer moving grip between them.
How the slips share the budget
When a tire has both a slip angle and a slip ratio, its tread elements are dragged diagonally through the contact patch. They deflect in the direction of that combined slip, start sliding when the total deflection force exceeds friction, and push back in roughly the direction of the combined slip.
The model in this lesson captures that with one idea: measure each slip as a fraction of the slip that gives peak force on its own,
The tire then produces the force it would make at a “total slip” of \(n\), shared between the two directions in proportion to \(n_x\) and \(n_y\). With only one kind of slip present, this reduces exactly to the pure curves of Lessons 2.1 and 2.2. At \(n = 1\) the forces land exactly on the friction ellipse.
Asking a tire for longitudinal force always costs lateral force, and vice versa. Braking or accelerating in a corner moves the operating point around the friction ellipse; it does not create extra grip.
- Start from Pure cornering and slowly add braking slip. Watch the red point slide around the ellipse: Fx grows while Fy shrinks.
- Choose Locked wheel, then move the slip angle from 2° to 15°. How much lateral force can a locked wheel make? What would the driver feel through the steering wheel?
- On the Trail braking preset, how much lateral force is lost compared with pure cornering? Now find a slip ratio that gives a 25% loss.
- Choose Too much power. If this were a rear tire mid-corner, which end of the car would slide first?
A tire with \(F_z = 3000\) N has \(\mu_x = 1.5\) and \(\mu_y = 1.4\), so \(F_{x,\max} = 4500\) N and \(F_{y,\max} = 4200\) N. While trail braking it supplies 3000 N of braking force. What is the most lateral force it can add?
\[ F_y = F_{y,\max}\sqrt{1 - \left(\frac{F_x}{F_{x,\max}}\right)^2} = 4200\sqrt{1 - \left(\tfrac{3000}{4500}\right)^2} = 4200 \times 0.745 = 3130\ \text{N} \]
Using two-thirds of the braking capacity still leaves three-quarters of the cornering capacity. The ellipse is “flat” near its top, which is exactly why gentle trail braking costs so little cornering force.
What combined slip does to the whole car
| Situation | What the tires are doing | What the car does |
|---|---|---|
| Front wheels locked | Front tires slide straight along their path, so almost no lateral force | Goes straight on regardless of steering: “locked-wheel understeer” |
| Rear wheels locked | Rear tires lose their lateral force while the fronts keep theirs | Rear swings round and the car spins. This is why brake bias is set forward. |
| Too much throttle (rear-drive) | Rear tires spend grip on Fx, so they have less for Fy | Power oversteer: the rear steps out |
| Too much throttle (front-drive) | Front tires spend grip on Fx | Power understeer: the car runs wide |
| Trail braking | Front tires share a little Fx with Fy, near the top of the ellipse | Small loss of front lateral force, more than repaid by load transfer onto the front (Module 3) |
“If I lock the brakes, I can still steer around the obstacle if I turn hard enough.” A locked tire's force opposes its sliding direction, which is almost straight backward. Turning the wheel changes almost nothing until the driver releases the brake enough for the tire to roll again. Avoiding this is the whole point of ABS.
Check your understanding
A tire at \(F_z = 2500\) N has \(\mu_x = 1.6\) and \(\mu_y = 1.5\). It is braking with 2400 N. Using the friction ellipse, what is the maximum lateral force it can produce at the same time?
Find \(F_{x,\max}\) and \(F_{y,\max}\) first, then use \(F_y = F_{y,\max}\sqrt{1 - (F_x/F_{x,\max})^2}\).
\(F_{x,\max} = 4000\) N and \(F_{y,\max} = 3750\) N. \(F_x/F_{x,\max} = 0.6\), so \(F_y = 3750\sqrt{1 - 0.36} = 3750 \times 0.8 = 3000\) N.
A rear tire carries 2500 N and obeys a friction circle with \(\mu = 1.4\). Mid-corner it must supply 3000 N of lateral force. What is the largest driving force the driver can add before the tire saturates?
For a circle, \(F_x^2 + F_y^2 \le (\mu F_z)^2\).
\(\mu F_z = 3500\) N, so \(F_x = \sqrt{3500^2 - 3000^2} = \sqrt{3.25 \times 10^6} = 1803\) N. Any more throttle than this and the rear tire starts to slide: power oversteer.
A driver brakes so hard that both rear wheels lock while the fronts keep rolling. What does the car most likely do?
- Stop in a straight line, just a little further away.That is closer to what happens with locked fronts. Locked rears lose their ability to hold the back of the car in line.
- Spin, because the rear tires can no longer make lateral force but the fronts still can.Any small yaw disturbance is resisted only by the front tires, which turns the car further. This is why brake bias is set toward the front.
- Understeer, because the rear tires are sliding.Losing rear lateral grip produces oversteer, not understeer.
Why does a small amount of trail braking cost only a little cornering force?
- The friction ellipse is nearly flat near pure cornering, so a modest Fx moves the operating point only slightly down in Fy.With \(F_x\) at 30% of its maximum, \(F_y\) can still reach \(\sqrt{1 - 0.09} = 95\%\) of its maximum.
- Braking increases the friction coefficient.Braking does not change \(\mu\). It changes where on the ellipse the tire operates.
- Lateral and longitudinal grip are independent.They are not — that is the whole point of the friction ellipse.
A rear-wheel-drive car is balanced mid-corner. The driver applies full throttle. Which end of the car is most likely to lose grip first, and why?
- The front, because acceleration moves load off the front tires.Load does move rearward, which helps the rear. But the rear tires are now also carrying a large Fx.
- The rear, because the driven tires spend part of their grip budget on Fx and have less left for Fy.This is power oversteer. With modest throttle, the rearward load transfer can outweigh it. Too much throttle and the rear slides.
- Neither, because the engine adds energy, not force, to the tires.The engine's torque appears as a longitudinal force at the driven contact patches.
Going further with Milliken
Read: RCVD Chapter 2, Tire Behavior, on combined lateral and longitudinal force (the friction ellipse or “friction circle” plots). Then revisit Chapter 9, “g-g” Diagram, and see how the car's envelope is built up from the four tires.
As you read, look for answers to these questions:
- How is combined-slip data usually plotted, and what do lines of constant slip angle look like on those plots?
- Is a real tire's combined-force boundary exactly an ellipse? Where does it differ?
- How do the authors relate a single tire's friction ellipse to the whole car's g-g diagram?
Summary
| Idea | Key relation |
|---|---|
| Friction ellipse | \((F_x/F_{x,\max})^2 + (F_y/F_{y,\max})^2 \le 1\) |
| Available lateral force under braking | \(F_y = F_{y,\max}\sqrt{1 - (F_x/F_{x,\max})^2}\) |
| Normalized combined slip | \(n = \sqrt{(\kappa/\kappa_{\text{peak}})^2 + (\alpha/\alpha_{\text{peak}})^2}\) |
| Locked or spinning tire | Force points along the sliding direction, so little lateral force |
Next: the last tire lesson looks at the conditions that set the size of the ellipse in the first place: camber, pressure and temperature.