Lap time is an acceleration problem
A race track fixes the path. The only way to get around it faster is to carry more speed, and the only way to carry more speed is to accelerate harder: get off the corner sooner, brake later, and go through the corner itself faster. Each of these is an acceleration — forward, backward or sideways.
Newton's second law says every acceleration needs a force, \(\Sigma \mathbf{F} = m\mathbf{a}\). Apart from aerodynamic forces (Module 4), every horizontal force on a race car comes from four rubber contact patches, each roughly the size of your hand. So the question “how fast can this car go around the lap?” becomes “how much force can those four patches produce, and how well can the driver use it?”
A race car is a machine for generating and controlling tire forces. Almost everything in this course — tires, load transfer, aerodynamics, suspension — matters because it changes how much tire force is available or how well it can be used.
Cornering is acceleration
A car driving around a curve of radius \(R\) at constant speed \(v\) is accelerating toward the center of the curve, even though its speed is not changing. From your dynamics course:
That lateral force \(F_y\) has to come from the tires. The simplest tire model is Coulomb friction: the tires can supply at most \(\mu\) times the normal force. On a flat road with no aerodynamic downforce the normal force is just the weight, \(mg\), so
Notice that the mass cancels. In this simple model a heavy car and a light car with the same tires corner at the same speed. (Real tires are not quite this simple — Module 2 shows why heavier cars actually lose a little grip, and Module 4 shows how downforce breaks the cancellation.)
- Double the radius from 50 m to 100 m. By what factor does the speed increase? Why \(\sqrt{2}\)?
- A 10% grip improvement gives only about 5% more corner speed. Use the equation to explain why.
- Compare a race car (\(\mu \approx 1.4\)) with the road-car curve in a 30 m hairpin and a 200 m sweeper.
A club racer on slick tires (\(\mu = 1.4\)) approaches a 50 m radius corner. What is the fastest it can take the corner?
\[ v_{\max} = \sqrt{1.4 \times 9.81\ \text{m/s}^2 \times 50\ \text{m}} = \sqrt{686.7}\ \text{m/s} = 26.2\ \text{m/s} \approx 94\ \text{km/h} \]
The lateral acceleration at that speed is \(v^2/R = 13.7\ \text{m/s}^2\), or 1.4 g — the same number as \(\mu\). That is no coincidence: with no downforce, the maximum lateral acceleration in g equals the friction coefficient.
The friction circle
A tire does not care which direction you ask it to push. It can produce braking force, driving force, cornering force, or a mix — but the total horizontal force is limited. For our simple friction model:
Divide through by \(mg\) and write accelerations in units of g, and the limit becomes a circle of radius \(\mu\):
The consequence is immediate and important: grip used for braking is not available for cornering. If a car with \(\mu = 1.5\) is braking at 0.9 g, the most lateral acceleration it can add is \(\sqrt{1.5^2 - 0.9^2} = 1.2\) g.
“Braking and turning are separate jobs, so the tires can do both at full strength.” They cannot — every contact patch has a single force budget. This is why a driver who brakes too hard into a corner runs wide.
The g-g diagram: a map of what the car can do
If you record a car's longitudinal and lateral acceleration throughout a lap and plot one against the other, the points fill a region bounded by the car's limits. Milliken calls the boundary of that region the car's performance envelope, and the plot itself the g-g diagram.
A real envelope is not a perfect circle:
- Bottom (braking): all four wheels brake, so braking is limited mainly by tire grip.
- Sides (cornering): limited by lateral grip — and, as Module 3 shows, by how evenly the front and rear tires share the work.
- Top (accelerating): usually limited by engine power or by the traction of only the driven wheels, so it is often much flatter and lower than the bottom.
We will model the envelope as an ellipse (lateral and braking grip as its semi-axes) cut off at the top by a drive limit.
- Put the dot at 0.9 g lateral and 0.6 g braking. How much of the available grip is being used?
- Hold the braking at 0.8 g and slide right until you reach the edge. Check the “max lateral” readout against the ellipse equation \(\left(\tfrac{a_y}{\mu_y}\right)^2 + \left(\tfrac{a_x}{\mu_x}\right)^2 = 1\).
- Reduce the drive limit to 0.3 g. Where on the diagram would a more powerful engine help, and where would it make no difference at all?
- Switch between the Novice and Expert traces. Which one spends more of the corner at the edge of the envelope?
Using the envelope: where the driver comes in
The envelope says what the car can do. Lap time depends on how much of the lap the driver spends at its edge. A novice tends to do one thing at a time: brake in a straight line, release the brake, turn, straighten, then accelerate. Their trace runs out along the axes and back to the middle, leaving large parts of the envelope unused.
An expert blends the phases: they keep some braking while turning in (trail braking), arrive at peak cornering exactly at the apex, and feed in throttle as they unwind the steering. Their trace sweeps around the boundary. Every moment spent inside the envelope is grip — and time — left on the table.
This is why Milliken treats the driver and vehicle as one system. A car with a large envelope that is unpredictable near the limit may be slower than a car with a smaller envelope that the driver can use with confidence. Lesson 2 of this module develops that idea of stability and control.
Check your understanding
A Formula SAE car with \(\mu = 1.2\) enters a 40 m radius corner. Assuming no downforce, what is its maximum cornering speed?
Start from \(m v^2 / R \le \mu m g\). Does the mass matter?
\(v_{\max} = \sqrt{\mu g R} = \sqrt{1.2 \times 9.81 \times 40} = \sqrt{470.9} = 21.7\ \text{m/s}\) (about 78 km/h). The mass cancels, so it is not needed.
A car's tires obey a friction circle with \(\mu = 1.5\). The driver is braking at 0.9 g. What is the maximum lateral acceleration the car can sustain at the same time?
The total acceleration cannot exceed the circle's radius: \(a_x^2 + a_y^2 \le \mu^2\).
\(a_y = \sqrt{\mu^2 - a_x^2} = \sqrt{1.5^2 - 0.9^2} = \sqrt{2.25 - 0.81} = \sqrt{1.44} = 1.2\ \text{g}\).
A car's g-g envelope is an ellipse with 1.4 g lateral and 1.3 g braking semi-axes. The driver is trail-braking at 0.65 g. What lateral acceleration is available?
Use \(\left(\tfrac{a_y}{1.4}\right)^2 + \left(\tfrac{a_x}{1.3}\right)^2 = 1\) and solve for \(a_y\).
\(a_x/\mu_x = 0.65/1.3 = 0.5\), so \(a_y = 1.4\sqrt{1 - 0.5^2} = 1.4 \times 0.866 = 1.21\ \text{g}\). Using half the braking capacity still leaves 87% of the cornering capacity — one reason trail braking is fast.
Using the simple friction model with no aerodynamics, a team adds 30 kg of ballast to their car. What happens to its maximum steady cornering speed?
- It decreases, because a heavier car needs more lateral force.It does need more lateral force — but its tires also get proportionally more normal force, so the two effects cancel in this model.
- It stays the same, because the required force and the available grip both scale with mass.\(m v^2/R \le \mu m g\): the mass cancels. Real tires are load-sensitive (Module 2), so in practice the heavier car is slightly slower.
- It increases, because more weight pushes the tires into the road.More weight does increase the normal force, but it increases the force needed to turn the car by exactly the same proportion.
On a typical rear-wheel-drive race car's g-g diagram, which part of the envelope is most likely to be limited by engine power rather than tire grip?
- The bottom (braking).All four wheels brake and brakes are very powerful, so braking is grip-limited.
- The left and right sides (pure cornering).Pure cornering needs almost no engine power at all — it is grip-limited.
- The top (forward acceleration), especially at higher speeds.Available thrust is roughly power ÷ speed, so at speed the engine runs out before the driven tires do. At low speed it is often the driven tires (traction) that limit instead.
Two drivers take the same corner in identical cars. Driver B's g-g trace follows the envelope boundary, while Driver A's trace runs along the axes. Why is Driver B faster?
- Driver B reaches a higher peak lateral acceleration.Both can reach the same peak — they have the same car. The difference is how long they stay near the limit.
- Driver B spends more of the corner using the full grip available, so the average acceleration is higher.Lap time depends on the time spent at the edge of the envelope, not just on its size.
- Driver B uses less grip, so the tires stay cooler.Tire management matters in long races, but it is not why a blended trace is faster through a single corner.
Going further with Milliken
Read: RCVD Chapter 1, The Problem Imposed by Racing, and Chapter 9, “g-g” Diagram.
As you read, look for answers to these questions:
- How do the authors divide up the “racing problem” between the tires, the vehicle and the driver?
- What does measured g-g data from a real car look like compared with our idealized ellipse? What explains the differences?
- How can a g-g plot be used to judge a driver, not just a car?
Summary
| Idea | Key relation |
|---|---|
| Centripetal acceleration | \(a_y = v^2 / R\) |
| Maximum corner speed (no aero) | \(v_{\max} = \sqrt{\mu g R}\) |
| Friction circle | \(\sqrt{a_x^2 + a_y^2} \le \mu\) (in g) |
| Friction ellipse / envelope | \((a_y/\mu_y)^2 + (a_x/\mu_x)^2 \le 1\), capped by the drive limit |
Next: Lesson 1.2 asks whether a driver can actually use the envelope, which brings in stability and control. After that, Module 2 replaces the single fixed friction coefficient \(\mu\): we open up the tire and find that the force it produces depends on slip angle, load, and more.