Aero balance and the speed-dependent car

One car, two corners

A driver comes in from practice: “Perfect in the hairpin, but it won't turn in the fast sweeper.” Or, more worrying: “Fine in the slow stuff, but the rear feels light through Turn 9 at 220.” Nothing on the car changed between those corners except the speed. In Module 3 the balance between the axles was set by weight, CG height and roll stiffness, none of which depend on speed. Downforce does, and it doesn't land on the axles in the same proportion as the weight.

Center of pressure and aero balance

Downforce is spread over the wings, the floor and the body, but for the axle loads all that matters is how it is shared between the front and rear. Call the front share the aero balance \(\lambda\):

\[ F_{Lf} = \lambda\,F_L, \qquad F_{Lr} = (1-\lambda)\,F_L \]

Just as the weight acts at the CG, the total downforce can be drawn as one force acting at the center of pressure (CoP). Taking moments about the front axle, \(F_L\,x_{CP} = F_{Lr}\,L\), so the CoP sits a distance

\[ x_{CP} = (1-\lambda)\,L \]

behind the front axle. The aero balance plays the same role for downforce that the static front weight \(b/L\) plays for weight. The GT car has \(\lambda = 42\%\) and \(L = 2.70\) m, so its CoP is \(0.58 \times 2.70 = 1.57\) m behind the front axle, while its CG (48% front) is 1.40 m behind it. The CoP is 16 cm behind the CG.

Aero balance. The total downforce acting at the center of pressure (left) loads the axles exactly as its two shares do (right): \(\lambda F_L\) on the front and \((1-\lambda)F_L\) on the rear. Arrows are to scale for the GT car (\(\lambda = 42\%\)). Teams move the CoP with the front-wing flap angle, the rear-wing angle and the car's rake (the difference between rear and front ride height).

Why the balance changes with speed

In a steady corner each axle has to supply its share of the cornering force. The yaw moments about the CG must balance (\(\Sigma M_z = 0\), Lesson 0.1), which makes the front axle's share equal to the static front weight fraction \(b/L\). The front axle's grip, though, comes from its share of the weight plus its share of the downforce. Dividing what each axle can supply by what it must supply gives each axle's own lateral limit, in g:

\[ \begin{aligned} a_{y,f} &= \mu\left(1 + \frac{F_L}{W}\,\frac{\lambda}{b/L}\right) \\[4pt] a_{y,r} &= \mu\left(1 + \frac{F_L}{W}\,\frac{1-\lambda}{a/L}\right) \end{aligned} \]

Whichever is smaller saturates first and sets both the car's limit and its balance (Lesson 3.1). At low speed \(F_L \approx 0\) and both limits are \(\mu\): the balance is whatever the mechanical set-up makes it. As speed rises, the comparison is between \(\lambda\) and \(b/L\):

\(\lambda = b/L\) (CoP at the CG): both axles gain grip in proportion, and the balance is the same at every speed.
\(\lambda < b/L\) (CoP behind the CG): the rear gains more. The front saturates first, and the car understeers more as speed rises.
\(\lambda > b/L\) (CoP ahead of the CG): the front gains more. The rear saturates first, and the car oversteers more as speed rises, exactly where Lesson 1.2 showed oversteer turns into instability.

Mechanical set-up (Module 3) decides the balance in slow corners; aero balance decides how it changes in fast ones. Most race cars run the CoP slightly behind the CG, so the car gains a little understeer, and stays stable, as speed rises.

Derivation: each axle's lateral limit Show

1. What each axle must supply

In a steady turn the yaw moments about the CG balance: \(F_{yf}\,a = F_{yr}\,b\). With \(F_{yf} + F_{yr} = m a_y\), the front axle must supply

\[ F_{yf} = m\,a_y\,\frac{b}{L} \]

2. What it can supply

Its vertical load is its share of the weight plus its share of the downforce, so its maximum lateral force is

\[ F_{yf,\max} = \mu\left(W\,\frac{b}{L} + \lambda F_L\right) \]

3. Set them equal and solve for \(a_y\)

\[ \begin{aligned} m\,a_y\,\frac{b}{L} &= \mu\left(W\,\frac{b}{L} + \lambda F_L\right) \\[4pt] \frac{a_y}{g} &= \mu\left(1 + \frac{F_L}{W}\,\frac{\lambda}{b/L}\right) \end{aligned} \]

The rear axle is the same with \(a/L\) and \(1-\lambda\). The two limits are equal when \(\lambda/(b/L) = (1-\lambda)/(a/L)\), which simplifies to \(\lambda = b/L\). This simple model leaves out load transfer and load sensitivity, which set the low-speed balance; the widget below adds them back as a single “mechanical balance” setting.

With the GT car's defaults, is the car understeering or oversteering at 250 km/h? By how much?
Set the aero balance equal to the static front weight. What happens to the right-hand plot?
Move the aero balance forward to 55%. At high speed the car now oversteers. Why is that more dangerous than high-speed understeer (Lesson 1.2)?
Give the car a little low-speed understeer (front grip −3%) and an aero balance of 52%. At what speed does the balance flip? In which corners would the driver complain, and about what?
Switch to the Formula SAE car. Why does aero balance change its handling more than the club racer's?

The GT car (\(\mu = 1.5\), \(b/L = 0.48\), \(\lambda = 0.42\)) corners at 250 km/h, where its downforce is 0.533 times its weight. Find each axle's limit.

Front: \(1.5\left(1 + 0.533 \times \dfrac{0.42}{0.48}\right) = 1.5 \times 1.466 = 2.20\) g.

Rear: \(1.5\left(1 + 0.533 \times \dfrac{0.58}{0.52}\right) = 1.5 \times 1.594 = 2.39\) g.

The front saturates first, so the car understeers and its limit is 2.20 g; the rear tires still have about 9% in hand. Moving the aero balance to 48% would bring both axles to 2.30 g, but the team may keep some of that understeer to give the driver a stable car at 250 km/h.

“Rear downforce makes a car stable, so the more of it the better.” Moving the center of pressure rearward does make a car more stable at speed, but past a point it just makes it understeer: the front saturates early and the rear tires' extra grip is wasted. The aim is a center of pressure slightly behind the CG, not as far back as possible.

Pitch, ride height and the aero platform

The aero balance is not fixed. Front wings and floors work close to the ground, and the closer they run, the more downforce they make, up to the point where the flow underneath chokes and the downforce collapses. So downforce, and its split, depends on the car's ride height and pitch:

Under braking the nose dives. The front runs closer to the ground and gains downforce, so the balance moves forward and the car tends to oversteer on corner entry.
Under acceleration the nose rises, the balance moves rearward, and the car tends to understeer on corner exit.
At high speed the downforce itself compresses the springs and lowers the car, which changes the downforce again.

Engineers call the car's attitude to the air its aero platform. High-downforce cars run stiff springs (Module 7) mainly to keep that platform steady, so the downforce and its balance stay predictable from braking to exit.

The g-g-V diagram

In Lesson 1.1 the g-g diagram drew a car's performance envelope as a single shape. With downforce and drag it becomes a family of shapes, one for each speed:

Cornering: \(\mu\,(1 + F_L/W)\), growing with speed.
Braking: \(\mu\,(1 + F_L/W) + F_D/W\). Drag helps the brakes, so the braking edge grows fastest.
Accelerating: the drive force (limited by traction at low speed and by power at high speed, Lesson 4.3) minus drag. It shrinks to zero at top speed.

Stacked on a speed axis, the envelopes form a three-dimensional shape called the g-g-V diagram. It is the main input to the lap-time simulations of Module 9.

With the GT car, compare the envelopes at 100 and 250 km/h. Which edge grows the most, and why?
Set the downforce to zero. What still changes with speed?
Raise the power. Which part of the envelope changes, and at which speeds?
Switch to the Formula SAE car and highlight a speed near its top speed. What does the top of the envelope look like, and why?

Find the GT car's envelope at 250 km/h, where \(F_L/W = 0.533\) and \(F_D/W = 0.232\), with 370 kW at the wheels.

Cornering: \(1.5 \times 1.533 = 2.30\) g. Braking: \(2.30 + 0.23 = 2.53\) g.

Accelerating: the power can push with \(P/V = 370\,000/69.4 = 5330\) N, or \(0.418\,W\). Drag takes \(0.232\,W\), leaving \(0.19\) g. At 100 km/h, by contrast, the same car can brake at about 1.7 g and accelerate at about 1.0 g.

Check your understanding

A car with a 2.5 m wheelbase has an aero balance of 40% front. How far behind the front axle is its center of pressure?

\(x_{CP} = (1 - \lambda)\,L\).

\(x_{CP} = 0.60 \times 2.5 = 1.5\) m.

A 750 kg car with 52% static front weight makes 2000 N of downforce with a 40% aero balance. What is the load on its front axle?

Front share of the weight plus front share of the downforce.

\(0.52 \times 750 \times 9.81 + 0.40 \times 2000 = 3826 + 800 = 4626\) N.

A car with \(\mu = 1.4\) and 45% static front weight has an aero balance of 50%. At a speed where its downforce is 0.8 times its weight, what is its lateral limit?

Work out both axle limits; the car's limit is the smaller one.

Front: \(1.4(1 + 0.8 \times 0.50/0.45) = 2.644\) g. Rear: \(1.4(1 + 0.8 \times 0.50/0.55) = 2.418\) g.

The rear saturates first, so the limit is 2.42 g and the car oversteers. The CoP is ahead of the CG (50% vs 45% front).

A car's aero balance is further forward than its static weight distribution (\(\lambda > b/L\)). How does its handling change as speed rises?

It understeers more.That happens when the CoP is behind the CG, giving the rear more of the extra grip.
It oversteers more, which can make it unstable in fast corners.The front gains grip faster than the rear, so the rear saturates first, and an oversteering car loses stability as speed rises (Lesson 1.2).
Nothing changes: downforce adds grip to both axles.It adds grip to both, but not in proportion to what each axle has to supply.

Under hard braking the car's nose dives toward the ground. What usually happens to the aero balance, and to the handling on corner entry?

The front gains downforce, the balance moves forward, and the car tends to oversteer on entry.Front wings and floors make more downforce closer to the ground.
The rear gains downforce and the car understeers on entry.Dive lowers the front, not the rear.
Nothing: aero balance depends only on the wing angles.Ride height and pitch change the downforce and its split.

On the g-g-V diagram, why does the braking limit grow faster with speed than the cornering limit?

Because the brakes get hotter.Brake temperature matters, but not in this model.
Because drag also decelerates the car, adding \(F_D/W\) on top of the tires' grip.Downforce helps both edges equally; drag only helps braking.
Because load transfer to the front is bigger at speed.Load transfer shares the load between the axles; with ideal brake bias it doesn't change the total.

Going further with Milliken

Read: RCVD Chapter 15, Applied Aerodynamics, on downforce distribution and aero balance, with Chapter 3 for the background.

As you read, look for answers to these questions:

How does the book describe the fore–aft distribution of downforce, and how is it measured?
How do ride height and pitch change downforce, and what does that mean for set-up?
Look back at Chapter 9, “g-g” Diagram. Which features of a car's g-g diagram would change with speed once it has wings?

Summary

Idea	Key relation
Aero balance	\(F_{Lf} = \lambda F_L,\quad F_{Lr} = (1-\lambda)F_L\)
Center of pressure	\(x_{CP} = (1-\lambda)L\) behind the front axle
Axle lateral limits	\(a_{y,f} = \mu[1 + (F_L/W)\,\lambda/(b/L)],\quad a_{y,r} = \mu[1 + (F_L/W)(1-\lambda)/(a/L)]\)
Speed-independent balance	\(\lambda = b/L\): CoP at the CG
Usual set-up	CoP slightly behind the CG: a little more understeer, and stability, at speed
g-g-V envelope	Lateral \(\mu(1+F_L/W)\); braking adds \(F_D/W\); acceleration is drive minus drag

Next: drag. What it costs on the straights, how power fights it, and how to choose a wing setting for a circuit.