Grip without weight
In Lesson 1.1 the maximum corner speed was \(V_{\max} = \sqrt{\mu g R}\). The mass cancelled out: a heavier car presses its tires harder into the road, but it also needs more force to turn, and the two effects balance. To go faster through a corner of a given radius, you seem to need stickier tires.
There is another way. If something other than gravity pushes the car onto the road, the tires get more load, and therefore more grip, without the car getting any harder to turn. That something is the air. A wing turned upside down, or a shaped floor, makes downforce: an aerodynamic load that presses the car down. Since Jim Hall's Chaparral race cars of the 1960s, downforce has been the biggest single source of cornering performance in top-level motorsport. It has a price, though: the same air that pushes down also pushes back, as drag.
Dynamic pressure
Every aerodynamic force on the car scales with the dynamic pressure of the oncoming air:
where \(\rho\) is the air density (1.225 kg/m³ at sea level and 15 °C) and \(V\) is the car's speed through the air. It is the pressure you would feel on a flat plate held square to the flow, and it grows with the square of speed: double the speed and every aerodynamic force is four times as large.
| Speed | Dynamic pressure \(q\) | GT car downforce (\(C_L A = 2.3\ \text{m}^2\)) |
|---|---|---|
| 50 km/h | 118 Pa | 272 N, the weight of 28 kg |
| 100 km/h | 473 Pa | 1087 N (111 kg) |
| 200 km/h | 1890 Pa | 4348 N (443 kg) |
| 300 km/h | 4254 Pa | 9783 N (997 kg) |
At town speeds aerodynamic forces are tiny. At 300 km/h the GT car's wing and floor push down with the weight of a tonne. A car with downforce is, in effect, a different car at every speed.
Drag and downforce coefficients
The shape of the car decides how much force it turns that pressure into. Engineers write the drag force \(F_D\) and the downforce \(F_L\) as
where \(A\) is a reference area (usually the frontal area) and \(C_D\) and \(C_L\) are the dimensionless drag and downforce coefficients. Because teams define \(A\) differently, the products \(C_D A\) and \(C_L A\) (“CDA” and “CLA”, in m²) are the numbers that actually get compared. Their ratio \(C_L/C_D\), the lift-to-drag ratio or efficiency, says how much downforce you get for each newton of drag.
| Car | \(C_D A\) (m²) | \(C_L A\) (m²) | \(C_L / C_D\) |
|---|---|---|---|
| Road car | 0.6–0.7 | about 0 (often slight lift) | — |
| Club racer (splitter and small wing) | 0.75 | 0.5 | 0.7 |
| GT car | 1.0 | 2.3 | 2.3 |
| Formula SAE with wings | 1.2 | 2.8 | 2.3 |
| Top single-seater (approximate) | ≈ 1.3 | ≈ 4–5 | ≈ 3–4 |
The middle three rows are the example cars used throughout this module; real values vary with set-up, and teams rarely publish theirs. Many references, RCVD included, use a lift coefficient that is negative for downforce. This course quotes downforce as a positive number.
Air density matters too. Density falls on hot days and at altitude (\(\rho = p/(R_{\text{air}}T)\)), so at Mexico City, 2240 m up, a car makes roughly 20% less downforce and 20% less drag at the same speed than at sea level.
Cornering with downforce
Return to the steady corner of Lesson 1.1. The tires must still supply \(mV^2/R\), but they can now push against the weight plus the downforce:
Solving for the speed at which the two sides are equal gives the grip-limited corner speed:
With no downforce \(R^*\) is infinite and this is the familiar \(\sqrt{\mu g R}\). With downforce, the gain depends on how the corner compares with \(R^*\), the aero radius:
- Tight corners (\(R \ll R^*\)) are taken slowly, so there is little dynamic pressure and little downforce. The gain is small.
- Fast corners (\(R\) approaching \(R^*\)) are taken quickly, where downforce is large. The gain grows rapidly.
- Beyond \(R^*\), downforce grows with speed faster than the force needed to turn, and this simple model says grip never runs out. In reality tire load sensitivity (Lesson 2.1) and the power needed to overcome drag (Lesson 4.3) set the limit. This is why the fastest cars take some fast corners flat out.
At any speed, the lateral acceleration the tires can sustain is \(a_y = \mu\,(1 + F_L/W)\) in g. Downforce equal to half the weight means 50% more cornering grip.
Downforce adds tire load without adding mass. The grip goes up but the inertia does not, so the car corners faster, and the benefit grows with the speed of the corner.
Derivation: corner speed with downforce Show
1. Force needed and force available
In a steady turn the tires must supply \(mV^2/R\). With friction coefficient \(\mu\) they can supply \(\mu\) times their total vertical load, the weight plus the downforce. At the limit the two are equal:
2. Collect the \(V^2\) terms
3. Divide through by \(m/R\)
The group \(2m/(\rho\,\mu\,C_L A)\) has units of length. Calling it \(R^*\) gives
4. Check the result
With \(C_L A = 0\), \(R^* \to \infty\) and \(V^2 = \mu g R\), as in Lesson 1.1. As \(R \to R^*\) the denominator goes to zero: the required speed rises without limit, because each extra km/h brings enough extra downforce to pay for itself. The ratio of corner speeds with and without downforce is \(1/\sqrt{1 - R/R^*}\), which depends only on how big the corner is compared with \(R^*\).
A heavier car has a larger \(R^*\), so the same wing helps it less: downforce has to share its benefit with more mass.
- With the Formula SAE car, compare the gain from downforce in a 15 m hairpin and a 60 m sweeper. Why are they so different?
- Set the downforce to zero. Which Lesson 1.1 formula does the left plot now show?
- Move the corner radius toward \(R^*\). What happens to the predicted speed, and what does the message below the readouts say limits the car instead?
- Lower the air density to 0.98 kg/m³ (Mexico City). How much corner speed do you lose in the 60 m corner?
- Switch to the GT car. Its \(C_L A\) is close to the Formula SAE car's, yet the gain in a 60 m corner is far smaller. Use \(R^*\) to explain why.
The Formula SAE car (\(m = 280\) kg, \(\mu = 1.6\), \(C_L A = 2.8\) m²) takes a 30 m radius corner at sea level. Find its corner speed with and without downforce.
Aero radius: \(R^* = \dfrac{2 \times 280}{1.225 \times 1.6 \times 2.8} = 102.0\) m.
With downforce: \(V^2 = \dfrac{1.6 \times 9.81 \times 30}{1 - 30/102.0} = \dfrac{470.9}{0.706} = 667\) m²/s², so \(V = 25.8\) m/s \(= 93\) km/h.
Without: \(V = \sqrt{1.6 \times 9.81 \times 30} = 21.7\) m/s \(= 78\) km/h. Downforce adds 19%.
Check: at 25.8 m/s, \(q = 408.5\) Pa and \(F_L = 1144\) N, which is 42% of the car's 2747 N weight. The lateral acceleration is \(1.6 \times 1.42 = 2.27\) g, and indeed \(V^2/(gR) = 667/(9.81 \times 30) = 2.27\) g.
“Downforce makes the car heavier, so it must be slower to accelerate and brake.” Downforce adds load, not mass. The force needed to accelerate the car is still \(ma\), so its inertia is unchanged, and the extra load actually lets the tires brake harder. The real costs are drag, which slows the car on the straights, and tire load sensitivity, which means the extra load buys a little less grip than you might hope.
Real-world complications
- Load sensitivity: doubling a tire's load less than doubles its grip (Lesson 2.1), so downforce is worth less than the simple formula suggests. Lesson 4.3 includes this in a lap-time model.
- Ride height and pitch: downforce changes as the car squats, dives and rolls, and the split between front and rear moves with it (Lesson 4.2).
- Dirty air: a car following closely sits in another car's wake and loses downforce, which is why it is hard to follow through fast corners.
- Drag: every wing that adds downforce also adds drag, which costs top speed (Lesson 4.3).
Check your understanding
What is the dynamic pressure at 180 km/h at sea level (\(\rho = 1.225\) kg/m³)?
Convert to m/s first, then \(q = \tfrac{1}{2}\rho V^2\).
\(V = 180/3.6 = 50\) m/s, so \(q = 0.5 \times 1.225 \times 50^2 = 1531\) Pa.
The GT car has \(C_L A = 2.3\) m². How much downforce does it make at 200 km/h at sea level?
\(F_L = \tfrac{1}{2}\rho V^2 C_L A\), with \(V\) in m/s.
\(V = 55.56\) m/s, \(q = 0.5 \times 1.225 \times 55.56^2 = 1890\) Pa, so \(F_L = 1890 \times 2.3 = 4348\) N, about a third of the car's weight.
The GT car (\(m = 1300\) kg, \(\mu = 1.5\), \(C_L A = 2.3\) m²) takes a 150 m radius corner at sea level. What is its grip-limited corner speed?
Find \(R^* = 2m/(\rho\mu C_L A)\) first, then \(V = \sqrt{\mu g R/(1 - R/R^*)}\).
\(R^* = 2600/(1.225 \times 1.5 \times 2.3) = 615\) m. \(V^2 = 1.5 \times 9.81 \times 150/(1 - 150/615) = 2207/0.756 = 2919\) m²/s², so \(V = 54.0\) m/s \(= 194.5\) km/h, 15% faster than the 169 km/h it could manage without downforce.
A car's speed rises from 100 km/h to 200 km/h. What happens to its downforce and drag?
- Both double.Aerodynamic forces scale with dynamic pressure, which goes with \(V^2\), not \(V\).
- Both become four times larger.\(q = \tfrac{1}{2}\rho V^2\): doubling \(V\) multiplies \(q\), and every aerodynamic force, by four.
- Downforce quadruples but drag only doubles.Both are \(q\) times a coefficient and an area, so they scale together.
Why does a wing raise the cornering speed much more in a fast sweeper than in a slow hairpin?
- Because the tires are hotter in fast corners.Temperature matters (Lesson 2.4), but the effect here comes from the aerodynamics alone.
- Because downforce grows with the square of speed: in a fast corner there is much more of it compared with the car's weight.The gain factor \(1/\sqrt{1 - R/R^*}\) is close to 1 when \(R \ll R^*\) and grows quickly as \(R\) approaches \(R^*\).
- Because the car is lighter in fast corners.The mass never changes; only the load on the tires does.
At 200 km/h a car's wings make downforce equal to half its weight. Compared with the same car without wings, how much force does it take to accelerate it at 0.5 g?
- 50% more, because the car is effectively heavier.Downforce is a load, not a mass. It doesn't add inertia.
- The same \(ma\), plus whatever extra force is needed to overcome the extra drag.Inertia depends on mass alone. The wings' cost when accelerating is their drag.
- Less, because the tires grip better.Better grip raises the force the tires can supply, not the force the acceleration requires.
Going further with Milliken
Read: RCVD Chapter 3, Aerodynamic Fundamentals, and the opening sections of Chapter 15, Applied Aerodynamics.
As you read, look for answers to these questions:
- How does Bernoulli's equation lead to the dynamic pressure \(\tfrac{1}{2}\rho V^2\) and to pressure coefficients?
- What sign convention and reference area does the book use for lift and drag coefficients?
- How does a wing's behavior change when it works close to the ground?
Summary
| Idea | Key relation |
|---|---|
| Dynamic pressure | \(q = \tfrac{1}{2}\rho V^2\); \(\rho = 1.225\) kg/m³ at sea level |
| Drag and downforce | \(F_D = q\,C_D A,\quad F_L = q\,C_L A\) |
| Lateral grip with downforce | \(a_y = \mu\,(1 + F_L/W)\) in g |
| Corner speed with downforce | \(V = \sqrt{\mu g R / (1 - R/R^*)},\quad R^* = 2m/(\rho\mu C_L A)\) |
| Where downforce helps most | Fast corners: the gain \(1/\sqrt{1 - R/R^*}\) grows as \(R\) approaches \(R^*\) |
Next: downforce does not land evenly on the two axles, and the split decides how the car handles at speed.