Twice the power, not twice the speed
Give a car twice the engine power and its top speed rises by only about a quarter. Give it a bigger wing and it laps one circuit faster and another slower. Both facts come from the same place: drag, and the power it takes to push the car through the air. This lesson puts power, traction and drag together, and uses them to choose a wing setting for a circuit.
Power is force times speed
An engine delivers power \(P\), not a fixed force. At speed \(V\) the driving force it can put through the wheels is
where \(P\) is the power reaching the driven wheels (the engine's output less drivetrain losses, typically 10–15%). This curve falls as speed rises, and at low speed it rises without limit. The tires cannot pass on unlimited force, though: the drive force is also capped by the traction limit of the driven axle. For a rear-wheel-drive car, accelerating moves load onto the rear tires (Lesson 3.1), which raises their limit:
The drive force available is the smaller of the two. Below the crossover speed where they cross, acceleration is traction-limited, and more power would only spin the wheels. Above it, acceleration is power-limited, and more grip would not help.
Derivation: the rear-wheel-drive traction limit Show
1. Moments about the front contact patch
Take drag to act at CG height, and add the inertial load \(m a_x\) at the CG as in Lesson 0.1. Moments about the front contact patch give the rear axle load:
2. Use Newton's second law
The net forward force is \(m a_x = F_x - F_D\), so the drag terms cancel and the rear load depends only on the drive force:
3. Set the drive force to the tires' limit
Collect the \(F_x\) terms:
The \(1 - \mu h/L\) in the denominator is load transfer helping: the harder the car accelerates, the more load the driven tires carry. It is the mirror image of the rear axle's lock-up limit under braking in Lesson 3.3.
Drag and top speed
What accelerates the car is the drive force minus the drag, \(m a_x = F_x - F_D\). Drag grows with \(V^2\) while the power-limited drive force shrinks with \(1/V\), so they must meet. Where they do, the car can go no faster. Setting \(P/V = \tfrac{1}{2}\rho V^2 C_D A\):
The power needed to overcome drag grows with the cube of speed, so top speed grows only with the cube root of power:
- 10% more power gives 3.2% more top speed.
- 10% less drag (\(C_D A\)) gives 3.6% more top speed: a slightly better deal.
- Twice the power gives 26% more top speed.
This lesson ignores rolling resistance, a force of roughly 1–2% of the tire load. It matters at low speed but is small next to drag at top speed.
- With the GT car, at what speed does acceleration switch from traction-limited to power-limited?
- Double the power. How much does the top speed rise? Why does the time to 100 km/h barely change?
- Raise the tire friction instead. Which part of the drive-force curve moves?
- Compare 10% more power with 10% less drag. Which gives the higher top speed?
- Add downforce. What happens to the traction limit, and to the top speed?
The GT car has 370 kW at the wheels and \(C_D A = 1.0\) m². Find its top speed, the power it would need for 320 km/h, and roughly where it stops being traction-limited.
Top speed: \(V_{\text{top}} = \left(\dfrac{2 \times 370\,000}{1.225 \times 1.0}\right)^{1/3} = 84.5\) m/s \(= 304\) km/h.
Power for 320 km/h (88.9 m/s): \(P = 0.5 \times 1.225 \times 1.0 \times 88.9^3 = 430\) kW. That is 16% more power for 5% more speed.
Crossover: ignoring downforce, the traction limit is \(\dfrac{1.5 \times 0.52 \times 12\,753}{1 - 1.5 \times 0.45/2.70} = \dfrac{9947}{0.75} = 13\,263\) N, about 1.04 g. Power takes over when \(P/V\) falls to that: \(V = 370\,000/13\,263 = 27.9\) m/s \(\approx 100\) km/h. (With downforce the traction limit is a little higher, so the widget shows 93 km/h.)
Downforce or drag? Choosing a wing for the circuit
A wing that makes more downforce also makes more drag. Its lift-to-drag ratio \(C_L/C_D\) says how much downforce each newton of drag buys; for a rear wing it is typically around 3 to 5. More wing means:
- Faster corners (Lesson 4.1), and later braking, because downforce adds grip.
- Slower straights: lower top speed and weaker acceleration once the car is power-limited.
Which effect wins depends on the circuit. On a tight circuit, most of the lap is spent in or near corners, so teams run the most downforce the car can carry; Monaco is the classic example. On a fast circuit with long straights, such as Monza, they trim the wings back to an optimum somewhere in between.
There is no “best” wing setting, only a best setting for a circuit. It balances the time downforce gains in the corners against the time its drag costs on the straights.
- With the GT car, find the fastest wing level on each of the three circuits. How do they compare?
- On the Fast circuit, compare the speed traces at the fastest wing level and at ×2.5. Where does the extra wing gain time, and where does it lose it?
- On the Mixed circuit, note the lap times at ×0 and ×1 with load sensitivity on, then off. By how much does the constant-\(\mu\) model overstate what the wing is worth, and why (Lesson 2.1)?
- Raise the wing efficiency L/D. How does the fastest wing level on the Fast circuit move?
- Try the Formula SAE car on its own circuits. Why do Formula SAE teams usually run large wings?
“The car with the highest top speed will set the fastest lap.” Top speed is reached for a moment at the end of the longest straight. On most circuits far more time is spent braking, cornering and accelerating out of corners, where downforce pays. A car trimmed for top speed is often slower over a lap.
Real-world complications
- Gearing and rev limits can cap top speed before drag does. Formula SAE cars are usually geared for acceleration, not for their theoretical top speed.
- Slipstreaming: a car close behind another sits in its low-pressure wake and has less drag, so it can pass on the straights, at the cost of downforce in the corners (Lesson 4.1).
- Adjustable aero: some series let drivers open a slot in the rear wing on the straights to cut drag, partly escaping the trade-off.
- Cooling: radiators and brake ducts add drag of their own, and hot races need bigger openings.
Check your understanding
A Formula SAE car has 60 kW at the wheels and \(C_D A = 1.2\) m². What is its drag-limited top speed at sea level?
\(V_{\text{top}} = (2P/(\rho C_D A))^{1/3}\) in m/s, then multiply by 3.6.
\(V_{\text{top}} = (2 \times 60\,000/(1.225 \times 1.2))^{1/3} = 81\,633^{1/3} = 43.4\) m/s \(= 156\) km/h. In practice gearing usually limits it below that.
The club racer has \(C_D A = 0.75\) m². How much power does it take just to overcome drag at 200 km/h?
\(P = F_D V = \tfrac{1}{2}\rho C_D A V^3\).
\(V = 55.56\) m/s. \(P = 0.5 \times 1.225 \times 0.75 \times 55.56^3 = 78\,800\) W \(= 78.8\) kW.
The club racer (no downforce, \(\mu = 1.4\), 52% static front weight, \(h = 0.40\) m, \(L = 2.45\) m) is rear-wheel drive. What is its traction-limited acceleration from a standstill?
At a standstill there is no drag or downforce, so \(a_x = F_{x,\text{trac}}/W = \mu(a/L)/(1 - \mu h/L)\), with \(a/L\) the rear weight fraction.
\(a/L = 0.48\) and \(\mu h/L = 1.4 \times 0.40/2.45 = 0.229\). \(a_x = 1.4 \times 0.48/(1 - 0.229) = 0.672/0.771 = 0.871\) g.
An engine upgrade adds 10% more power. Roughly how much does the drag-limited top speed rise?
- 10%.The power needed to overcome drag grows with \(V^3\), not \(V\).
- 5%.That would follow from a \(V^2\) law. Power is force times speed, so it goes as \(V^3\).
- About 3%.\(V_{\text{top}} \propto P^{1/3}\), and \(1.1^{1/3} = 1.032\).
A team races at a tight street circuit one weekend and a fast circuit with long straights the next. How should their rear-wing setting change?
- Keep it the same; the best wing setting is a property of the car.The best setting depends on how the lap time splits between corners and straights.
- Run more wing at the street circuit and trim it back for the fast circuit.Tight circuits reward corner speed; long straights punish drag.
- Run less wing at the street circuit, because the speeds are low there.Low speeds make the wing's drag cheap, and the corners are where the lap time is.
Why does adding downforce do almost nothing for a car's 0–100 km/h time?
- Because downforce adds mass.It adds load, never mass.
- Because at low speed there is very little downforce, and that is where the car is traction-limited.Downforce raises the traction limit only once speed, and dynamic pressure, have built up, by which time power is usually the limit.
- Because downforce only acts on the front axle.It acts on both, split by the aero balance.
Going further with Milliken
Read: RCVD Chapter 15, Applied Aerodynamics, on drag, downforce and how they are traded off, with Chapter 3 for the fundamentals.
As you read, look for answers to these questions:
- Where does a race car's drag come from: which parts of the car make most of it?
- How does the book weigh the value of downforce against the cost of drag?
- How does running in another car's wake change both drag and downforce?
Summary
| Idea | Key relation |
|---|---|
| Power-limited drive force | \(F_x = P/V\), with \(P\) at the driven wheels |
| Traction limit (rear-wheel drive) | \(F_{x,\text{trac}} = \mu(W a/L + F_{Lr})/(1 - \mu h/L)\) |
| Net acceleration | \(m a_x = \min(F_{x,\text{trac}},\,P/V) - F_D\) |
| Top speed | \(V_{\text{top}} = (2P/(\rho C_D A))^{1/3}\); power to overcome drag \(\propto V^3\) |
| Wing choice | More wing on tight circuits, less on fast ones; the optimum depends on the circuit |
That completes aerodynamics. Module 5 turns to the car's steady-state cornering behavior with the bicycle model, where the understeer gradient of Lesson 1.2 is finally derived.