Superspeedways: aka Power-Drag Limited Tracks

The term ‘power-drag’ turns up in my article over at NBC Sports. Here’s a little more on what exactly that means.

Power-Drag

Physicists and engineers have a historic rivalry. Engineers think physicists are too removed from reality and physicists think engineers oversimplify things by ignoring the details.

In my NBC Sports piece, I interviewed engineer (and Cornell grad) Nick Fishbein. Formerly of Chip Ganassi Racing, he’s now head of GM Racing’s Performance Engineering Group. I wanted to understand more about how teams are approaching the newly repaved Atlanta Motor Speedway.

‘Superspeedway’ is not a well-defined term. Initially, any track more than a mile in length was called a superspeedway. More recently, the term meant Daytona and Talladega, tracks where drafting was important. But drafting plays a role at Michigan and Fontana, so are they superspeedways?

That’s why I liked Nick’s response: Engineer look at a track based on what physics limits speed. All tracks fall somewhere between two extremes.

• The power/drag relationship limits the car’s speed
• Tire grip limits the car’s speed

Grip-Limited

When friction between the tires and the track limits speed, the track is grip-limited. If you’ve ever taken an off-ramp from a highway a little too fast and then had to brake as your car moved toward the outside of the curve, that’s exactly what I’m talking about. A driver can’t keep their foot on the gas because the tires loose traction. At these tracks, engineers focus on finding a setup that lets the driver take the corners as fast as possible without crashing.

Power-Drag Limited

I’m going to be pedantic physics professor here for a moment and note that the expression ‘power-drag’ drives me a little nuts. Power and force (drag is a force) are two very different quantities. Power is force times speed: basically how much energy the car’s engine outputs in a given time. So directly comparing the two rankles my physicist senses.

However, I can see why engineers use the term. Car engines are spec’ed by horsepower, not the force they put out. The actual force reaching the wheels depends on a series of things like friction loses in the drivetrain, gearing, etc. Engine horsepower is a pretty easy thing to measure. Engine force isn’t.

Anyway, a power-drag track is one where the drag becomes large enough that the engine doesn’t have enough power to overcome it. A driver can keep their foot on the gas all the way around the track without worrying about the tires losing traction.

Newton’s Laws Vis-a-Vis Race Cars

As I explained in my piece for NBC Sports, we can model a racecar as subject to two forces: the force from the engine and the drag force. Since the drag force always acts in the direction opposite the car’s motion, these two force will always act in opposition to each other.

\Sigma F=ma
F_{engine}-F_{drag}=ma

It’s a tug of war: The engine force pushes forward, the drag force pulls backward. Because of the quadratic dependence of the drag force on speed, you can reach a condition where there simply isn’t enough engine force to overcome the drag. You don’t stop going, but you stop accelerating.

You reach terminal velocity: the maximum velocity possible given the engine power and the aerodynamic profile of the car.

The Math of the Force-Drag Relationship

For those more technical, here are the mathematical details. The drag force is proportional to the square of the velocity

F_{drag} \propto v^{2}

This means that the maximum drag is proportional to the terminal velocity squared. Power, on the other hand, depends on velocity cubed.

P \propto v^{3}

The terminal velocity is proportional to the ratio of power to drag.

v_{t} \propto \frac{P}{D}

Using the fact that the maximum drag is proportional to the terminal velocity squared:

Which means that the terminal velocity ends up depending on the cube root of the power!

Practically, What Does that Mean?

If engine power doubles (increases by a factor of 2), the terminal velocity only increases by the cube root of two, which is 1.26.

Let’s say, for the sake of round numbers, that we assume 450-hp engine can produce a terminal velocity of 200 mph. If we replaced the 450-hp engine with a 900-hp engine, the terminal velocity would only be 252 mph.

Thatâ€™s pretty surprising â€“ you double the engine power and you only get 50 mph more.  Such is the power of cube roots.

But this equation is why minimizing drag is so important at power-drag limited tracks. Drag is in the denominator of the equation for terminal velocity, so the smaller the drag, the bigger the terminal velocity.

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