Of Power and Torque

Of Power and Torque

I propose to demonstrate the following points:

(1) Power directly determines a car's top (unlimited) track speed.
(2) Power doesn't directly determine a car's acceleration.
(3) Torque determines a car’s acceleration.
(4) How gearing can take advantage of high-rpm torque.
(5) Using the power curve as an indication of a car's acceleration capability over-exaggerates the effect of high-rpm torque.

First, let’s accept that the standard formula P = C * Q * x, where P is power, Q is torque, x is engine rpm, and C is a constant, is correct. This formula is derived by looking at the force F on the edge of an imaginary disc of radius R driven by the engine. F = Q / R, and a point on the edge is moving at a linear speed of V = R * x. By the formula P = F * V, we get P = Q * x. C is there to convert between units.

Second, in what follows, we assume no drive-train power loss. This is not a big handicap, because power loss can be approximated by a constant percentage, and we can just assume that the engine makes proportionally less power overall.

(1)

At constant top speed V, all the engine’s power is used to overcome the air resistance force Fa. Per unit time dt, the (negative) work air resistance does on the car is Fa * V, and it is equal to the work the engine does, because there is no change in kinetic energy. Therefore P = Fa * V.

Here, if the car is in equilibrium state, P must be equal to Pmax, and Fa is determined by V and the car’s drag coefficient, so ultimately V can be solved as a function of Pmax.

Corollary: When driven at unlimited max track speed, the engine must be performing at the peak of the power curve.

(2)

For acceleration, we first assume that no gear change is involved. For high-end cars like Ferraris or Z06, this covers 0-60 mph already. For more vanilla cars, you can still reach almost 40 mph in first gear.

Having a slow speed allows us to ignore air resistance, so work done by the engine is entirely converted into the car’s kinetic energy:

Ek = \integral P(t) dt.

Taking derivatives with t, remembering that Ek = 0.5 * m * V^2, we get:

d(Ek)/dt = m * V * dV/dt = P(t).

Or, m * V * A = P(t), where A is acceleration. Now you see it plainly: As the vehicle’s speed increases, the same amount of power generates less and less acceleration!

So what directly affects acceleration? Here’s a hint: If you divide by V, you get A = P(t) / V, and remembering that V is linearly proportional to x (rpm) in one gear, then from the formula we proved first, we see that A should be directly proportional to Q, or torque.

(3)

Here’s another way to prove this fact. Start with the simple formula

A = Fw / m,

where Fw is the driving force measured at the wheels. Fw = Qw / Rw, where Qw is the torque at the wheels and Rw is the wheel radius. And Qw = K * Q, where Q is engine torque and K is transmission gear ratio (times final drive ratio). From this we get:

A = (K / Rw / m) * Q,

where the parentheses bracket a constant. This proves that acceleration is directly determined by torque.

Corollary: Peak acceleration in any gear occurs precisely at the peak of the torque curve. It has nothing to do with where peak power occurs.

This formula is for instantaneous acceleration. To compute 0-60 times, we need to set up and solve an ordinary differential equation:

A = dV/dt, Q is Q(x), but x = V * K / Rw, therefore

dV/dt = (K / Rw / m) * Q(K / Rw * V).

Separating variables V and t:

dV / Q(K / Rw * V) = (K / Rw / m) * dt,

and we can integrate:

\integrate_0^V (1 / Q(K / Rw * V)) dV = (K / Rw / m) * t.

The inverse function of the left-hand side gives the total terminal speed after a period of time. As you can see, this formula involves the torque curve only, nowhere does x (rpm) enter into it.

(4)

Back to the formula:

A = (K / Rw / m) * Q.

Obviously, if you can increase K (gear ratio), you can increase acceleration. This comes at the expense of using up the engine’s rpm range faster. Therefore, if you have an engine with lower torque levels, you better make it rev very high. Then another side effect kicks in: fuel economy suffers when you actually want to get performance out of the engine.

If you look at the effective gear ratios for most family cars, you’ll see that they don’t vary that much. So there’s not much engineers can play with gear ratios to make up for low torque.

(5)

If you market the power curve as the indicator of vehicle acceleration, you are multiplying the high-rpm end of the torque curve by the high rpm, so of course you get impressive numbers. But I’ve demonstrated that it’s not the power curve that determines acceleration, it’s the torque curve plus gearing.

So we have here a concept, power, that can’t be directly measured. Its direct effect, on kinetic energy, can’t be felt by the driver. It doesn’t have a direct effect on acceleration, which can be felt and measured. Conclusion: The power curve isn’t relevant.

You may ask why the power concept got introduced in car benchmark in the first place. Historically, performance was measured more by top speed than acceleration. From section (1), we saw that maximum power was indeed the relevant measure for that. Nowadays, it just serves a marketing purpose.

Ferdinand Piech (Chairman of VW): “People buy horsepower, but they drive torque.”