# How much does the resistance force realistically change when driving a car?

When driving a car, one accelerates by having the driving force greater than resistance forces (friction, air resistance, etc). Then, to move at a constant velocity, the accelerator has to be in the right spot so that the forces are equal in magnitude. How much does this equal spot change when moving at different velocities (e.g. 10 mph to 100 mph). In other words, how much does the resistance force increase as you go faster?

• Drag from air resistance scales to the squared of speed, at least. Don't know about how gearbox friction scales with speed, and car transmission in particular with their viscous torque converters and such. May 9 at 2:04
• How much it would exactly change would depend on the exact numbers of the specific car road friction, aerodynamic parameters, engine drive train details. So, asking how much it would change by is not very fruitful in this forum. It is better to ask what are the different factors and how do they change with velocity ? May 9 at 5:19
• If you could compare the gas mileage you get at the two speeds you would get precisely the information you're looking for. Not sure how easy that info is too come by though... May 9 at 10:51
• May 9 at 13:23
• Stick your hand out of the window and feel it yourself. May 9 at 13:58

So you have the very basic math model of a car with thrust and resistance, and you are trying to balance the two: Resistance $$F$$ is a combination of aerodynamic drag, tire rolling resistance, driveline resistance, and the hill grade you are climbing.

At low speeds all of the above are important, but because the pressure drag part of the aerodynamic forces increases with speed squared, this is what dominates on higher speed.

You can calculate the drag force from the frontal area of a car $$A$$, the air density $$\rho$$, speed $$v$$ and the coefficient of drag $$c_d$$

$$F_{\rm air} = \tfrac{1}{2} \rho A c_d v^2$$

If you know the top speed of a car, you can estimate the general coefficient $$F_{\rm air} = \beta v^2$$ by balancing with the thrust provided (by means of power $$P=F_{\rm air} v$$).

Qualitatively think of the following form and you can do some basic estimations.

A car with mass $$m=1500 \;\mathrm{kg}$$, peak power output $$P=216\;\mathrm{kW}$$ and top speed of $$v_{\rm top} = 60\;\mathrm{m/s}$$

• Find the general air resistance coefficient, $$\beta$$, from $$P/v = \beta v^2$$ $$\beta = \frac{P}{v^3} = \frac{216000}{60^3} = 1$$
• Build the mathematical model for acceleration $$a$$ and engine power provided $$P$$ $$a = \frac{1}{m} \left( \frac{P}{v} - \beta\,v^2 \right)$$ and note that the power provided is a function of speed also. You cannot provide full power at zero speed. Based on the gearing or electric motor characteristics, peak power is available only at a few speeds.

You can build a table of aerodynamic resistance at various speeds now

Speed [mph] Speed [m/s] Air Resistance [N]
10 4.48 20.1
40 17.9 321.2
60 26.9 722.6
80 35.8 1284.6
100 44.8 2007.3
120 53.8 2890.5

In summary, aerodynamic resistance changes a lot with speed.

While driving at velocity $$v$$, a car of mass $$m$$ is subject to two kinds of friction: rolling resistance and air resticance. The former is given by $$F_r = \mu_r m g \cos{\theta}$$, where $$\mu$$ is a constant and $$\theta$$ is the angle between the rolling surface and the gravitational force, which is always directed towards the center of Earth. The latter, on the other hand, has various possible formulations but, leaving aside any aerodynamical consideration, the easiest to deal with mathematically is $$F_a = -bv$$, where $$b$$ is a constant.

Note that, assuming the angle $$\theta$$ is constant while driving, the rolling resistance is constant, while the air resistance depends on velocity and therefore varies anytime the car accelerates.

To find the motion of such car, we need to solve the following differential equation, where $$F_m$$ is the motor force

$$F_m - \mu_rmg-b\dot x= m\ddot x$$

Even though it's a second order differential equation, just plug $$v=\dot x$$ into the equation and integrate using separation of variables.