# Terminal Velocity of a Car

When explaining how a car reaches its terminal velocity, is it solely down to how aerodynamic the car is? My thoughts are that if you have a fixed mass $m$, then the contact frictional force will be constant,$F=\mu{R}$, and because the thrust due to the engine will be much greater than this value, then surely only air resistance will increase the total value of drag? And this is why as the speed increase the acceleration begins to tend to 0 as Drag ∝$v^2$.

• There is more than just air friction and surface contact friction that acts to resist acceleration in a car. – JMac May 8 '17 at 13:09
• Rolling resistance is not constant, but it depends on many factors. The formula $F = \mu R$ is only valid for sliding dry friction (with slipping). – John Alexiou May 8 '17 at 13:40
• @ja72 does it follow though that if the mass decreases, is the rolling resistance decreases too? – math111 May 8 '17 at 14:04
• Not necessarily. Resistance of internal drivetrain parts for example may depend on engine torque more than vehicle mass. Many rotating parts follow the stribeck curve where resistance is high at low speeds, drops and the rises again. I think tires follow a similar pattern. – John Alexiou May 8 '17 at 14:58
• Note that $F=\mu R$ is not happening here. This is the formula for kinetic friction, while a car is influenced by static friction. Static friction is a variable force unlike kinetic friction, and does not stay at a fixed value according to that formula. – Steeven Feb 5 '18 at 14:51

It is just a simplification to say that total drag is $\propto v^2$ and to ignore all other forms of friction. This applies because the power to overcome aerodynamic drag is $\gg$ power loss on the drivetrain and rolling resistance. If you included all the other terms the final speed will decrease only very slightly.

So a simplified model of acceleration (as a function of speed $v$) is

$$a = \frac{P(v)}{m v} - \beta v^2$$

Where $P(v)$ is the engine power at speed $v$, $m$ is the mass and $\beta$ is some drag constant.

Top speed occurs when $a=0$ and hence $$v_{top} \approx \sqrt[3]{ \frac{P_{max}}{m \beta} }$$

This applies only the the car is drag limits (and not gear limited) and that it is designed with getting peak power at the rpm corresponding to top speed. Some cars with 6 speeds, may only achieve top speed in 5th gear, and 6th gear is just an "overdrive".

• Isn't the top speed $\sqrt[3]{\frac{P_{max}}{\beta}}$? – swineone Mar 16 '20 at 3:27
• @swineone - what happened to the $m$? You can rewrite the condition $a=0$ as $$\frac{P_{\rm max}}{m} - \beta v^3 =0$$ which has the solution shown above. – John Alexiou Mar 16 '20 at 23:51
• I made the comment because I tried the formula with the known power, top speed, and drag coefficient of a car, and it only matched if I took $m$ out. Maybe it’s a difference in definitions. – swineone Mar 16 '20 at 23:53
• @swineone Or maybe you didn't use consistent units. Try $m$ in kg, $P$ in W, $v$ in m/s and $\beta$ in 1/m. Or you used an incorrect value for $\beta$. or .. I don't know. Unless you show your work I don't know. I know the formula is correct though. I have used it many times. – John Alexiou Mar 17 '20 at 1:47
• That's exactly the issue. I considered the drag equation $F_{drag} = \frac{1}{2} \rho C_d A v^2$ and assumed $\beta = \frac{1}{2} \rho C_d A$. However, $\beta v^2$ is an acceleration, not a force. – swineone Mar 17 '20 at 1:51

The car reaches terminal velocity when power in = power out. The power that an engine can deliver depends on the gear ratio and the rpm.

The car has several sources of power dissipation - rolling friction and air drag are the two most important ones. The power these take is a function of velocity - so more power (but a smaller fraction) goes into rolling friction as the car goes faster.

Simple way of formulating this:

\begin{align}P_{engine}&=f(v,...)\\ F_{air}&=\frac12\rho v^2 A\cdot C_D\\ F_{roll}&=\mu_r F_n\\ P_{resistance}&=(F_{air}+F_{roll})v\end{align}

• Floris, rolling resistance is not just a frictional thing. At normal inflation pressure, the bottom surface of the tires of an automobile (the part that contacts the road) deform as they rotate. This deformation is continuous while the car is moving, it heats the tires, and it is a very substantial portion of total power consumption (approximately 50%) at normal highway speeds (e.g., 60 miles per hour). – David White Jan 1 '18 at 20:41
• @DavidWhite - I have no argument with your statement. Would you prefer that I change "friction" to "resistance"? Or are you suggesting that I formulate a more complex set of equations to describe this? I was trying to keep it simple - maybe I overdid it. – Floris Jan 1 '18 at 20:43
• Floris - no, I'm not advocating that you change your statement. I merely want the OP to realize that there is more involved than the rolling resistance typically associated with a static friction coefficient. – David White Jan 5 '18 at 2:41

Answer is "it depends" (on speed).

Forces acting against engine force $$F_e$$ are:

Air friction: $$F_a = - \frac 1 2 \rho C_D A v^2$$

Rolling friction: $$F_r = -m g \mu_{rr}$$

Equation for terminal velocity (balance of forces):

$$F_e - F_a - F_r = 0$$

$$F_e = F_a + F_r$$

Expressing them in terms of power (P = F*v):

$$P_e = F_a * v + F_r * v$$

$$P_e = \frac 1 2 \rho C_D A v^3 + m g \mu_{rr} v$$

If you plot separately the two factors, you can see that for speeds lower than $$V_{threshold}$$ the $$m g \mu_{rr} v$$ factor will prevail, hence at low speed the rolling friction prevails on air friction:

Y = Power [W]

X = Speed [m/s]

• mass = 1000 kg
• $$\mu_{rr} = 0.01$$
• $$\rho = 1.225 kg/m3$$
• $$g = 9.81 \frac m{s^2}$$
• Cd = 0.3
• $$A = 2.2 m^2$$

$$F_e - F_a - F_w = m * \frac {dv}{dt}$$
$$\frac {P_e} v - \frac 1 2 \rho C_D A v^2 - m g \mu_{rr} = m * \frac {dv}{dt}$$
$$m \frac {dv}{dt} = \frac {K_1} v + K_2 v^2 + K_3$$