I've always wanted to know how a car's top speed could be determined. Usually, I thought it was horsepower and aerodynamic drag. I learned that weight can affect top speed, but I've recently found out it only changes at a very small rate; not enough to actually consider weight as a factor.

For example, Car A can reach 60 mph in 3.8 seconds and reach a speed of 160 mph with 470 hp. Car B can reach 60 in 3.5 seconds with 300 hp, AWD, and reach a speed of 124.3 mph. From this, if you give the two cars the same hp, Car A would still be faster.

However, I'm unsure about the actual factors that affect top speed, apart from horsepower and drag.

car A car B
Horsepower 470hp 300 hp
0-60mph 3.8 sec 3.5 sec
Top speed 160 mph 124.3 mph
drive RWD AWD
car model 2009 Mercedes-Benz AMG C-Class DTM Citroën Hymotion4 WR
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    $\begingroup$ assuming the gearing allows application of power, those are the 2 factors. $\endgroup$
    – Tiger Guy
    Commented Aug 31, 2021 at 4:00
  • 1
    $\begingroup$ For those in the record-setting class, ability not to go airborne matters as well. $\endgroup$ Commented Aug 31, 2021 at 12:43
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    $\begingroup$ In the example cars you have picked there, you are comparing apples and oranges. The DTM car is a circuit racer and has to tackle long (perhaps almost a Km) tarmac straights, while the WRC car will rarely get close to it's max speed, being used on mainly twisty rough surfaces, and it is geared accordingly. $\endgroup$
    – KevInSol
    Commented Sep 1, 2021 at 13:22

3 Answers 3


The factors that affect a car's top speed - IMHO - can be grouped to the power train factors :

  • engine power
  • gear box and differential ratios
  • size of the wheel

and to the losses:

  • overall aerodynamic coefficient of drag
  • friction related loses (e.g. wheel resistance), (however for the top speed it has a negligible effect).

Usually regarding the losses the top speed is mainly affected by the aerodynamic drag force $F_D$ which is $$F_D =\frac{C_D}{2} A \rho u^2 $$ where:

  • $F_D $ is the aerodynamic force in N
  • $A$ is the frontal crosssectional area of the car
  • $\rho$ is the density of the air (changes only with altitude and temperature)
  • $u$ is the velocity of the car
  • $C_D $ is drag coefficient

Engine power

The engine power output has a significant effect. Below is a graph that shows the torque and the power. Observe that the maximum torque and the maximum power occur at different rpm. E.g. in this engine the max power is at approximately 5500rpm, (while max torque is at 3500 rpm)

enter image description here

Figure : engine power and torque curves for a car ICE engine of about 150 hp (source carthrottle.com )

The top speed is theoretically reached at the maximum power output, and the highest gear.

The top speed can be approximated by equating the car engine power and the losses (i.e. drag) using the following equation:

$$P = \color{red}{F_D} \cdot u = \color{red}{\frac{C_D}{2} A \rho u^2} \cdot u $$

so the top theoretical (see below why this is stressed) speed that engine can reach is:

$$u = \sqrt[3]{\frac{2 P}{C_D A \rho }}$$

if the data in the original question presented is revisited (in tabular format)

car A car B
Horsepower 470hp 300 hp
0-60mph 3.8 sec 3.5 sec
Top speed 160 mph 124.3 mph
drive ? AWD

and the equation is applied, then you see because the ratio of power is about 1.56 ($=\frac{470}{300}$), then the ratio of the theoretical speeds should be $1.16 = \sqrt[3]{1.56}$. So (setting as reference car A), then the theoretical speed of car B should have been about 137 mph (which about 10% greater).

This difference can be attributed to A) the AWD which has greater frictional losses in the drivetrain, B) the gearbox setting (usually) AWD vehicles are load carrying so the acceleration is more important than the top speed.

UPDATE: Additionally a third reason (thanks to TigerGuy's comment), that the theoretical ratio is different is that these two cars might have dissimilar Coefficient of drag $C_D$ And different crosssectional areas. As a result the aerodynamic losses are not directly comparable.


Although, theoretically the top speed is determined by power and aerodynamic drag, the gearbox plays an important role. First of all, (my pardon for being so pedantically detailed), the gearbox determines the compromise between the acceleration and top speed.

For simplicity purposes I will consider the differential as part of the gear box (so I will consider the ratio of the entire drivetrain). I will denote the entire drivetrain as $i_d$, and its defined as:

$$i_d = \frac{n_{engine}}{n_{wheels}}$$

To make more clear what I'm saying, is that if in the above example of the car with 150hp at 5500 rpm, the $i_d$ was equal to 1, and the car had a tire diameter of $d_W=0.5m$, then at the max power the speed would have been:

$$u= \frac{2\cdot \pi}{60} \cdot \frac{n_{engine}}{i_d} \frac{d_W}{2} = 904 \frac{m}{s}$$

so it would be supersonic (speed of sound is 340 m/s). So why don't we do that?

The problem is that if the gear ratio is too low, then the torque (and therefore the force applied from the tires is too small). On the other had if the drivetrain ratio is high, then the forces on the tires would be too high (and the tires would lose traction - think accelerating in a manual car to hard in 1st gear).

So there is an optimal range, that determines the drivetrain gear ratio.

For example, for Car A (we know the max power of 470hp, and the max speed is 160pmh) if we assume that the max power is reached at 6000 rpm, and the tire diameter is 0.5m, then the overall theoretically optimal drivetrain can be calculated to be $$i_d= \frac{\pi \cdot n_{engine} \cdot d_w}{ u}$$ $$i_d= 13.8$$

(there is a small asterisk here, which has to do with the exact definition of the car engine rpm (see this, but this is not pertinent to the discussion).

tire size

Tire size is in effect an extension to the gearbox and the differential. By reducing the size of the wheel you can have better acceleration at the expense of top speed (and the inverse is true).

So if you are interested in the top speed, you should select larger wheels. The downside is that it will take you longer to reach a velocity (e.g. 60mph). In general though, the size of the tires, and the differential and the gearbox are set to optimal values.

UPDATE: Calculation of theoretical top speed ratio

The theoretical velocity for car A is $uA = \sqrt[3]{\frac{2 P_A}{C_{D,A} A_A \rho }}$ and for B is $u_B = \sqrt[3]{\frac{2 P_B}{C_{D,B} A_B \rho }}$, then the ratio $$\frac{u_A}{u_B} = \sqrt[3]{\frac{P_A}{P_B}\cdot \frac{C_{D,B} A_{B} }{C_{D,A} A_A }}$$

  • $\begingroup$ The theoretical speed ratio really only works if it's two engines in the same vehicle since air resistance will be different between the two vehicles. I would have included tire size in the gear ratios discussion, but this is a well thought out and complete answer. $\endgroup$
    – Tiger Guy
    Commented Aug 31, 2021 at 15:11
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    $\begingroup$ I agree they are very similar. I opted to put them separately (while point out that its an extension), because it might not be be obvious to some this particular detail. As to the theoretical speed ratio, you are right that I should also include the air resistance as a cause for the discrepancy. $\endgroup$
    – NMech
    Commented Aug 31, 2021 at 15:18
  • $\begingroup$ I don't if I should have mentioned this but car A is RWD. Anyway, thanks for your answer. I did know something about the gearing and tire sizes but just had a hard time trying to figure out how much it would affect the top speed. But just to be clear, theoretical speed is how fast the car would be in general right? $\endgroup$ Commented Aug 31, 2021 at 23:39
  • $\begingroup$ Also, how did you plug in the numbers to get the horsepower speed? The example I've given above were just 2 of 11 race cars, but I want to see if I can get the theoretical speeds to the other cars instead of asking you. I've seen formulas like these before but just wasn't sure how to calculate it the right way. $\endgroup$ Commented Aug 31, 2021 at 23:42
  • $\begingroup$ @HoshikoMizudori the theoretical velocity for car A is $uA = \sqrt[3]{\frac{2 P_A}{C_{D,A} A_A \rho }}$ and for B is $u_B = \sqrt[3]{\frac{2 P_B}{C_{D_B} A_B \rho }}$, then the ratio $\frac{u_A}{u_b} = \sqrt[3]{\frac{P_A}{P_B}\cdot \frac{C_{D,B} A_{B} }{C_{DA} A_A }}$. Since, I didn't have the values for $C_D, A$ for both cars, I assumed that they have the same. A photo of car A, B would have been useful if you can post it. $\endgroup$
    – NMech
    Commented Sep 1, 2021 at 2:00

What limits top speed is technically how high the wheel RPM can go. because even with infinite torque and traction for the wheel, the car isn't going to go any faster than the linear velocity the edge of the wheel is turning at.

But what limits the car's ability to get to theoretical top speed is the torque output to overcome the ever increasing losses as you go faster on top of the decreasing engine torque past a particular RPM.

So to achieve any given speed, the engine has to be able to turn the wheel at an RPM that will actually move the car that fast but it also has to be able to produce enough torque to overcome the losses at that RPM to keep the wheel spinning.

One source of loss is aerodynamic drag. Another source of loss is the weight of the car since it causes the wheels to deform more which produces more rolling friction.

The engine also has a torque speed curve at beyond some speed the torque starts to drop off. The engine is cycling going so fast that the valves don't stay open long enough for the engine to breath. I suspect this last factor is the one you are looking for.

  • $\begingroup$ 'torque @ RPM' <-- that curve only flattens out for ICE. Try a BEV :-) $\endgroup$ Commented Aug 31, 2021 at 12:44
  • $\begingroup$ @CarlWitthoft I don't know what your point is since I am not talking about low RPM nor did I ever say flatten. I am talking about high RPMs where torque drops off to zero for both engines and electric motors, both synchronous and induction. Has nothing to do with torque to accelerate at low RPM. $\endgroup$
    – DKNguyen
    Commented Aug 31, 2021 at 13:19
  • Horsepower -weight -aerodynamic drag
  • Most of all aerodynamic lift......... As a car body as it goes faster will siphon air below it's underbelly acting as a lifting force.

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