How To Calculate Maximum Car Speed + Examples (Mercedes C-180, Bugatti Veyron)

How do you determine the maximum possible speed your car can go? Well, one rather straight-forward option is to just get into your car, go on the Autobahn and push down the pedal until the needle stops moving. The problem with this option is that there’s not always an Autobahn nearby. So we need to find another way.

Luckily, physics can help us out here. You probably know that whenever a body is moving at constant speed, there must be a balance of forces in play. The force that is aiming to accelerate the object is exactly balanced by the force that wants to decelerate it. Our first job is to find out what forces we are dealing with.

Obvious candidates for the retarding forces are ground friction and air resistance. However, in our case looking at the latter is sufficient since at high speeds, air resistance becomes the dominating factor. This makes things considerably easier for us. So how can we calculate air resistance?

To compute air resistance we need to know several inputs. One of these is the air density D (in kg/m³), which at sea level has the value D = 1.25 kg/m³. We also need to know the projected area A (in m²) of the car, which is just the product of width times height. Of course there’s also the dependence on the velocity v (in m/s) relative to the air. The formula for the drag force is:

F = 0.5 · c · D · A · v²

with c (dimensionless) being the drag coefficient. This is the one quantity in this formula that is tough to determine. You probably don’t know this value for your car and there’s a good chance you will never find it out even if you try. In general, you want to have this value as low as possible.

On you can find a table of drag coefficients for many common modern car models. Excluding prototype models, the drag coefficient in this list ranges between c = 0.25 for the Honda Insight to c = 0.58 for the Jeep Wrangler TJ Soft Top. The average value is c = 0.33. In first approximation you can estimate your car’s drag coefficient by placing it in this range depending on how streamlined it looks compared to the average car.

With the equation: power equals force times speed, we can use the above formula to find out how much power (in W) we need to provide to counter the air resistance at a certain speed:

P = F · v = 0.5 · c · D · A · v³

Of course we can also reverse this equation. Given that our car is able to provide a certain amount of power P, this is the maximum speed v we can achieve:

v = ( 2 · P / (c · D · A) )1/3

From the formula we can see that the top speed grows with the third root of the car’s power, meaning that when we increase the power eightfold, the maximum speed doubles. So even a slight increase in top speed has to be bought with a significant increase in energy output.

Note the we have to input the power in the standard physical unit watt rather than the often used unit horsepower. Luckily the conversion is very easy, just multiply horsepower with 746 to get to watt.

Let’s put the formula to the test.


I drive a ten year old Mercedes C180 Compressor. According the Mercedes-Benz homepage, its drag coefficient is c = 0.29 and its power P = 143 HP ≈ 106,680 W. Its width and height is w = 1.77 m and h = 1.45 m respectively. What is the maximum possible speed?

First we need the projected area of the car:

A = 1.77 m · 1.45 m ≈ 2.57 m²

Now we can use the formula:

v = ( 2 · 106,680 / (0.29 · 1.25 · 2.57) )1/3

v ≈ 61.2 m/s ≈ 220.3 km/h ≈ 136.6 mph

From my experience on the Autobahn, this seems to be very realistic. You can reach 200 Km/h quite well, but the acceleration is already noticeably lower at this point.

If you ever get the chance to visit Germany, make sure to rent a ridiculously fast sports car (you can rent a Porsche 911 Carrera for as little as 200 $ per day) and find a nice section on the Autobahn with unlimited speed. But remember: unless you’re overtaking, always use the right lane. The left lanes are reserved for overtaking. Never overtake on the right side, nobody will expect you there. And make sure to check the rear-view mirror often. You might think you’re going fast, but there’s always someone going even faster. Let them pass. Last but not least, stay focused and keep your eyes on the road. Traffic jams can appear out of nowhere and you don’t want to end up in the back of a truck at these speeds.


The fastest production car at the present time is the Bugatti Veyron Super Sport. Is has a drag coefficient of c = 0.35, width w = 2 m, height h = 1.19 m and power P = 1200 HP = 895,200 W. Let’s calculate its maximum possible speed:

v = ( 2 · 895,200 / (0.35 · 1.25 · 2 · 1.19) )1/3

v ≈ 119.8 m/s ≈ 431.3 km/h ≈ 267.4 mph

Does this seem unreasonably high? It does. But the car has actually been recorded going 431 Km/h, so we are right on target. If you’d like to purchase this car, make sure you have 4,000,000 $ in your bank account.


This was an excerpt from the ebook More Great Formulas Explained.

Check out my BEST OF for more interesting physics articles.



    1. I’d like to see a formula that takes what speed a car/bike has gone to determine how aerodynamic the car/bike is.

  1. Pingback: 2014 NEW BUGATTI
    1. Hi Jumpjack, thanks a lot for commenting, I’ll gladly answer your questions.

      1) Note that tanh(F/(mVt) * t) approaches 1 as t approaches infinity. So the above equation tells us how fast the terminal speed is approached, but not how to calculate Vt itself. The greater F and the smaller m and Vt, the faster the the maximum speed is reached.

      2) You can use P = F * v to convert from power to force and vice versa.

      3) Yes, you can include ground friction. Set up the balance equation: F(car) = F(air resistance) + F(ground friction). Problem is though, the equation can’t be solved anymore analytically. You’d have to solve it numerically given all the inputs except the terminal speed.

      Hope that was helpful!

      1. Thanks for your reply.
        I was indeed looking for a formula to calculate time needed to go from 0 to 60 MPH, to compare this value to manufacturer-given one and so check its plausiblity.
        But I don’t understand if Torque or Power, usually provided with cars data, affect maximum speed and/or time-to-60.

  2. המון נהגים עוברים את המהירות המותרת בחוק הישראלי. נהגים כאלה נקראים עברייני תנועה. ישנם מצלמות שמצלמות את תנועת הרכבים בכבישים ויש מכמונות מהירות שקולטות את המהירות של הרכב הנוסע במהירות גבוה שאינה מותרת על החוק.

  3. Hey Jumpjack, I just remembered that I recently did a regression regarding the time-to-60. It depends mainly on the ratio of vehicle mass m (in kg) to maximum power P (in hp = horsepower). The larger it is, the longer it takes to get to the desired speed. You can use this formula as a rough estimate:

    t = 0.87 * m / P

    The equation however does produce an error of around plus/minus 10 % and is even less accurate for cars with 200 hp and more. From a physical point of view this relationship is to be expected, as t is the ratio of energy needed in total (kinetic energy, proportional to m) and energy provided per unit time (power). Hope this is what you were looking for!

  4. Obviously this is a way to calculate the potential top speed of the car, how do you calculate the actual top speed of the car in its present setup?

    Am I right in thinking it would involve the maximum RPM of the engine at the output shaft, multiplied by the ratio of the final gear, multiplied by the circumference of the wheels?

    That would presumably only be theoretical though, given the resistance from the drivetrain and, as you’ve pointed out, air resistance also.

  5. Cool article! I was trying to figure this out by myself, but kept tripping up on how to get from power to force. This was super helpful, and the examples were awesome. In particular, it’s amazing how accurate these calculations can be. There are, however, a couple things I’d like to point out, particularly in regards to the Veyron example.

    Firstly, you listed the power of the Bugatti Veyron Super Sport as 1200 HP, but it’s actually 1200 PS (metric horsepower) = 1184 HP = 882 kW. Secondly, I don’t think you can just multiply height by width to get frontal area, because cars aren’t exactly rectangles (seen from the front). I found a spec sheet from Car and Driver ( for the regular Veyron, not the Super Sport; they have a more tuned engine but the same basic shape and size, and you can see there that the view from the front is not a rectangle. They calculate frontal area at 22.3 sqft or 2.07 square meters. They also list Cd at .39. If we accept all of these changes, the top speed is a remarkably similar 433.5 km/h.

    Also, I disagree with the claim that you can’t solve for top speed numerically if you include ground friction. I ended up with a third-degree polynomial, which is totally nasty but nonetheless analytically solvable.

  6. a car accelerates uniformly from rest for 20s with an acceleration of 1.5m/s ^ .it then travels at a constant speed of 60s before slowing with a uniform deceleration to come to rest in a further 10s. calculate.
    a. the maximum speed in km/h. distance travel..

    i need help ti answer this plz

  7. Mass is not relevant?
    If you could load a car with 5000kg of bricks, will it achieve same top speed? I think not.
    Engine power is allways flywheel rated and available power for pushing the car by wheels is lower because transmission loss! Sometimes by a whole 30% in 4×4.

    1. It matters, but not as much as you think.

      As speed increases air drag gets huge compared to the wheel losses caused by the weight of the car.

      But it depends on the speed, on a veyron at 400km/h the engine power used to contrast drag is (let’s say) 1000kW, and the power used to contrast wheel friction is probably in the order of tens on kW let’s say 25kW. For a total of 1025kW

      Wheel friction and losses are proportional to the weight so if you double the weight you will only need 25 more kw to reach the same top speed for a total of 1050.

      This is because drag does not depends on mass of the car and drag is immensely bigger than wheel losses at those speeds.

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