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 ecomodder.com 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.





Space Shuttle Launch and Sound Suppression

The Space Shuttle’s first flight (STS-1) in 1981 was considered a great success as almost all the technical and scientific goals were achieved. However, post flight analysis showed one potentially fatal problem: 16 heat shield tiles had been destroyed and another 148 damaged. How did that happen? The culprit was quickly determined to be sound. During launch the shuttle’s main engine and the SRBs (Solid Rocket Boosters) produce intense sound waves which cause strong vibrations. A sound suppression system was needed to protect the shuttle from acoustically induced damage such as cracks and mechanical fatigue. But how do you suppress the sound coming from a jet engine?

Let’s take a step back. What is the source of this sound? When the hot exhaust gas meets the ambient air, mixing occurs. This leads to the formation of a large number of eddies. The small-scale eddies close to the engine are responsible for high frequency noise, while the large-scale eddies that appear downstream cause intense low-frequency noise. Lighthill showed that the power P (in W) of the sound increases with the jet velocity v (in m/s) and the size s (in m) of the eddies:

P = K * D * c-5 * s2 * v8

with K being a constant, D the exhaust gas density and c the speed of sound. Note the extremely strong dependence of acoustic power on jet velocity: if you double the velocity, the power increases by a factor of 256. Such a strong relationship is very unusual in physics. The dependence on eddy size is also significant, doubling the size leads to a quadrupling in power. The formula tells us what we must do to effectively suppress sound: reduce jet velocity and the size of the eddies. Water injection into the exhaust gas achieves both. The water droplets absorb kinetic energy from the gas molecules, thus slowing them down. At the same time, the water breaks down the eddies.

During the second Space Shuttle launch (STS-2) a water injection system was used to suppress potentially catastrophic acoustic vibrations. This proved to be successful, it reduced the sound level by 10 – 20 dB (depending on location), and accordingly was used during every launch since then. But large amounts of water are needed to accomplish this reduction. The tank at the launch pad holds about 300,000 gallons. The flow starts at T minus 6.6 seconds and last for about 20 seconds. The peak flow rate is roughly 15,000 gallons per seconds. That’s a lot of water!

The video below shows a test run of the sound suppression system:

Sources and further reading:

Click to access art09.pdf


Click to access CAE_XUYue_Investigation-of-Flow-Control-with-Fluidic-injection-for-Jet-Noise-Reduction.pdf

Intensity: How Much Power Will Burst Your Eardrums?

Under ideal circumstances, sound or light waves emitted from a point source propagate in a spherical fashion from the source. As the distance to the source grows, the energy of the waves is spread over a larger area and thus the perceived intensity decreases. We’ll take a look at the formula that allows us to compute the intensity at any distance from a source.

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First of all, what do we mean by intensity? The intensity I tells us how much energy we receive from the source per second and per square meter. Accordingly, it is measured in the unit J per s and m² or simply W/m². To calculate it properly we need the power of the source P (in W) and the distance r (in m) to it.

I = P / (4 · π · r²)

This is one of these formulas that can quickly get you hooked on physics. It’s simple and extremely useful. In a later section you will meet the denominator again. It is the expression for the surface area of a sphere with radius r.

Before we go to the examples, let’s take a look at a special intensity scale that is often used in acoustics. Instead of expressing the sound intensity in the common physical unit W/m², we convert it to its decibel value dB using this formula:

dB ≈ 120 + 4.34 · ln(I)

with ln being the natural logarithm. For example, a sound intensity of I = 0.00001 W/m² (busy traffic) translates into 70 dB. This conversion is done to avoid dealing with very small or large numbers. Here are some typical values to keep in mind:

0 dB → Threshold of Hearing
20 dB → Whispering
60 dB → Normal Conversation
80 dB → Vacuum Cleaner
110 dB → Front Row at Rock Concert
130 dB → Threshold of Pain
160 dB → Bursting Eardrums

No onto the examples.


We just bought a P = 300 W speaker and want to try it out at maximal power. To get the full dose, we sit at a distance of only r = 1 m. Is that a bad idea? To find out, let’s calculate the intensity at this distance and the matching decibel value.

I = 300 W / (4 · π · (1 m)²) ≈ 23.9 W/m²

dB ≈ 120 + 4.34 · ln(23.9) ≈ 134 dB

This is already past the threshold of pain, so yes, it is a bad idea. But on the bright side, there’s no danger of the eardrums bursting. So it shouldn’t be dangerous to your health as long as you’re not exposed to this intensity for a longer period of time.

As a side note: the speaker is of course no point source, so all these values are just estimates founded on the idea that as long as you’re not too close to a source, it can be regarded as a point source in good approximation. The more the source resembles a point source and the farther you’re from it, the better the estimates computed using the formula will be.


Let’s reverse the situation from the previous example. Again we assume a distance of r = 1 m from the speaker. At what power P would our eardrums burst? Have a guess before reading on.

As we can see from the table, this happens at 160 dB. To be able to use the intensity formula, we need to know the corresponding intensity in the common physical quantity W/m². We can find that out using this equation:

160 ≈ 120 + 4.34 · ln(I)

We’ll subtract 120 from both sides and divide by 4.34:

40 ≈ 4.34 · ln(I)   

9.22 ≈ ln(I)

The inverse of the natural logarithm ln is Euler’s number e. In other words: e to the power of ln(I) is just I. So in order to get rid of the natural logarithm in this equation, we’ll just use Euler’s number as the basis on both sides:

e^9.22 ≈ e^ln(I)

10,100 ≈ I

Thus, 160 dB correspond to I = 10,100 W/m². At this intensity eardrums will burst. Now we can answer the question of which amount of power P will do that, given that we are only r = 1 m from the sound source. We insert the values into the intensity formula and solve for P:

10,100 = P / (4 · π · 1²)

10,100 = 0.08 · P

P ≈ 126,000 W

So don’t worry about ever bursting your eardrums with a speaker or a set of speakers. Not even the powerful sound systems at rock concerts could accomplish this.


This was an excerpt from the ebook “Great Formulas Explained – Physics, Mathematics, Economics”, released yesterday and available here: http://www.amazon.com/dp/B00G807Y00.