Name: Car crash calculation counsel
Rating: 3.3 (4732 reviews)
Author: peateasea

Originally published on peateasea.de.

A trip to the library ended up being more eventful and instructive than planned. This is the story of how a bit of physics led me to look deeper into a car crash.

On my way to the library recently, something interesting came across my path. Namely, a car. On the bike path. And partially on the footpath. Fortunately, it didn’t come across my path while I was on the bike path; from what I could tell, the crash must have happened the night before.

Because the danger of being run over by an angry, speeding motorist is, unfortunately, omnipresent on German roads, my first thought was something along the lines of: “Bloody idiot! Was probably speeding and lost control!” That’s when my physicist brain kicked in and I thought “hrm, just how fast_was_ the car going?”. I took a couple of photos, used my shoe for an approximate scale, and went merrily on my way.

My expectation for the car’s speed was well over the speed limit because this particular bit of road is often used for races.¹ Even when racing isn’t involved, the road has its fair share of accidents.

The thing is, expectations have a habit of being wrong. Thus, it’s best to do the calculations and see if the expectation makes any sense.

Constructing a plan of action

Let’s look at the scene.

We see that the car hit two concrete bollards (the half-spheres in the image; one on the bike path, and one at the front right-hand side of the car) and moved them a few metres along the pathway. It’s safe to say that the energy required to move the bollards from where they originally stood is going to be approximately equal to the kinetic energy of the car when it hit them. If we know the kinetic energy of the car, then we can work out how fast it was going and we can test my speeding car hypothesis. With this plan in hand, the task is now a simple case of working out the details and piecing them together.

Speed from energy

To move the bollards, the car applied a force to them. Ignoring the force to get the bollard moving, the force the car applied will be equal to the (kinetic) frictional force of the bollard sliding across the pavement. This diagram hopefully makes the situation clearer:

If we assume that this force is roughly constant (it’s not, but we’re approximating a lot here anyway), then the energy to move the bollards is equal to the work done against the frictional force over the distance they were moved. I.e.

W = F_\mathrm{friction} d

where $W$ is the work, $F_\mathrm{friction}$ is the frictional force and $d$ is the collective distance the bollards travelled.

As mentioned above, this work will be equal to the kinetic energy of the car as it hit the bollards. In other words:

E_k = \frac{1}{2} m v^2 = F_\mathrm{friction} d

where $E_k$ is the kinetic energy of the car, $m$ is the mass of the car and $v$ is its speed.

Since we want to know the speed of the car, we solve for $v$ :

v = \sqrt{\frac{2 F_\mathrm{friction} d}{m}}

Thus we just have to work out $F_\mathrm{friction}$ , $d$ and $m$ , plug in the numbers and we’ll get the speed.

What’s the force, Kenneth?

The frictional force will be the weight of the bollard multiplied by its coefficient of kinetic friction (because the bollard is moving as it’s slowing down the car), i.e.

F_\mathrm{friction} = \mu_k F_g

where $\mu_k$ is the coefficient of kinetic friction and $F_g$ is the bollard’s weight. The coefficient of friction can be looked up in a table and I’m going to use the maximum for concrete on a surface similar to what we have here, i.e.:

Cement concrete on dry rock rock: 0.60 - 0.70

In other words, I’ll choose $\mu_k = 0.7$ in our calculation here. My reasoning is that there are other forces that I’ve not taken into account, so I think choosing the high value here is ok.

The weight of the bollard is its mass times the acceleration due to gravity:

F_g = m_b g

where $g = 9.81 \mathrm{m/s^2}$ and $m_b$ is the mass of the bollard.

The mass we can work out from the density of concrete and the bollard’s volume. As with the coefficient of friction, we can look up a reference value for the density of concreteand find that it is

around 2,400 kilograms per cubic metre.

The volume of the bollard is the volume of half a sphere, which is

V = \frac{2}{3} \pi r^3

where $r$ is the radius. The thing is, what’s the radius? Fortunately, I brought my shoe along.

Ok, now we know how many pixels long my shoe is, but how long is it really? After I got home, I measured my shoe’s length and found that it was 32 cm. We’ll round this 0.3 m, because that’s a nicer number and we’re making lots of approximations already. Because the bollard’s diameter is twice as long as my shoe, we know that the bollard’s radius is as long as my shoe, i.e. 0.3 m. Now we can work out the bollard’s volume:

V = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (0.3)^3 \approx 0.188 \mathrm{m}^3

The bollard’s mass is its density multiplied by its volume, i.e.

m_b = \rho_\mathrm{concrete} V = \rho_\mathrm{concrete} \frac{2}{3} \pi r^3 \approx 2400 \times 0.188 \approx 452 \mathrm{kg}

These things are heavy!

Now we can work out the frictional force:

F_\mathrm{friction} = \mu_k m_b g \approx 0.7 \times 452 \times 9.81 \approx 3106 \mathrm{N}

How far can you go?

Now that we know what the frictional force is, we just need to work out the collective distance that the bollards travelled. To find these distances we measure the distance in pixels from the image and knowledge of some appropriate physical lengths to be able to convert pixels into metres. Thus, I spent a fair bit of time in Gimp cropping parts of the image to get the appropriate pixel measurements. Then I found out the length and width of the car in real life via the internet. This information, along with the diameter of the bollard (0.6 m), is sufficient for us to calculate a relationship between pixels and metres and thus determine how far each of the bollards moved.

Why do we need the length and width of the car? Well, I took the photo from an angle to the car, hence the car’s diagonal will be more perpendicular to the camera’s perspective and hence the relationship between pixels and metres for the diagonal will be more reliable.

Here are the values I measured on the full-resolution image:

We know that the car is a Mercedes Benz CLK 320 which has a length of 4.567 m and a width of 1.722 m, thus its diagonal is:

diagonal = \sqrt{length^2 + width^2} \approx 4.881 m

Since the length of the car’s diagonal from the image is 1230 pixels, we can say that the relationship between metres and pixels is:

\frac{4.881}{1230} \approx 0.00397 \mathrm{m/pixel}

From the diagram, we know that the second bollard (the one still stuck under the car’s right-hand front wheel) travelled 1330 pixels, hence, using the factor we just calculated we can work out this value in metres which is:

d_{b2} = \frac{4.881}{1230} \times 1330 \approx 5.278 \mathrm{m}

where $d_{b2}$ means the distance travelled by the second bollard.

Does this make sense? Well, it’s roughly one bollard’s diameter longer than the car, which, when squinting at the image, seems about right, but also seems a bit underestimated. If we consider that the measurement of how far the second bollard travelled in pixels is measured across the image, but in reality, the bollard travelled a bit into the image, that means that the measurement for the bollard’s distance is underestimated. Oh well. Let’s run with this value anyway and see what we end up with.

The first bollard (the one closest to the camera) is 490 pixels across, which we know corresponds to 0.6 m. We know that it travelled 1120 pixels in the image, hence the distance it travelled in reality is

d_{b1} = \frac{0.6}{490} \times 1120 \approx 1.371 \mathrm{m}

which is a little bit more than the width of two bollards, which again seems a bit underestimated when looking at the image.

The total collective distance travelled by the bollards is then:

d = d_{b1} + d_{b2} \approx 6.650 m

Quickly calculating the speed

The last thing we need to know is the mass of the car, which we can find by looking up the relevant spec sheet and is 1,495 kg.²

Now we’re in a position to calculate the car’s velocity. Using the equation we derived earlier:

v = \sqrt{\frac{2 F_\mathrm{friction} d}{m}}

and plugging in the values, we get

v = \sqrt{\frac{2 \times 3106 \times 6.650}{1,495}} \approx 5.248 \mathrm{m/s} \approx 18.893 \mathrm{km/h}

(insert here needle-scratching-across-record sound, Ally McBeal-style)

Hang on? What??? The car was only travelling at (roughly) 20 km/h?

Even if we increase the distance the bollards travelled to 7 m and increase the coefficient of friction to the (unrealistic) value of 1, we still only get a speed of approximately 23 km/h. What’s happening here? Why is the value so low?

Stand back a bit, take a breath, and think

After making this calculation, and being rather confused, I returned to the crash site again, this time with a different perspective.

Look at the scene from the back

and from the front.

Notice that there aren’t any skid marks. If this were a high-speed crash, usually the driver has slammed on the brakes and there would be skid marks leading up to the car wreck. Here they’re missing. This suggests that the crash didn’t happen at a high speed and hence my initial hypothesis is wrong.

I also noticed that the airbag hadn’t deployed. This also suggests a slow crash.

Thinking about this further, I realised that there could be a deeper, possibly much more tragic, story behind what looked like just another car crash. What if the driver had become unconscious while at the wheel? If the car had been going at the speed limit (50 km/h) and the driver became unconscious it’s certainly possible for the car to be travelling at approximately 20 km/h when it hit the bollards. Considering that the car remained where it was for five days, one can imagine that the driver couldn’t organise to have the car towed because they were now in hospital due to reasons unrelated to the crash (or worse).

It definitely made me think.

Conclusions

What can we take away from this?

It’s important not to jump to conclusions: one can be prejudiced from previous experience and a knee-jerk judgement of a situation probably isn’t correct.

It’s nice that a bit of high-school physics and simple digging around on the internet was enough to solve this puzzle (if somewhat imprecisely). It’s also interesting that “listening” to the physics can help correct incorrect expectations and lead to deeper understanding.

I should take photos from the correct perspective in the future and thus avoid parallax problems. 🤦

These bollards provide scarily little protection for cyclists and pedestrians. Perhaps this is something that the city can consider when planning or improving cycling and pedestrian infrastructure in the future. For instance, consider that we can rearrange the equation for the speed mentioned above to calculate the distance a bollard would move for a given speed:

d = \frac{m v^2}{2 F}

Using 50 km/h (13.9 m/s) as the input speed, with the mass of the car
$m$ and the frictional force $F$ as before, we get

d = \frac{1495 \times 13.9^2}{2 \times 3106} \approx 46.571 \mathrm{m}

At that speed, the car would have landed in the river, taking any hapless cyclist or pedestrian with it. Even at low speeds cars carry an enormous amount of energy.

Food for thought!

I used to live near there and one could hear the cars clearly on Friday and Saturday nights. ↩
That’s 1.5 tonnes! Just to move a 80-90 kg person from place to place! Doesn’t that seem a bit excessive? ↩