# A Brief Look At Car-Following Models

Recently I posted a short introduction to recurrence relations – what they are and how they can be used for mathematical modeling. This post expands on the topic as car-following models are a nice example of recurrence relations applied to the real-world.

Suppose a car is traveling on the road at the speed u(t) at time t. Another car approaches this car from behind and starts following it. Obviously the driver of the car that is following cannot choose his speed freely. Rather, his speed v(t) at time t will be a result of whatever the driver in the leading car is doing.

The most basic car-following model assumes that the acceleration a(t) at time t of the follower is determined by the difference in speeds. If the leader is faster than the follower, the follower accelerates. If the leader is slower than the follower, the follower decelerates. The follower assumes a constant speed if there’s no speed difference. In mathematical form, this statement looks like this:

a(t) = λ * (u(t) – v(t))

The factor λ (sensitivity) determines how strongly the follower accelerates in response to a speed difference. To be more specific: it is the acceleration that results from a speed difference of one unit.

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Before we go on: how is this a recurrence relation? In a recurrence relation we determine a quantity from its values at an earlier time. This seems to be missing here. But remember that the acceleration is given by:

a(t) = (v(t+h) – v(t)) / h

with h being a time span. Inserted into the above car-following equation, we can see that it indeed implies a recurrence relation.

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Our model is still very crude. Here’s the biggest problem: The response of the driver is instantaneous. He picks up the speed difference at time t and turns this information into an acceleration also at time t. But more realistically, there will be a time lag. His response at time t will be a result of the speed difference at an earlier time t – Λ, with Λ being the reaction time.

a(t) = λ * (u(t – Λ) – v(t – Λ))

The reaction time is usually in the order of one second and consist of the time needed to process the information as well as the time it takes to move the muscles and press the pedal. There are several things we can do to make the model even more realistic. First of all, studies show that the speed difference is not the only factor. The distance d(t) between the leader and follower also plays an important role. The smaller it is, the stronger the follower will react. We can take this into account by putting the distance in the denominator:

a(t) = (λ / d(t)) * (u(t – Λ) – v(t – Λ))

You can also interpret this as making the sensitivity distance-dependent. There’s still one adjustment we need to make. The above model allows any value of acceleration, but we know that we can only reach certain maximum values in a car. Let’s symbolize the maximum acceleration by a(acc) and the maximum deceleration by a(dec). The latter will be a number smaller than zero since deceleration is by definition negative acceleration. We can write:

a(t) = a(acc) if (λ / d(t)) * (u(t – Λ) – v(t – Λ)) > a(acc)
a(t) = a(dec) if (λ / d(t)) * (u(t – Λ) – v(t – Λ)) < a(dec)
a(t) = (λ / d(t)) * (u(t – Λ) – v(t – Λ)) else

It probably looks simpler using an if-statement:

a(t) = (λ / d(t)) * (u(t – Λ) – v(t – Λ))

IF a(t) > a(acc) THEN
a(t) = a(acc)
ELSEIF a(t) < a(dec) THEN
a(t) = a(dec)
END IF

This model already catches a lot of nuances of car traffic. I hope I was able to give you some  insight into what car-following models are and how you can fine-tune them to satisfy certain conditions.

# Hollywood and Physics

We’ve all seen these kinds of movies. After a fast and dramatic chase, the bad guy jumps out of the car, determined to end the good guy once and for all. His evil plans have been thwarted for the last time! In self-defense, the good guy is forced to take a shot and when the bullet hits, the evildoer is thrown violently backwards as a result of the impact and through the nearest shop window. Once the hero is reunited with the love of his life, the credits roll and we are left to wonder if that’s really how physics work.

In a previous example we calculated the momentum of a common 9 mm bullet (p = 5.4 kg m/s). Suppose the m = 75 kg evildoer gets hit by just this bullet. Since the bullet practically comes to a halt on impact, this momentum has to be transferred to the unlucky antihero for the conservation of momentum to hold true. Accordingly, this is the speed at which the bad guy is thrown back:

5.4 kg m/s = 75 kg · v’

v’ ≈ 0.07 m/s ≈ 0.26 km/h ≈ 0.16 mph

This is not even enough to topple a person, let alone make him fly dramatically through the air. From a kinematic point of view, the impact is not noticeable. The same is true for more massive and faster bullets as well as for a series of impacts. The only thing that can make a person fall instantly after getting shot is a sudden drop in blood pressure and the resulting loss of consciousness. But in this case, the evildoer would simply drop where he stands instead of being thrown backwards.

This is not the only example of Hollywood bending the laws of physics. You’ve probably heard the weak “fut” sound a Hollywood gun makes when equipped with a silencer. This way the hero can take out an entire army of bad guys without anyone noticing. But that’s not how pistol silencers work. At best, they can reduce the the sound level to about 120 dB, which is equivalent to what you hear standing near a pneumatic hammer or right in front of the speakers at a rock concert. So unless the hero is up against an army of hearing impaired seniors (which wouldn’t make him that much of a hero), his coming will be noticed.

This was an excerpt from my Kindle book: Physics! In Quantities and Examples

For more interesting physics, check out my Best of Physics selection.

# Wavelength (And: Why Is The Sky Blue?)

A very important type of length is wavelength, usually symbolized by the Greek letter λ (in m). It is defined as the distance from crest to crest (one complete cycle) and can easily be calculated for any wave by dividing the speed of the wave c (in m/s) by its frequency f (in Hz):

λ = c / f

What are typical wavelengths for sound? At room temperature, sound travels with a speed of c = 343 m/s. The chamber pitch has a frequency of f = 440 Hz. According to the equation, the corresponding wavelength is:

λ = 343 / 440 ≈ 0.8 m ≈ 2.6 ft

Are you surprised? I bet most people would greatly underestimate this value. Bass sounds are even longer than that. The lowest tone on a four-string bass guitar has a frequency of f = 41.2 Hz, which leads to the wavelength:

λ = 343 / 41.2 ≈ 8.3 m ≈ 27 ft

So the wave coming from the open E string of a bass guitar doesn’t even fit in a common room. In the case of light, the situation is very different. As noted in the introduction, the wavelength of light ranges between 4000 A (violet light) and 7000 A (red light), which is just below the size of a bacterium.

Wavelength plays an important role in explaining why the sky is blue. When light collides with a particle, parts of it are deflected while the rest continues along the initial path. This phenomenon is known as scattering. The smaller the wavelength of the light, the stronger the effect. This means that scattering is particularly pronounced for violet and blue light.

Unless you are looking directly at the Sun, all the light you see when looking at the sky is scattered light coming from the particles in the atmosphere. Since blue light tends to scatter so easily, the sky ends up in just this color. But why not violet? This is a legitimate question. After all, due to its smaller wavelength, violet light is even more willing to scatter. While this is true, it is also important to note that the sun’s rays don’t contain all the colors in the same ratio. In particular, they carry much less violet than blue light. On top of that, our eyes are less sensitive to violet light.

(This is an excerpt from my Kindle book: Physics! In Quantities and Examples)

# 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.

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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.

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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.

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

Check out my BEST OF for more interesting physics articles.

Sources:

http://ecomodder.com/wiki/index.php/Vehicle_Coefficient_of_Drag_List

http://www.mercedes-benz.de/content/germany/mpc/mpc_germany_website/de/home_mpc/passengercars/home/_used_cars/technical_data.0006.html

http://www.carfolio.com/specifications/models/car/?car=218999

# Another Home Experiment – Wind Speed and Sound Level

Recently I told you about my home experiment regarding impact speed and sound level. I did another experiment with my sound level meter, this time I was interested in finding out how the sound level varies with the wind speed. So I took my anemometer (yep, that’s a thing) to measure the wind speed and at the same time noted the sound level. I collected some data points and plotted them. Here’s the result:

As you can see the fit is not that bad (the adjusted r-square is 0.91).

So the sound level grows with the wind velocity to the power of 0.22, meaning that if the wind speed increases by a factor of twenty-five, the sound level doubles. According to the empirical formula, the noise from the wind inside a category 1 and 2 hurricane is comparable to the sound level at a rock concert. This is of course assuming that the formula holds true past the 12 m/s range over which it was determined (which is not necessarily the case, but for now the best guess).

# Home Experiment – Impact Speed and Sound Level

A while ago I got my hands on a sound level meter and pondered what to do with it. Sound level versus distance from source? Too boring, there’s already a formula for that (see here: Intensity: How Much Power Will Burst Your Eardrums?). What I noticed though is that I’ve never seen a formula relating impact height or speed to sound level, that seemed interesting. So I bought a small wooden sphere at a local store and dropped it from various heights, at each impact recording the maximum sound level. I dropped the sphere from 8 different heights and to reduce the effect of random fluctuations 20 times from each height. So in total I collected 160 data points. I’m not so sure if my neighbors were happy about that.

I calculated the impact speed v from the drop height h using the common v = sqrt (2 * g * h). As you might know, this formula neglects air resistance. However, I’m not concerned about that. The wooden sphere was small and massive and only dropped from heights below about 1 ft. The computed impact speed shouldn’t be off by more than a few percent.

Here’s the resulting plot of impact speed versus sound level (in decibels):

The fit turned out to be fantastic and implies that if you increase the impact speed by a factor of five, the sound level doubles. What’s the point of this? I don’t know, but it’s a neat graph and that’s good enough for me.

# Physics: Free Fall and Terminal Velocity

After a while of free fall, any object will reach and maintain a terminal velocity. To calculate it, we need a lot of inputs.

The necessary quantities are: the mass of the object (in kg), the gravitational acceleration (in m/s²), the density of air D (in kg/m³), the projected area of the object A (in m²) and the drag coefficient c (dimensionless). The latter two quantities need some explaining.

The projected area is the largest cross-section in the direction of fall. You can think of it as the shadow of the object on the ground when the sun’s rays hit the ground at a ninety degree angle. For example, if the falling object is a sphere, the projected area will be a circle with the same radius.

The drag coefficient is a dimensionless number that depends in a very complex way on the geometry of the object. There’s no simple way to compute it, usually it is determined in a wind tunnel. However, you can find the drag coefficients for common shapes in the picture below.

Now that we know all the inputs, let’s look at the formula for the terminal velocity v (in m/s). It will be valid for objects dropped from such a great heights that they manage to reach this limiting value, which is basically a result of the air resistance canceling out gravity.

v = sq root (2 * m * g / (c * D * A) )

Let’s do an example.

Skydivers are in free fall after leaving the plane, but soon reach the terminal velocity. We will set the mass to m = 75 kg, g = 9.81 (as usual) and D = 1.2 kg/m³. In a head-first position the skydiver has a drag coefficient of c = 0.8 and a projected area A = 0.3 m². What is the terminal velocity of the skydiver?

v = sq root (2 * 75 * 9.81 / (0.8 * 1.2 * 0.3) )

v ≈ 70 m/s ≈ 260 km/h ≈ 160 mph

Let’s take a look how changing the inputs varies the terminal velocity. Two bullet points will be sufficient here:

• If you quadruple the mass (or the gravitational acceleration), the terminal velocity doubles. So a very heavy skydiver or a regular skydiver on a massive planet would fall much faster.
• If you quadruple the drag coefficient (or the density or the projected area), the terminal velocity halves. This is why parachutes work. They have a higher drag coefficient and larger area, thus effectively reducing the terminal velocity.

This was an excerpt from the Kindle ebook: Great Formulas Explained – Physics. Mathematics, Economics. Check out my BEST OF for more interesting physics articles.