# 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.

# Car Dynamics – Sliding and Overturning

In this post we will take a look at car performance in curves. Of central importance for our considerations is the centrifugal force. Whenever a body is moving in a curved path, this force comes into play. You probably felt it many times in your car. It is the force that tries to push you out of a curve as you go through it.

The centrifugal force C (in N) depends on three factors: the velocity v (in m/s) of the car, its mass m (in kg) and the radius r (in m) of the curve. Given these quantities, we can easily compute the centrifugal force using this formula:

C = m · v² / r

Note the quadratic dependence on speed. If you double the car’s speed, the centrifugal force quadruples. With this force acting, there must be a counter-force to cancel it for the car not to slide. This force is provided by the sideways friction of the tires. The frictional force F (in N) can be calculated from the so called coefficient of friction μ (dimensionless), the car mass m and the gravitational acceleration g (in m/s²).

F = μ · m · g

The coefficient of friction depends mainly on the road type and condition. On dry asphalt we can set μ ≈ 0.8, on wet asphalt μ ≈ 0.6, on snow μ ≈ 0.2 and on ice μ ≈ 0.1. At low speeds the frictional force exceeds the centrifugal force and the car will be able to go through the curve without any problems. However, as we increase the velocity, so does the centrifugal force and at a certain critical velocity the forces cancel each other out. Any increase in speed from this point on will result in the car sliding.

We can compute the critical speed s (in m/s) by equating the expressions for the forces:

m · s² / r = μ · m · g

s = sqrt (μ · r · g)

This is the speed at which the car begins to slide. Note that there’s no dependence on mass anymore. Since both the centrifugal as well as the frictional force grow proportionally to the car’s mass, it doesn’t play a role in determining the critical speed for sliding. All that’s left in terms of variables is the coefficient of friction (lower friction, lower critical speed) and the radius of the curve (smaller radius, more narrow curve, smaller critical speed).

However, sliding is not the only problem that can occur in curves. Under certain circumstances a car can also overturn. Again the centrifugal force is the culprit. Assuming the center of gravity (in short: CG) of the car is at a height of h (in m), the centrifugal force will produce a torque T acting to overturn the car:

T = h · C = m · v² · h / r

On the other hand, there’s the weight of the car giving rise to an opposing torque T’ that grows with the width w (in m) and mass m of the car:

T’ = 0.5 · m · g · w

At low speeds, the torque caused by the centrifugal force will be lower than the one caused by the gravitational pull. But at a certain critical speed o (in m/s), the torques will cancel each other and any further increase in speed will result in the car overturning. Equating the above expressions, we get:

m · o² · h / r = 0.5 · m · g · w

o = sqrt (0.5 · r · g · w / h)

Aside from the coefficient of friction, the determining factor here is the ratio of width to height. The larger it is, the harder it will be for the centrifugal force to overturn the car. This is why lowering a car when intending to go fast makes sense. If you lower the CG while keeping the width the same, the ratio w / h, and thus the critical speed for overturning, will increase.

Let’s look at some examples before drawing a final conclusion from these truly great formulas.

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According to caranddriver.com the center of gravity of a 2014 BMW 435i is h = 0.5 m above the ground. The width of the car is about w = 1.8 m. Calculate the critical speed for sliding and overturning in a curve of radius r = 300 m on a dry asphalt road (μ ≈ 0.8).

Nothing to do but to apply the formulas:

s = sqrt (0.8 · 300 m · 9.81 m/s²)

s ≈ 49 m/s ≈ 175 km/h ≈ 108 mph

So with normal driving behavior you certainly won’t get anywhere near sliding. But note that sudden steering in a curve can cause the radius of the your car’s path to be considerably lower than the actual curve radius.

Onto the critical overturning speed:

o = sqrt (0.5 · 300 m · 9.81 m/s² · 3.6)

o ≈ 73 m/s ≈ 262 km/h ≈ 162 mph

Not even Michael Schumacher could bring this car to overturn.

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How would the critical speeds change if we drove the 2014 BMW 435i through the same curve on an icy road? In this case the coefficient is considerably lower (μ ≈ 0.1). For the critical sliding speed we get:

s = sqrt (0.1 · 300 m · 9.81 m/s²)

s ≈ 17 m/s ≈ 62 km/h ≈ 38 mph

So even this sweet sport car is in danger of sliding relatively quickly under these conditions. What about the overturning speed? Well, it has nothing to do with the friction of the tires, so it will still be at 73 m/s.

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This was an excerpt from More Great Formulas Explained. Interested in more car dynamics? Take a look at my post on How to Compute Maximum Car Speed. For other interesting physics articles, check out my BEST OF. I hope you enjoyed and drive safe!

# The Time Value of Money and Inflation

To make a point, I’ll start this blog entry in an unusual way, that is, by talking about vectors. A vector is basically an ordered row of numbers. Consider this expression for example:

(12, 3, 5)

This vector could represent a lot of things. For example a point in a three dimensional coordinate system, with the vector components being the x-, y- and z-values respectively. Or for a company offering three products, it could stand for the sales of these products in a certain year.

Why this talk about vectors? You were probably very surprised when you heard grandma say that she paid only 150 \$ for her first car. It seems so amazingly cheap. But it is not. Your dear grandma is talking about 1950’s money, while you are thinking of today’s money. These two have a very different value.

If you want to specify the costs of a good precisely, merely giving an amount of money will not be sufficient. The value of money changes over time and thus to be absolutely precise, you should always couple this amount with a certain year. For example, this is what grandma’s car really cost:

(150 \$, 1950)

This is far from (150 \$, 2012), which is what you were thinking of when grandma shared the story with you. Using an online inflation calculator, we can conclude that this is actually what the car would cost in today’s money:

(1410 \$, 2012)

Not an expensive car, but certainly more than 150 \$ in today’s money. Now you can see why I started this chapter using vectors. They allow us to easily and clearly couple an amount with a year. A true pedant would even ask for one more component since we are still missing the respective months. But let’s not get too pedantic.

How can we justify saying that 150 \$ in 1950’s money is the same as 1410 \$ in today’s money? We can look at how much of a certain good these amounts would buy in the given year. With 150 \$ in 1950 you could fill your basket with about as many apples as you can with 1410 \$ today. The same goes for most other common goods: oranges, potatoes, water, cinema tickets, and so on.

This is inflation, goods get more expensive each year. At a later point we will take a look at what reasons there are for inflation to occur. But before that, let’s define the rate of inflation and see how it is measured …

This was an excerpt from the ebook “Business Math Basics – Practical and Simple”, available for Kindle here: http://www.amazon.com/dp/B00FXB8QSO.