Wave Analysis of Traffic Flow

Although a series of cars on the highway appears to be a bunch of particles, wave analysis can be used to understand traffic congestion. Here we will show how the formula for waves is the same as the formula for traffic. We will examine the implications.

Written 1999

Formatted 2010

  cars as waves

The basic formula is: v = 2df

For a wave:

For Traffic Flow:

v represents the velocity of the wave

d represents to distance between peaks and troughs ( 1/2 wavelength)

f represents the frequency of the waves per unit time (typ: second)

v represents the average car velocity

d represents the average distance between drivers

f represents the average number of cars that pass a point per unit time (typ: minute)

Now, the basic wave equation applies to traffic. So, if the average velocity drops, say, due to construction, for the cars to pass at the same frequency the distance d between the cars must decrease. Is anything learned yet?

When will highway congestion occur? If we let b represent the typical bumper to bumper length of each car, we can determine when congestion will occur. When a drop in velocity forces d<b congestion will occur.

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Normal Highway:

  • v = 60 mph = 88 ft/sec
  • frequency: f = 1 (1 cars every 1 seconds)
  • distance: d = v/2f = 44 ft between drivers
  • car size: b = 12 ft bumper to bumper

Construction Delay:

  • Velocity: v = 16 mph = 24 ft/sec
  • frequency (no change) f=1
  • distance: d = 12 ft between drivers
  • car size: (no change) b= 12 ft


In this example, the construction delay forces the distance between the drivers down to the same size as the cars. That means the cars are now touching. To eliminate the collisions from this example will require lowering the frequency of passing cars, that means to create a delay.

cars as waves

To sum: the wavelength must remain larger than the particle size to prevent collisions or a drop in frequency.

Is there an analogy to this approach that applies to quantum particle theory?


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