Fundamentals of Transportation/Shockwaves

Shockwaves

Visualization

While you have probably experienced plenty of traffic congestion first hand, it is useful to see it systematically from three different perspectives. That of the driver (with which you are familiar), a birdseye view, and a helicopter view. Some excellent simulation are available here, please see the movies:

Visualization with the TU-Dresden 3D Traffic Simulator

This movie shows traffic jams without an "obvious source" such as an on-ramp, but instead due to randomness in driver behavior: Shockwave traffic jams recreated for first time

Analysis of shockwaves

Shockwaves can be seen by the cascading of brake lights upstream along a highway. They are often caused by a change in capacity on the roadways (4 lanes drops to 3) or an incident, or a traffic signal on an arterial, or a merge on freeway. As seen above, just heavy traffic flow (flow above capacity) can also induce shockwaves. In general, it must be remembered that capacity is a function of drivers rather than just being a property of the roadway and environment. As the capacity (maximum flow) drops from $C_1$ to $C_2$, optimum density also changes. Speeds of the vehicles passing the bottleneck will of course be reduced, but the drop in speed will cascade upstream as following vehicles also have to decelerate.

The figures illustrate the issues. On the main road, far upstream of the bottleneck, traffic moves at density $k_1$, below capacity ($k_{o}pt$). At the bottleneck, density increases to accommodate the most of the flow, but speed drops.

Shockwave Math

Shockwave speed

If the flow rates in the two sections are $q_1$ and $q_2$, then $q_1=k_1{v}_1$ and $q_2=k_2{v}_2$.

$v_w=\frac{q_2-q_1}{k_2-k_1}$

Relative speed

With $v_1$ equal to the space mean speed of vehicles in area 1, the speed relative to the line $w$ is:

$v_{r1}=v_1-v_w$

The speed of vehicles in area 2 relative to the line w is $v_{r2}=v_2-v_w$

Boundary crossing

The number of vehicles crossing line 2 from area 1 during time period $t$ is $N_1=v_{r1}{k}_{1}t=(v_1-v_w)k_{1}t$

and similarly

$N_2=v_{r2}{k}_{2}t=(v_2-v_w)k_{2}t$

By conservation of flow, the number of vehicles crossing from left equals the number that crossed on the right

$N_1=N_2$

so:

$v_2{k}_2-v_1{k}_1=v_w(k_2-k_1)$

or

$q_2-q_1=v_w(k_2-k_1)$

which is equivalent to

$v_w=\frac{q_2-q_1}{k_2-k_1}$

Example

Problem

The traffic flow on a highway is $q_1=2000veh/hr$ with speed of $v_1=80km/hr$. As the result of an accident, the road is blocked. The density in the queue is $k_2=275veh/km$. (Jam density, vehicle length = 3.63 meters).

(A) What is the wave speed ($v_w$)? (B) What is the rate at which the queue grows, in units of vehicles per hour ($q$)?

Solution

(A) At what rate does the queue increase?

1. Identify Unknowns:

$k_1=q/v_1=2000/80=25veh/km$

$v_2=0,q_2=k_2{v}_2=0$

2. solve for wave speed ($v_w$)

$v_w=\frac{q_2-q_1}{k_2-k_1}=\frac{0-2000}{275-25}=-8km/hr$

Conclusion: the queue grows against traffic

(B) What is the rate at which the queue grows, in units of vehicles per hour?

$\begin{array}{c}{N}_1=\left(v_1-v_w\right)k_{1}t=\left(v_2-v_w\right)k_{2}t=N_2\\ droppingt\left(lett=1\right)\\ v_1{k}_1-v_w{k}_1=v_2{k}_2-v_w{k}_2\\ q_1-v_w{k}_1=q_2-v_w{k}_2\\ 2000-(-8)*25=0-(-8)*275\\ 2200veh/hr=2200veh/hr\end{array}$

Application

• Problem ( Solution)

Key Terms

• Shockwaves
• Time lag, space lag

Variables

• $q$ = flow
• $c$ = capacity (maximum flow)
• $k$ = density
• $v$ =speed
• $v_r$ = relative speed (travel speed minus wave speed)
• $v_w$ = wave speed
• $N$ = number of vehicles crossing wave boundary