Speed of shock waves

Let a stream flowing under condition A (with $u_A$, $k_A$, and $q_A$) meet another stream flowing under condition B (with $u_B$, $k_B$, and $q_B$). Assume that the speed of the resultant shock wave is $u_{sw}$. Then relative to the shock wave vehicles in condition A are moving at a speed of $u_A-u_{sw}$ and vehicles in condition B are moving at a speed of $u_B-u_{sw}$.

Now, recall that the shock wave is a demarcation between the two conditions. Hence, it can be said that in a time duration of $\Delta t$ the number of vehicles crossing over the shock wave from condition A is $(u_A-u_{sw})\Delta t \times k_A$. Similarly, it can be said that in a time duration of $\Delta t$ the number of vehicles crossing over the shock wave from condition B is $(u_B-u_{sw})\Delta t \times k_B$. Note that, physically, some vehicles are crossing over the shock wave from one condition to the other. Since vehicles are neither created nor destroyed in the process of crossing over, the number of vehicles crossing over the shock wave from the perspectives of conditions A and B must be equal. Therefore,

$$(u_A-u_{sw})\Delta t \times k_A = (u_B-u_{sw})\Delta t \times k_B$$

Rearranging the terms and substituting $u_ik_i$ by $q_i$, the following expression can be obtained:

$$u_{sw} = \frac{q_A - q_B}{k_A-k_B} = \frac{q_B - q_A}{k_B-k_A}$$ (1)

It should be noted that the above equation for speed of shock wave can also be obtained using principles of geometry from a distance-time plot of the type shown in Figure 1. Further, the above equation also has a simple graphical interpretation; it states that the speed of the shock wave is given by the slope of the line joining the points representing the two conditions (on a $q-k$ graph) whose confluence gives rise to the shock wave. Figure 2 shows this graphically.

図 2: Illustration of the speed of shock waves on a $q-k$ plot.

There can be three types of shock waves: (i) the forward moving shock wave, i.e., speed of shock wave is positive (see Figure 3 (a)), (ii) the stationary shock, i.e., speed of shock wave is zero (see Figure 3 (b)), and (iii) the backward moving shock wave, i.e., speed of shock wave is negative (see Figure 3 (c)). As can be seen from the figure, the first type of shock wave will occur when a stream with lower flow and lower density meets a stream with higher flow and higher density or when a stream with higher flow and higher density meets a stream with lower flow and lower density. Stationary shock waves will occur when the streams meeting have the same flow value but different densities. The third kind of shock waves will occur when a stream with higher flow and lower density meets a stream with lower flow and higher density or when a stream with lower flow and higher density meets a stream with higher flow and lower density.

図 3: Illustration of different types of shock waves on a $q-k$ plot.

In the following an example is worked out in order to show how knowledge of shock waves can be used to obtain different traffic flow parameters of interest and to also illustrate how one can obtain information about where a shock wave starts and where it ends (which are the other two parameters related to the description of a shock wave).

Example

Traffic is moving on a one way road at $q_A=1000$ vph, and $k_A=16$ vpkm. A truck enters the stream at a point P (which is at a distance of 1 km from an upstream benchmark point BM) at a speed of $u_B=16$ kmph. Due to the decreased speed the density behind the truck increases to 75 vpkm. After 10 minutes the truck leaves the stream. The platoon behind the truck then releases itself at capacity conditions $q_C=1400$ vph and $k_C=44$ vpkm. Determine (i) the speed of all shock waves that form, (ii) the starting point of the platoon (behind the truck) forming shock wave, (iii) the starting point of the platoon dissipating shock wave, (iv) the ending points of the platoon forming and platoon dissipating shock waves, (v) the maximum length of the platoon, and (vi) the time it takes for the platoon to dissipate, and also plot the (vii) location of the front of the platoon and the rear of the platoon versus time, and (viii) length of the platoon versus time.

Solution

Consider the distance time diagram shown in Figure 4 plotted for the scenario described in the problem. This diagram is shown here to help the reader understand the problem better; strictly speaking the complete diagram is not necessary for solving the problem. However, an understanding of the physical scenario definitely helps.

図 4: Distance time diagram illustrating the example problem on shock waves.

(i) Speeds of the various shock waves (shock wave $i$ is denoted as SW$i$) can be obtained directly by using Equation 1 as follows:

$U_{SW1}$	=	$\frac{q_B-q_A}{k_B-k_A}$	=	$\frac{1200-1000}{75-16}$	=	3.39 kmph
$U_{SW2}$	=	$\frac{q_C-q_B}{k_C-k_B}$	=	$\frac{1400-1200}{44-75}$	=	-6.45 kmph
$U_{SW3}$	=	$\frac{q_C-q_A}{k_C-k_A}$	=	$\frac{1400-1000}{44-16}$	=	14.29 kmph
$U_{SW4}$	=	$\frac{q_B-q_D}{k_B-k_D}$	=	$\frac{1200-0}{75-0}$	=	16 kmph
$U_{SW5}$	=	$\frac{q_C-q_D}{k_C-k_D}$	=	$\frac{1400-0}{44-0}$	=	31.8 kmph

(ii) Shock wave 1 is the platoon forming shock wave. It starts at Point P and at the time when the truck enters the stream.

(iii) Shock wave 2 is the platoon dissipating shock wave. It starts at Point Q (i.e., where the truck leaves the stream) and 10 minutes (note the truck remains in the stream for 10 minutes) after the truck entered the traffic stream. Point Q is $16 \times (10/60) = 2.67$ km downstream of Point P.

(iv) Both shock waves 1 and 2 will end if the platoon condition (i.e., condition B) ends. This condition will end whenever Shock waves 1 and 2 meet. Say they meet at time $t$ hours after the start of Shock wave 1. Assuming that these two shock waves meet, their position at time $t$ must be the same. Their positions at time $t$ can be determined from their starting positions and distances by which these travel by time $t$. Thus, knowing that P is 1 km from BM and Q is 3.67 km from BM, one can write the following

$$1+3.39t = 3.67 - 6.45\{t-(10/60)\}$$

or,

$$t=\frac{3.745}{9.84}=0.381 \;\;\mbox{hr}$$

or, $t= 22.84$ minutes.

Hence, the two shock waves end 22.84 minutes after the start of Shock wave 1 and at a distance of $1+3.39\times 0.381=2.29$ km downstream of BM.

(v) The maximum length of the platoon will be at the instant where Shock wave 2 is just about to start. The platoon at any given time is defined by the length between the front of the platoon (Shock wave 4) and the rear of the platoon (Shock wave 1). Hence, the length of the platoon grows at a speed of $16-3.39=12.61$ kmph. The length is maximum at 10 minutes after the platoon starts forming. Hence the maximum length is equal to $12.61 \times (10/60) = 2.1$ km. In terms of number of vehicles the maximum length of the platoon is $k_B \times 2.1$ = $75 \times 2.1 = 157.5$ $\approx 158$ vehicles.

(vi) In part (iv) it was determined that the platoon ceases to exist 22.84 minutes after the start of platoon formation. Out of this for the first 10 minutes the platoon only grows (and there is no dissipation). Hence it takes 12.84 minutes for the platoon to dissipate.

(vii) Figure 5 (a) shows the required plot. In the plot, time is assumed to be zero when Shock wave 1 starts; the distances are as measured from BM. In the figure, Lines 1 and 2 represent the front of the platoon and Line 3 represent the rear of the platoon. Further, the slope of Line 1 is equal to $U_{SW4}$, the slope of Line 2 is equal to $U_{SW2}$, and the slope of Line 3 is equal to $U_{SW1}$.

(viii) Figure 5 (b) shows the required plot. In the plot, time is assumed to be zero when Shock wave 1 starts. Slope of Line 1 is equal to the rate of growth of the platoon. This value is, as determined in part (v), 12.61 kmph. The slope of Line 2 is basically the rate of dissipation of the platoon; however, this need not be calculated since one knows the maximum length of the platoon and when the platoon completely dissipates.

図 5: (a) Plot of platoon front and rear locations versus time for the example, and (b) Plot of length of platoon versus time for the example.