Greenshields model and shock waves at a grade crossing: 16-Civ-A6 Highway Design, Construction, and Maintenance, December 2013, Q4

Question 4: Greenshields model and shock waves at a grade crossing (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A one‑lane approach obeying the Greenshields linear speed–density model, with a 6‑minute gate closure at a grade crossing.

Given data
Quantity	Symbol	Value
Capacity	$q_{\max}$	1,500 veh/h
Jam density	$k_j$	100 veh/km
Normal volume	$q_A$	1,260 veh/h
Normal density	$k_A$	30 veh/km
Gate closure	$t_c$	6 min = 0.1 h

Find. (a) free‑flow speed $u_f$ and density at capacity $k_m$; (b) the platoon length when the gate opens; (c) the time for the platoon to dissipate.

Greenshields flow–density parabola. State A is the approaching (normal) flow, state B the stopped (jam) condition behind the gate, and state C the capacity discharge after the gate opens. The stopping wave runs from A to B and the starting wave from B to C.

Approach. Fit Greenshields from capacity and jam density; identify the three traffic states (A approach, B jam, C capacity discharge); then apply the shock‑wave speed $u_w=\Delta q/\Delta k$ to the stopping and starting waves.

Free‑flow speed and density at capacity. Greenshields gives $q_{\max}=u_f k_j/4$, so $$u_f=\frac{4q_{\max}}{k_j}=\frac{4(1500)}{100}=\boxed{60\ \text{km/h}},\qquad k_m=\frac{k_j}{2}=\boxed{50\ \text{veh/km}}.$$ Check: the normal state gives $u_A=q_A/k_A=1260/30=42$ km/h $=60(1-30/100)$, confirming A lies on the curve.
Stopping shock wave A→B. Behind the closed gate the flow is stopped: $k_B=k_j=100$, $q_B=0$. The back‑of‑queue wave speed is $$u_{AB}=\frac{q_B-q_A}{k_B-k_A}=\frac{0-1260}{100-30}=-18\ \text{km/h},$$ i.e. the queue front propagates upstream at 18 km/h.
Platoon length at gate opening. The stopping wave acts for the full 6‑min closure, $t_c=0.1$ h: $$L=|u_{AB}|\,t_c=18\times0.1=\boxed{1.8\ \text{km}},$$ containing $N=L\,k_j=1.8\times100=\boxed{180\ \text{vehicles}}$.
Starting shock wave B→C. When the gate opens, vehicles discharge at capacity: $k_C=k_m=50$, $q_C=1500$. The starting wave speed is $$u_{BC}=\frac{q_C-q_B}{k_C-k_B}=\frac{1500-0}{50-100}=-30\ \text{km/h}.$$
Dissipation time. The starting wave (30 km/h upstream) chases the still‑advancing stopping wave (18 km/h upstream); the platoon is gone when it overtakes it: $$\tau=\frac{L}{|u_{BC}|-|u_{AB}|}=\frac{1.8}{30-18}=0.15\ \text{h}=\boxed{9\ \text{min}}.$$

Final results — Question 4
Quantity	Result
Free-flow speed $u_f$	60 km/h
Density at capacity $k_m$	50 veh/km
Stopping wave speed $u_{AB}$	−18 km/h (upstream)
Platoon length at opening	1.8 km (180 vehicles)
Starting wave speed $u_{BC}$	−30 km/h (upstream)
Dissipation time after opening	9 min

Topic. Greenshields speed–density model and shock‑wave (kinematic‑wave) analysis of a stop‑and‑release event.

Key relations. $q=u_f k(1-k/k_j)$; $q_{\max}=u_f k_j/4$ at $k_m=k_j/2$; shock speed $u_w=(q_2-q_1)/(k_2-k_1)$; platoon length $=|u_{AB}|t_c$; dissipation $\tau=L/(|u_{BC}|-|u_{AB}|)$.

Why this works. A shock is the boundary between two traffic states; its speed is the slope of the chord joining them on the $q$–$k$ diagram. The queue keeps growing upstream during the closure at the stopping‑wave speed; once released, the faster starting wave eats into it from the front, and the platoon vanishes when the two waves meet.

Common pitfall. Using the free‑flow speed (60) instead of the starting‑wave speed for dissipation, or forgetting that the back of the queue is still moving upstream so the closing speed is the difference of the two wave speeds, not just $|u_{BC}|$.

Source. Garber & Hoel, Traffic and Highway Engineering, ch. 6 (traffic flow, Greenshields and shock waves).

Question 4 of 7: Greenshields model and shock waves at a grade crossing

Question 4: Greenshields model and shock waves at a grade crossing (20 marks)