16-Civ-A6 Highway Design, Construction, and Maintenance · May 2016

Question 4 of 7: Greenshields' Model and Shock Waves at a Railway Grade Crossing

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: 98-Civ-A6 Transportation Planning & Engineering, National Examination May 2016. Seven questions, all of equal value (20 marks); any five constitute a complete examination. Closed book, one two-sided aid sheet permitted. Three hours. All seven questions are solved below as a study resource.

Reference texts. Mannering, Washburn & Kilareski, Principles of Highway Engineering and Traffic Analysis (Wiley) — queueing, shock waves and traffic-stream models; Papacostas & Prevedouros, Transportation Engineering and Planning (Prentice Hall) — the four-step demand model; Ortuzar & Willumsen, Modelling Transport (Wiley) — trip generation, distribution, mode choice and assignment; Roess, Prassas & McShane, Traffic Engineering (Pearson) — signalised-intersection delay.

Question 4: Greenshields' Model and Shock Waves at a Railway 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 to a railway grade crossing whose traffic stream obeys the Greenshields linear speed–density relationship, interrupted by a three-minute gate closure.

Given data
Quantity	Symbol	Value
Capacity	$q_{\max}$	2 000 veh/h
Jam density	$k_j$	160 veh/km
Normal (approaching) volume	$q_A$	1 500 veh/h
Normal (approaching) density	$k_A$	40 veh/km
Gate closure duration	$t_c$	3 min = 0.05 h

Find. The free-flow speed and the density at capacity; the physical length of the stopped platoon at the moment the gate reopens; and the elapsed time from reopening until the platoon has completely dissipated.

Time–space (shock-wave) diagram for the grade crossing. The stopping wave leaves the gate at 12.5 km/h upstream when the gate closes; the faster starting wave (25 km/h upstream) leaves when the gate opens and overtakes it 1.25 km upstream, at which instant the platoon has dissipated.

Approach. Fix the Greenshields parameters from capacity and jam density, then treat each abrupt change of traffic state as a shock wave whose speed is the slope of the chord joining the two states on the flow–density diagram, $u_w = \Delta q/\Delta k$.

Recover the free-flow speed from the capacity relationship. Greenshields' linear speed–density law $u = u_f(1-k/k_j)$ gives the parabolic flow–density curve $q = u_f k (1-k/k_j)$, whose maximum occurs at $k = k_j/2$ with value $q_{\max}=u_f k_j/4$. Inverting, $$ u_f = \frac{4\,q_{\max}}{k_j} = \frac{4 \times 2\,000}{160} \quad\Longrightarrow\quad \boxed{u_f = 50\ \text{km/h}} $$
Obtain the density and speed at capacity. The optimum density is half the jam density and the optimum speed is half the free-flow speed: $$ k_m = \frac{k_j}{2} = \frac{160}{2} = \boxed{80\ \text{veh/km}}, \qquad u_m = \frac{u_f}{2} = 25\ \text{km/h} $$ As a check, $q_{\max}=u_m k_m = 25 \times 80 = 2\,000$ veh/h, which reproduces the given capacity.
Verify that the stated normal condition lies on the same Greenshields curve. At $k_A = 40$ veh/km the model predicts $$ u_A = u_f\!\left(1-\frac{k_A}{k_j}\right) = 50\!\left(1-\frac{40}{160}\right) = 37.5\ \text{km/h} $$ and hence $q_A = u_A k_A = 37.5 \times 40 = 1\,500$ veh/h, exactly the volume given. The approaching state A is therefore consistent, and it lies on the uncongested branch of the curve.
Compute the stopping shock wave. When the gate closes, state A meets the jam state B, at which $k_B = k_j = 160$ veh/km and $q_B = 0$. The shock speed is the chord slope $$ u_{AB} = \frac{q_B-q_A}{k_B-k_A} = \frac{0-1\,500}{160-40} = -12.5\ \text{km/h} $$ The negative sign means the boundary between moving and stopped traffic travels upstream, away from the gate, at 12.5 km/h.
Convert the wave travel to a platoon length — part (b). The gate is closed for $t_c = 3\ \text{min} = 0.05\ \text{h}$, during which the stopping wave sweeps back a distance $$ L = |u_{AB}|\,t_c = 12.5 \times 0.05 \quad\Longrightarrow\quad \boxed{L = 0.625\ \text{km} = 625\ \text{m}} $$ Since every vehicle inside that length is at jam density, the platoon contains $$ N = L\,k_j = 0.625 \times 160 = \boxed{100\ \text{vehicles}} $$
Compute the starting shock wave. When the gate opens, the head of the queue discharges at capacity, so the jam state B is replaced by the capacity state C with $k_C = k_m = 80$ veh/km and $q_C = q_{\max}= 2\,000$ veh/h. The boundary between them moves at $$ u_{BC} = \frac{q_C-q_B}{k_C-k_B} = \frac{2\,000-0}{80-160} = -25\ \text{km/h} $$ so the starting wave also travels upstream, but at twice the speed of the stopping wave.
Let the faster starting wave overtake the stopping wave — part (c). At the instant the gate opens, the two waves are 0.625 km apart, and the stopping wave keeps retreating because vehicles still arrive at 1 500 veh/h and join the back of the queue. The gap therefore closes at the relative speed $$ |u_{BC}| - |u_{AB}| = 25 - 12.5 = 12.5\ \text{km/h} $$ and the time to close it is $$ t_d = \frac{L}{|u_{BC}|-|u_{AB}|} = \frac{0.625}{12.5} = 0.05\ \text{h} \quad\Longrightarrow\quad \boxed{t_d = 3.0\ \text{minutes}} $$
Report the extent of the queue at the moment it vanishes. During those three minutes the stopping wave travels a further $12.5 \times 0.05 = 0.625$ km, so the two waves meet 1.25 km upstream of the crossing, having admitted a further 100 vehicles to the queue. Beyond that point and after that instant the road returns to a moving state, and the residual boundary between the capacity discharge and the approaching normal flow moves back downstream at $(2\,000-1\,500)/(80-40) = +12.5$ km/h.

Final results — Question 4
Quantity	Symbol	Value
Free-flow speed	$u_f$	50 km/h
Density at capacity	$k_m$	80 veh/km
Speed at capacity	$u_m$	25 km/h
Speed of the approaching stream	$u_A$	37.5 km/h
Stopping shock wave	$u_{AB}$	−12.5 km/h (upstream)
Platoon length when the gate opens	$L$	625 m (100 vehicles)
Starting shock wave	$u_{BC}$	−25 km/h (upstream)
Time for the platoon to dissipate	$t_d$	3.0 min after the gate opens
Maximum queue extent	$L_{\max}$	1.25 km upstream of the crossing

Topic. The Greenshields linear traffic-stream model and shock-wave analysis of a temporary complete blockage.

Key relations. $u = u_f(1-k/k_j)$; $q = uk$; $k_m = k_j/2$, $u_m = u_f/2$, $q_{\max}= u_f k_j/4$; shock-wave speed $u_w = (q_2-q_1)/(k_2-k_1)$, positive downstream; platoon length $L = |u_{AB}|\,t_c$; vehicles in the platoon $N = L\,k_j$; dissipation time $t_d = L/(|u_{BC}|-|u_{AB}|)$.

Why this works. A shock wave is the trajectory of the boundary between two uniform traffic states, and conservation of vehicles across that boundary forces its speed to equal the slope of the chord joining the two states on the $q$–$k$ diagram. Because a complete blockage sends the stream to the jam state and a release sends it to the capacity state, the two chords are fixed by the geometry of the parabola alone and no differential equation is needed. The starting wave is always steeper than the stopping wave for an uncongested approach, which guarantees that the queue is eventually consumed — and the time-space diagram above makes that overtaking explicit as the intersection of two straight lines.

Common pitfall. Measuring the platoon dissipation from the moment the gate closed rather than from the moment it opened, or dividing the platoon length by the starting-wave speed alone (giving 1.5 min here). The stopping wave does not stop when the gate opens — vehicles keep arriving and joining the back of the queue — so the correct denominator is the relative speed of the two waves. A second error is using the approaching density $k_A$ rather than $k_j$ to count the vehicles in a stopped platoon.

Source. Mannering, Washburn & Kilareski, Principles of Highway Engineering and Traffic Analysis, Ch. 5 (traffic-stream models and shock waves); Roess, Prassas & McShane, Traffic Engineering, Ch. 5; Papacostas & Prevedouros, Ch. 4.

← Question 3 Next: Question 5 →

Quantity	Symbol	Value
Capacity	\(q_{\max}\)	2 000 veh/h
Jam density	\(k_j\)	160 veh/km
Normal (approaching) volume	\(q_A\)	1 500 veh/h
Normal (approaching) density	\(k_A\)	40 veh/km
Gate closure duration	\(t_c\)	3 min = 0.05 h

Quantity	Symbol	Value
Free-flow speed	\(u_f\)	50 km/h
Density at capacity	\(k_m\)	80 veh/km
Speed at capacity	\(u_m\)	25 km/h
Speed of the approaching stream	\(u_A\)	37.5 km/h
Stopping shock wave	\(u_{AB}\)	−12.5 km/h (upstream)
Platoon length when the gate opens	\(L\)	625 m (100 vehicles)
Starting shock wave	\(u_{BC}\)	−25 km/h (upstream)
Time for the platoon to dissipate	\(t_d\)	3.0 min after the gate opens
Maximum queue extent	\(L_{\max}\)	1.25 km upstream of the crossing