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

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, described by a Greenshields speed–density relationship, interrupted by a six-minute gate closure.

Find. The jam density and the density at capacity; the platoon length at the instant the gate opens; the speed of the front of the platoon; and the time for the platoon to dissipate.

Approach. Fit the Greenshields parabola from the capacity and free-flow speed, place the three traffic states (approach, jam, capacity discharge) on it, and take each shock-wave speed as the chord slope \(w = \Delta q / \Delta k\) between the two states it separates.

1. Fit the Greenshields parabola and read off the two densities asked for in part (a). Greenshields assumes a linear speed–density law \(u = u_f(1 - k/k_j)\), so \(q = uk = u_f k (1 - k/k_j)\) and the maximum of that parabola is \(q_{\max} = u_f k_j / 4\) at \(k = k_j/2\). Inverting for the jam density, \[k_j = \frac{4 q_{\max}}{u_f} = \frac{4 \times 1500}{60} = 100\ \text{veh/km}, \qquad k_c = \frac{k_j}{2} = 50\ \text{veh/km}.\] \[\boxed{k_j = 100\ \text{veh/km},\qquad k_c = 50\ \text{veh/km}}\] The speed at capacity follows as \(u_c = u_f/2 = 30\) km/h.
2. Confirm that the printed operating point lies on the fitted curve. The approach density is \(k_A = q_A/u_A = 1260/42 = 30\) veh/km, and the fitted law returns \[u = 60\left(1 - \frac{30}{100}\right) = 42\ \text{km/h},\] which is exactly the printed speed. The four data the examiner supplies are mutually consistent, so no reconciliation of a contradictory reading is needed — a check worth making explicitly, because this question type is often over-determined.
3. Define the three traffic states. State A is the arriving stream, \((k_A, q_A, u_A) = (30, 1260, 42)\). State B is the stopped queue behind the gate, which is at jam density with zero flow, \((k_B, q_B) = (100, 0)\). State C is the discharge after the gate opens; because the road downstream is uncongested, the queue discharges at capacity, \((k_C, q_C, u_C) = (50, 1500, 30)\).
4. Compute the stopping shock wave and hence the platoon length (part b). The boundary between the arriving stream and the stopped queue moves at the chord slope \[w_{AB} = \frac{q_B - q_A}{k_B - k_A} = \frac{0 - 1260}{100 - 30} = -18\ \text{km/h},\] the negative sign meaning it travels upstream, against the direction of travel. Over the six-minute closure it sweeps back \[L = |w_{AB}|\,t_b = 18 \times 0.10 = \boxed{1.8\ \text{km}}\] so the platoon is 1.8 km long at the instant the gate opens.
5. Cross-check the platoon length against the vehicles it must store. At jam density the platoon holds \(k_j L = 100 \times 1.8 = 180\) vehicles. Independently, the number of vehicles the shock overtakes in six minutes is \[N = (u_A - w_{AB})\,k_A\,t_b = (42 + 18)(30)(0.10) = 180\ \text{veh},\] and the two agree. Note that this is not the 126 vehicles that would pass a fixed point in six minutes at 1,260 veh/h: the queue grows faster than the arrival rate alone would suggest, because its tail is itself moving upstream to meet the arriving traffic.
6. Compute the starting shock wave and answer part (c). When the gate lifts, the boundary between the jam and the capacity discharge moves at \[w_{BC} = \frac{q_C - q_B}{k_C - k_B} = \frac{1500 - 0}{50 - 100} = -30\ \text{km/h},\] also upstream. Under Greenshields this is always equal in magnitude to the discharge speed, since \(|w_{BC}| = q_{\max}/(k_j - k_c) = u_f/2 = u_c\). The phrase “speed of the front of the platoon” is therefore safe under either reading: the release boundary travels upstream at 30 km/h, and the lead vehicles accelerate away downstream at the discharge speed of 30 km/h. \[\boxed{|w_{BC}| = u_C = 30\ \text{km/h}}\]
7. Close the two waves on each other for the dissipation time (part d). The stopping wave does not stop when the gate opens — state-A traffic is still arriving and still joining the back of the queue — so it continues upstream at 18 km/h while the starting wave chases it upstream at 30 km/h. The queue is consumed at the closing speed of the two boundaries: \[t_{\text{diss}} = \frac{L}{|w_{BC}| - |w_{AB}|} = \frac{1.8}{30 - 18} = 0.15\ \text{h} = \boxed{9\ \text{minutes}}\]
8. Locate the dissipation point and confirm it from both ends. The starting wave has run \(30 \times 0.15 = 4.5\) km upstream when it catches the tail, and the stopping wave has run \(18 \times (0.10 + 0.15) = 4.5\) km upstream over the whole episode. The queue therefore vanishes 4.5 km upstream of the crossing, 15 minutes after the gate first closed, having released \(1500 \times 0.15 = 225\) vehicles at capacity. From that instant a recovery shock separates the capacity discharge from the normal stream, travelling downstream at \[w_{CA} = \frac{q_A - q_C}{k_A - k_C} = \frac{1260 - 1500}{30 - 50} = +12\ \text{km/h}.\]