Question 4 of 7: Greenshields Model and Shock Wave Analysis

Question 4: Greenshields Model and Shock Wave Analysis (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 single lane on which the approaching stream and the Greenshields curve are both specified, interrupted by a total blockage of five minutes.

Find. The jam density and the density at capacity, the length of the stopped platoon at the instant the blockage is cleared, the speed at which its front moves once released, and the further time required for the platoon to disappear.

Approach. Fix the Greenshields curve from the free-flow speed and the capacity, identify the three traffic states involved — approaching, stopped, and discharging at capacity — and obtain each boundary between them as a shock wave whose speed is the slope of the chord joining the two states on the flow-density curve.

Steps 1 and 2 answer part (a), the jam density and the density at capacity; steps 3 and 4 answer part (b), the length of the platoon; step 5 answers part (c), the speed of the front of the platoon; and step 6 answers part (d), the time the platoon takes to dissipate.

1. Part (a) — fix the Greenshields curve from the capacity. The Greenshields relation $u = u_f(1 - k/k_j)$ gives the parabolic flow-density curve $q = u_f(k - k^2/k_j)$, whose maximum sits at $k = k_j/2$ with value $q_{\max} = u_f k_j / 4$. Inverting for the jam density, $$k_j = \frac{4 q_{\max}}{u_f} = \frac{4(1400)}{50} = \boxed{112 \ \text{veh/km}}$$ and the density at capacity is half of it, $$k_c = \frac{k_j}{2} = \boxed{56 \ \text{veh/km}}$$ with the corresponding speed $u_c = u_f/2 = 25$ km/h.
2. Establish the three traffic states. State A is the approaching stream, given directly as $u_A = 40$ km/h and $k_A = 25$ veh/km, hence $$q_A = u_A k_A = (40)(25) = 1000 \ \text{veh/h}$$ State B is the stopped condition behind the blockage: $u_B = 0$, $k_B = k_j = 112$ veh/km, $q_B = 0$. State C is the discharging condition after the blockage is cleared, which is capacity flow into an empty road downstream: $q_C = 1400$ veh/h, $k_C = 56$ veh/km, $u_C = 25$ km/h.
3. Part (b) — compute the stopping shock wave. The speed of the boundary between two states is the slope of the chord between them on the flow-density curve, $$\omega_{AB} = \frac{q_B - q_A}{k_B - k_A} = \frac{0 - 1000}{112 - 25} = \boxed{-11.49 \ \text{km/h}}$$ The negative sign means the wave travels upstream, against the direction of traffic, at 11.49 km/h — this is the back of the queue moving toward oncoming vehicles.
4. Part (b) — convert the wave travel into a platoon length. The stopped vehicle blocks the lane for $t_b = 5$ min $= 1/12$ h, during which the stopping wave has moved upstream by $$L = |\omega_{AB}|\, t_b = \frac{11.494}{12} = 0.958 \ \text{km}$$ Since the front of the platoon is held at the blockage, that distance is the whole extent of the stopped queue, $$\boxed{L = 0.958 \ \text{km} \approx 958 \ \text{m}}$$ As a check on the physics, the number of vehicles stored in the platoon can be obtained two ways and must agree. Counting the vehicles that overtake the shock wave, $(u_A - \omega_{AB})k_A t_b = (40 + 11.494)(25)/12 = 107.3$ vehicles; storing them at jam density, $k_j L = (112)(0.958) = 107.3$ vehicles. They agree, which confirms the wave speed. Note that this is not the same as the 83.3 vehicles that would pass a fixed point in five minutes at 1000 veh/h — more vehicles join the queue than that, because the queue is itself advancing to meet them.
5. Part (c) — find the speed of the front of the platoon. When the obstruction is removed, the vehicles at the head of the queue accelerate away into an uncongested road and discharge at capacity, so the leading vehicles move forward at the capacity speed $$u_C = \frac{q_{\max}}{k_c} = \frac{1400}{56} = \boxed{25 \ \text{km/h}}$$ The release itself propagates backward through the queue as the starting shock wave between the jam state and the discharge state, $$\omega_{BC} = \frac{q_C - q_B}{k_C - k_B} = \frac{1400 - 0}{56 - 112} = \boxed{-25 \ \text{km/h}}$$ that is, 25 km/h upstream. The coincidence of the two magnitudes is not an accident: under Greenshields, $u_c = u_f/2$ and $\omega_{BC} = q_{\max}/(k_c - k_j) = -u_f/2$ always, so the front of the platoon moves forward at exactly the speed at which the release wave moves back through it.
6. Part (d) — close the two waves to find the dissipation time. After the blockage clears, the stopping wave continues upstream at 11.49 km/h because vehicles are still arriving in state A and still joining the back of the queue, while the starting wave chases it upstream at 25 km/h. The stopped region therefore shrinks at the difference of the two speeds, $25 - 11.494 = 13.506$ km/h, from its length of 0.958 km: $$t_d = \frac{L}{|\omega_{BC}| - |\omega_{AB}|} = \frac{0.9579}{13.506} = 0.07092 \ \text{h}$$ $$\boxed{t_d = 4.26 \ \text{minutes after the vehicle was removed}}$$ which is 9.26 minutes after the blockage began. The two wave lines meet on the time-space diagram 1.77 km upstream of the incident, and beyond that point no vehicle is ever brought to a stop.