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Lec 12, Ch.6, pp.191-197: Shock Waves in Traffic Streams (Objectives)

Lec 12, Ch.6, pp.191-197: Shock Waves in Traffic Streams (Objectives). Understand how a shock wave can be created in traffic stream Be able to explain the creation of a shock wave using the fundamental diagram of traffic flow Be able to derive the formula to compute the speed of a shock wave

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Lec 12, Ch.6, pp.191-197: Shock Waves in Traffic Streams (Objectives)

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  1. Lec 12, Ch.6, pp.191-197: Shock Waves in Traffic Streams (Objectives) • Understand how a shock wave can be created in traffic stream • Be able to explain the creation of a shock wave using the fundamental diagram of traffic flow • Be able to derive the formula to compute the speed of a shock wave • Be able to compute shock wave velocities for special cases (like a shock wave after a slow truck, at a traffic light, etc.)

  2. What we cover today in class… • Seeing traffic flow as water flow • Connecting the fundamental diagram of traffic flow and shock wave • Obtaining the shock wave formula • Special cases of shock wave propagation (we can apply the basic formula – no need to use special case formulas)

  3. Have you seen this phenomenon in water flow? Backwater Weir When the capacity of the weir opening is smaller than the upstream flow, what would happen? Backwater is created. This is like a queue created upstream of a bottleneck. The tail of the queue moves upstream until the upstream in flow and the outflow from the opening become equal. This phenomenon is called “shock wave.” So, the formation of queue upstream of a bottleneck is like the formation of backwater caused by a weir.

  4. Fundamental diagram of traffic flow and shock wave For upstream q1 Slope gives velocity uw of shock wave for q1 Flow (q) q2 Work zone For bottleneck k2 k1 kj Density (k) Queue forms upstream of the bottleneck. So we use the diagram of the upstream section

  5. By the way, what would happen if upstream flow is still lower than bottleneck capacity? For upstream For bottleneck q1 Flow (q) q2 Speed decreases & density increases but no queue. k1 k2 kj Density (k)

  6. Derive uw from the conservation of flow principle (p. 192) u1 uW k1 P w Q u2 k2 Speed relative to w No. of veh’s crossing line w during t Conservation of flow says N1 = N2. Hence, (Review Example 6-4)

  7. Special cases (a simpler approach) Stopping waves (like “stopped at the red light”: uw q2 =0 q1 k2 = kj Flow (q) (Use this to solve Example 6-5) q2 =0 k1 k2 kj Density (k)

  8. Special cases (cont) Starting waves (like “leaving the stop line given green light”: uw q1 =0 q2= qmax k1 = kj Flow (q) k2 = ko q1 =0 k2 k1 kj

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