Solution Chapter 26

# Solution Chapter 26 - 1 Traffic Engineering, 4th Edition...

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Unformatted text preview: 1 Traffic Engineering, 4th Edition Roess, R.P., Prassas, E.S., and McShane, W.R. Solutions to Problems in Chapter 26 Problem 26‐1 Offset, Assuming a Moving Platoon: 1000 L t= = = 17.0 s. S 40 * 1.47 Offset, Assuming a Standing Platoon: L t = l 1 + = 2.0 + 17.0 = 19.0 s S Offset with Queue of 3 Vehicles Downstream: L t = − (Qh + l 1 ) = 17.0 − (3 * 2 − 2) = 13.0 s S Reverse Offset: The sum of the offsets in each direction must be equal to an even multiple of the cycle length. In this case, vehicles can make a “round trip” within one cycle. Thus, the offset in the opposite direction is 60.0 – 17.0 = 43.0 s. This would result in platoons arriving most likely during the red interval, and high delays would occur. Effect of a Poor Speed Estimate: If the actual desired speed were 45 mi/h, the ideal offset should have been 1000/(45*1.47) = 15.1 s. The lead vehicle of the platoon would, therefore, arrive 17.0‐15.1 = 1.9 s “early,” i.e., before the green is initiated. This would add 1.9 s of delay to each vehicle in the platoon. 2 Problem 26‐2 Distance (ft) SIG 4 3000 SIG 3 2500 2000 1500 SIG 2 1000 50 0 SIG 1 0 0 BW = 10 s BW = 30 s Progression Speed = 3,000 ft/60 s = 50 ft/s or 34.0 mi/h 30 60 90 120 Time (secs) Time‐Space Diagram for Problem 26‐2 3 Problem 26‐3 I would select a single‐alternating signal, as the block lengths are fairly long. For a single‐alternating progression, the cycle length is given by: CL = 2S C 750 = 2 30 * 1.47 2 * 750 C= = 34 s. ⇒ 35 s 30 * 1.47 Problem 26‐4 Time‐Space Diagram for Problem 26‐4 Northbound Vehicle At 50 ft/s, the lead NB vehicle will arrive at the first signal (700 ft) in 700/50 = 14 s. It will stop, and will leave when the green begins at 30 s, compiling 16 s of delay. It stops again at the second signal (1400 ft) at 30 + 14 = 44 s, and departs on the green at 55 s, compiling another 11 s of delay. The vehicle will not have to stop at the third signal. Result: 2 Stops; 25 s of delay. 4 Southbound Vehicle At 50 ft/s, the lead SB vehicle does not stop at the third signal nor at the second. It does stop at the first signal, however. Since the green at signal 3 starts at 5 s, and it takes the SB vehicle 2000/50 = 40 s to get to the first signal, it will stop at 45 s, and leave at the green, which begins at 60 s. Result: 1 stop; 15 s of delay. These answers assume that the lead vehicle is moving when it hits the initial green phase. ...
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## This note was uploaded on 01/18/2011 for the course PROJECT MA PM 587 taught by Professor Lee during the Spring '10 term at Keller Graduate School of Management.

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