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Solution Chapter 20

# Solution Chapter 20 - 1 TrafficEngineering,4thEdition...

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Traffic Engineering, 4 th  Edition Roess, R.P., Prassas, E.S., and McShane, W.R. Solutions to Problems in Chapter 20 Problem 20-1 The easiest way to plot the data is to first compute the average headway for each queue  position.  The average headway is then plotted against queue position, and the best curve  (by eye) is drawn through the data.  In computing the average headway for each position,  note that all queue positions do not have the same number of observations.  Queue Cycle Number Total No. of Ave. Position 1 2 3 4 5 6 7 8 9 10 Headway Headways Headway 1 3.6 3.4 3.2 3.5 3.5 3.3 3.6 3.5 2.4 3.5 33.5 10 3.4 2 2.8 2.7 2.6 2.7 2.5 2.6 2.9 2.6 2.7 2.8 26.9 10 2.7 3 2.2 2.4 2.3 2.1 2.5 2.4 2.4 2.4 2.6 2.4 23.7 10 2.4 4 2.0 2.2 2.1 2.1 2.3 2.1 2.0 2.2 2.2 2.2 21.4 10 2.1 5 2.1 1.9 2.0 2.2 2.1 2.0 2.1 1.8 1.9 1.8 19.9 10 2.0 6 1.9 2.0 2.1 2.0 1.8 2.1 2.0 1.8 2.0 1.7 19.4 10 1.9 7 1.9 2.0 1.8 2.1 1.9 1.9 2.1 1.9 2.0 2.0 19.6 10 2.0 8 x 2.1 1.8 1.9 2.0 2.0 2.0 1.8 x 1.9 15.5 8 1.9 9 x 1.8 x 2.0 x 2.0 1.9 x x 1.8 9.5 5 1.9 10 x 1.9 x 1.8 x x 2.0 x x 1.8 7.5 4 1.9 AVG HEADWAY vs QUEUE POSTION 1.5 2.0 2.5 3.0 3.5 0 2 4 6 8 10 Queue Position Average Headway (s) From the figure, the saturation headway is the extension of the flat portion of the curve,  which, in this case, starts with the 5 th  queue position.   From the figure, the saturation  headway is:   1.95 s. 1 1

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The first four headway involve some component of start-up lost time.  The start-up lost  time is the difference between the actual headway and the saturation headway (1.95 s) in  each case.  Reading the actual headways from the figure, these values are: Position 1: 3.40 – 1.95 = 1.45   s Position 2: 2.70 – 1.95 = 0.75   s Position 3: 2.35 – 1.95 = 0.40 s Position 4: 2.10 – 1.95 = 0.15 s SUM           2.75 s The start-up lost time is, then, 2.75 s. The saturation flow rate is computed from the saturation headway as: ln / / 846 , 1 95 . 1 3600 3600 h veh h s = = = Problem 20-2 In general, the capacity of an intersection approach is given by: = C g s c In this case, C = 75 s (given), s = 3600/h = 3600/2.48 = 1,452 veh/hg, g = G+Y-t = 40 +  4 – (2.3+1.1) = 40.6   s.  Then: ( 29 ln / / 786 75 6 . 40 1452 h veh c = = As there are three lanes on this approach, the total capacity of the approach is 3*786 =  2,358   veh/h.   Problem 20-3 The equation suggests a start-up time of 2.04 s/veh and a saturation headway of 2.35  s/veh, which translates to a saturation headway of 3600/2.35 = 1,532   veh/hg. Problem 20-4 2 2
The maximum sum of critical lane volumes is computed as follows: h veh V V C t N h V c c L c / 334 , 1 35 . 2 6 . 135 , 3 100 3600 * 3 . 4 * 3 600 , 3 35 . 2 1 3600 600 , 3 1 = = - = - = Problem 20-5

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Solution Chapter 20 - 1 TrafficEngineering,4thEdition...

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