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Solution Chapter 22

# Solution Chapter 22 - 1 TrafficEngineering,4thEdition...

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1 Traffic Engineering, 4 th Edition Roess, R.P., Prassas, E.S., and McShane, W.R. Solutions to Problems in Chapter 22 Problem 22 1 Detector Placement Detectors are to be placed to produce a minimum green of 6 s. Then: ft d d Int d Int d Int G 00 . 40 01 . 20 0 . 2 2 2 6 20 20 0 . 2 0 . 2 6 20 0 . 2 1 min = = = + = + = l Passage Time As a point detector is in use, the passage time (PT) should be set to the minimum allowable headway (MAH), which is generally set at 3.0 s. The passage time must be at least long enough to allow a vehicle to traverse the distance between the front edge of the detector and the intersection line: OK s PT 09 . 1 25 * 47 . 1 40 0 . 3 = = Yellow and All Red Times All semi actuated signals are two phase. Thus, there are two sets of yellow and all red times, one for each street. For the minor street: s S L w ar s G a S t y 1 . 3 ) 5 25 ( * 47 . 1 20 ) 6 * 12 ( 47 . 1 2 . 3 2 . 2 0 . 1 10 * 2 ) 5 25 ( * 47 . 1 0 . 1 ) 01 . 0 * 4 . 64 ( 2 47 . 1 15 85 = + = + = = + = + + = + + = 1

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2 For the major street: s ar s y 7 . 0 ) 5 45 ( * 47 . 1 20 ) 2 * 12 ( 7 . 4 7 . 3 0 . 1 10 * 2 ) 5 45 ( * 47 . 1 0 . 1 = + = = + = + + = Note that because pedestrian flows are “low,” the all red is timed to allow vehicles to clear the width of the intersection, w, NOT the distance to the far edge of the crosswalk. Assuming that the standard default values for start up lost time and encroachment are in use, the total lost time in the cycle is equal to the sum of the yellow and all red intervals, or (3.2+3.1) + (4.7+0.7) = 6.3 + 5.4 = 11.7 s. Maximum Green Time (Minor) and Minimum Green Time (Major) To time the critical cycle for a semi actuated signal, the Maximum Green Time for the minor street and the Minimum Green Time for the major street must be determined. To do this, all volumes must be converted to tvu’s/ln using the equivalents shown in Tables 18.1 and 18.2 of the textbook. All left and right turns are permitted. Approach Mvt Volume (veh/h) Equiv Volume (tvu/h) Volume (tvu/h/ln) EB L T R 2 1200 5 10.00 1.00 1.21 20 1200 6 1226/3 = 409 WB L T R 3 1000 6 15.00 1.00 1.21 45 1000 7 1052/3 = 351 NB L T R 10 100 8 1.94* 1.00 1.21 20 100 10 130/1 = 130 SB L T R 14 120 12 1.80* 1.00 1.21 25 120 15 160/1 = 160 *interpolated; boldface = critical lane volume. 2