Question

3- Consider the third example solved in the lecture videos. Two additional 12-ft through lanes are added to Vine Street (the street in the intersection of the first example solved in the lecture videos), one lane in each direction. If the peak-hour traffic volumes are unchanged but the Vine Street left-turn saturation flow rates increase by 120 veh/h because of the added through lanes, what would the revised effective green time, yellow time, and all-red time be for each phase? Assume minimum cycle length and a critical intersection v/c of 0.95 is desired.

4- Consider Problem 3. Calculate the new minimum pedestrian green time, assuming the effective crosswalk width is 6 ft and the maximum number of crossing pedestrians in any phase is 20.

Answer 1

ANSWER :
© 390 NBTR = 3.18 NBTR = 0.108. we know that formula Yc ='EBL + WBTR + NBL 50.171+0.338 +0.157 Yc = = 0.666 we have L=12 give x = 0.90 calculating cycle length We know (min= LXC ų - Yc 2 12 X090 0.90* 0.66 10.8 0.234 Cmin = 46.06 : 2 Cmin: = 50 times effective green calielate the we know Xc = Ye Emin Cmin-c:

& 50-666.X50 50-12 33.3 38 XE 0.876 (min o .EBL. بالا go8sec 8, (min WBTR 2 xa -0,338 X 50 0.876 19.3dec or (min NBL IC 0.157 X 50 0.870 Oz -8.9 sec check the values 0,+ O2 + O₂ + 2 = 50 vine street 35 x 5280 Vane 3600 Vune 51333 Ats

we assumed that tr=10, 2=20, a = 10 ft/s2 Whale boft we have, Y vine ert Vuire 009 quine 3.567 Ivine = 40 sec 4 I AR vine = W moplet Vuire = 1.558" Arvine 2.0 sec maple street: YOX 5280 Vnaple 3600 maple = 58.667 we assumes that tre1.0, l=20, a 10ft/32- Wvine 60 ftly wecalculate the yellow and red times:

4 Imaple = txt maple 2.9. = 3.993 20 Imaple = 4.0 sec we know that Almaple Wrine +1 maple 1.364. Armaple = 1.5sec Here given data to determine the Revises effective green times, yellow times and all red we times for each phase we knowi EBL = 0.171 EBTR=0.324 WEL = 0.143 WBTR = 0.338 calculating new vis Ratios. SBL 70 450+.100

8 70 550 SBL=0.127 SBTR = 370 /2018 SBTR = 0.103 go NBL = = 0.157 ㄷ 90 475 + 100 575 NBL = 0.157