Question

2. A class 1, 30mile long rural two-lane highway segment is in rolling terrain. Peak demand is 900 vph (total for both directions), with a 70:30 directional split, and consists of 20% trucks (no rv's or busses). The PHF is 0.80. There are 80% no passing zones, and the base free-flow speed is 55 mi/h. Lane and shoulder widths width are 11 and 2 ft, respectively. There are 6 access points per mile. a) Estimate the average travel speed in the critical direction only for the given conditions, and determine the LOS based on this criterion only. b) Identify and compare (supported by calculations) two alternatives for improving the LOS.

Answer 1

Arman) Given data Peak demand = 900 uph derectional splet = 70:30 Py = 20% PHF = 0.8 No passing zones = 80%. BFFS - 55 mph = 88.51 knuph Lame width = 1H shoulder width=2ft Accen points = 6 pee mile Grade Adjustment factor f = 1.00 foregiven volwn Hem Hy= - find fun for Average travel speed HP, CE, -1) + PR (FR-1) E for Average Hand speed is E,=1.2 f = 1 0.96 to.2 (1.2-7) to PHP * fg* for 900 = 1171.875 Pull 0.8 1*0.96 • Use Up = 1172 Palh.

Calculate highest derection flowrate up=0.7*1172 = 82o. 4 pulu Use Up = 820 peln Check the highest directional flow rate and two way flow rate against capacity values of 1000 Pun, and 3200 pula respectively. . 1192m 7ooRelh LLLLLLLLLLLLLLLLLLL 821Pc Doup/ R pethan 112 pch (3200 pch -ok. - compute Ifs ffs= Bffs-his-ta - 88.5 Fus-ta tis - 4.9 for came width =11H & Shouldewidth =oft fA - 4.0 for 6 Access points pukn Theofore, UU AFS = BB.5 4.9-4 = 79.6 kmph Ifs = 661 79.6 kuph ffs= 79.6 km/u compute the average travel speed ATS = Ffs- 0.0125 Up top tnp = 3.08 for No palsing zone = 80 & Up=1172 uph 1111111 Therefore, ATS=79.6-0.0115 (112)-3.08 = 61.87 kmph ATS= 61.87 kup

Sind for for time spent following fu=10 = 1.0 fuu - 0.2 (1-1) + 900 = 1125 peln 0.8x1x1 PILIPPILLARRRRRRRRRRRAA highest dinational How = 0.0x1125 = 287.5 = 788 belh 787.5 1200 peth 1125 L3200 peln I check Ok. compute base percent time spent . BRISE- lonci -0.ooo8o9up) BPTSF = 100 (ie -0.000809*1125) BPTSF = 62.8% = 62.8% compute percent time spent following PTSF = BPI SF + feje Falue = 10.70 for directinal Sower 70/30 & up=1125 pun Thermore, PTSF = 62.8+ 10.7 = 73.5% : PiSG = 73.5% P

Thirutore for ATS = 61. 87 kontn 4 PTSF = 73.57. The Lovel of service in 'D'. (a) ATS = 61.87 kmph, and LOS 'D'. . To improve Los if Adjustment for lane widthd shoulde width (ks) and Adjustment for Access points density is a) in changed then we can improve the average travel speed for example, if we increase Lane width 3.6m and shoulder width 21.8m then t=0 like wise if we de reduce access points per kon e it Access points per km so then. Faro Thenfore, FES= BEFS - tis - fA = 88.5 -0.0 11 FF s = 88.5 krough AE

ve change no passing zone to 07 tuen. tupro ATS- IfS-00125 Up top -88-5-0-0125*1122 = 73.85 ATS - 73.85 km/h if No-passing zone 2014 then fand tale =0 for percent time spent following thenfore, PTSF = BPT SF + tapes = 62.8to = 62.8y. PT SF=62.8%. Therefore, for ATS= 73.85 km/h, PT SF = 62-87. The level of service in 'C'. Initial Changing ATS= 61.87 km/h PTSF = 23.5%. ATS = 73.85 kelly PTSG - 62.8%. for No passing zone - 0% bane wilth 23.6. Shoulde, width 31.8m for No passing zone= 80% Lame width =W2t Shoulder width=240 Los got 'D' Los got 'e Cimproved).