Question

What are number of lanes required to meet the criterion of LOS B for a new 3.5 km segment of undivided multilane highway, if design hour volume is 70000 v/h, 85 km/h speed limit, peak hour volume is 20% of daily traffic, the terrain is level, PHF= 0.8, and 2.2 m lateral , 15 access points in the segment, SRV 5% trucks, 3%clearance from the right edge.

Answer 1

daily traffic = 70000 vehicle/day - I think your question has a typo, design hour volume cannot be 70000 veh/hour (this is not possible). It must be daily traffic is 70000

so i assume one direction daily traffic is 70000 veh/day

so design hour volume in one direction = 20% of 70000 = 14000

Calculate HEAVY VEHICLE FACTOR

Assume rolling terrain so ET= 1.5 and ER = 1.2

We know PT = 3% = 0.03, PR = 5% = 0.05

This is calculated using the formula

fHV and it is calculated using the formula

So fHV = 1/(1+.03*.5 + .05*.2) = 1/1.025 = 0.9756

=======================================================================================

So Calculate vp based on the formula

where

vp = 15 mt passenger car equivalent flow rate in pc/h/ln

V = hourly volume = let this be an unknown value

PHF = 0.80

N = Number of lanes in one direction = N (unknown)

fHV = heavy vehicle factor calculated earlier to be equal to 0.9756

fp = driver population factor. Assume equal to 1 because all are familiar drivers

vp = V/(0.8 * N * .9756 * 1) = V/(N*.7805) pc/hr = 17937/N pc/hr/ln

Calculate FFS

FFS = BFFS - fLW - fLC -fN -fID

where

DO all calcs in mph and then convert back to kmph at the last

BFFS = base free flow speed = 85kmph = 52.816 mph

fLW = adjustment for lane width = 0 mph (assume lanes are greater than 12ft)

fLC = adjustment for right-shoulder lateral clearance = 0 mph (because 2.2m i.e 7.2 ft clearancesee table 2 below since 6 ft shoulders)

fN = adjusment for number of lanes = 1.5 mph (assume this to be 1.5 mph i.e corresponding to 4 lanes, we dont know number of lanes, but assume it to be four for the purpose of calculating reduction)

fID = adjustment for interchange density = 7.5 mph (15 access pts/3.5km or 6.8971 access pts/mile. so from HCM tables, reduction can be found)

Therfore FFS = 52.816 - 0 - 0 - 1.5 - 7.5 = 50 - 12 = 43.816 mph = 70.5148 kmph

=================================================================================== So density = vp/S = (17937/N)/70.5148 = 127.187/N pc/km/ln

But for LOS B maximum density should be 11 pc/km/ln

so we have 254.372/N = 11 - So N = 23.12

so 24 lanes are required.