Module 5 : Uninterrupted Flow
Lecture 23 : Multilane Highways
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Numerical example 2

A segment of an east-west five-lane highway with two travel lanes in each direction separated by a two-way left-turn lane (TWLTL) on a level terrain has- 83.0-km/h 85th-percentile speed ,3.6-m lane width, 1,500-veh/h peak-hour volume, 6 % trucks and buses, 8 access points/km (WB), 6 access points/km (EB), 0.90 PHF, 3.6-m and greater lateral clearance for westbound and eastbound. What is the LOS of the highway on level terrain during the peak hour?

Solution

The solution steps are enumerate below:
  1. Data given:  Level terrain, 85th-percentile speed is 83.0 km/h , lane width is 3.6 m, peak-hour volume, v=1500 veh/h percent of trucks and buses PT=0.06, 8 access points/km in WB, 6 access points/km in EB, PHF = 0.90, and lateral clearance for westbound and eastbound is more than 3.6 m.
  2. Determination of flow rate(VP):  LOS can be calculated by knowing flow rate and free flow speed. Flow rate (Vp) is calculated from the equation
    $\displaystyle Vp$ $\displaystyle =$ $\displaystyle \frac{V}{(PHF\times~N\times~fHV\times~fp)}$  

    where, Vp = 15-min passenger-car equivalent flow rate (pc/h/ln), V = hourly volume (veh/h), PHF = peak-hour factor, N = number of lanes, fHV = heavy-vehicle adjustment factor, and fp = driver population factor Since $ f_{HV}$ is unknown it is calculated from the equation
    $\displaystyle f_{HV}$ $\displaystyle =$ $\displaystyle \frac{1}{(1+PT(ET-1)+PR(ER-1)}$  

    where, ET and ER = passenger-car equivalents for trucks and buses and for recreational vehicles (RVs), respectively PT and PR = proportion of trucks and buses, and RVs, respectively, in the traffic stream (expressed as a decimal fraction) Assume no RVs, since none is indicated.
    $\displaystyle f_{HV}$ $\displaystyle =$ $\displaystyle \frac{1}{1+0.06(1.5-1)+0}$  
      $\displaystyle =$ $\displaystyle 0.970.$  


    $\displaystyle Vp$ $\displaystyle =$ $\displaystyle \frac{1500}{(0.90\times~2\times~0.970\times~1)}$  
      $\displaystyle =$ $\displaystyle 858~\mathrm{pc/h/lane}$  

  3. Determination of free flow speed(S):  BFFS is approximately equal to 62.4 km/h when the 85 th percentile speed is 64 km/h, and it is 91.2 km/h when the 85 th percentile speed is 96 km/h and the in between speed values is found out by interpolation. Hence, BFFS = 80 km/h. Now, compute east bound and west bound free-flow speeds
    $\displaystyle FFS$ $\displaystyle =$ $\displaystyle BFFS-f_{LW}-f_{LC}-f_{A}-f_{M}$  
      $\displaystyle =$ $\displaystyle 80-0-0-4-0$  
      $\displaystyle =$ $\displaystyle 76~\mathrm{kmph~(WB)}$  
      $\displaystyle =$ $\displaystyle 80-0-0-5.3-0$  
      $\displaystyle =$ $\displaystyle 74.7~\mathrm{kmph~(EB)}$  

  4. Determination of LOS:  LOS determined from the speed-flow diagram. LOS = C (for EB) LOS = C (for WB)