Module 5 : Uninterrupted Flow
Lecture 23 : Multilane Highways
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Numerical example 1

A segment of undivided four-lane highway on level terrain has field-measured FFS 74.0-km/h, lane width 3.4-m, peak-hour volume 1,900-veh/h, 13 percent trucks and buses, 2 percent RVs, and 0.90 PHF. What is the peak-hour LOS, speed, and density for the level terrain portion of the highway?

Solution

The solution steps are given below:
  1. Data given:  Level terrain, field measured FFS = 74 km/h, lane width is 3.4 m, peak-hour volume = 1900 veh/h, percent trucks and buses pt = 0.13, percent RVs PR = 0.02, and PHF=0.90.
  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
    $\displaystyle Vp$ $\displaystyle =$ $\displaystyle \frac{V}{(PHF\times~N\times~fHV\times~fp)}$  

    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 are passenger-car equivalents for trucks and buses and for recreational vehicles (RVs) respectively PT and PR are proportion of trucks and buses, and RVs, respectively, in the traffic stream (expressed as a decimal fraction)
    $\displaystyle f_{HV}$ $\displaystyle =$ $\displaystyle \frac{1}{1+0.13(1.5-1)+0.02(1.2-1)}$  
      $\displaystyle =$ $\displaystyle 0.935.$  

    Assume no RVs, since none is indicated.
    $\displaystyle Vp$ $\displaystyle =$ $\displaystyle \frac{1900}{(0.90\times~2\times~0.935\times~1)}$  
      $\displaystyle =$ $\displaystyle 1129~\mathrm{pc/h/ln}.$  

  3. Determination of free flow speed(S):  In this example the free flow speed (FFS) measured at the field is given and hence no need to compute free flow speed by indirect method. Therefore, $ FFS = S = 74.0 km/h$.
  4. Determination of density(D):  The density of flow is computed from the equation $ D = Vp/S=15.3 $
  5. Determination of LOS:  LOS determined from the speed-flow diagram. LOS = C.