Module 5 : Uninterrupted Flow
Lecture 23 : Multilane Highways
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Numerical example 3

A 10 km long 4 lane undivided multilane highway in a suburban area has a segment 1 km long with a 3% upgrade and a segment 1 km long with a 3% downgrade. The section has a volume of 1900 vehicles/hr in each direction with 13% trucks and buses and 2% recreational vehicles. The 85 th percentile speed of passenger car is 80 km/hr on upgrade and 86km/hr on downgrade. There are total of 12 access points on both sides of the roadway. The lane width is 3.6 m, PHF is 0.90 and having a 3m lateral clearance. Determine the LOS of the highway section (upgrade and downgrade) during the peak hour? From HCM, For a 3% upgrade and 1 km length( ET=1.5 , ER=3) For a 3% downgrade and 1 km length( ET=1.5 , ER=1.2 )

Solution

  1. Data given:  3%upgrade and 3%downgrade No of lanes = 4, N = 4, 80.0 km/h 85th-percentile speed for upgrade, 86 km/h 85t h-percentile speed for downgrade, 3.6-m lane width, 1,900-veh/h peak-hour volume, (V =1900 veh/h) 13 % trucks and buses, (PT =0.13) 2 % Recreational vehicles, ( Pr=0.02 ) 12 access points/km, PHF = 0.90 lateral clearance = 3 m
  2. Determination of flow rate(VP):  LOS can be calculated by knowing flow rate and free flow speed.

    For upgrade:  Flow rate is calculated from the equation

    $\displaystyle Vp = \frac{V}{(PHF\times~N\times~fHV\times~fp)}$      

    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 fHV 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.
    $\displaystyle f_{HV}$ $\displaystyle =$ $\displaystyle \frac{1}{1+0.13(1.5-1)+0.02(3-1)}$  
      $\displaystyle =$ $\displaystyle 0.905.$  
    $\displaystyle Vp$ $\displaystyle =$ $\displaystyle \frac{1900}{(0.90\times~2\times~0.905*1)}$  
      $\displaystyle =$ $\displaystyle 1166~{\mathrm pc/h/ln}$  

    For downgrade: 

    $\displaystyle f_{HV}$ $\displaystyle =$ $\displaystyle \frac{1}{1+0.13(1.5-1)+0.02(1.2-1)}$  
      $\displaystyle =$ $\displaystyle 0.935$  
    $\displaystyle Vp$ $\displaystyle =$ $\displaystyle \frac{1900}{(0.90\times~2\times~0.935\times~1)}$  
      $\displaystyle =$ $\displaystyle 1128~\mathrm{pc/h/ln}$  

  3. Determination of free flow speed(S):  For upgrade: 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 for 86 km/hr $ 85_{th}$ percentile speed from interpolation we get, BFFS = 77.0 km/h Now, Compute east bound and west bound free-flow speeds
    $\displaystyle FFS$ $\displaystyle =$ $\displaystyle BFFS-f_{LW}-f_{LC}-f_{A}-f_{M}$  
      $\displaystyle =$ $\displaystyle 77-0-0.6-8.0-2.6$  
      $\displaystyle =$ $\displaystyle 65.8~\mathrm{km/h}$  

    For downgrade: 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 for 86 km/hr 85th percentile speed from interpolation we get, BFFS= 82.0 km/h Now, Compute the free-flow speed
    $\displaystyle FFS$ $\displaystyle =$ $\displaystyle BFFS-f_{LW}-f_{LC}-f_{A}-f_{M}$  
      $\displaystyle =$ $\displaystyle 82-0-0.6-8.0-2.6$  
      $\displaystyle =$ $\displaystyle 71~\mathrm{km/h}$  

  4. Determination of LOS LOS determined from the speed-flow diagram. LOS = D (for upgrade) LOS = D (for downgrade)