Module 5 : Uninterrupted Flow
Lecture 25 : Ramp Metering
1 2 3 4 5 6 7 8 9 10 11
 

Numerical example

Consider a single lane on-ramp to a six-lane freeway. The length of the acceleration lane is 150 m. What is the LOS during the peak hour for the first on-ramp? Given that the peak hour factor is 0.95, the heavy vehicle adjustment factor is 0.976, the driver adjustment factor is 1.0 and proportion of approaching freeway flow remaining is 55.5%? The freeway volume is 3000 veh/hr and the on-ramp volume is 1800 veh/hr.

Solution

  1. Convert volume to flow rate:  Convert volume in (veh/hr) to flow rate (pc/hr) using

    $\displaystyle v_i= \frac{V_i}{PHF\times F_{hv}\times F_p}$    

    where, vi is the flow rate in pc/hr for direction i, Vi is the hourly volume in veh/hr for direction i, PHF is the peak hour factor, and Fhv is the adjustment factor for heavy vehicles, and Fp is the adjustment factor for driver population.
    $\displaystyle V_{F}$ $\displaystyle =$ $\displaystyle 3236~\mathrm{pc/hr}~~(F_{hv}=0.976,~~Fp= 1.000)$  
    $\displaystyle V_{R}$ $\displaystyle =$ $\displaystyle 1941~\mathrm{pc/hr}~~(F_{hv}=0.976,~~Fp= 1.000)$  

  2. Compute $ V_{12}$ as:
    $\displaystyle V_{12}$ $\displaystyle =$ $\displaystyle V_{F}\times P_{FM}$  
      $\displaystyle =$ $\displaystyle 3236\times0.555 = 1796~\mathrm{pc/hr}.$  

  3. Compute density at ramp influence area using equation:
    $\displaystyle D_{R}$ $\displaystyle =$ $\displaystyle a~+~b~V_{R}~+~c~V_{12}~+~d~L_{A}$  
      $\displaystyle =$ $\displaystyle 3.402~+~0.00456~V_{R}~+~0.0048~V_{12}~-~0.01278~L_{A}$  
      $\displaystyle =$ $\displaystyle 3.402+0.00456\times1941+0.0048\times1796-0.01278\times150$  
      $\displaystyle =$ $\displaystyle 18.96~\mathrm{pc/km/ln}.$  

  4. Compute LOS For $ D_{R}$=18.96 pc/km/ln, the LOS = D from the LOS table above.