A new section of Richmond Highway is being designed as a six-lane facility (three in each direction) with a two-way left-turn lane. Determine the peak hour LOS. Traffic data include directional design hourly volume = 3600 veh/h, PHF = 0.94, assumed base free flow speed = 55 mi/h. Geometric data include: urban setting, rolling terrain, lane width = 11 ft, shoulder widths = 4 ft (right side) and 1 ft (left side), and average access point spacing = 12 points per mile on each side.
What will be an ideal response?
Step 1: Compute passenger car equivalent flow rate for peak 15-minute period
using Equation 9.4.
Determine fp (driver population factor); assume mainly commuter traffic, fp = 1.0
Determine ET (PCE for trucks), using Table 9.3; ET = 2.5
Assume percentage of trucks is ???????? ???? 0.05
Determine fHV (heavy vehicle adjustment factor), using Equation 9.5
Step 2: Compute free flow speed using Equation 9.7
FFS = BFFS – fLW – fLC – fM – fA
Determine fLW using Table 9.1, fLW = 1.9
Determine fLC using Table 9.7, fLC = 1.5
Determine fM using Table 9.8, fM = 0.0
Determine fA using Table 9.9, fA = 3.0
FFS = 55 – 1.9 – 1.5 – 0.0 – 3.0 = 48.6 mi/h
Step 3: Compute average passenger car speed and density to determine LOS.
For FFS = 48.6 mi/h, greater than 47.5 and less than 52.5, the 50 mi/h speed-flow
curve will be used for this analysis. It was calculated that vp = 1373 pc/h/ln.
The expected speed of the traffic stream will be 50mi/h, as vp <1,400 pc/h/ln.
The density will be:
Density = (1373 pc/h/ln) / 50 mi/h = 27.46 pc/mi/ln
This corresponds to LOS D (Table 9.10).
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