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A four-lane freeway with two lanes in each direction has a measured free-flow speed of 55 mi/hr. ...

A four-lane freeway with two lanes in each direction has a measured free-flow speed of

55 mi/hr. The peak-hour factor is 0.80 and there are 6% trucks, 2% buses, and 2%

recreational vehicles in the traffic stream. One upgrade is 5% and 0.5 miles long. A

traffic engineer has determined that the freeway is operating at capacity on this

upgrade during the peak hour. If the peak-hour traffic volume is 3900 vehicles, what

value for the driver population adjustment factor was used?

engineering   Civil-Engineering   
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Answer #1

Given, Free flow speed (FFS) = 55 mph

PHF = 0.80

Trucks = 6 %

Buses = 2 %

RVs = 2 %

One upgrade is 5% and 0.5 miles long

fHV = 1/ (1 + 0.06 (2.5-1) + 0.02 (3.0-1) + 0.02 (4.0-1)) = 1/1.19 = 0.84

V = Hourly volume = 3900 veh/hr

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP)

Since, freeway is operating at capacity on this upgrade during peak hour,

For FFS = 55 mph, Maximum Vp = 1700 + 10 FFS = 1700 + 10 * 55 = 2250 veh/hr/ln

2250 = 3900 / (0.80*2*0.84*fP)

fP = 1.29

Driver population adjustment factor = 1.29

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