Question

9) An existing rural freeway in level terrain is to be analyzed to determine LOS using the following information: number of lanes in each direction equals 2, peak hour volume of 2140 veh/h (in the peak direction), 15% trucks, 2% recreational vehicles, PHF = 0.90, lane width = 12 ft, lateral clearance = 10 ft, average interchange spacing is 3 mi.

Answer 1

Assume Base free flow speed (BFFS) = 75 mph

Lane width = 12 ft

Reduction in speed corresponding to lane width, f_LW = 0 mph

Lateral Clearance = 10 ft

Reduction in speed corresponding to lateral clearance, f_LC = 0 mph

Interchanges/Ramps = 1/3 mile = 0.33 /mile

Reduction in speed corresponding to Interchanges/ramps, f_ID = 0 mph

No. of lanes = 2

Reduction in speed corresponding to number of lanes, f_N = 4.5 mph

Free Flow Speed (FFS) = BFFS – f_LW – f_LC – f_N – f_ID= 75 – 0 – 0 – 4.5 – 0 = 70.5 mph

Peak Flow, V = 2140 veh/hr

Peak-hour factor = 0.90

Trucks = 15 %

RVs = 2 %

Level Terrain

f_HV = 1/ (1 + 0.15 (1.5-1) + 0.02 (1.2-1)) = 1/1.079 = 0.927

f_P = 1.0

Peak Flow Rate, V_p = V / (PHV*n*f_HV*f_P) = 2140/ (0.90*2*0.927*1.0) = 1282.5 ~ 1283 veh/hr/ln

V_p < 3400 - 30 FFS

S = FFS

S = 70.5 mph

Density = V_p/S = (1283) / (70.5) = 18.2 veh/mi/ln

Hence Level of Service is C

Density of LOS C should lie between 18 - 26 veh/mi/ln