Question

A four-lane divided highway in suburban area has the following criteria:

• Average grade is less than 2% (level Terrain)

• Posted speed 55 MPH

• 11 ft lanes, 4 ft left and 4 ft right clearance

• 9% heavy vehicle

• Access spacing is 500 ft

• Directional Design Hour Volume = 2600 vph

• PHF = 0.92

Find the level of service for the highway

Answer 1

Given, Base free flow speed (BFFS) = 55 mph

Lane width = 11 ft

Reduction in speed corresponding to lane width, f_LW = 1.9 mph

Total Lateral Clearance = 8 ft

Reduction in speed corresponding to lateral clearance, f_LC = 0.9 mph

Access Points = 5280 / 500 mile = 10.56 /mile

Reduction in speed corresponding to Access Points, f_A = 2.5 mph

No. of lanes = 4

Reduction in speed corresponding to divided highway, f_M = 0 mph

Free Flow Speed (FFS) = BFFS – f_LW – f_LC – f_M – f_A = 55 – 1.9 – 0.9 – 0 – 2.5 = 49.7 mph

Directional Design Hour Volume, V = 2600 veh/hr

Peak-hour factor = 0.92

Heavy Vehicles = 9 %

Level Terrain

f_HV = 1/ (1 + 0.09 (1.5-1)) = 1/1.045 = 0.957

f_P = 1.0

Peak Flow Rate, V_p = V / (PHV*n*f_HV*f_P) = 2600/ (0.92*2*0.957*1.0) = 1476.53 ~ 1477 veh/hr/ln

S = FFS = 49.7 mph

Density = V_p/S = (1477) / (49.7) = 29.72 veh/mi/ln

Hence Level of Service is D

Density of LOS D should lie between 26 – 35 veh/mi/ln