Question

By using HCM Find the level of service LOS for the peak direction of a 2-mile section of a rural highway with the following characteristics: • Rural, two-lane highway, Class I • Rolling terrain • Design speed = 55 mph • 12 ft travel lanes, 6 ft shoulders • 60% no-passing zones (both directions) • No access points • AADT (in design year) = 8,000 vpd • 10% trucks, 5% RVs • K = 12% • D = 55% (in direction of increasing station) • PHF = 0.9

Answer 1

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

Lane width = 12 ft

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

Total Lateral Clearance = 6 ft

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

Access Points = 0

Reduction in speed corresponding to access points, f_ID = 0 mph

No. of lanes in each direction = 1

Reduction in speed corresponding to undivided highway, f_M = 1.6 mph

Free Flow Speed (FFS) = BFFS – f_LW – f_LC – f_M – f_ID = 55 – 0 – 0 – 1.6 – 0 = 53.4 mph

Heavy Vehicle Adjustment factor:

Trucks = 10%

Recreational Vehicles = 5 %

Rolling Terrain

f_HV = 1/(1+0.10(2.5-1)+0.05(2-1)) = 1/1.2 = 0.833

Flow rate:

AADT = 8000 vpd

K = 12 %

D = 55%

Peak Flow, V = AADT * K * D = 8000 * 0.12 * 0.55 = 528 veh/hr

Peak-hour factor = 0.9

f_P = 1.0

Peak Flow Rate, V_p = V / (PHV*n*f_HV*f_P) = 528/ (0.9*1*0.833*1.0) = 704.282 ~ 705 veh/hr/ln

Level of service:

Density = V_p/S = (705) / (53.4) = 13.2 veh/mi/ln

Hence Level of Service is B

Density of LOS B should lie between 11 – 18 veh/mi/ln