Question

A divided multilane highway is a recreational area (fp = 0.90) has four lanes (two lanes in each direction) and is on rolling terrain. The highway has 10-ft lanes with a 6-ft right-side shoulder and a 3-ft left-side shoulder. The posted speed is 50 mi/h. Formerly there were 4 access points per mile, but recent development has increased the number of access points to 12 per mile. Before the development, the peak-hour factor was 0.95 and the directional hourly volume was 2,300 vehicles with 10% large trucks and buses and 3% recreational vehicles. After the development, the peak-hour directional flow is 2,700 vehicles with the same vehicle percentage and peak-hour factor. What is the level-of-service before and after the development?

Answer 1

Before the development:

Base free flow speed (BFFS) = 50 mph

Lane width = 10 ft

Reduction in speed corresponding to lane width, f_LW = 6.6 mph

Total Lateral Clearance = 6 + 3 = 9 ft

Reduction in speed corresponding to lateral clearance, f_LC = 0.6 mph

Total Access points = 4 /mile

Reduction in speed corresponding to access points, f_{​​​​A} = 1 mph

no. of lanes in each direction = 2

Reduction in speed corresponding to divided highway, f_M = 3 mph

Free Flow Speed (FFS) = BFFS – f_LW – f_LC – f_M – f_A = 50 – 6.6 – 0.6 – 0 – 1 = 41.8 mph

Peak Flow, V = 2300 veh/hr

Peak-hour factor = 0.95

Trucks = 10%

RVs = 3%

Rolling Terrain

f_HV = 1/ (1 + 0.10 (2.5-1) + 0.03 (2 - 1)) = 1/1.18 = 0.847

f_P = 0.9

Peak Flow Rate, V_p = V / (PHV*n*f_HV*f_P) = 2300/ (0.95*2*0.847*0.9) = 1587.99 ~ 1588 veh/hr/ln

Average speed of vehicles, S = FFS = 41.8 mph

Density = V_p/S = 1588/41.8 = 37.99 veh/mi/ln

Level of service E

Density of LOS E lies between 36 - 45 veh/mi/ln

After the development:

Assume Base free flow speed (BFFS) = 50 mph

Lane width = 10 ft

Reduction in speed corresponding to lane width, f_LW = 6.6 mph

Total Lateral Clearance = 9 ft

Reduction in speed corresponding to lateral clearance, f_LC = 0.6 mph

Total Access points = 12 /mile

Reduction in speed corresponding to access points, f_{​​​​A} = 3 mph

no. of lanes = 2

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

Free Flow Speed (FFS) = BFFS – f_LW – f_LC – f_M – f_A = 50 – 6.6 – 0.6 – 0 - 3 = 39.8 mph

Peak Flow, V = 2700 veh/hr

Peak-hour factor = 0.95

Trucks = 10%

RVs = 3%

Rolling Terrain

f_HV = 1/ (1 + 0.10 (2.5-1)+ 0.03 (2 - 1)) = 1/1.18 = 0.847

f_P = 0.9

Peak Flow Rate, V_p = V / (PHV*n*f_HV*f_P) = 2700/ (0.95*2*0.847*0.9) = 1864.16 ~ 1865 veh/hr/ln

Average speed of vehicles, S = 39.8 mph

Density = V_p/S = 1865/39.8= 46.83 veh/mi/ln

Level of service F

Density of LOS F is greater than 45 veh/mi/ln