Question

A four-lane freeway (two lanes in each direction) operates at capacity during the peak hour. It has 11-ft lanes, 4-ft shoulders, and there are 3 ramps within 3 miles upstream of the segment midpoint and 4 ramps within 3 miles downstream of the segment midpoint. The freeway is on rolling terrain and has 8% heavy vehicles with a peak-hour factor of 0.85. It is known that 12% of the AADT occurs in the peak hour and that the directional factor is 0.6. What is the freeway's AADT?

((Final and should be 48,573))

((Final and should be 48,573))

((Final and should be 48,573))

((Final and should be 48,573))

((Final and should be 48,573))

Answer 1

Assume Base free flow speed (BFFS) = 75 mph

Lane width = 11 ft

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

Lateral Clearance = 4 ft

Reduction in speed corresponding to lateral clearance, f_LC = 1.2 mph

Interchanges/Ramps = 7 / 6 miles = 1.17 / mile

Reduction in speed corresponding to Interchanges/ramps, f_ID = 3.7 mph

No. of lanes in each direction = 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 – 1.9 – 1.2 – 4.5 – 3.7 = 63.7 mph

Peak-hour factor = 0.85

Heavy vehicles = 8 %

Rolling Terrain

f_HV = 1/ (1 + 0.08 (2.5-1)) = 1/1.12 = 0.893

f_P = 1.0

Peak Flow Rate, V_p = V / (PHV*n*f_HV*f_P) = V/ (0.85*2*0.893*1.0) = V/1.5181 veh/hr/ln

S = FFS

S = 63.7 mph

Density = V_p/S

Assume Level of Service is E

Density of LOS E should lie between 36 – 45 veh/mi/ln

Assume density = 36 veh/mi/ln

Peak flow rate = V/1.5181 = 36 * 63.7

V = 3481.3 veh/hr ~ 3482 veh/hr

V = AADT * K * D = AADT * 0.12 * 0.6

AADT = 3482/(0.12 * 0.6) = 48361 veh/day

Freeway AADT = 48361 veh/day