Question

Freeway highway analysis

• six-lane freeway, three -lane in each direction

• Lane width = 11 ft; Right-side lateral clearance = 4 ft

• Peak hour direction demand volume = 2,500 veh/h

• Traffic composition: 11% trucks and 6% RVs

• PHF = 0.97, Level Terrain; 24 ramps in six miles segment

• Commuter traffic

Determine the level of service (LOS)

Answer 1

Ans) We know,

V_p = V / ( PHF x N x f_HV x f_p)

where, Vp = 15 min passenger car equivalent

PHF = peak hour factor

N = number of lanes in each direction

f_HV = heavy vehicle adjustment factor

f_p = driver population factor

Also, f_HV = 1 / [1 + P_T ( E_T -1) + P_R(E_R -1)]

where P_T and P_R are proportion of buses and RVs respectively

E_T and E_R are passenger car equivalent for trucks and RVs respectively

For level terrain E_T = 1.5 and E_R = 1.2

=> f_HV = 1 / [ 1 + 0.11(0.5) + 0.06(0.2)]

= 0.937

=> Vp = 2500 / ( 0.97 x 3 x 0.937 x 1)

=916 pc/ph/lane

Now, we know ,

FFS = BFFS - f_{LW -} f_LC - f_N - f_ID

where, FFS = free flow speed

BFFS = Basic free flow speed (assume 75 mph)

f_{LW =} adjustment factor for lane width = 0 for 11 ft lane

f_LC = adjustment factor for lateral clearance = 1.80 mph for 4ft lateral clearance

f_N = adjustment factor for number of lanes = 3

f_ID = adjustment factor for interchange density = 2 mph for 4 access point per mile

=> FFS = 75 - 0 - 1.85 - 3 - 2  = 68.15 mph

According to speed flow curve from highway manual, for Vp = 916 pcphpl and FFS = 68.15 mph, LOS is B