Question

A six-lane divided highway (three lanes in each direction) is on rolling terrain with two access points per mile and has 10- ft lanes, with a 5-ft shoulder on the right side and a 3-ft shoulder on the left side. The peakhour factor is 0.9, and the directional peak-hour volume is 3500 vehicles per hour. There are 7% single-unit trucks and 3% tractortrailer trucks in the traffic stream. No speed studies are available, but the posted speed limit is 55 mi/h. Determine the level of service.

Answer 1

Answer

given,

6 lanes divided highway 3 lanes in each direction

rolling terrain

lane width = 10'

shoulder on right = 5'

PHF = 0.9

shoulder on the left direction = 3'

peak hour volume = 3500 veh/hr

large truck = 7 %

tractor trailer = 3 %

speed = 55 mi/h

LOS is determined based on V p

10' lane weight ;  f_{Lw}=6.6 mi/h

5' on right   ;    f_{Lc} = 0.4 mi/hr

3' on left   ;      no adjustment

3 lanes in each direction    f n = 3 mi/h

$v_p =\dfrac{V}{f_{HV}\times N\times f_p\times PHF}$

$f_{HV}=\dfrac{1}{1+P_T(E_T-1)+P_R(E_R-1)}$

$f_{HV}=\dfrac{1}{1+0.08(2.5-1)+0.02(2-1)}$

          = 0.877

$v_p =\dfrac{3500}{0.877\times 3\times 0.95\times0.9}$

       = 1,555 veh/hr/lane

$FFS = BFFS - F_{Lw}-F_{Lc}-F_{N}-F_{ID}$

      = (55 + 5) - 6.6 - 0.4 -3 -0

      = 50 mi/h

$D = \dfrac{V_P}{s}$

$D = \dfrac{1555}{55} =28.27$

level of service is D using speed flow curves and LOS for basic free moving of vehicle