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

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 =(V)/(f_(HV)* N* f_p* PHF)`

`f_(HV)=(1)/(1+P_T(E_T-1)+P_R(E_R-1))`

`f_(HV)=(1)/(1+0.08(2.5-1)+0.02(2-1))`

= 0.877

`v_p =(3500)/(0.877* 3* 0.95*0.9)`

= 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 = (V_P)/(s)`

`D = (1555)/(55) =28.27`

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