Respuesta :
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
[tex]v_p =\dfrac{V}{f_{HV}\times N\times f_p\times PHF}[/tex]
[tex]f_{HV}=\dfrac{1}{1+P_T(E_T-1)+P_R(E_R-1)}[/tex]
[tex]f_{HV}=\dfrac{1}{1+0.08(2.5-1)+0.02(2-1)}[/tex]
= 0.877
[tex]v_p =\dfrac{3500}{0.877\times 3\times 0.95\times0.9}[/tex]
= 1,555 veh/hr/lane
[tex]FFS = BFFS - F_{Lw}-F_{Lc}-F_{N}-F_{ID}[/tex]
= (55 + 5) - 6.6 - 0.4 -3 -0
= 50 mi/h
[tex]D = \dfrac{V_P}{s}[/tex]
[tex]D = \dfrac{1555}{55} =28.27[/tex]
level of service is D using speed flow curves and LOS for basic free moving of vehicle
