Respuesta :
Answer:
The answer is " [tex]4,071 \ \frac{veh}{h}[/tex]".
Explanation:
Compute the Free-Flow Speed estimation (FFS)
The formulation:
[tex]FFS=75.4-f_{LW}-f_{LC}-3.22 TRD^{0.84}[/tex]
The lane width change is 11 ft. The width of the lane.
[tex]F_{LW} \ is \ 19 \ \frac{m}{h}[/tex]
Adaptation for lateral clearing for both the night shoulder
Right side shoulder equivalent 4 ft
In one way clearance or 3 lanes.
[tex]f_{LC} \ is \ 0.8 \ \frac{m}{h}[/tex]
[tex]FFS = 75.4 -1.9 -0.8 -3.22(\frac{3}{6})^{0.84}[/tex]
[tex]= 75.4 -1.9 -0.8 - 1.799\\\\= 70.901\ \frac{mi}{h}\\\\= 70 \ \frac{mi}{h}[/tex]
Adjustment factor (f_{HV})
The eqlivdents for the passenger vehicles (PCEs). Any corresponding segment of its moving field. Take the [tex]E_T \ value \ 2.5[/tex] and the [tex]E_R \ value \ 2.0[/tex]
[tex]It \ replace \ 10\% \ with \ p_r, \ 2.5 \ with\ E_r, \ 0 \ with \ P_R, \ E_R's \ and \ 2.0:[/tex]
[tex]f_{(HV)}= \frac{1}{1 + \frac{10}{100} (2.5-1)+(0)(2.0-1)}[/tex]
[tex]= 0.869[/tex]
Compute volume(V) hourly:
Please take 15 Minute passenger car eqvialent flow rate for the LOS Parameters for the LOS C and FFS of [tex]70 \frac{mi}{h}[/tex] So, [tex]v_p[/tex] is worth [tex]1,735 \frac{pc}{\frac{h}{In}}[/tex]
Consider that[tex]f_p[/tex] is 1.00 for commuters.
Replace[tex]v_p[/tex] with [tex]1,735 \frac{pc}{\frac{h}{In}}[/tex]
Thus, the value of[tex]v_p[/tex], is [tex]1,735 \frac{pc}{\frac{h}{In}}[/tex]
Consider for commuters the value of f_p, is 1.00.
Substitute for v_p,
[tex]0.90 \ for \ PHF\ 3 for \ N,\ 0.869 \ for\\F_{HV} \ and \ \ 1.00 \ for\ f_v \\\\1,735 = \frac{V}{0.9 \times 3 \times 0.869 \times 1} \\\\V= 1,735 \times 0.9 \times 0.869 \\= 4,070.83 = 4,071 \ \ \frac{veh}{h}[/tex]
