Question

1. A six lane freeway (three lanes in each direction) currently operates at maximum LOS C conditions. The lanes are 11 ft wide, the right-side shoulder is 4 ft wide, and there are 2 ramps within 3 miles upstream of the segment midpoint and 1 ramp within 3 miles downstream. The highway is on rolling terrain with 10% heavy vehicles, and the peak hour factor is 0.90. Determine the hourly volume for these conditions.

Answer 1

Answer:

The answer is " $4,071 \ \frac{veh}{h}$".

Explanation:

Compute the Free-Flow Speed estimation (FFS)  

The formulation:  

$FFS=75.4-f_{LW}-f_{LC}-3.22 TRD^{0.84}$

The lane width change is 11 ft.  The width of the lane.

$F_{LW} \ is \ 19 \ \frac{m}{h}$

Adaptation for lateral clearing for both the night shoulder  

Right side shoulder equivalent 4 ft  

In one way clearance or 3 lanes.

$f_{LC} \ is \ 0.8 \ \frac{m}{h}$

$FFS = 75.4 -1.9 -0.8 -3.22(\frac{3}{6})^{0.84}$

        $= 75.4 -1.9 -0.8 - 1.799\\\\= 70.901\ \frac{mi}{h}\\\\= 70 \ \frac{mi}{h}$

Adjustment factor (f_{HV})

The eqlivdents for the passenger vehicles (PCEs).  Any corresponding segment of its moving field.  Take the $E_T \ value \ 2.5$ and the $E_R \ value \ 2.0$  

$It \ replace \ 10\% \ with \ p_r, \ 2.5 \ with\ E_r, \ 0 \ with \ P_R, \ E_R's \ and \ 2.0:$

$f_{(HV)}= \frac{1}{1 + \frac{10}{100} (2.5-1)+(0)(2.0-1)}$

         $= 0.869$

Compute volume(V) hourly:

Please take 15 Minute passenger car eqvialent flow rate for the LOS Parameters for the LOS C and FFS of $70 \frac{mi}{h}$ So, $v_p$ is worth $1,735 \frac{pc}{\frac{h}{In}}$

Consider that$f_p$ is 1.00 for commuters.  

Replace$v_p$ with $1,735 \frac{pc}{\frac{h}{In}}$

 Thus, the value of$v_p$, is $1,735 \frac{pc}{\frac{h}{In}}$

Consider for commuters the value of f_p, is 1.00.

Substitute for v_p,

$0.90 \  for \ PHF\ 3 for \ N,\ 0.869 \ for\\F_{HV} \ and \ \ 1.00 \ for\ f_v \\\\1,735 = \frac{V}{0.9 \times 3 \times 0.869 \times 1} \\\\V= 1,735 \times 0.9 \times 0.869 \\= 4,070.83 = 4,071 \ \ \frac{veh}{h}$