Question

(10 pts) A four-lane freeway (two lanes in each direction) is located on rolling terrain and has 12-ft lanes, no lateral obstructions within 6 ft of the pavement edges, and there are 2 ramps within 3 miles upstream of the segment midpoint and 3 ramps within 3 miles downstream of the segment midpoint. A weekday directional peak-hour volume of 1800 vehicles (familiar users) is observed, with 700 arriving in the most congested 15-min period. Ifa level of service no worse than C is desired, determine the maximum number of heavy vehicles that can be present in the peak-hour traffic stream.

Answer 1

Answer:

309

Step-by-step explanation:

To determine the estimated free-flow speed

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

From the table "Adjustment for lane width," which corresponds to a lane width of 12 feet. As a result, $f_{LW}$ equals 0 mi/h.

Take the meaning from the table "Adjustment for right-shoulder lateral clearance," which corresponds to 6 feet of right shoulder lateral clearance and two lanes in one direction.

The $f_{LC} = 0 mi/h$

$FFS = 75.4 -0-0-3.22 ( \dfrac{5}{6})^{0.84}$

$= 72.64 \ mi/h$

$\simeq 73 \ mi/h$

Determine the peak-hour factor

$PHF = \dfrac{V}{V_{15}\times 4} \\ \\ =\dfrac{1800}{700\times 4}\\ \\ = 0.6429$

Now, Find the heavy-vehicle adjustment factor.

$v_P = \dfrac{V}{PHF\times N \times f_{HV}\times f_P} --- (1)$

Take the value for the 15-minute passenger car equivalent flow rate from the table "LOS requirements for freeway segments" for the FFS and LOS C conditions. The free-flow speed is estimated to be 73 miles per hour

$v_p = 1735+ \dfrac{73-70}{75-70}(1775-1735)$

$v_p = 1759 \ pc/h/In$

Think about for familiar users the value of f_p = 1.00

Replace all of the values obtained in (1)

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

$1759 = \dfrac{1800}{0.6429\times 2 \times f_{HV}\times 1}$

$f_{HV} = \dfrac{1800}{0.6429\times 2 }\times \dfrac{1}{1759}$

= 0.795

Calculate the percentage of trucks and buses in the flow of traffic stream.

$f_{HV} = \dfrac{1}{1+P_{\tau}(E_{\tau}-1) + P_R(E_R-1)}$

Take the values from the table "Passenger car equivalent for extended highway segments" referring to trucks and buses and rolling terrains for passenger car equivalent for trucks and buses and recreational vehicles. As a result, $E_T$ has a value of 2.5 and $E_R$ has a value of 2.0.

Since $P_R$ is zero and there is no recreational vehicle.

Then;

$0.759 = \dfrac{1}{1+ P_T(2.5 -1) +0}$

$P_T = 0.1719$

Finally, to estimate the maximum number of large trucks and buses; we have:

$= V\times P_T = 1800 \times 0.1719$

Maximum number of large trucks and buses = 309