Question

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. If a 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.

Consider the freeway and traffic conditions described in Problem 6.3. If 180 of the 1800 vehicles observed in the peak hour were heavy vehicles (assume a 70%/30% SUT/TT split), what would the level of service of this freeway be on a 5-mi, 6% downgrade?

Answer 1

Answer:

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Explanation:

FFS = Free-Flow Speed (km/h) BFFS = the Base Free-Flow Speed (km/h) fLS = the Adjustment for lane and shoulder widths less than 3.65 m and 1.80 m, respectively. fAPD = Adjustment for access points density (km/h) fm = Adjustment for proportion  (km/h) 3.0 Results

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Answer 2

Determining the anticipated free-flow speed.

$\to FFS = 75.4-f_{LW}-f_{Lc}-3.22TRD^{0.84}$

• A table "Adjustment regarding lane width" relates to a $12 \ ft$span length. As a result, $f_{LW}\ is\ 0 \ \frac{mi}{h}$.
• Use the lateral clearances correction value from the table "Alteration for lateral aspect clearance" which corresponds to 6 feet of shoulder lateral clearance and two lines in one way. As a corollary, $f_{LC} \ is\ 0\ \frac{mi}{h}.$

             $\to FFS = 75.4-0-0-3.22 (\frac{5}{6})^{0.84} = 72.64\ \frac{mi}{h} \approx 73\ \frac{mi}{h}$

• Evaluating the peak-hour factor.

$\to PHF = \frac{V}{ V_{15} \times 4}=\frac{1,800}{700 \times 4} = 0.6429$

• Evaluating the heavy-vehicle adjustment factor.

$\to V_p= \frac{V}{PHF \times N \times f_{HV} \times f_{p}}...................(1)$

• Calculate the 15-minute customer car equivalent mass flow figure from the table "LOS criterion for motorway portions" that match FFS and LOS C circumstances. Therefore, estimating the flow of free speed is $73 \ \frac{mi}{h}$. that takes the values for $70\ \frac{mi}{h}\ \ and \ \ 75\ \frac{mi}{h}$, then interpolate for $73 \ \frac{mi}{h}$

$\to v_p=1,735+ \frac{(73-70)}{ (75-70)}(1,775 - 1,735)\ =1,759 \frac{\frac{pc}{h}}{In}$

• Considering the familiar users so, the value is $f_p$=1.00.  Substituting all the obtaining value in (1).  

$\to V_p =\frac{V}{PHF \times N \times f_{HV} \times f_{p}} \\\\\to 1,759= \frac{1,800}{0.6429 \times 2 \times \times f_{HV} \times 1}\\\\ \to f_{HV}= \frac{1,800}{0.6429 \times 2} \times \frac{1}{1,759} = 0.795$

• Determine the frequency of commercial vehicles inside the traffic flow.

$\to f_{HV}=\frac{1}{1+ P_{T}(E_{T} -1)+P_{R}(E_{R}-1)}$

• Take any value for regular car equivalent for commercial vehicles and passenger car equivalent for recreational vehicles from the tables "Passenger vehicle equivalent for extended freeway segments" for buses and trucks on rolling terrain. As a consequence, the number of $E_T =2.5$, whereas the value of $E_R =2.0$. Since there is no leisure car, $P_R =0$.

$\to 0.759=\frac{1}{1+P_T(2.5-1)+0}\\\\ \to 1.5P_T=\frac{1}{0.759}-1 \\\\\to P_T = 0.1719$

• Find the maximum number of heavy buses and trucks which can be used.

$\to \binom{\text{Maximum Number of large}}{\text{ trucks and buses}} = V\times P_r= 1,800 \times 0.1719 = 309$

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