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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.

1 Answer

5 votes

Answer:

The answer is "
4,071 \ (veh)/(h)".

Step-by-step 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 \ (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 \ (m)/(h)


FFS = 75.4 -1.9 -0.8 -3.22((3)/(6))^(0.84)


= 75.4 -1.9 -0.8 - 1.799\\\\= 70.901\ (mi)/(h)\\\\= 70 \ (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))= (1)/(1 + (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 (mi)/(h) So,
v_p is worth
1,735 (pc)/((h)/(In))

Consider that
f_p is 1.00 for commuters.

Replace
v_p with
1,735 (pc)/((h)/(In))

Thus, the value of
v_p, is
1,735 (pc)/((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 = (V)/(0.9 * 3 * 0.869 * 1) \\\\V= 1,735 * 0.9 * 0.869 \\= 4,070.83 = 4,071 \ \ (veh)/(h)

User John Mee
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