Question

A new suburban freeway is being designed. Estimated AADT is 40,000 veh/day, 60% of the peak flow traffic travels in the peak direction, the PHF is 0.85, and the terrain is rolling. There will be 3 ramps per mile, and traffic is expected to include 18% heavy vehicles (50% SUTs and 50% TTs). Lanes will be 12 feet wide with a lateral clearance of 6 feet. One segment of the road of the road has a 0.75-mile-long section with a 3.5% grade.

a) How many lanes are needed to provide level of service C during the peak hour using the 30th highest hourly volume (see Figure 6.8 in text and in lecture notes on design hourly volumes) on the road and how many lanes are required on the segment with the 3.5% grade?

b) Does the number of lanes change if you design for LOS D?

Answer 1

AADT = 40000 veh/day

directionality factor = 0.6

assume a k factor (which converts daily flow into peak hourly flow) = 0.12

so Design Hourly Volume = V = 40000 * .6 * .12 = 2800 veh/hr

Its rolling terrain with 3.5% upgrade over 3/4mile

18% heavy vehices of which 50% are trucks + buses and 50% are RV's

so we can say PT = Percentage of Trucks + Buses = 9% = 0.09

PR = Percentage of RV's = 9% = 0.09

So we can take ET = 2.5 and ER = 2.0

So

So we can calculate Heavy vehicle factor = fHV and it is calculated using the formula

So fHV = 1/(1+(.09 * 1.5) + .(09*1 ) = 1/1.225= 0.8163

so we get

vp is calculated by the formula

where

vp = 15 mt passenger car equivalent flow rate in pc/h

V = hourly volume per direction = 2800 veh/hr

PHF = 0.85 given to us

N = Number of lanes in one direction = 3

fHV = heavy vehicle factor calculated earlier to be equal to 0.8163

fp = driver familiarity factor - assume it to be 1 and that most are familiar drivers

So vp = 2800/(0.85 * 3 * .8163 * 1) = 2800/2.08156 = 1345.14 pc/hr/lane

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now calculate FFS which is based on Base Free Flow Speed

FFS = BFFS - fLW - fLC -fN -fID

BFFS = base free flow speed = 70mph assume this since urban area

fLW = adjustment for lane width = 0 mph (see table in HCM for factor with 12ft lanes)

fLC = adjustment for right-shoulder lateral clearance = 0 mph (see table in HCM for 6ft lateral clearance).

fN = adjusment for number of lanes = 3 mph (see table in HCM for number of lanes in one direction = 3)

fID = adjustment for interchange density = 7.5mph since TRD = 3/mi

FFS = BFFS - fLW - fLC -fN -fID

Therfore FFS = 70 - 0 -0 - 3 - 7.5 = 70 - 10.5 = 59.5 mph

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Density = vp/S = 1345.14/59.5 = 22.607 pc/mi/ln = 14.0476 pc/km/ln.

so based on the density guildelines table this corresponds to LOS C

In order to provide LOS C, then density can go as high as 16 pc/km/ln = 25.749 pc/mi/ln

so calculate the vp required to maintain a density of 25.749 pc/mi/ln

25.749 =vp/ 59.5 and so vp = 59.5 * 25.749 = 1533 pc/hr/ln

So vp = 2800/(0.85 * N * .8163 * 1) = 1533

So solving for N, we get N = 2800/1064 = 2.6315

So 3 lanes are required to maintain LOS C on the segment which has a 3.5% grade

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For the segment without the 3.5 % grade and rolling terrain with 9% trucks+ buses and 9% RV's

we know So we can take ET = 2.5 and ER = 2.0

so these are identical to the situation when there was no grade. So nothing elsewill change inthe calculations

so number or lanes required for segment with 3.5% upgrade = 3

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Now if you are allowing LOS D, that means you increase density upto a maximum of 22 pc/km/ln = 35.4055 pc/mi/ln

so density = vp/59.5 = 35.4055

So vp = 35.4055 * 59.5 = 2107 pc/hr/ln

But vp = 2800/(0.85 * N * .8163 * 1) = 2107

Therefore solving for N, we get N = 1.915

So for maintaining LOS D you only need 2 lanes in each direction

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