Question

A section of a multilane highway is to be reconstructed to improve the level of service. The section being considered is on a 5.0% upgrade that is 3/4 mile long. The highway currently has 4-lanes (2 in each direction – all are 12-ft lanes) with a two-way left-turn lane in the middle and 4 foot shoulders on the right side. It is to be reconstructed into a 6-lane facility (3-lanes in each direction) undivided facility but, due to commercial development surrounding the highway, must remain in the current 72 foot right of way. There are currently 35 access points per mile and the free flow speed is determined to be 50 mi/h. It is known that the road currently operates at capacity with 420 trucks/buses (no recreational vehicles) during the peak hour, a peak hour factor of 0.95 and all-commuter traffic.

The redesign is to reduce the number of access points per mile to 10 and to reduce the grade to 4.0% for 3/4 mile. It is estimated that the new design will increase traffic by 13%.

1. Determine the lane width and shoulder width combination that will maximize capacity given that the 3 lanes (each direction) must fit within 36ft?
2. For the maximum-capacity lane/shoulder combination chosen above, determine the new design's level of service and density.

Answer 1

We are given a current design section, but dont know the flow rate.

So we have to calculate the vp

Assume one way flow in the peak hour = V veh/hour

Currently we have

Number of lanes in one direction = 2 - all 12 foot lanes

PHF = 0.95

Percentage of Trucks + Buses = PT  = ?? = 420/V

Since it is 5% upgrade over a 3/4 length segment, we can assume ET = 4

Percentage of RV's = PR  = 0

Roadway is at capacity so LOS = E and maximum possible density = 28pc/km/ln = 45.061 pc/mi/ln

Lets calculate the heavy vehicle factor using the formula

fHV and it is calculated using the formula

We have assume ET  = 4.0 and PT = 420/V and PR = 0

So fHV = 1/(1+420/V * 3) = 1/ (1+ 1260/V) = V/(V + 1260)

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So Calculate vp based on the formula

where

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

V = hourly volume = let this be an unknown value

PHF = 0.95

N = Number of lanes in one direction = 2

fHV = heavy vehicle factor calculated earlier to be equal to V/(V + 1260)

fp = driver population factor. Assume equal to 1 because all are familiar drivers

vp = V/(0.95 * 2 * V/(V + 1260) * 1) = V/ (1.9 * V/(V + 1260)) = (V + 1260)/1.9

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Base Free Flow Speed = 50mph and we know that

FFS = BFFS - fLW - fLC -fN -fID

where

DO all calcs in mph and then convert back to kmph at the last

BFFS = base free flow speed = 50mph for an urban freeway

fLW = adjustment for lane width = 0 mph (see HCM tabe for corresponding to 12 ft lanes)

fLC = adjustment for right-shoulder lateral clearance = 1.2 mph (see HCM tabe for corresponding to  4 ft shoulders)

fN = adjusment for number of lanes = 4.5 mph (see HCM tabe for corresponding to  number of lanes = 2)

fID = adjustment for interchange density = 7.5 mph (see HCM tabe for corresponding to 35 interchanges per mile)

So FFS = 50 - 0 - 1.2 - 4.5 - 7.5 = 50 - 13.2 = 36.8 mph

we know that density = vp/FFS = vp/36.8 = 45.061 pc/mi/ln

Therefore vp = 45.061 * 36.8 = 1658.24 pc/hr/ln

so (V + 1260)/1.9 = 1658.24 so solving for V, we get V = 1891 veh/hr

so percentage of trucks = 420/1891 = .2221

Therfore fHV =.6001

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ROAD IMPROVEMENT

Now we are improving the road. So we have

N = 3 now

also V is increased by 13%, so new V = 1891 * 1.13 =2137 veh/hr

ET = 3 now because its 4% over 3/4 mile upgrade

PT = .2221 same as before (we are assuming that when traffic increases by 13%, trucks also increase in same proportion, so that in new increaed traffic percentage of trucks is kept the same)

we have to calculate fHV using the formula

So fHV = 1/(1+.2221*2) = 1/ 1.4442 = 0.6924

vp = 2137/(.95 * 3 * .6924 * 1) = 2137/1.9733 = 260) * 1) = 1083 pc/hr/ln

density = vp/FFS so capacity will be maximized when density is highest, but it should not touch 28 pc/kn/ln because at 28pc/km/ln it will be come LOS E

So for maximizing density, FFS should be lower

FFS = 50 - FFS = BFFS - fLW - fLC -fN -fID

fLW = adjustment for lane width = 0 mph (see HCM tabe for corresponding to 12 ft lanes)

fLC = adjustment for right-shoulder lateral clearance = 1.2 mph (see HCM tabe for corresponding to  4 ft shoulders)

fN = adjusment for number of lanes = 3.0 mph (see HCM tabe for corresponding to  number of lanes = 2)

fID = adjustment for interchange density = 7.5 mph (see HCM tabe for corresponding to 10 interchanges per mile)

So FFS = 50 - fLW- fLC - 3 - 7.5 = 39.5 - fLW- fLC = 39.5 - (fLW + fLC)

So FFS will be lowest when the combination of (fLW + fLC) will be highest. We only have 36ft lane width to play

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Options

Option 1 - for 3 lanes at 12ft each + 4ft lateral clearance,(fLW + fLC) = 0 + 0.8 = 0.8

Option 2 - for 3 lanes at 11ft each + 7ft lateral clearance (fLW + fLC) = 2 + 0 = 2

Option 3 - for 3 lanes at 10.5ft each + 8.5 ft lateral clearance (fLW + fLC) = 4.25 + 0 = 4.25

Option 4 - for 3 lanes at 10ft each + 10 ft lateral clearance (fLW + fLC) = 6.5 + 0 = 6.5 - but this option means lane widths are too narrow, so you might have car side scraping etc

So lets calculate density for all three options

Option 1 - FFS = 39.5 - 0.8 = 38.7 mph, so density in pc/km/ln = 1083/38.7/1.60934 = 17.3888 pc/km/ln - LOS D

Option 2 - FFS = 39.5 - 2 = 37.5 mph, so density in pc/km/ln = 1083/37.5/1.60934 = 17.9452 pc/km/ln - LOS D

Option 3 - FFS = 39.5 - 4.25= 35.25 mph, so density in pc/km/ln = 1083/35.25/1.60934 = 19.091 pc/km/ln - LOS D

Option 4 - FFS = 39.5 - 6.5 = 33 mph, so density in pc/km/ln = 1083/33/1.60934 = 20.39232 pc/km/ln - LOS D

So the best option for re-design is to have

3lanes in each direction, each is 10.5 ft wide and right-shoulder clearance = 8.5 ft.

The new designs LOS will be D and density will be 19.091 pc/km/ln