Question

CEN 334-Traffic Engineering Spring 2019 Beaudry Due HCM Chapter 12: Two-Lane Highways A 3-mile segment of Class I two-lane highway has the following conditions: 1. 1700 veh/hr (two-way volume) .10 access points per mile 6% trucks, 4% RVs . 30% no-passing zones BFFS- 50 mph 70/30 directional split Rolling terrain - Typical weekday commuters · PHF 0.90 11 ft lane widths 4-ft right shoulder lateral clearance . a. Using the HCM procedures for two-lane highways and SHOWING ALL WORK, calculate i) Average travel speed, ii) iii) Percent time spent following, and LOS of the intersection. b. Repeat part a) using the HCS7 software, and compare final results. Repeat Question 1 assuming that the road is a Class II two-lane highway (all other details remain 2. identical) using HCS7 SOFTWARE ONLY. Are there any differences in the final calculated LOS? NOTE: Linear interpolation may be required when using HCM tables. HCM Chapter 13: Freeway Weaving Segments

Answer 1

Solution:

1 a)

Trial value of Vp = V / PHF = 1700 /0.9

= 1888.889 pc/h

heavy vehicle adjustment factor = fHV

fHV = 1 / [ 1 + Pt(Et -1 ) + Pr(Er -1)]

for rolling terrain,

Et = 1 , Er = 1

fHV = 1 / [ 1 + 0.06(1 -1 ) + 0.04(1 -1)]

= 1

fG = grade adjustment factor = 1

Vp = V / ( PHF * fG * fHV)

= 1700 / 0.9

= 1888.889 pc/h

Base percent time spent following:

BPTSF = 100 * [ 1 - e^-0.000879*Vp ]

= 100 * [ 1 - e^-0.000879*1888.889 ]

= 81%

PTSF = BPTSF + f(d/np)

we have to interpolate to find f(d/np)

Directional split = 70/30 , no-passing zones = 30%

the value will be 4.7 when Vp = 1400 ,,

= 3.15 when Vp = 2000

So, when Vp = 1888.889

f(d/np) = 3.437

PTSF = 81 + 3.44 = 84.44%

Average travel speed (ATS) = FFS - 0.0776*Vp - fnp

fnp for no passing zones

for Vp = 1888.889 pc/h , 30% no passing zones,

interpolation,

fnp for 30% no passing zones

1800 , 0.6

2000 ,0.55

for 1888.889 it will be 0.578

ATS = FFS - 0.00776*1888.889 - 0.578

FFS = BFFS - fLS - fA

= 50 - 1.7 - 2.5

= 45.8 mi/h

ATS = 45.8 - 0.00776*1888.889 - 0.578

= 30.56 mi/h

PTSF = 84.44%

So, for the above data LOS E