Use the data and results obtained in Problem 9-18 to determine the level of service of a two-lane section if a passing lane 1.5 mi long is added. The passing lane begins 0.75 mi from the starting point of the analysis segment.
What will be an ideal response?
To determine the level of service, lengths of the regions in the segment, PTSFpl
and ATSpl, must first be determined.
From Problem 9-16, PTSFd = 76.83% and ATSd = 37.88 mi/h
Step 1: Determine region lengths.
Region I: Lu = 0.75 mi
Region II: Lpl = 1.5 mi
Region III:
For PTSF, from Table 9.27, Lde = 5.8 mi; from Table 9.29, fpl = 0.62
For ATS, from Table 9.27, Lde = 1.7 mi; from Table 9.30, fpl = 1.11
Region IV:Use Equation 9.21.
Ld = Lt - (Lu +Lpl + Lde )
For PTSF, Ld = 4.5 – 0.75 –1.5 – 5.8 = –3.55; use Ld = 0
and L’de = 4.5 – 0.75 – 1.5 = 2.25
For ATS, Ld = 4.5 – 0.75 – 1.5 – 1.7 = 0.55
Step 2: Compute PTSFpl using Equation 9.23
Step 4: Determine level of service from Table 9.11 (for Class I highway): LOS E.
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