Question

A six-lane recreational freeway has the following characteristics: 12 ft lanes, 6 ft lateral clearances, 2 ramps/mi, 10% trucks (standard mix), peak-hour factor=0.95, level terrain, peak-hour demand=4,000 veh/h. Because the facility serves recreational users to a nearby ski area, the peak period occurs on weekends, and light to moderate snow is the normal weather pattern. If virtually all users of the facility are unfamiliar with the roadway, what is the expected LOS during the peak period?

Answer 1

Ans) Calculate 15 min peak hourly passanger car equivalent, vp

vp = V / (PHF x N x fHV x fP)

where, V = hourly volume = 40000 veh/hr

PHF = peak hour factor = 0.95

N = number of lanes in each direction = 3

fHV = Heavy vehicle adjustment factor

fp = Driver population factor = 1

Also, fHV = 1 / [1 + PT(ET -1) + PR(ER - 1)]

where, PT and PR are proportions of trucks and RVs respectively

ET and ER are passenger car equivalent for trucks and RVs respectively

For level terrain and trucks F, ET = 1.5

For level terrain and RVs , ER = 1.2

=> fHV = 1 / [1 + 0.1(1.5 - 1) + 0(1.2 -1)]

=> fHV = 0.952

=> vp = 4000 / ( 0.95 x 3 x 0.952 x 1)

=> vp = 1475 pc/h/lane

Now,calculate free flow speed (FFS),

FFS = BFFS - fLW - fLC - fN - fAD

where, BFFS = base free flow speed = 70 mph

fLW = adjustment for lane width = 0 for 12 ft lane

fLC = adjustment for lateral clearance = 0 mph for 6 ft

fN = adjustment for number of lane = 3 mph for 3 lanes in each direction

fAD = adjustment for access point density = 7.5 mph for 2 interchange per mile

=> FFS = 70 -- 3 - 7.5 = 59.5 mph

Now, according to speed flow curve, for vp = 1475 pc/ph/pl and FFS = 59.5 mph , Level of service (LOS) is C