Question

[HCM for Basic Freeway Segments] A four-lane freeway (two lanes in each direction) has an observed demand volume of 2400 veh/h (one direction) during the peak hour. The freeway is on rolling terrain and the traffic stream consists of passenger cars and trucks only. The peak-hour factor is 0.75 and the traffic is all commuters. If the peak 15-min demand flow rate under equivalent base conditions is estimated to be 2000 pc/h/ln, what is the percentage of trucks in this traffic stream?

Answer 1

Ans) We know, 15 min hourly passanger car equivalent, vp

vp = V / (PHF x N x fHV x fP)...................................(1)

where, V = hourly volume = 2400 veh/hr

PHF = peak hour factor = 0.75

N = number of lanes in each direction = 2

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 rolling terrain and trucks F, ET = 2.5

For rolling terrain and RVs , ER = 2

Since traffic consist of only passenger car and trucks, proportion of RVs = 0

=> fHV = 1 / [1 + PT(2.5 - 1 ) + 0.0(2 -1)]

=> fHV = 1 / (1 + 1.5 PT)

Given, vp = 2000 pc/h/ln

Putting values in equation 1,

=> 2000 = 2400 / [0.75 x 2 x (1 / (1 + 1.5 PT)) x 1]

=> [0.75 x 2 x (1 / (1 + 1.5 PT)) x 1] = 1.2

=> 1.5 / (1 + 1.5 PT) = 1.2

=> 1 + 1.5PT = 1.25

=> 1.5 PT = 0.25

=> PT = 0.1667 or 16.67%

Hence, percent of trucks in traffic stream is 16.67%