Question

A two-lane highway is currently operating with flow rate at 3000 pc/h in rolling terrain. The traffic stream consists of cars and trucks only. A recent traffic count revealed 600 vehicles (total, both directions) arriving in the most congested 15-min interval. What is the percentage of trucks in the traffic stream based on the ATS service measure?

Answer 1

Step 1: Determine grade adjustment factor

Its Rolling Terrain and two way volume = 3000 pc/hr so clearly greater than 1200 veh/hour

So f_G (grade adjustment factor) based on HCM figure 20-7 = 0.99

GRADE ADJUSTMENT FACTOR (CG) TO DETERMINE SPEEDS ON TWO-WAY AND DIRECTIONAL SEGMENTS Type of Terrain Range of Two-Way Flow Range of Directional Flow Level Rolling Rates (pc/h) Rates (pc/h) 0-600 0-300 1.00 0.71 >600-1200 > 300-600 1.00 0.93 > 1200 > 600 1.00 0.99

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Step 2: Compute f_HV and ATS in the anlaysis direction

Percent Trucks = P_T = ?? this is unknown, we need to find it

Percent RV's = P_R = 0%

for Rolling terrain E_T = 2.5 and E_R = 2.0

Vehicle Level Rolling Mountainous ET 1.5 2.5 4.5 Es 1.2 2.0 4.0

So calculate heavy vehicle factor f_HV using the formulae

1 fuv 1 + P (E, - 1) + P (Ex - 1)

So f_HV = 1/(1+(P_T* 1.5) + (0 * 1) =  = 1/(1+ 1.5P_T)

Two way volume V = 3000 veh/hr,

So in analysis direction V = 1500 veh/hr

v_ps = adjusted hourly demand and is calculated by the formula

v_ps = V/(PHF * N * f_hv* f_p * f_G)

where

v_ps = adjusted flow rate in pc/hr

PHF = 300*4/1500 = = 0.80

V = 1500 veh/hr, PHF = 0.8 , N = Number of lanes in one direction = 1, f_HV = 1/(1+ 1.5P_T) , f_G = 0.99

assume driver familiarity factor f_p =1

Therefore v_ps = 1500/(.8 * 1/(1+ 1.5P_T) * 1 * 0.99) = 1500 * (1+ 1.5P_T)/0.792 = 1893.93 * (1+ 1.5P_T)

But were are given that peak 15 minute flow = 600 pc/hr

So in our our peak flow = 600*4 = 2400

so we have 2400 = 1893.93 * (1+ 1.5P_T)

So solving we get PT = .17855 = 17.85%

So percentage of trucks = 17.85%

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