Question

A two-lane highway is currently operating at its two-way capacity in
rolling terrain. The traffic stream consists of cars and trucks only. A recent traffic
count revealed 720 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?
Given:
ATS:
fG=0.99 when it is rolling terrain, vp>1200
ET= 1.5
when it is rolling terrain, vp>1200

Answer 1

We are given fG  = 0.99

E_T = 1.5 and E_R = 1.1, Assume % of trucks = P_T (expressed as a decimal)

So calculate heavy vehicle factor f_HV using the formulae

So f_HV = 1/(1+(P_T *0.5)  

v_p is calculated by the formula

where

vp = 15 mt passenger car equivalent flow rate in pc/h

V = measures in veh/hour in one direction so in our case V = 720 *4/2 (assuming that directionality is 50-50 split)

So V = 1440 veh/hour

So v_p = 1440*(1 + 0.5P_T) this is the v_p in each direction

We are given roadway is at capacity to LOS = E

Based on the following graph (taken from HCM), for LOS, the maximum possible ATS = 30 mi/hour

but we know that

ATS = FFS - .0076*V_ps - f_np

Average travel speed
where ATS = average travel speed

FFS = base free flow speed = assume it as 75kmph = 46.603mph

V_ps = volume of passenger cars for speed = 1440/(1 + 0.5P_T) (calculated earlier)

f_np = no passing zone correction to be taken from HCM - so for a two way demand of approximately 2880 pc/hr and a assume 50% no passing zones,

so we get fnp = 0.65

So ATS = 46.603 - .0076 * 1440*(1 + 0.5P_T) - 0.65 = 30

so we get .0076 * 1440*(1 + 0.5P_T) = 46.603 - 0.65 - 30 = 15.953

So 1440*(1 + 0.5P_T) = 2099.08

so we get P_T = (2099.08/1440 -1)/0.05 = .9153

So % of trucks in the traffic = 91.53%