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A Class III two-lane highway is on level terrain, has a measured free-flow speed of 45 mi/h, and has 100% no-passing zones. During the peak hour, the analysis direction flow rate is 150 veh/h, the opposing direction flow rate is 100 veh/h, and the PHF-0.95. There are 5% large trucks and 10% recreational vehicles. Determine the level of service.

Sagot :

batolisis

Answer:

LOS = A

Explanation:

Given all the parameters the level of service as seen from the attached graph

is LOS =  A

To determine the LOS from the attached graph

calculate the trial value of Vp

Vp = V / PHF

     = (100 + 150) / 0.95  =  263 pc/h

since the trial value of Vp = ( 0 to 600 ) pc/h . hence E.T = 1.7 , ER = 1

next we will calculate the flow rate

flow rate = 1 / [ ( 1 + PT(ET - 1 ) + PR ( ER - 1 ) ]

             Fhr  = 1 / 1.035 = 0.966 ≈ 1

next calculate the real value of Vp

Vp = V / ( PHF * N * Fhr * Fp )

     = ( 100 + 150 ) / ( 0.95 * 2 * 1 * 1 )

Vp ≈ 126 pc/h/In

Next calculate the density

D = Vp /  S  =  126 / ( 45 * 1.61 )  = 1.74 pc/km/In

View image batolisis
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