Answered

Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Connect with a community of experts ready to provide precise solutions to your questions on our user-friendly Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

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
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.