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13 October, 23:48

A single lane highway has a horizontal curve. The curve has a super elevation of 4% and a design speed of 45 mph. The PC station is 105+00 and the PI is at 108+75.

What is the station of the PT?

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Answers (1)
  1. 14 October, 01:42
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    Answer: 112 + 19.27

    Explanation:

    Super elevation is an inward transverse slope provided through out the length of the horizontal curve which ends up serving as a counteract to the centrifugal force and checks tendency of overturning. It changes from infinite radius to radius of a transition curve.

    Super curve elevation (e) = 4%

    4/100 = 0.04

    e = V^2/gR

    Make R the subject of the formula.

    egR = V^2

    R = V^2/eg

    V = 45mph

    =45 * 0.44704m/s

    =20.1168m/s

    g (force due to gravity) = 9.81

    Therefore,

    R = (20.1168) ^2/9.81 * 0.04

    = 1031.31m

    Tangent Length (T) = PI - PC

    Tangent Length = 10875 - 10500

    =375m

    T = R Tan (I/2)

    375 = 1031.31 * Tan (I/2)

    I = 39.96

    Also,

    L = πRI/180

    = 719.27m

    Station PT = Stat PC + L

    10500 + 719.27

    =11219.27

    =112 + 19.27
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