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Math Help - applying math to surveying

  1. #1
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    Apr 2009
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    applying math to surveying

    Hi there,

    Dont know if have posted this question to the rite forum.

    Have attached a question on solving of vertical curves.
    can any help or give me a start of where to begin solving the attched question.

    many thanks
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  2. #2
    Super Member
    Joined
    Jan 2009
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    Quote Originally Posted by masiboy View Post
    Hi there,

    Dont know if have posted this question to the rite forum.

    Have attached a question on solving of vertical curves.
    can any help or give me a start of where to begin solving the attched question.

    many thanks
    Almost all vertical curves have equal forward and backward tangents.
    In some rare special cases the horizontal distance from G1 to the Vertical Intersection Point and from VIP to G2 will be different.

    The length of the VC is 160 m, so the length from G1 to VIP is 80 m.
    Station of G1 is 4+50.00
    Station of G2 is 6+10.00

    Typically, you would calculate elevation points ON THE TANGENTS, and then compute the vertical offsets to the curve.

    Elevation of VIP 621.00:
    G1 elevation: 621.00 - 80 * 0.04 = 617.80
    G2 elevation: 621.00 + 80 * -0.021 = 619.32

    The RATE OF CHANGE of the curve is uniform through the curve, but the offsets from the tangent are quadratic.

    Almost any surveying/civil engineering book will have elaborate explanations for handling vertical curves.

    If your book lacks such, you can always use GOOGLE.

    Vertical curves


    The following is from here:
    Vertical Curves
    One you have located the VPI, VPC, and VPT, you are ready to develop the shape of your curve. The equation that calculates the elevation at every point along an equal-tangent parabolic vertical curve is shown below.
    Y = VPCy + B*x + (A*x2)/(200*L)
    Where:
    Y = Elevation of the curve at a distance x from the VPC (ft)
    VPCy = Elevation of the VPC (ft)
    B = Slope of the approaching roadway, or the roadway that intersects the VPC
    A = The change in grade between the disjointed segments (From 2% to -2% would be a change of -4% or -4)
    L = Length of the curve (ft)
    x = Horizontal distance from the VPC (ft) (Varied from 0 to L for graphing.)
    At this point, you have everything that you need to develop the shape of a simple equal-tangent vertical curve. The procedure above will work for both sag and crest vertical curves.
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  3. #3
    Newbie
    Joined
    Apr 2009
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    Quote Originally Posted by aidan View Post
    Almost all vertical curves have equal forward and backward tangents.
    In some rare special cases the horizontal distance from G1 to the Vertical Intersection Point and from VIP to G2 will be different.

    The length of the VC is 160 m, so the length from G1 to VIP is 80 m.
    Station of G1 is 4+50.00
    Station of G2 is 6+10.00

    Typically, you would calculate elevation points ON THE TANGENTS, and then compute the vertical offsets to the curve.

    Elevation of VIP 621.00:
    G1 elevation: 621.00 - 80 * 0.04 = 617.80
    G2 elevation: 621.00 + 80 * -0.021 = 619.32

    The RATE OF CHANGE of the curve is uniform through the curve, but the offsets from the tangent are quadratic.

    Almost any surveying/civil engineering book will have elaborate explanations for handling vertical curves.

    If your book lacks such, you can always use GOOGLE.

    Vertical curves


    The following is from here:
    Vertical Curves
    One you have located the VPI, VPC, and VPT, you are ready to develop the shape of your curve. The equation that calculates the elevation at every point along an equal-tangent parabolic vertical curve is shown below.
    Y = VPCy + B*x + (A*x2)/(200*L)
    Where:
    Y = Elevation of the curve at a distance x from the VPC (ft)
    VPCy = Elevation of the VPC (ft)
    B = Slope of the approaching roadway, or the roadway that intersects the VPC
    A = The change in grade between the disjointed segments (From 2% to -2% would be a change of -4% or -4)
    L = Length of the curve (ft)
    x = Horizontal distance from the VPC (ft) (Varied from 0 to L for graphing.)
    At this point, you have everything that you need to develop the shape of a simple equal-tangent vertical curve. The procedure above will work for both sag and crest vertical curves.
    Hi aidan,

    thanks for the solution..

    it would be great if you explain on more how you got the values for:

    Station of G1 is 4+50.00
    Station of G2 is 6+10.00

    and how does the chainage of 5+ 30.00 in question fits in for G1 4+50 & G2 6+10.

    cheers
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