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Determining the distance difference of towers of a bridge

  1. Sep 27, 2017 #1
    1. The problem statement, all variables and given/known data
    Huge suspension bridges are build with the earth's roundness in mind. The two towers are plumb line straight up and down and yet, because of their colossal size, they are a bit further apart at their tops than they are at their base. So, how can we calculate what this difference would be?

    Here is the input data:

    If we know the earth's radius; the location of the base of the towers above sea level; The distance (from the center of each base of the tower) between the bases; and the height of the tower, how would we calculate the distance from the tops of the towers.

    2. Relevant equations

    I have seen a similar question answered once using something called "the law of cosines" where, if you know an angle and the length of two vectors, you can calculate the distances between the two vectors? I hope that helps and gives us a clue.
    3. The attempt at a solution

    Here I am not taking a stab at solving this on my own. The first thing I noticed in trying to solve this is that the lines forming the towers of the bridge can be thought of forming two right triangles back to back:

    bridge%2001.jpg

    Now, this is going way back for me -- I am talking decades -- since I last used this stuff. But, as I recall, Sine of an angle equals the Opposite side over the Hypotenuse. So, since we know what the Hypotenuse is. The angle is something we can figure out by taking the amount of the circle is taken up by the vectors.

    bridge%2002.png
    (this shows those helpful square thingies that indicate a right angle)

    Now, I am going to actually put the mathematics to use with an actual example: the Golden Gate Bridge.

    In order to get this as precise as possible, I am not going to google what the radius of the earth is and then, separately, google what the circumference of the earth is. If I do this, I am going to have messed up data because both numbers are going to be rounded. So in in order to get more precise data, I am going to use a given radius for the earth and then calculate the circumference from that data.

    radius = 6371000

    We have to use corresponding radius and circumference in this equation in order for the results to be accurate since we are dealing with very small changes of measurements

    Circumference = 2 π r
    r is the radius of the circle
    For π, I think this degree of precision will be enough 3.1415926535897932
    radius is 6371000
    So Circumference is 40030173.59201743

    Now to get to work on a real life example. Take the golden gate bridge into consideration

    Height of the tower
    Height of tower above water: 746 ft = 227 m
    the distance between the support is 1,280 m
    since, on that small scale in comparison to the globe, there is little curvature, I am just going to use 1280 meters as a percentage of the circumference of the globe
    40.075 million meters
    40075000
    So:
    1280 x
    -------- = -------
    40030173.59201743 360
    So:
    40030173.59201743 x = 460800
    So:
    460800
    x = -----------
    40030173.59201743

    0.0115113165557666 degrees
    That is the whole angle. We have to split it in half now to get our angle for each of our right angle triangles:
    0.0057556582778833 degrees
    so, now for our Hypotenuse. Our hypotenuse is the radius of the earth plus the height of the tower (6,371,000 meters plus 227) = 6371227
    That is our H in "Sine 0.0057556582778833 = O/h"
    With X being the unknown value (also known as "O" -- not zero, by the way -- it is the oposite side, the half distance between the tops of the towers ):

    so
    (0.000100342822) * 6371227 = (half the distance between the tops of the towers)
    640.0228
    doubled is
    1,280.0456

    So the difference in size between the top of the towers from the base is uh, 4 millimeters. In reality, it is a bit more. Where did I go wrong?
     
  2. jcsd
  3. Sep 27, 2017 #2

    Nidum

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    Science Advisor
    Gold Member

    This problem can be solved in one line . Think about similar triangles and a scaling factor .
     
  4. Sep 27, 2017 #3
    I think I am going to try some C++ code and use "long double" variables
     
  5. Sep 27, 2017 #4

    Nidum

    User Avatar
    Science Advisor
    Gold Member

    The one line calculation gives 45.6 mm . You got the right answer actually - you just slipped a place in reading the . 0456

    All you essentially needed to solve the problem though was the scaling factor : Earths radius plus the column height / Earths radius
     
  6. Sep 28, 2017 #5
    I solved this.

    And I wrote some software in C++ to verify it.

    I will share it here later

    :)
     
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