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B The curvature of the Earth

  1. Mar 6, 2017 #1
    I propose a calculation of the drop X in the horizon that an observer sees due to the curvature of the Earth surface the parameters are:

    h: the elevation of the observer from the surface of the Earth in km
    H: the length of the horizon at which the drop is perceived in km
    R: the radius of the spherical Earth ≈ 6371 km
    d: the distance from the observer to the horizon in km

    The following ( see figure 1) sets the layout for the calculation, the observer field of view intersects the globe in the circle (C).

    upload_2017-3-6_19-7-39.png
    Computing d:
    We have: (R+h)2=d2+R2

    d=√(2hR+h2)

    Computing e: e is the radius of (C)

    we have sin(b)=cos(a)⇒ e/R=√(1-(e/d)2)⇒e=Rd/√(R2+d2)

    Now switching to the plane of the circle (C) (see figure 2):

    upload_2017-3-6_19-6-59.png
    Computing the drop X(h,H):

    X=e-√(e2-H2/4)

    Now under the reasonable approximation that h<<R:

    X(h, H)=H2/2√(Rh)⇒ X(h, H)=0.00626×H2/√h (km)

    For example; if H=1 km and h=2 m (human height)⇒ X(1, 0.02)= 44.3 m .
     
  2. jcsd
  3. Mar 6, 2017 #2

    Drakkith

    User Avatar

    Staff: Mentor

    Hi Axion! Welcome to PF!

    That's an interesting calculation, but may I ask what the purpose of posting it here is? Just to share? Or did you have a question about it?
     
  4. Mar 6, 2017 #3
    Hi,
    Just to share may be someone will need it.
     
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