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Homework Help: Distance to the horizon

  1. May 28, 2006 #1
    My problem is to determine how far one can see to the horizon, depending on how far above the water the person is and to create a formula that determines the distance as a function of the height.

    To start with, the hypotenuse is the distance from the core to the persons eyes. (x+e). The cathetus being the distance from the core to the waterlevel (x) and the 2nd cathetus being the distance to the horizon (d).
    This is easily calculated with Pythagoras (x+e)^2 - (x)^2 = (d)^2.
    With the fact that a cathetus to a spheric object has an angle of 90degrees.

    How shall I proceed to create this formula?
     
  2. jcsd
  3. May 29, 2006 #2

    Tide

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    You're almost there. x is the radius of the Earth so just solve for d. Noting that [itex]a^2-b^2 = (a-b)(a+b)[/itex] may help.
     
  4. May 30, 2006 #3
    Okay, to put this formula into situations depending on the height above the water level in meters:
    (x+e)^2 - (x) = "the distance to the horizon" ^2
    When I keep my eyes just above the water:
    (6378150+0.02)^2 - (6378150)^2 = 255126.
    I can then theoretically see 505.1 meters untill I can only see the horizon.

    A person of 1.85 meters have his eyes at a height of 1.65 meters above the water. (Roughly)
    (6378150+1,65)^2 - (6378150)^2 = 21047898
    The person is capable of seeing 4587.8 meters until the horizon.

    A person in a lighthouse looking at the horizon from a height of 20 meters:
    (6378150+20)^2 - (6378150)^2 = 255107020,1
    The person is then looking at a distance of 15972 meters to the horizon.

    Are these numbers probable?
    How do you advice me to solve the problem to make a formula giving the distance as a function of e (the height from my eyes to the water) ?
     
  5. May 31, 2006 #4

    Tide

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    Alec,

    Those numbers look pretty good!
     
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