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Math Help - Calculating total height of object (above horizon plus below horizon)

  1. #1
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    Calculating total height of object (above horizon plus below horizon)

    Hi all,

    Hoping that someone can come up with a simple explanation or solution to my problem ... yes, I have Googled extensively and haven't come across anything that helps.

    Basically, I'm looking for a way to calculate the TOTAL height of an object that includes the height component that can be seen above your personal horizon PLUS the height component that cannot be seen because it lies below your personal horizon.

    Picture a very high tower that is located 100's of kilometres away. Naturally the base of the tower and a percentage of it's total height will not be visible to you because of the Earth's curvature. Obviously it's trivial to measure the height component that you can see above the horizon ... but how do you calculate the height component that lies below the horizon ?

    Thanks in advance to all those who may be able to shed some light on this.
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  2. #2
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    Tower below the visible horizon

    Hello stevek
    Quote Originally Posted by stevek View Post
    Hi all,

    Hoping that someone can come up with a simple explanation or solution to my problem ... yes, I have Googled extensively and haven't come across anything that helps.

    Basically, I'm looking for a way to calculate the TOTAL height of an object that includes the height component that can be seen above your personal horizon PLUS the height component that cannot be seen because it lies below your personal horizon.

    Picture a very high tower that is located 100's of kilometres away. Naturally the base of the tower and a percentage of it's total height will not be visible to you because of the Earth's curvature. Obviously it's trivial to measure the height component that you can see above the horizon ... but how do you calculate the height component that lies below the horizon ?

    Thanks in advance to all those who may be able to shed some light on this.
    In the attached diagram:
    O is the centre of the earth, whose radius is r

    PS = h_1 = height of observer above surface of earth

    QT = h_2 = height of part of tower not visible below the horizon. It is h_2 that we need to calculate.

    R is a point on the horizon visible from P; therefore PRQ is a tangent to the circle at R, and \angle PRO = \angle QRO = 90^o

    \theta_1 and \theta_2 are the angles (in radians) subtended at O by the arcs SR, RT; therefore the arc lengths SR and RT are r\theta_1 and r\theta_2 respectively. If s is the total length of the arc ST, then:
    s = r\theta_1+r\theta_2

    \Rightarrow \theta_2 = \frac{s}{r}-\theta_1
    ...(1)
    Now I am assuming that we are given
    r (the radius of the earth),

    h_1 (the height of the observer) and

    s (the distance measured around the surface of the earth from the foot of the observer to the foot of the tower).
    Therefore, in \triangle PRO:
    \cos\theta_1 = \frac{r}{r+h_1}

    \Rightarrow\theta_1 = \arccos\left(\frac{r}{r+h_1}\right)...(2)
    In \triangle QRO:
    \sec\theta_2 = \frac{r+h_2}{r}

    \Rightarrow h_2 = r(\sec\theta_2-1)
    This, then, together with equations (1) and (2) will give you the height of the part of the tower below the visible horizon.

    I've done a calculation on a spreadsheet (see second attachment) given:
    r = 6378000 (radius of earth in m)

    s = 100000 (distance 100 km away)

    h_1 = 50 (height of observer in m)
    which gives
    h_2\approx 438 m.
    Seems OK to me.

    Grandad
    Attached Thumbnails Attached Thumbnails Calculating total height of object (above horizon plus below horizon)-untitled.jpg   Calculating total height of object (above horizon plus below horizon)-untitled2.jpg  
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  3. #3
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    Grandad,
    Many thanks to you for that very clear and usefull explanation ... it's exactly what i was after. Amazingly enough, even though I struggled to come up with a solution myself, after following your explanation it immediately became one of those "of course ..." moments when it all fell into place for me :-)

    Thanks again ...
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  4. #4
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    It's the difference in perspective from space,
    compared to standing on the surface itself!
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