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Re: Distance off with Chicago buildings

From: Bill (no email)
Date: Thu Oct 06 2005 - 23:43:30 EDT

  • Next message: Bill: "Re: Side error etc (Re: Beginner)"

    > So instead of using
    > the altitudes directly, use their difference: 8.7 minutes of arc.

    Ah, now I see. Very clever. If of you are 24 sm from the Hancock, and on
    the beach, you are either in Gary Indiana, or Highland Park, IL. The
    Michigan City cooling tower would not be visible from Highland park, so it
    must be Gary, nominally N 41d 37.7', W 87d 20.0'. Given a horizon 4-5
    miles away, height of eye would be approx. 14 feet.

    It looks like you're walking the beach in Gary Indiana. At a steel mill
    perhaps? Interesting choice for a beach stroll. <G>

    This is where I become deeply confused. Where does the 10.7' figure for the
    361' cooling tower come from? The above location is approx. 23 sm from the
    Michigan City cooling tower. (Reference NOAA Chart 14905.) Therefore the
    land base and shore would be below your horizon at 14 ft height of eye. If
    I use Bowditch table 15, "vertical angle between top of object and sea
    horizon," you would be nominally 13 nm/15 sm away from the cooling tower
    with an angle of 11'. Not enough. If use plane trig with the assumption
    one could see the base/waterline, it works out within a < 1 mile.

    I have not checked out the separation angle yet, as I am very iffy about my
    building lat/lon (derived from Street Atlas). What have I missed?

    Bill


  • Next message: Bill: "Re: Side error etc (Re: Beginner)"



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