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    Re: Almanac for Computers
    From: Karl Hahn
    Date: 1996 Aug 28, 17:36 EDT

    John H Henderson wrote:
    >
    > >                  Computing Position w/o a DR Position
    > >
    > > In principle, taking a sight reduces the possible positions you
    > > might be at in 3-dimensional space to a single plane in space.
    >
    > Taking one sight put you on a cone in space (which, when it intersects
    > the surface of the earth, gives you a circle.)
    >
    > > Taking two separate sights reduces it still further to the
    > > intersection of two planes, which is, of course, a line. You
    > > know, a priori, that your position is also on the surface of the
    > > earth.  So the intersection of that line with the surface of the
    > > earth yields exactly two points at which you might be.  Usually
    > > one of them is ridiculous, and so you choose the nonridiculous one.
    >
    > Two sights give you two intersecting cones, which give you
    > two lines at their intersection, and thus two points at the surface
    > of the earth, one of which you can rule out as unreasonable.
    >
    > :
    > :
    > > The equations for the two planes are:
    > >
    > >   x_1 * x  +  y_1 * y  +  z_1 * z   =   sin(theta_1)
    > >
    > > and...
    >
    > I could understand this plane approach if the planes were tangent to the
    > cones at your location, but there is no information from the other
    > sight in each equation, so I don't know how these equations express
    > the one correct plane.  Also, these planes are not tangent to the
    > "sight cone" since they do not pass through the center of the earth.
    > (Unless theta=0: set x=y=z=0). (Although any two planes that pass through
    > the observer's location would work.)
    >
    > I just need help understanding how this expression for these planes
    > is derived.
    The plane of the equation above is normal to the vector connecting
    the center of the earth to the celestial object sighted.  That
    plane's closest approach to the center of the earth is
    sin(theta) * radius_of_the_earth.  It intersects the earth's
    surface along a circle, and that circle is your LOP (so the plane
    is tangent iff you sighted the object at 90 degrees above the
    horizon).
    The derivation of the equation is this.  You have two vectors, one
    that connects the center of the earth with the celestial object,
    and one that connects the center of the earth with your unknown
    position.  Let both vectors be of unit length.  Then their dot
    product is the cosine of the angle between them.  The cosine of
    that angle is, by simple geometry, the same as the sine of the
    angle you see between the object and the horizon.  Hope that helps.
    Take care,
    Karl
    <PRE>
    --
    |         (V)              |  "Tiger gotta hunt.  Bird gotta fly.
    |   (^    (`>              |   Man gotta sit and wonder why, why, why.
    |  ((\\__/ )               |   Tiger gotta sleep.  Bird gotta land.
    |  (\\<   )   der Nethahn  |   Man gotta tell himself he understand."
    |    \<  )                 |
    |     ( /                  |                Kurt Vonnegut Jr.
    |      |                   |
    |      ^  hahn{at}XXX.XXX/~hahn
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