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    Re: Almanac for Computers
    From: Karl Hahn
    Date: 1996 Aug 28, 09:56 EDT

    Jacky Wong wrote:
    >         Your work should be very accurate. I had compare your example with my
    > JavaScript program, which can be find at
    > http://www.speednet.net/~himarj/navscript/sunsight.htm
    I noticed that Mr. Wong's program requires a DR position in order
    to do sight-reduction.  When a computer is available to do the
    grunt work for you, a programmer can take advantage of the computer's
    talent for doing grunt work in enormous volumes.  This includes
    things you would never want to do by hand at sea, such as solving
    matrices and quadratic equations, and taking trig and inverse
    trig functions.
    What I am leading up to here is that there is an algebraic solution
    to reducing a pair of sights that does not need a DR position.  For
    the mathematically inclined, I will outline it here.
                     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 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.
    In more detail: If you imagine the sky as being the surface of a
    unit sphere, convert the GHA and declination of the object(s) sighted
    to Cartesian coordinates.  So for each object (1 and 2) you would
    have a point:
      (x_1, y_1, z_1)  and  (x_2, y_2, z_2)
    You also have the corrected angles that you read from your sextant,
    theta_1 and theta_2.  Take their sines.
    The equations for the two planes are:
      x_1 * x  +  y_1 * y  +  z_1 * z   =   sin(theta_1)
      x_2 * x  +  y_2 * y  +  z_2 * z   =   sin(theta_2)
    Now solve those two equations as far as you can using Gaussian
    elimination.  You (or actually, your computer program) should be
    able to get linear expressions for x in terms of z and y in terms
    of z.  Your final equation is the equation of the surface of a sphere:
       x^2  +  y^2  +  z^2   =   r^2
    Assume r to be 1.  Substitute into this equation from the other
    two, and you get a quadratic, from which you can get two solutions
    using the quadratic formula.  From those, you can establish two
    Cartesian points that lie on the sphere at which you might be.
    Convert those points back to spherical coordinates (ie lat & lon)
    using the standard formula, and you're done.
    I don't know how many math-nerds like myself read this group,
    but of those who are out there, I hope you found this interesting.
    Take care,
    |         (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@XXX.XXX/~hahn
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