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    Re: Clarification of Question regarding LAN
    From: Joel Jacobs
    Date: 2004 Aug 4, 18:02 -0400

    Hello Jim,
    I had a distraction and came back to your message and now I'm more confused
    than I was.
    I thought Chuck was talking about using equal altitudes to calculate LON. If
    that's correct, then after halving the time, isn't it a simple time to arc
    conversion to get the point at which the sun crossed the ship's meridian.
    All you do is plot a vertical line at this point on a previously determined
    LAT which was the middle of the three sight's Chuck said to take.
    There is no E or W prime vertical involved, no azimuths that are N or S of
    the ship's position. Yes, there can be some adjustments to make, but the
    aren't key to the procedure.
    So I ask you where did I go wrong in understanding what Chuck was offering?
    Joel Jacobs
    ----- Original Message -----
    From: "Jim Thompson" 
    Sent: Wednesday, August 04, 2004 5:14 PM
    Subject: Re: Clarification of Question regarding LAN
    > Jim Thompson
    > jim2{at}jimthompson.net
    > www.jimthompson.net
    > Outgoing mail scanned by Norton Antivirus
    > -----------------------------------------
    > > -----Original Message-----
    > > From: Navigation Mailing List
    > > [mailto:NAVIGATION-L{at}LISTSERV.WEBKAHUNA.COM]On Behalf Of Robert Gainer
    > >
    > > Chuck said,
    > > >The more traditional way of determining longitude was
    > > >to use a time sight at the time the sun crosses the
    > > >Prime Vertical (i.e., the time at which the sun is due
    > > >east or due west of you).  This procedure is described
    > > >in Bowditch and elsewhere.  It requires that you know
    > > >your latitude, which you can get from a noon sight or
    > > >from an observation of Polaris.
    > >
    > > Chuck,
    > > I don't understand how that will work. The magnetic variation and the
    > > latitude must be problems in that method. If you are at 23 degrees north
    > > latitude or greater the sun is never due east or west.  If you do not
    > > the magnetite variation with some degree of accuracy wont that have a
    > > large effect on the method? Is this practical at all?
    > > All the best,
    > > Robert Gainer
    > Robert, I really got stuck on that too when I was learning this stuff last
    > year.  The key is to realize that the Prime Vertical Circle is not in the
    > terrestrial coordinate system.  Let's see if my explanation stands the
    > of this posting:
    > See the text and diagrams at
    > zonCoordinates
    > for a discussion of the Horizon Coordinate System, and
    > ed
    > for an explanation of how the various coordinate systems are linked
    > the PRINCIPAL vertical circle (not PRIME).
    > The intersection of the horizon coordinate system's PRINCIPAL vertical
    > circle with the celestial (and so terrestrial) equator defines the north
    > cardinal point in the horizon coordinate system.  If you then go 90
    > clockwise from that point around the horizon, you reach the horizon
    > coordinate system's West cardinal point.  See also Figure 1527a in
    > The vertical circle in the horizon coordinate system that goes through
    > point is the PRIME vertical circle.
    > Now here's the tricky part, requiring a lot of reflection until it comes
    > clear in the mind: When a body is on the PRIME vertical circle relative to
    > your position, it is by definition also exactly west of you in the
    > terrestrial coordinate system.  Bowditch' definition of the Prime Vertical
    > Circle includes this statement: "The intersections of the prime vertical
    > circle with the horizon define the east and west points of the horizon.".
    > For the practical application referred to by Chuck, see Figure 3 on
    > http://jimthompson.net/boating/CelestialNav/NoonSunSight.htm
    > titled, "Figure 3. Sun sight when the sun is on its prime vertical".
    > Jim

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