|
|
|
|
|||||
|
||||||
From: Joel Jacobs (no email)
Date: Wed Aug 04 2004 - 18:02:28 EDT
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" <>
To: <>
Sent: Wednesday, August 04, 2004 5:14 PM
Subject: Re: Clarification of Question regarding LAN
> Jim Thompson
>
> www.jimthompson.net
> Outgoing mail scanned by Norton Antivirus
> -----------------------------------------
>
> > -----Original Message-----
> > From: Navigation Mailing List
> > [mailto:]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
know
> > the magnetite variation with some degree of accuracy wont that have a
very
> > 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
test
> of this posting:
>
> See the text and diagrams at
>
http://jimthompson.net/boating/CelestialNav/CelestNotes/Coordinates.htm#Hori
> zonCoordinates
> for a discussion of the Horizon Coordinate System, and
>
http://jimthompson.net/boating/CelestialNav/CelestNotes/Coordinates.htm#Link
> ed
> for an explanation of how the various coordinate systems are linked
through
> 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
degrees
> clockwise from that point around the horizon, you reach the horizon
> coordinate system's West cardinal point. See also Figure 1527a in
Bowditch.
> The vertical circle in the horizon coordinate system that goes through
that
> 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
|