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I'm sorry, Russell, I thought I'd sent the following message but it
doesn't show up in the mailing list.
In addition, the reason to draw a LOP perpendicular to the azimuth of
the sun at the time of obs is that it is an approximation to the
altitude circle that you're supposed to be on. It's just the same as
in the case of a time sight; the difference is in the calculation.


Here is the missing message again:
-------------------------------------------------------------------

>   You said...
>   Surely, the latitude that comes out of the
>   calculation is dependent on an assumed longitude, because the meridian
>   angle t is.
>   -----
>   I don't really see why the latitude is dependant on an assumed longitude.
>   Typically time of LAN would be calculated and the altitude measured. DR
>   Longitude would be used to anticipate the time of LAN .
>   The longitude would then be dependant on the time of the observation, then
>   using arc to time (or GHA = long for east observer, 360-LHA = long for west
>   observer), it could be worked out.
>   ---
I'd rather say that the time of observation depends on the longitude.
Reason: it's easier to wait until the sun is near the meridian, than
to adjust your position to the same effect. Further, the time
difference between the actual MerPass and the actual time of
observation translates (time-to-arc) into the meridian angle t,
subject to the condition that your own motion over the earth is
sufficiently slow. You need your assumed (DR) longitude to calculate
the time of MerPass, so an error in DR longitude has effect on the
angle t and thus on  the reduction to the meridian and thus on the
latitude derived.
>
>
>   you also wrote...
>   >
>   Please note that the latitude you get is the latitude at the time
>   when you took the sight, not at the time of merpass. The tables only
>   correct for the apparent motion of the body, not for the motion of the
>   observer.
>   ------
>   Yes, I certainly understand this, in fact that is my point, because what I
>   would do is to work out my latitude for the time of LAN. On a small boat I
>   may assume that the vessel's movement was negligible. If say, I had been 15
>   minutes later than LAN in observing and suppose that LAN was 1200 UT; then
>   using an ex-meridian sight (taken at 1215) I would calculate my latitude for
>   what it should have been at 1200 -- is this not the point of the ex-meridan
>   altitude??
Agreed. If you're doing 8 knots in N-S direction the error will
be only 2', sufficient in mid-ocean.

>   But then that directional offset of the LOP is what I (and I
>   think you) are puzzled about
>   Russell
>
I found this in my 1984 Bowdich Volume I, art. 2104:

  "If reasonable doubt exists regarding the longitude, the azimuth of
  the body at the time of observation should be determined, and the
  line of position drawn perpendicular to it (through the point
  defined by the 'observed' latitude and the assumed longitude),
  rather than as a latitude line. There are alternative methods available."

You calculate the azimuth for the assumed longitude. The whole
procedure is then just slightly simpler than the usual HO249 method
that I prefer to use. Personally I want to get as much practice as
possible using one method that serves all so as to reduce the risk of
mistakes. I even hardly ever take a meridian altitude because I don't
like waiting for the sun to culminate, loosing approx. the same time
gained from the shorter calculation. Besides, the best place for the
sextant is below deck in its box.

Regards, _Steven.

   

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