A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Peter Hakel
Date: 2009 Oct 1, 12:34 -0700
In my method I needed the latitude to translate the E/W miles into change in longitude (and change in moment of meridian transit) in the E/W-DR portion of the algorithm. I chose the latitude calculated from the fit. This seems to be adequate for examples given by Jeremy and Antoine. As you move faster, you cover larger distances, the curvature of the Earth becomes even more visible and this approximation of picking a fixed latitude will progressively break down. In Gary's example, however, the course is along a meridian, so this ambiguity should not be an issue for the moment of transit determination - though at some speed the geometry will become more complicated.
Perhaps Gary will tell us how he constructed this example and what therefore the "exact" solution is. My longitude result is in the neighborhood of his DR value. If one changes the speed, the Ho's themselves will change accordingly. However, the moment of meridian transit should come out independent of speed, as long as the vessel moves along the same meridian.
I do get the latitude from an Hmax; but this Hmax is calculated from a curve fit through N/S-DR/declination-change prorated altitudes (advanced to a common parallel of latitude), not directly from the given Ho's.
From: Andres Ruiz <firstname.lastname@example.org>
Sent: Thursday, October 1, 2009 2:53:01 AM
Subject: [NavList 10014] Re: Time of meridian passage accuracy
if I have not done anything wrong, the possible difference could come from this:
Jim Wilson correction for time depends on latitude, B:
Dt = 3600.0*180.0/PI * (dBdt - dDecdt) * (TAN( B ) - TAN( Dec )) / SQ( dLHAdt )
In this case, high N/S component of the velocity, the difference between the time of maximum altitude and LAN is very important, and also the associated altitude.
If you take estimation for latitude using the Hmax, de solution becomes erroneous.
[mailto:email@example.com] En nombre
de P H
Enviado el: jueves, 01 de octubre de 2009 8:20
Asunto: [NavList 10012] Re: Time of meridian passage accuracy
UT of LAN: 17:17:35
Latitude at LAN: N 22 deg 40.8'
Longitude at LAN: W 79 deg 42.4'
Sent: Wednesday, September 30, 2009 9:19:19 PM
Subject: [NavList 10009] Re: Time of meridian passage accuracy
Jim Wilson wrote on Sept.
"I compiled a table in my Navigation paper showing differences between maximum alititude and meridian passage ranging from 10 seconds to 37 minutes. My trial which led me to write the paper had a difference of a minute and a half, and that was at a moderate latitude and sailboat speeds."
I am curious how well it works at higher speeds. I have created a data
set and I would like you to evaluate it and tell me the time of meridian
passage, the longitude and the latitude at meridian passage.
This hypothetical takes place on December 22, 2009 and the sights are taken at five minute intervals on the exact five minute mark and all sextant corrections have been applied so the values stated are the observed altitudes, Ho. They cover a two hour period from 1700 Z to 1900 Z.
17:00:00 42º 14'
:05 42º 44'
:10 43º 13'
:15 43º 39'
:20 44º 04'
:25 44º 26'
:30 44º 47'
:35 45º 06'
:40 45º 22'
:45 45º 36'
:50 45º 49'
:55 45º 59'
18:00:00 46º 06'
:05 46º 12'
:10 46º 15'
:15 46º 16'
:20 46º 14'
:25 46º 10'
:30 46º 04'
:35 45º 56'
:40 45º 45'
:45 45º 33'
:50 45º 18'
:55 45º 01'
19:00:00 44º 41'
The D.R. at 1700 Z is 24º 25' N, 79º 50' W.
The course is 180º True.
The ground speed is 300 knots.
James N Wilson wrote:
You state that "the effect is very small as to be of no practical concern to practical navigators."
I compiled a table in my Navigation paper showing differences between maximum alititude and meridian passage ranging from 10 seconds to 37 minutes. My trial which led me to write the paper had a difference of a minute and a half, and that was at a moderate latitude and sailboat speeds.
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