NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: No Two-Body Fix Problems
From: Peter Hakel
Date: 2009 Nov 6, 11:02 -0800
In general there are two latitudes, one from A+B and the other from A-B. Both correspond to genuine solutions.
For either latitude, there are two LHA1 (and hence longitude) possibilities, hence the four candidates. This was the original "caveat" that started the discussion thread. Each of the two latitudes has one candidate LHA1 that matches the Ho2 (within some round-off), while the other one I called "spurious." That is how we get our two LOP intersections, which I call "genuine" solutions. The spurious ones are "solutions" for the GP1-Pole-Ship triangle only, not for the whole problem.
If both bodies are on the same meridian, then our four candidates degenerate into two, and both are genuine solutions. Imagine gradually bringing the two circles (with fixed radii) onto the same meridian. The two parallels of latitude merge into one. The spurious "solution" for A+B overlaps with the genuine solution for A-B, and vice versa. Here my spreadsheet threw me another numerical curveball. All four candidates now yield the correct Ho2 but with unpredictable differences due to round off. As a result I saw some cases in which the same longitude was chosen for both the A+B and A-B branch of the code - thus returning one of the genuine solutions twice and throwing out the other. I fixed this by detecting the "same meridian" condition and flipping one of the longitudes to the other side of that meridian.
Peter
From: George Huxtable <george@hux.me.uk>
To: navlist@fer3.com
Sent: Fri, November 6, 2009 9:25:57 AM
Subject: [NavList 10473] Re: No Two-Body Fix Problems
[first part deleted by PH]
John is correct in maintaining that there's no actual "ambiguity" about
this, that needs to be resolved. Only one of the two alternative solutions
for arc-cos LHA1 can actually apply, but to discover which, the
corresponding value of (A-B) or (A+B), whichever is the chosen one, has to
be inspected.
John's equations may provide what looks like a perfect mathematical
solution. But to get the right answer, it seems to me that a real-life
navigator needs a bit more.
Or am I misunderstanding the matter, rather seriously? Could be...
George.
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From: Peter Hakel
Date: 2009 Nov 6, 11:02 -0800
George, in response to the second part of your posting, I hereby repeat a message that I had sent to John off-the-list.
It seems to me that by now we have addressed the questions quite thoroughly.
Peter Hakel
========================================================================
John,
It all makes perfect sense, I just wasn't precise enough in the wording of my previous message.
It seems to me that by now we have addressed the questions quite thoroughly.
Peter Hakel
========================================================================
John,
It all makes perfect sense, I just wasn't precise enough in the wording of my previous message.
In general there are two latitudes, one from A+B and the other from A-B. Both correspond to genuine solutions.
For either latitude, there are two LHA1 (and hence longitude) possibilities, hence the four candidates. This was the original "caveat" that started the discussion thread. Each of the two latitudes has one candidate LHA1 that matches the Ho2 (within some round-off), while the other one I called "spurious." That is how we get our two LOP intersections, which I call "genuine" solutions. The spurious ones are "solutions" for the GP1-Pole-Ship triangle only, not for the whole problem.
If both bodies are on the same meridian, then our four candidates degenerate into two, and both are genuine solutions. Imagine gradually bringing the two circles (with fixed radii) onto the same meridian. The two parallels of latitude merge into one. The spurious "solution" for A+B overlaps with the genuine solution for A-B, and vice versa. Here my spreadsheet threw me another numerical curveball. All four candidates now yield the correct Ho2 but with unpredictable differences due to round off. As a result I saw some cases in which the same longitude was chosen for both the A+B and A-B branch of the code - thus returning one of the genuine solutions twice and throwing out the other. I fixed this by detecting the "same meridian" condition and flipping one of the longitudes to the other side of that meridian.
Peter
From: George Huxtable <george@hux.me.uk>
To: navlist@fer3.com
Sent: Fri, November 6, 2009 9:25:57 AM
Subject: [NavList 10473] Re: No Two-Body Fix Problems
[first part deleted by PH]
John is correct in maintaining that there's no actual "ambiguity" about
this, that needs to be resolved. Only one of the two alternative solutions
for arc-cos LHA1 can actually apply, but to discover which, the
corresponding value of (A-B) or (A+B), whichever is the chosen one, has to
be inspected.
John's equations may provide what looks like a perfect mathematical
solution. But to get the right answer, it seems to me that a real-life
navigator needs a bit more.
Or am I misunderstanding the matter, rather seriously? Could be...
George.
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NavList message boards: www.fer3.com/arc
Or post by email to: NavList@fer3.com
To , email NavList+@fer3.com
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