NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Two-body fix caveat
From: Peter Hakel
Date: 2009 Nov 4, 18:43 -0800
On the second point:
The three parameters of the quadratic fit are given in Meeus's "Astronomical Algorithms" (2nd edition) as Eqs. 4.5 and 4.6 on page 43. They can be derived as the determinantal solution of the 3x3 linear system arising from the least-squares minimization conditions with a quadratic fit. As such, the equations are perfectly correct in the pure world of mathamatics; in real life they have numerical stability issues due to round-off errors, as I have found out. I have been using Excel 2004 on a Mac.
Peter Hakel
From: George Huxtable <george@hux.me.uk>
To: navlist@fer3.com
Sent: Wed, November 4, 2009 3:48:10 PM
Subject: [NavList 10453] Re: Two-body fix caveat
I'm a bit puzzled by Peter Hakel's "illustration" in the two pages attached
to [10414], although the more-recent mathematical talk is quite beyond my
grasp.
The discussion seems to be about whether various mathematical techniques
produce a "fix". I would not describe the position circles in that example,
which are osculating rather than intersecting, as producing a "fix". Both
are tangential to a North-South line passing through 20º West. Although the
point on the equator at 20º West may be the only point that actually
satisfies the two altitudes, other positions, to its North and South, come
so infinitesimally close to meeting that requirement that the result is
really a "line-fix". It's an example of the degenerate situation, of two
bodies with the same azimuth, that navigators know to avoid. In reality,
they would choose another body, or wait a bit, in time.
In those circumstances, it's no surprise to me that difficulties arise in a
mathematical solution. A real navigator, working from a simple plot, would
realise what was going wrong immediately.
===================
On another matter altogether, Peter has experienced some problems in
quadratic fitting with Excel. So have I, using Excel 2000, and in doing the
same job, fitting a quadratic for longitude-around-noon. Excel will
cheerfully plot a quadratic fit to such a set of points, and apparently a
good fit too, and if asked to, will display the resulting coefficients, as
numbers printed on top of the fitted graph. And the only way I've found to
extract those coefficients, to use for further Excel calculation, is to read
them off that graph, and then type them back in again. Frustrating, but
acceptable. But, having done just that, and then used those numbers to
regenerate the quadratic curve once again, for checking, the resuting plot
can differ significantly from the quadratic that Excel had plotted before.
The reason is that in displaying those parameters, they are truncated by
rounding-off in such a way as that significant digits have been lost. I've
got round it by renumbering the divisions along the x-axis, so its numerical
values increase by 10, or perhaps 100, accordingly.
If Peter can offer any insight into how to get at those coefficients
directly, or how to alter their rounding-off, it would be helpful.
George.
contact George Huxtable, at george@hux.me.uk
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
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Or post by email to: NavList@fer3.com
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From: Peter Hakel
Date: 2009 Nov 4, 18:43 -0800
On the first point:
George, you are quite right about the impractical nature of this artificial example. Mathematics and geometry consist of idealized models, in which spheres are perfectly round, lines are infinitely thin, etc. It is useful and often necessary to look at singular cases to check how general or robust a method is, both in its fundamentals and its imperfect computerized implementation. I constructed a bare-bones degenerate case with osculating LOPs in order to better isolate the ambiguity in question (LHA1) for the purposes of illustration. I did not want to distract the debate from LHA1 by having another ambiguity (two separate "fixes" from intersecting LOPs) on top of all that. I could have done it (e.g., by having GHA1 = 15 instead of 10) as the problem in question is unrelated to LOPs crossing or just touching. I guess I should not have called the ship's position a "fix" in this case; the correct terminology is the intersection of the two sets ( LOP1 ∩ LOP2 ), which in this case contains only one point instead of the usually distinct two.
George, you are quite right about the impractical nature of this artificial example. Mathematics and geometry consist of idealized models, in which spheres are perfectly round, lines are infinitely thin, etc. It is useful and often necessary to look at singular cases to check how general or robust a method is, both in its fundamentals and its imperfect computerized implementation. I constructed a bare-bones degenerate case with osculating LOPs in order to better isolate the ambiguity in question (LHA1) for the purposes of illustration. I did not want to distract the debate from LHA1 by having another ambiguity (two separate "fixes" from intersecting LOPs) on top of all that. I could have done it (e.g., by having GHA1 = 15 instead of 10) as the problem in question is unrelated to LOPs crossing or just touching. I guess I should not have called the ship's position a "fix" in this case; the correct terminology is the intersection of the two sets ( LOP1 ∩ LOP2 ), which in this case contains only one point instead of the usually distinct two.
On the second point:
The three parameters of the quadratic fit are given in Meeus's "Astronomical Algorithms" (2nd edition) as Eqs. 4.5 and 4.6 on page 43. They can be derived as the determinantal solution of the 3x3 linear system arising from the least-squares minimization conditions with a quadratic fit. As such, the equations are perfectly correct in the pure world of mathamatics; in real life they have numerical stability issues due to round-off errors, as I have found out. I have been using Excel 2004 on a Mac.
Peter Hakel
From: George Huxtable <george@hux.me.uk>
To: navlist@fer3.com
Sent: Wed, November 4, 2009 3:48:10 PM
Subject: [NavList 10453] Re: Two-body fix caveat
I'm a bit puzzled by Peter Hakel's "illustration" in the two pages attached
to [10414], although the more-recent mathematical talk is quite beyond my
grasp.
The discussion seems to be about whether various mathematical techniques
produce a "fix". I would not describe the position circles in that example,
which are osculating rather than intersecting, as producing a "fix". Both
are tangential to a North-South line passing through 20º West. Although the
point on the equator at 20º West may be the only point that actually
satisfies the two altitudes, other positions, to its North and South, come
so infinitesimally close to meeting that requirement that the result is
really a "line-fix". It's an example of the degenerate situation, of two
bodies with the same azimuth, that navigators know to avoid. In reality,
they would choose another body, or wait a bit, in time.
In those circumstances, it's no surprise to me that difficulties arise in a
mathematical solution. A real navigator, working from a simple plot, would
realise what was going wrong immediately.
===================
On another matter altogether, Peter has experienced some problems in
quadratic fitting with Excel. So have I, using Excel 2000, and in doing the
same job, fitting a quadratic for longitude-around-noon. Excel will
cheerfully plot a quadratic fit to such a set of points, and apparently a
good fit too, and if asked to, will display the resulting coefficients, as
numbers printed on top of the fitted graph. And the only way I've found to
extract those coefficients, to use for further Excel calculation, is to read
them off that graph, and then type them back in again. Frustrating, but
acceptable. But, having done just that, and then used those numbers to
regenerate the quadratic curve once again, for checking, the resuting plot
can differ significantly from the quadratic that Excel had plotted before.
The reason is that in displaying those parameters, they are truncated by
rounding-off in such a way as that significant digits have been lost. I've
got round it by renumbering the divisions along the x-axis, so its numerical
values increase by 10, or perhaps 100, accordingly.
If Peter can offer any insight into how to get at those coefficients
directly, or how to alter their rounding-off, it would be helpful.
George.
contact George Huxtable, at george@hux.me.uk
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
--~--~---------~--~----~------------~-------~--~----~
NavList message boards: www.fer3.com/arc
Or post by email to: NavList@fer3.com
To , email NavList+@fer3.com
-~----------~----~----~----~------~----~------~--~---
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