NavList:

   

Reply

Chuck

THANK YOU for your summary. As a novice in cel nav without an education
(that I can recall for any practical use) in higher mathematics, the
running-gun-battle-in-a-corner regarding averaging has been at times
difficult to follow.  It has been educational for which I thank all
involved.

I too think Alex's argument is as simple as taking several sights of one
body over a short period of time, averaging them, and using that average of
altitude, time (and related craft position if applicable) for plotting an
LOP.  Given the resolution of the system/operator, this is good enough. "In
theory, practice and theory are the same, in practice they are not the
same."--Yogi Berra

There are other methods of creating slope, fitting to pre-calculated slope,
averaging azimuths and intercepts before plotting etc. from a "run" of one
body.  This was at the heart of the discussion that Alex and I had off-list,
the averaging aspect being what Alex posted, resulting in the highly-rated
on-list debate.  Again I recommend David Burch's web site.

Lacking an understanding of mathematics on the plane that Alex and Herbert
dwell, I would not bet the farm on on my assumption of the their arguments
or your summary.

From my (limited) vantage point, Herbert has brought in many other variables
above and beyond Alex's original query, or for that matter, my question to
Herbert regarding worst-case scenarios.  One that lies there in the brush
like a baited trap is plotting the intercepts and LOPs.  Working from
memory, in an earlier posting Herbert spoke of the cocked hat and bisecting
the angles, and alluded to problems therein. Not averaging one run, rather
plotting. In a recent post, Herbert gave an example regarding averaging
where (by coincidence?) all the azimuths were within 180d of each other.  If
I understand Dutton, when all azimuths of observed bodies are within 180d
the fix may well be outside the cocked hat. Herbert's interjection
obfuscates Alex's original premise and fails to address the question on the
table--in my mind.

Last. As a neophyte:

Would someone, anyone, PLEASE define "Overdetermined Celestial Fix" in
layman's terms?

Thanks

Bill

> Alexandre Eremenko and Herbert Prinz have been
> discussing the averaging of celestial sights.  The
> following is an attempt to summarize:
>
> Consider the following hypothetical situation (the
> actual numbers here are made up):
>
> Body       Time        Altitude
>
> Spica      20-00-00    25d 10.5'
> Spica      20-01-00    25d 05.1'
>
> Arcturus   20-03-00    35d 30.0'
> Arcturus   20-04-00    35d 25.0'
> Arcturus   20-05-00    35d 20.0'
> Arcturus   20-06-00    35d 15.0'
> Arcturus   20-07-00    35d 10.0'
>
> Dubhe      20-09-00    42d 00.0'
>
> We have 2 observations of Spica, 5 of Arcturus, and 1
> of Dubhe, for a total of 8 observations.  How should
> we proceed in obtaining a fix?
>
> One obvious solution is to simply choose one
> observation of each body, then reduce and plot.  The
> fix would be plotted somewhere within the "cocked
> hat".
>
> Alex would argue that we could do better than that by
> averaging the two sights of Spica (treating the
> average time and the average altitude as a one sight),
> averaging the 5 sights of Arcturus, and taking the
> single sight of Dubhe. (The motivation for averaging
> is the expection that random errors will cancel out in
> the process of averaging.)
>
> If I understand Herbert's objection correctly, he
> would claim, "Now wait a minute.  You have 5 sights of
> Arcturus, but only 2 of Spica and 1 of Dubhe.  How can
> you give the same amount of weight to the single sight
> of Dubhe as you do to the 5 sights of Arcturus?  There
> is more information in the 5 sights of Arcturus than
> in the 1 sight of Dubhe, yet you are giving them the
> same weights."  (I hope Herbert will forgive me for
> putting words in his mouth.  If I have misunderstood
> him, I apologize in advance.)
>
> In theory you can only justify giving equal weights to
> each set of observations if each set contains an equal
> number of sights.  In practice, you will probably not
> go far wrong anyway.  Nor will you go far wrong in
> practice by selecting one sight from each set ("run")
> of sights.
>
> I happen to have a copy of the paper, "On the
> Overdetermined Celestial Fix", by Thomas and Frederic
> Metcalf.  (Tom was once a contributor to this list.
> The last I heard, he was associated with the Institute
> of Astronomy, University of Hawaii, Honolulu.) I also
> have a copy of a follow-on paper, "An Extension to the
> Overdetermined Celestial Fix" by Tom Metcalf (alone).
> I obtained my copies of these papers some years ago
> directly from Tom.
>
> At the risk of gross oversimplification, I present the
> following analogy:
>
> Suppose you have two equations in two unknowns, each
> representing a line in a plane.  Assuming the two
> lines are not parallel, the two lines intersect at a
> single point, and that point can be found by solving
> those two equations simultaneously. The solution can
> be by one of several methods, including Gaussian
> elimination, Gauss-Jordon elimination, or by the use
> of matrix algebra.
>
> Now suppose that we have 3 equations in two unknowns.
> We observed above that a unique solution is determined
> by two equations in two unknowns.  Now we have an
> "overdetermined" set of equations with no unique
> solution.  This is where the method of least squares
> enters in.  We find the point such that the squared
> distance between that point and each of the lines in
> minimized.  As Herbert pointed out, we can do this
> using matrix algebra.
>
> Now, let us return to the example of our sights of
> Spica, Arcturus, and Dubhe.  Herbert's argument is
> that we ought to compute 8 lines of position (2 for
> Spica, 5 for Arcturus, and 1 for Dubhe) and find that
> point which minimizes the sum of the squared distances
> between it and each of the 8 lines of position.  That
> would be solving an overdetermined celestial fix by
> the method of least squares.  With this method, each
> observation from each body is given equal weight.
>
> Herbert, have I stated your argument fairly?
>
> The Metcalf papers actually address a slightly
> different form of overdetermined celestial fix.
> Quoting from the earlier of the two papers, "... using
> these methods, an accurate fix can quite often be
> determined from 10 to 20 observations of a single
> celestial body spanning only 10 to 20 min of time."
> The algorithm uses a Lagrange multiplier and the
> simultaneous solution of a nonlinear equation for the
> Lagrange multiplier and a number of linear equations.
> The solution is iterative.  It's an interesting
> concept.  Tom tried it out with "real observations of
> the Moon taken on land with an inexpensive plastic
> sextant, and with a pan of vegetable oil serving as an
> artifical horizon."  He took 18 sights over
> approximately 35 minutes, and was able to confirm his
> known position within about 2.5 nautical miles. He
> programmed his algorithm on an HP-48SX calculator.
> The paper offers to make the code available to
> interested parties who send a stamped, self-addressed
> envelope.  I'm not sure whether there was any
> expiration date on the offer, as it appeared in
> "Navigation: Journal of The Institute of Navigation",
> Vol 38, No. 1, Spring 1991. The follow-on paper
> appeared in Vol. 39, No. 4 (Winter 1992-1993).
>
> Best regards to all,
>
> Chuck Taylor

   

Reply