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There seems to be continuing interest in the question of directly computing
a position, from the intersection of two position circles, without any
chartwork or prior estimations.

In navlist 3438, John Karl wrote-

 "My book presents a direct calculation (no
iterations) at the bottom of page 77.  It uses the same two equations
of the St. Hilaire method, five times.  Whereas a two-body St. Hilaire
fix uses four of these equations.  So with just one more equation, of
the same type, we eliminate the straight-line approximation, using no
iterations."

and I responded, in navlist 3443-

"That is indeed of great interest. It's at the bottom of page 78, by the
way,
not 77. It takes the simultaneous altitudes of two bodies, together with the
decs and LHAs of those bodies, and deduces the position (actually, two
alternative positions) of the observer to satisfy those requirements. It
requires no estimations, and no chartwork. I haven't come across it before,
and wonder whether it's an entirely new piece of work, from John. Certainly
very different from any St. Hilaire procedure.

But I wonder whether there are shortcomings that might show up on closer
investigation. A few questions come to mind-

How does that method handle a situation in which there's a distance-run, or
just an interval, between two sights?
As equations 7.5b, 7.5c, and 7.5e use an arc-cos, how well do they manage
when that angle nears zero degrees, and its cos is changing hardly at all?
Are there problems of putting angles into the right quadrant?"

===============

Andes Ruiz has offered a procedure, and d walden has provided a spreadsheet
using algorithms from a van Allen paper. That spreadsheet seems to get
angles using an arctan2 function. That's is a powerful way to derive angles
from two components, without ambiguity or regions of insensitivity , usually
much better in those respects than using arc sine or arc cos.

I have provided a routine to work on a pocket calculator, with an
explanation of how it works, which has finally got across, as a .doc
attachment. And there's another paper, that nobody has mentioned yet, by
George Bennett, with the unpromising title "General Conventions and
Solutions - their use in Celestial Navigation" in Navigation, vol 26 no. 4,
Winter 1979-80. This provides another set of equations to get the two
crossings of a pair of position circles (using arc-tan). For all I know,
there may be many more such ways of doing the job.

But let's go back to the set of equations that John Karl provides in his new
book "Celestial navigation in the GPS age", referred to above. You will need
that book at hand to follow what comes next (and it's worth acquiring). They
are equations 7.5a to 7.5f. Equations 7.5a to 7.5d take you to the latitude
of the crossing; indeed, the two latitudes, that you have to choose between,
for the two crossings. No analysis will choose for you which of those two
you are at; you have to do that for yourself, and it's hardly ever a problem
to do so. And until that point, the thing appears to have been done by John
Karl in a rather succinct way, much more elegantly that shown in the
attachment to my earlier email.

But let's leave that choice aside, for now, and settle on just one of those
two positions, defined by using the angle (A-B) in eq. 7.5d (the other would
come from using (A+B). Equation 7e now looks for the longitude which
corresponds to that latitude and one of the two circles. Unfortunately, any
such line of latitude crosses a circle at two points (or not at all), so
there are two solutions to equation 7.5e, which correspond to equal positive
and negative values for LHA1, the result of finding it from its cosine. And
I don't see a simple way for the mathematics to discover which of those two
longitudes is the correct one, the one associated with our crossing
position. It corresponds to deciding whether that position is East or West
of the meridian of GP1, which is not obvious, especially when GP1 is
anywhere near North or South of the observer. Those two values for LHA1 give
(in 7.5f) rise to two different longitudes for our chosen circle-crossing,
to go with our known latitude.  That's an extra degree of ambiguity, which
it is possible to resolve, but John's equations don't do it. That process
forms a large, and rather clumsy part, of my attachment on intersecting
circles. I wonder if it can be done a bit more elegantly than that.

Next, we need to look at the other alternative latitude, obtained from (A+B)
in 7.5d, and we find again another ambiguity in the resulting long, which
also needs resolving in the same way.

George.

contact George Huxtable at george@huxtable.u-net.com
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.





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