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Subject: Noon by least squares.
From: Herbert Prinz (hprinz@XXX.XXX)
Date: Thu Apr 11 2002 - 02:22:02 EDT
What I don't like about Walter Guinon's procedure is that it assumes the DR
position as an absolute given and derives local noon as a function of it and some
observed altitudes. What is the application of this to the real world? In the
general case it forces an error into t (= local time) to compensate for any error
in assumed latitude. The beauty of the combined altitude method, however, is that,
correctly applied, it will give a simultaneous solution for local time AND
latitude.
If one really must have a least square fit, it still can be done. After all,
solving for latitude and local time from two altitudes is the same problem as
solving for latitude and longitude from two altitudes, except that the last step,
i.e. converting local time to longitude via GMT is omitted. So we can use St.
Hilaire on the celestial sphere instead of on the globe of the Earth. The least
square algorithm is given in the explanatory pages of the N.A. Just replace all
GHAs (and your longitude) by SHAs.
Rod Deyo holds that "You can determine the circles of constant altitude for three
or more celestial bodies (assuming a spherical earth - something quite reasonable
for practical navigation) and find their best-fit intersection numerically. A
problem arises if you need to plot the circles of constant altitude on a Mercator
chart to find the intersection [...]. Then you really do want something like the
Marcq St. Hilarie intercept method [...]."
Passing over the mistake about having to assume a spherical earth, which George
Huxtable has already pointed out, it must be emphasized that St. Hilaire is not
just needed for plotting. It is vital to the best-fit solution. I would be curious
to know how Rod finds the best fit intersection of three circles, if not by first
linearizing the problem. This is exactly what the intercept method does. It enables
us to formulate the problem of the n-overdetermined fix as a system of n+2 linear
equations in 2 variables for the solution of which we have a well established
procedure. The penalty for the linear approximation is that the procedure is
iterative. Rod does not need to worry about changed azimuths or high altitudes: The
algorithm will converge.
In short, trying to somehow solve for a least square fit for noon and/or latitude
while bypassing St. Hilaire would be re-inventing the wheel or, worse, solving the
wrong problem.
Herbert Prinz
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