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Gary LaPook wrote:
> And I think that I remember this correctly, that the geodetic gratical used 
for map making places the landmarks where they would be placed by celestial 
navigation.

Actually, there's a difference between geodetic coordinates (the ones
shown on a chart or GPS receiver) and the astronomic coordinates derived
from celestial observations. I will later show that at one California
city, picked at random, the discrepancy is a quarter mile!

The reason is Earth's non-uniform gravity field. Of course this
phenomenon was unknown long ago. But as geodesy progressed, it became
necessary to express latitude and longitude with respect to a
mathematically perfect surface (an ellipsoid) instead of an imaginary
mean sea level surface (the geoid).

The ellipsoid's equatorial radius and its flattening factor (about 1
part in 300) are chosen to approximate the shape of the geoid. When you
specify the ellipsoid's position with respect to a set of points on the
surface of Earth, you have a geodetic datum.

Historically, multiple datums were created to fit the geoid well in
various parts of the world. I have no idea what the "local datum"
mentioned on the chart might have been. WGS84 is an attempt at a
worldwide best fit datum. Because it's a global compromise, the vertical
separation between its ellipsoid and the geoid is 100 meters in some places.

Celestial observations are independent of any datum since the instuments
are oriented to gravity, not an ellipsoid. But the astronomic
coordinates obtained thereby should be close to the geodetic coordinates
of a suitable datum. The discrepancy is due to the deflection of the
vertical, given by angles ξ (xi) and η (eta). The reference direction is
the normal (perpendicular) to the datum's ellipsoid. If a plumb line
intersects the celestial sphere north (east) of the ellipsoid normal, xi
(eta) is positive.

Another way to look at it: if astronomic latitude (longitude) is north
(east) of the geodetic coordinate, xi (eta) is positive by that difference.

In U.S. territory, a National Geodetic Survey online calculator can give
these angles. Let's look at Burbank, California, at N34 12 W118 22. The
calculator says xi = -11.07 and eta = -9.66 seconds. These fairly large
value are probably due to the mountains and sea close by.
http://www.ngs.noaa.gov/cgi-bin/GEOID_STUFF/deflec12A_prompt.prl

I haven't found an easy way to do that worldwide. However, there is a
utility to compute the height of the geoid with respect to the WGS84
ellipsoid:
http://earth-info.nga.mil/GandG/wgs84/gravitymod/egm96/intpt.html

By computing the slope of the geoid with respect to the ellipsoid, we
can estimate the deflection of the vertical.

"The coordinates found on the most recent chart now show it also at 0°
48' 00" north, 176° 37' west."

Get the geoid height (meters) at those coordinates, and one minute in
each cardinal direction:

----- 19.64 -----
19.68 19.66 19.63
----- 19.68 -----

Assume one minute of latitude or longitude is 1850 meters. Then the
extreme N-S and E-W points are 3700 m apart. Xi = arc tan ((19.68 -
19.64) / 3700) = 2″. Similarly, eta = 3″. A celestial fix would be that
much north and east, respectively, of the WGS 84 coordinates.
Essentially, there would be no discrepancy.

That method for estimating deflection of the vertical is not highly
accurate, as will be evident if you try it at Burbank. Part of the
problem is that the gravity field in that area has complex structure,
with significant changes in xi and eta even one minute away. However,
the approximation comes reasonably close to the correct values.

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