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What an interesting discussion. I've followed it closely and saved all
the messages in a dedicated folder. For the first time I've had to
really think out how our spherical Earth assumptions affect celestial
navigation.

It's my belief that the almanac and sight reduction tables are equally
valid on both a sphere and an ellipsoid. Let's explore this with a
thought experiment.

Imagine the inside of the celestial sphere is marked with parallels
and meridians, so an observer on Earth can look up and see them. For
simplicity's sake, assume there's no Earth rotation.

Initially, let's pretend Earth is a perfect sphere and you are at 30N
90W. At this location set up a surveyor's transit, level it, and
adjust the horizontal circle to read zero when the telescope is
sighted on the celestial pole. By doing this you have aligned the
instrument's vertical axis to the zenith, which is the intersection of
the 30N and 90W lines on the celestial sphere. You've also brought the
horizontal circle's zero point into the plane of the 90W celestial
meridian.

If I name a specific parallel and meridian on the celestial sphere,
can you compute the altitude and azimuth required to direct the
transit's scope to that spot in the sky? Yes, of course. This is
basically the same problem we solve during a sight reduction. Like our
imaginary lines in the sky, the coordinates in the almanac are pure
spherical coordinates and aren't based on any datum. If we assume
Earth is spherical too, computing altitude and azimuth is
straightforward.

But what if the world is ellipsoidal? Doesn't that make the
computation much more difficult? No, because of the way latitude is
defined on an ellipsoid. Mapping and navigation use geodetic latitude,
which is the angle between a normal to the ellipsoid (e.g., a plumb
line) at your location and the equatorial plane. (Meeus calls this
"geographical latitude".)

So if you're at 30N 90W on an ellipsoid, the 30N and 90W lines on the
celestial sphere intersect at your zenith, just as they did on the
spherical Earth. Leveling your transit and aligning it to north gives
it the same orientation relative to the celestial sphere that it had
on the spherical world.

To be sure, its location has changed. But since this movement is
negligible compared to the enormous radius of the celestial sphere,
the apparent positions of points on the sphere don't change. Thus on
an ellipsoidal Earth you can accurately compute Hc and Zn with
spherical trig.

But we're not quite done yet. The actual shape of the Earth, as sensed
by a transit or sextant, is the geoid. That's an imaginary sea level
surface perpendicular to gravity at all points. Its shape is extremely
complex, too hard to handle in routine computations. That's why datums
assume Earth's cross section through the poles is a perfect ellipse.

Unfortunately our instruments know nothing about this, and align
themselves to gravity. Their celestial observations produce astronomic
latitude and longitude, which are coordinates on the geoid and aren't
in any datum at all. (I guess you could call it "God's datum".) But
the geodesists who create a datum always try to make it a good overall
fit to the geoid. Astronomic coordinates should match the geodetic
coordinates quite well.

Exactly what "quite well" means, depends on the angle between a normal
to the ellipsoid and the local vertical at that spot. The angle is
called "deflection of the vertical", and is separated into north-south
and east-west components. By the standards of the celestial navigator,
deflection of the vertical is very small, but it can be greater than
some observational errors.

http://www.ngs.noaa.gov/GEOID/DEFLEC99/

Click the interactive computations link and try inputting some
coordinates. I tried that star atop the Texas capitol dome -- sum of
N-S and E-W deflections came to 4.08 seconds, about 400 feet on the
ground. That's how much a perfect celestial fix would differ from NAD
83.


Bottom line:

1. The tabulated coordinates in the almanac are independent of datum.

2. The correction for Moon's parallax is slightly (.2') affected by
the non-spherical shape of Earth. The almanac's text explains how to
compute an accurate value. The Moon correction tables don't bother
about this, though. All other bodies are so far away, Earth may be
considered a sphere as far as parallax is concerned.

3. Sight reduction tables and formulas lose no accuracy on an
ellipsoidal world. The plotting is slighly affected. But given the
geometric liberties taken by the intercept method, the accuracy of
sextant observations, and the small size of the intercepts, "one
minute equals one mile" is close enough. Sight reduction programs
which converge interatively on the solution, like the method described
in the almanac, can finish with very short intercepts. Here
non-spherocity should have virtually no effect.

4. The discrepancy between the astronomic coordinates obtained from
celestial observations and the geodetic coordinates on charts is
slight, at least in NAD 83. However, for a perfectionist applying all
possible corrections it can be significant.

--


paulhirose@earthlink.net (Paul Hirose)

   

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