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Re: spherical geometry

Regarding site checking site that Bob notes:

| As part of your answer find a formula for the heading angle of the
| geodesic connecting 2 arbitrary fixed points (of known latitude
| and longitude) on the earth's surface as measured by an observer
| a one of those points

Use this on-line calculator to test your answer:

Bob Sciamanda

I wonder if this calculator calculates things using a spherical
earth approximation (as my problem above assumed), or if it actually
uses the formula for calculating geodesic distances and headings for
a surface which is an oblate spheroid. In the latter case there are
two different versions depending on the definition of the latitude
angle used. The common geographic/astronomical latitude measures
the angle between the tangent plane at some point on the surface and
the plane tangent plane to the equator at the same longitude. But
the geodetic latitude measures the central angle at the center of
the earth between a ray from the center through the surface point of
interest and a second ray from the center to a point on the equator
at the same longitude. If an accuracy of about 1% or so is
required a spherical earth approximation is plenty adequate, but if
an accuracy to within 1/10 of a percent is required on some long
distance paths then the spherical earth approximation is not
sufficient because the earth has an oblateness of about 1/298.

BTW, it is substantially more challenging to do the problem of
finding formulae for heading angles and geodesic distances on an
oblate spheroid of revolution. Regardless of which definition of
latitude is used the geodesic heading/distance formulae can be
relatively simply converted into the corresponding formulae for
the other definition of latitude because there is a relatively
simple formula connecting these two different latitude angles to
each other.

Anybody up for doing some spheroidal geometry problems? Recall
the Cassini spacecraft orbiting Saturn is dealing with a planet
that has a 10% oblateness at the surface of the cloud deck and
has a gravitational field that possesses a significant
quadrupole moment contribution.

David Bowman