Re: spherical geometry (was Re: navigation riddle)

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding Jack's comments:

> I think that the interesting part of the original problem is that it
> requires the solver to realize that plane geometry is not involved.

I disagree.  The radius of curvature of the earth is over 6000 times
the length scale of the problem as it is stated.  A flat earth
approximation is quite sufficient.

Rather than considering non-planar geometry what is required is to
realize that the concepts of the cardinal directions of N, S, E, & W
do not refer to approximately Cartesian coordinate directions in
the vicinity of a geographic pole, but rather refer to *polar*
coordinate directions (with the attendant coordinate singularity
effects of such a coordinate system).

> The path has two corners and three equal legs, returning to the
> starting point.

The 3 legs have equal *lengths* but they are not equivalent.  This is
because the E-W direction is not necessarily a (geodesically)
straight direction (except in the vicinity of the equator).  OTOH the
N-S direction is always a geodesically straight direction no matter
where on earth one may be situated.  The 2nd leg of the trip can only
be geodesically straight when it is carried out on the equator.  For
all other locations the 2nd leg has a path that curves to either the
left or the right depending on hemisphere it is traversed on.  The
*only* effect of spherical geometry on the problem is the latitude
dependence of the degree of curvature of a path along the E/W
direction.

> On a plane surface, this must be an equilateral triangle.

It is not a triangle.  Triangles do not have geodesically curved legs
(even when they are inscribed on gently curved surfaces).  All
solutions to the problem have the 2nd leg of the path follow a
geodesically curved path on an effectively flat surface.

> But the problem, stated on a plane surface, would make two of the
> legs parallel - impossible on a plane surface.

It works just fine on a plane surface.  You only need to realize that
the solutions require that the E/W direction is a circumferential
direction in a polar coordinization of the plane and that the N/S
direction is the radial direction in such a coordinization.

> So we need a surface where either a "southern" path is not parallel
> to a northern path, or two of the corners are oincident - impossible
> on a planar surface.

A south-going path is never parallel to a north-going path some E/W
distance from it (unless one happens to be effectively on the
equator).  The typical case is that 2 such N/S paths are *not*
parallel, but the E/W distance from one to the other depends on
how far along the paths one goes before measuring the E/W (or
even the geodesic) distance between the paths. Of course the degree
of this dependence is itself latitude dependent. As examples of this
consider the states of Colorado and Wyoming.  Their northern corners
are significantly closer together than their southern corners are.
BTW the northern boundary of Colorado is about as long as the
southern boundary of Wyoming is (a good part of them being a shared
boundary).  Both states have also have the same distance from their
northern boundary to their southern boundary.  This is because both
states have their boundaries defined being about 4 degrees of
longitude wide from north to south and have their eastern boundary
about 7 degrees of longitude east of their western boundary.  So even
though both states have effectively the same angular dimensions the
aspect ratio of Colorado is obviously different than that of Wyoming
and Colorado is obviously wider from west to east than corresponding
parts of Wyoming is because Wyoming is north of Colorado and both are
in the Northern Hemisphere.

>                                 Regards,
>                                        Jack

David Bowman