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*From*: David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>*Date*: Sat, 14 Aug 2004 10:36:28 -0400

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

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