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Re: spherical geometry (was Re: navigation riddle)



Regarding John Denker's comments concerning the spherical equilateral
triangle problem:

I would argue that this "extra credit" part of the assignment
is far and away the most worthwhile part ... because if you do
this, you get the other parts practically for free.

and

For example, suppose we want to answer the very physically-
significant question of how much a pointer precesses if
flatlanders carry it along during a trip around the spherical
triangle. This involves "parallel transport" i.e. at each
step the new pointer orientation is aligned with (parallel to)
the old pointer orientation.

Lemma: To convince yourself that this is an interesting
question, consider the example of the spherical triangle
that has the following properties:
-- sides: length s = (pi/2) R
-- interior angles: three right angles
-- area: (pi/2) R^2
-- solid angle: one octant i.e. (pi/2)

Now we argue that the amount of precession is proportional to
the area of the triangle. ...
...
This tells us that the amount of precession is proportional
to area. The remaining work is to determine the constant of
proportionality. We can obtain this from the aforementioned
90-90-90 triangle. It has an area of (pi/2) R^2 and a
precession of (pi/2). So we conclude that the bottom-line
answer is incredibly simple:
precession = area / R^2

John's comments and argument noting and arguing for the
proportionality of the enclosed area and the precession (or
excess turning) angle are *extremely* useful in helping to do the
problem and its extra credit part in a relatively painless manner.
However, I seem to disagree with the first quote above assessing
what part of the problem is "free" and what part is the key to what
other part.

In my experience in solving the problem I found (and I think Carl can
back me up here) that it is relatively straightforward to find the
answer to the first part of the problem by a few simple techniques
such as the use of vectors in R^3 space (that contains the S^2
surface), an occasional vector projection, deducing angles from
vector dot products, etc. This allow us to find the interior angle
of the triangle in a relatively painless way. The interior angle is
very simply related to the excess turning angle or precession angle
that John discusses. Thus John's (nicely calibrated) proportionality
of the enclosed area to the excess turning angle allows us to get the
*area* for free once we found the expression for the interior angle
of the triangle by these other means. But if we wanted to calculate
the area of the triangle by a direct brute force integration the
problem quickly becomes *very* nasty. When I integrated it the only
way could figure out how to do it was to break up a messy expression
in an integrand into complex-valued partial fractions of monomials of
sinh() functions in the denominators of the partial fractions and
integrate the partial fractions separately (and this was after doing
a messy integration by parts). The resulting integrated form was a
very messy expression involving manifestly complex-valued functions
inside *and* outside of logarithm functions. The overall value was
manifestly real because the overall expression involved the sum of
a very complicated complex-valued expression and its complex
conjugate. I tested this messy result by sticking in the various
known special case values (tiny triangles, the 90-90-90 case, & the
180-180-180 case) to check for a math error. The result passed the
test. But simplifying this huge mess into a more friendly formula
proved to be a near nightmare.

Thus the use of the excess turning angle/area proportionality proves
to be an extremely valuable way of getting the triangle's area (for
free) from its interior angle. In my experience it didn't seem to
work well in getting the interior angle (for free) from the area
because of a lack of a general *simply-obtained* prior expression for
the triangle's area found by other means. The areas of a few
important special cases are easy to find (and do help calibrate the
proportionality that John notes) but it doesn't seem at all simple to
find the general area expression by some *other* means and to then
find the interior angle from the proportionality relationship (for
free).

Maybe John also has a neat trick for finding the triangle area by
some other means than by direct integration so he can use the
precession/area proportionality to find the interior angle? If so,
maybe he could share *that* secret with us as well? Or maybe he
has a trick for making a magic transformation to the integral
expressions that simplify them into an easily doable form?

BTW, *another* simple expression for the interior angle A of the
triangle is: sin(A/2) = sec(s/2)/2 where s is the side length in
units of the sphere's radius. Using John's observation of the
precession/area proportionality gives us a nice simple expression
for the triangle's area S in units of the sphere's squared radius
(i.e. the solid angle) which is:

S = 3*A - [pi] = 6*Arcsin(sec(s/2)/2) - [pi] .

This expression is *way* nicer than what comes unmassaged out of
the integral I encountered when finding S by brute force.

David Bowman