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From: Archimedes Plutonium on 29 Jan 2010 08:06


David Bernier wrote:
> Archimedes Plutonium wrote:
> >
> > Archimedes Plutonium wrote:
> > (snipped)
> >> That should read 18 by 18 degrees of latitude longitude of a sphere
> >> globe,
> >> and changed on the original post.
> >>
> >> Also, I suspect that the Upper Bound if it is 18 by 18 degree latitude
> >> and longitude
> >> that the Elliptic, Euclidean and Hyperbolic triangles involved that
> >> none of them are
> >> equilateral. Here again, my intuition is easy to go astray. This
> >> entire adventure of this
> >> geometry expedition is going on all in my head, my intuition and only
> >> with the aid of
> >> the globe as a visual aid. So in these "mind wanderings" there are
> >> alot of traps and
> >> pitfalls.
> >>
> >
> > Enrico may be correct from just looking at the problem on the globe of
> > say the 1/8 surface area of Greenwich to Equator to 90W.
> >
> > In my previous post I called it a "rod" from North Pole to 90W as the
> > line-segment underneath the surface of Earth that forms the line and
> > to
> > construct a unique Euclidean plane (Euclidean triangle for the "point
> > not on the chord" of the Greenwich intersect Equator vertex.
> >
> > So I am looking at it, trying to visualize whether when that Euclidean
> > triangle
> > is formed and thence I take the two longitude arcs and so to speak
> > rotate them
> > on their hinges at the poles and equator to rotate them as concave
> > inwards
> > forming a North Pole Hyperbolic triangle vertex. I am wondering if
> > they overlap.
> [...]
>
> From what I remember, one can get a model of elliptic
> two-dimensional geometry starting with the globe and
> identifying points that are antipodes (such as
> the North Pole and the South Pole points on
> the globe).
>
> It's easy to embed a Euclidean plane in 3D space,
> so going from a 90 degree, 90 degree, 90 degree
> spherical triangle on the globe to an associated
> one in a Euclidean plane that passes through the
> three vertices of the 90-90-90 spherical triangle
> is not hard.
>
> If you simply rotate the three arcs of the
> spherical triangle inwards (along the hinges),
> this may at first glance give the impression
> of having defined a triangle in the hyperbolic
> plane. But I see a problem: for the
> Euclidean triangle, there was an obvious choice
> for the Euclidean plane where the Euclidean triangle
> exists. If you want to properly define a hyperbolic
> triangle, you should specify a hyperbolic plane
> in some way; unfortunately, I don't think
> giving the three vertices and even the
> assumed sides is enough. Or, the question is:
> if there's a hyperbolic triangle, it's part of
> a hyperbolic plane. Which hyperbolic plane
> do you want to choose for the hyperbolic triangle?
>
> David Bernier

Yes, thanks David, this helps alot. I was struggling with finding the
Euclidean
analog to attach the hinges of the arcs.

So if I take a representative elliptic triangle and then find its
antipodal elliptic
triangle would thus induce me to find that plane which is orthogonal
to both
and then peg that plane with the three vertices. Then simply just
determine the
arc and arc length and transfer to the Euclidean triangles and see if
they can
produce a hyperbolic triangle.

Another way of doing this is to simply surgically cut the podal and
antipodal
elliptic triangles then paste them onto a Euclidean plane sandwiched
between
the two surgical cuts. Then poke a hole into the plane from the three
vertices
yielding the Euclidean triangle analog. Now the test comes as to
whether we
can surgically cut the arcs of one of the elliptic triangles and to be
able to
compose a hyperbolic triangle from the Euclidean triangle vertices. I
am assured
of contructing a elliptic 2D triangles with those arcs, but the
question I am wanting
is whether I can construct a hyperbolic triangle with those arcs, and
if not, at what
size of elliptic triangle does it stop?


Unless I am mistaken, I think that Enrico's
estimate of 1/8 surface area is not able to yield a hyperbolic
triangle and it
must be a smaller area than 1/8. From the looks of it, the sides of
the hyperbolic
overlapp. The elliptic triangle in the case of 1/8 is 90 by 90 by 90
degrees and from
the looks of it the Euclidean triangle analog would be 60 by 60 by 60
and it looks as
though the arcs overlapp when their concavity is reversed.

Thanks for the help, and getting much closer to a finale.

I do not understand your questions about chosing a hyperbolic plane.
All I want to
find out, David is what size of a Elliptic triangle on a sphere is the
upper bound for
producing a reverse concavity. That is, what size of elliptic triangle
where I reverse
the concavity and try to fit that onto the sides of the analog-
Euclidean-triangle, where
it prevents me from having a hyperbolic triangle since the arcs
overlapp. So I need only the elliptic triangles and the Euclidean
analog-triangles. I do not need a hyperbolic plane, at least I do not
think so.

David, once I find the answer, hopefully it is 10% of the hemisphere,
then I do need a hyperbolic plane, because I would like to know if
that 10% hemisphere surface area is
the maximum area which a sphere can rest-in-tangency into the
hyperbolic plane-analog.
In that case I would need to have a criterion for a **natural
analog**.

Working on this problem was more exciting for me tonight, rather than
going to sleep.


Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
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