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From: Archimedes Plutonium on 30 Jan 2010 02:42
Making some progress here. At least I am on some sure footing.

I can begin to see what a 1/8 surface area of sphere is going to be
too large.
That amounts to a 45 by 45 degree latitude longitude rectangle on the
one hemisphere of the sphere. And the reason that a elliptic triangle
of 1/8 fails
is because when we convert it into the representative Euclidean
triangle the
angles of 90 + 90 + 90 become 60 + 60 +60 and then when we transport
the
arc curves they overlapp into the hyperbolic triangle.

A 1/8 total surface area or 45 by 45 rectangle of hemisphere is far
too large
of an upper bound. The upper bound is smaller, and I believe it is 10%
of the
hemisphere, making a rectangle of 18 by 18 degrees latitude longitude.

As the triangles formed from the sphere are smaller and smaller, they
approach
that of euclidean-triangles and thus when we reverse the concavity of
the elliptic
triangle, we can form a hyperbolic triangle. And I think the upper
limit here is
a 10% of the hemisphere. A 10% is small enough of a triangle that by
reversing
the concavity of elliptic to hyperbolic, the sides do not overlapp.

Correct me if wrong, but throughout mathematics, I never recalled of
any important
number ending up at a cool 10%. It seems as though 1/10 was never a
important
feature of either Number theory or geometry. But maybe this is a first
for the number
1/10.

Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
From: David Bernier on 30 Jan 2010 03:34
Archimedes Plutonium wrote:
> Making some progress here. At least I am on some sure footing.
>
> I can begin to see what a 1/8 surface area of sphere is going to be
> too large.
> That amounts to a 45 by 45 degree latitude longitude rectangle on the
> one hemisphere of the sphere. And the reason that a elliptic triangle
> of 1/8 fails
> is because when we convert it into the representative Euclidean
> triangle the
> angles of 90 + 90 + 90 become 60 + 60 +60 and then when we transport
> the
> arc curves they overlapp into the hyperbolic triangle.
>
> A 1/8 total surface area or 45 by 45 rectangle of hemisphere is far
> too large
> of an upper bound. The upper bound is smaller, and I believe it is 10%
> of the
> hemisphere, making a rectangle of 18 by 18 degrees latitude longitude.
>
> As the triangles formed from the sphere are smaller and smaller, they
> approach
> that of euclidean-triangles and thus when we reverse the concavity of
> the elliptic
> triangle, we can form a hyperbolic triangle. And I think the upper
> limit here is
> a 10% of the hemisphere. A 10% is small enough of a triangle that by
> reversing
> the concavity of elliptic to hyperbolic, the sides do not overlapp.

I think I can help. If you take three quarters (25 cents, 3 of them)
and put them on a smooth surface like a table-top so that each
quarter just manages to be in contact with the other two quarters,
as:
OO
O <-- move left 1/2 a space

then the three circles are tangent to each other, and in the Poincar�'s
disk model of the hyperbolic plane, this gives three straight lines
of "infinite" length. Anyway, the inside figure of the three circles
has three points and cusps just like at the center of an eight
figure: '8' . At the center of the '8', the upper loop and the lower loop
are tangent, and you get two cusps, one left of center, the other right of
center. So with the three quarters/circles, the three interior
angles at each contact point between two circles is zero or
virtually zero. The centers of the three circles can form the
vertices of an equilateral Euclidean triangle.
Then by basic geometry, the three arcs measure 60 degrees,
in the same sense as equator to North pole is a 90 degree arc.

So you can imagine three points on a sphere, always at equal distance from
each other. Starting from a tiny spherical triangle, imagine moving
the points away from each other until the three arcs on the sphere
subtend an angle of 60 degrees as measured from the center of the
sphere. Then that size of spherical triangle can enclose
the shape of any equilateral triangle where the convex to concave
operation will keeps the arcs from intersecting.

So I think three 60 degree circular arcs is the smallest upper bound,
although it needs checking.

David Bernier

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