Re: closed curve question

"John Doe" <johndoe9485@xxxxxxxxx> wrote in message
news:op.stkacuku6sy38m@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx

> Yes, I am ruling out the perpendicular case. And I was interested in your
> thoughts on both parallel and perspective projections. The geometry is
> Euclidean and as far as closedness is concerned, the curve forms a region
> (area) on the plane separate from the rest of the plane. If you lack the
> time, I am always interested in book/article recommendations.

The type of curve you describe is a "Jordan curve".

"Just d' FAQs" already mentioned an example where perspective
projection maps a circle to a parabola, so a perspective map does
not necessarily preserve the topological property of Jordan curve.

For a parallel projection with the aforementioned constraints on
the planes, the Jordan curve property is preserved. You can
assign 2D coordinate systems to both planes. If the origin of
the second plane is chosen to be the projection of the origin of
the first plane, you obtain an invertible linear mapping of one
plane to the other. The mapping is necessarily a continuous
function. Continuous functions map connected sets to connected
sets and compact sets to compact sets. The parallel projection
has the additional property that it maps a simply connected sets
to a simply connected set, preserving the Jordan curve property.

A sketch of the proof (by contradiction). Let S be the simply
connected region bounded by the curve on the first plane. Let
F be the linear mapping to the second plane. Let F(S) be the
image of S. Suppose F(S) is not simply connected. There must
exist a curve C in F(S) and a point P for which P is inside the curve
but outside F(S). The inverse image of C is F^{-1}(C) and is
contained in S. The inverse image of P is Q = F^{-1}(P) and is
not contained in S. The inverse function F^{-1} is itself a parallel
projection. Such projections preserve point-in-curve property.
That is, P is inside C, so Q must be inside F^{-1}(C). Since S
is simply connected, Q must also be inside F^{-1}(C) and by
implication in S, which is the contradiction.

The point-in-curve preservation is easy to see geometrically.
The curve C is extruded in the direction of projection, forming
a generalized cylinder surface. A point Q in C is also extruded,
forming a line that is inside the cylinder surface. The intersection
of the line and projection plane must be a point inside the
intersection of the generalized cylinder surface and the projection
plane.

--
Dave Eberly
http://www.geometrictools.com


.

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