Re: Rotated square in space

If I understand right, what OP looks for
is as follows:
you have a point P inside the square
p1,p2,p3,p4.
You want to know where it is on the
quadrangle p1a, p2a, p3a, p4a
which is the picture of the square.
You choose a coordinate system
p4
|
p1--p2
where p1 is 0,0 and p2 coordinate x is set to 1
(and p4 coordinate y is set to 1).
Then you have the coordinates
Px, Py of your point within the square
and search for the coordinates Pax, Pay
in the quadrangle. You get the point Pa
as intersection of two lines within the
quadrangle, where the endings of
these lines lies on the edges, so, that
the endings split the edges in the same
proportion as in the original square i.e.
the proportion value is the value of
coordinates x,y in the choosen new
coordinate system.
I assume, that I can leave out here the
exact equations for getting the x,y
coordinates of Pa if the coordinates
of p1, p2, p3, p4, p1a, p2a, p3a, p4a
and P are given, because it's quite
simple to get them (but will have much
terms).

That's all.
Luiz Borges, am I right?

Claudio

"Luiz Borges" <luiz_borges@xxxxxxx> schrieb im Newsbeitrag
news:1118163699.394017.96140@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> Replying to Claudio, I didn't give those last details because I haven't
> tthought about them... but yesterday I thought that using angles would
> be simpler, would gave the exact same results, and I wont need to go
> into the 3d space...
> But I still haven't had time to proof it... otherwise I will have to
> use the projection plane aproach... in that case, i thought about
> working side by side, with projections of faces from a pyramid... the
> projection plane would intersect the piramid...
>



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