Re: Rotated square in space

Ok, than you don't need the Z-coordinates at all.

So your question was about transformation of
points of a square on a quadrangle which
results from 3D perspective transformation,
right?

This is another kind of question as your original
one:
> I have this projection:
> p1 0,5 p2 5,0 p3 15,5 p4 10,15
> and now I want p1,p2,p3,p4 in 3d world coordinates.

Conclusion?
It is a good idea to give details about what
one tries to achieve along with the question.
This helps to get better response and make
the responding people happier, because
they have better chances to provide what
is asked for.

Claudio

"Luiz Borges" <luiz_borges@xxxxxxx> schrieb im Newsbeitrag
news:1118141232.121924.213110@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> I think I found a way to solve my problems...
>
> 1) With the four projected corners, if I trace diagonals from them, the
> crossing of diagonals will be on the exact point in projection of where
> the center of the original square would lie.
> 2) Having the center C and the working with an edge at time (thus a
> triangle) with corners A and B, I can find any point I want on the line
> AB working with the angle ACB.
>
> Example: if i need the middle point of AB I will just bissect the angle
> ACB and pick the point where the bissection crosses AB. The same works
> for any other point in the border, if I need the point at 10% (I MUST
> work with percenteges) of side AB, I just pick 10% of the angle ACB
> from the line CA and make a line from C with that angle, the point
> where it intersects AB is my desired point.
>
> Can anybody correct me? I tried to be as clear as possible, but english
> isn't my first language.
> I will try to proof my results on paper before implement them, but I
> think this is one of the ways.
>
> Thanks,
> Luiz Borges
>


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