Re: Rotated square in space

> Any other suggestions or clarifications?

I have assumed, that you speak about parallel
projection, but you don't, so it is not known
what I assumed as given, i.e. the x, y coordinates
of the corner points :(

So if you want to do it for a picture created by
a camera, I _suppose_ you have to know the
appropriate parameter of the lens system, the
current setting of camera focusing and the
distance of two corners of the square from the
camera.
I _suppose_, that if you only know, that the
object is a square it is not enough to calculate
the transformation matrix.

I think, that if you ask this question in
comp.graphics.algorithms
you will get better and more qualified
response.

By the way:
distance in 3D between two points p1 p2 is:
sqrt( (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2 )
where
sqrt() means square root (exponent value of 0.5)

Claudio

"Luiz Borges" <luiz_borges@xxxxxxx> schrieb im Newsbeitrag
news:1118072652.226562.37970@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> Thanks all for your answers...
> Actually if I find ANY square in the 3d space the fits the projection,
> it will work... it doesn't matter if there are infinetely squares...
> today I was think on something alike Claudio said, but working with
> angles formed between the edges... but i guess that working with
> distances might work better.
>
> What I don't see is how to calculate that, my last geometry lesson was
> 5 years ago lol :)
>
> Explaining a bit, what I want to do is that: a program that takes a
> picture, the user marks the four corners of an square object in the
> picture, and then I have to mark some specifics points inside that
> object... if I have transformation matrix for the object in 3d space, I
> can plot those points...
>
> Any other suggestions or clarifications?
>
> Luiz Borges
>


.

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