Re: Too lazy to find an algorithm myself :-)

Might not be the perfect solution, but I suppose you could use
csBox3::ProjectBox() to project the box as a csBox2 on the view plane, then
multiply the camera distance by the ratio of bounding box dimension to view
box dimension. Whether dimension means width or height would depend on a
comparison between the ratio of bbox width and height and the ratio of vbox
width and height.

csBox2 bbox;
float minz, maxz;
if (input_box.ProjectBox (camera->GetTransform (), camera->GetFOV (), 0, 0,
bbox, minz, maxz))
{
  float dist = (input_box.GetCenter () - camera->GetTransform ().GetOrigin
()).Norm ();

  float bbox_girth = (bbox.MaxX () - bbox.MinX ()) / (bbox.MaxY () -
bbox.MinY ());
  float vbox_girth = (vbox.MaxX () - vbox.MinX ()) / (vbox.MaxY () -
vbox.MinY ());
  if (bbox_girth > vbox_girth)
  {
    return dist * (bbox.MaxX () - bbox.MinX ()) / (vbox.MaxX () - vbox.MinX
());
  }
  else
  {
    return dist * (bbox.MaxY () - bbox.MinY ()) / (vbox.MaxY () - vbox.MinY
());
  }
}
// not tested ;)


On 11 May 2012 15:21, Jorrit Tyberghein <[email protected]> wrote:

> So I just ask you guys.
>
> Given an axis aligned bounding box in world space and given a camera that
> is located
> at a certain spot in the world I would like to calculate at which the
> camera has to be
> so that the entire bounding box is exactly visible in the given view
> rectangle. So assume
> there is a line between the center of the box going through the current
> camera position.
> I want a new position on that line where the camera can see the bounding
> box (assuming
> it is oriented so that it looks at the center of the bounding box) and the
> box is just visible
> in the view rectangle.
>
> Any takers? :-)
>
> Thanks in advance,
>
> --
> Project Manager of Crystal Space (http://www.crystalspace3d.org)
> and CEL (http://cel.crystalspace3d.org)
> Support Crystal Space. Donate at
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> Personal page: http://users.telenet.be/jorritTyberghein/
>
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