Zoom to fit function implementation problem

Posted on 2006-06-29
Medium Priority
Last Modified: 2013-12-06
I think this is done/asked millions of time but I couldnt find any useful information about it.
I am rendering some points scattered through the scene. I want to implement zoom to fit function, for example when the user presses "F" program will automatically align the model so all of the model appears on the view and as maximum.

The steps I took:
1. I have calculated the bounding sphere
2. And I am trying to compute some parameters according to it, here the problem occurs.

Here are some code I have written (in MFC):
NOTE: m_xpos, m_ypos, m_zoom, m_xrot, m_yrot and m_zrot are variables that changes with user interaction.

void COrthographic::OnDraw(CDC *pDC)
      CspetrexDoc *doc = (CspetrexDoc*) GetDocument();

      // Position the camera
      glTranslatef( m_xpos, -m_ypos, -m_zoom );

      // Adjust viewport for md3 models which
      // use a different coordinate system -
      // 3DSMAX format.
      glRotatef( -90.0f, 1.0f, 0.0f, 0.0f );
      glRotatef( -90.0f, 0.0f, 0.0f, 1.0f );

      // Rotate the camera
      glRotatef( m_xrot, 1.0f, 0.0f, 0.0f );
      glRotatef( m_yrot, 0.0f, 1.0f, 0.0f );
      glRotatef( m_zrot, 0.0f, 0.0f, 1.0f );

void COrthographic::OnSize(UINT nType, int cx, int cy)
      COpenGLWnd::OnSize(nType, cx, cy);
      if ( 0 >= cx || 0 >= cy || nType == SIZE_MINIMIZED )

      // Change the orthographic viewing volume to
      // reflect the new dimensions of the window
      // and the zoom and position of the viewport.
      glViewport( 0, 0, cx, cy );
      glMatrixMode( GL_PROJECTION );
      glOrtho( (float)(cx)/(float)(cy)*-m_zoom-m_xpos, (float)(cx)/(float)(cy)*m_zoom-m_xpos,      
            -m_zoom+m_ypos, m_zoom+m_ypos, -200.0f, 200.0f );
      glMatrixMode( GL_MODELVIEW );

// HERE!! this is zoomToFit function!!!
void COrthographic::zoomToFit()
      Sphere boundingSphere;
      Point3D center;
      CspetrexDoc *doc = (CspetrexDoc*) GetDocument();
      CRect cr;
      GetClientRect( &cr );

      boundingSphere = doc->getBoundingSphere();
      center = boundingSphere.getCenter();

      m_xpos = -center.getX();
      m_ypos = center.getY();
      m_zoom = (double)cr.Height()/cr.Width()*boundingSphere.getDiameter();

      OnSize( SIZE_MAXIMIZED, cr.Width(), cr.Height() );

I will be glad if anyone helps.
Question by:bitkidoku
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Author Comment

ID: 17031306
I think this is a hard question, lets make it more attractive (raised value to 300)
LVL 25

Expert Comment

ID: 17132615
What output are you actually getting?

Expert Comment

ID: 17165636
Observation: Since you're using an Orthographic projection, it is impossible to change the dispursal of the points, except if you change their location. Moving the camera will do no good (no 'zoom' in Orthographic projection). What you can do, is move the points to the center of the screen (as much as possible) by creating a bounding box/sphere and setting the look-at vector of the camera to the center of it (you should actually move the points using glTranslate).
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Author Comment

ID: 17170130
@nargov: Do you mean it is impossible to implement "zoom to fit" function using glOrtho() ? If you mean that, i dont think so.
LVL 25

Expert Comment

ID: 17170299
You could scale the scene using glScale*() by a factor of w/(2d). Where w is the screen width, and d is the distance of the furthest horizontal point within the scene.
LVL 18

Accepted Solution

Jose Parrot earned 1200 total points
ID: 17195805
Hi, bitkidoku,

As far I am understanding the objective, I see the following procedures:

1. Find the model's projection center.
Finding the bounding sphere is just an intermediate approach, because give us the model's center, not the projection center. The model can have points so far from the center, that force the radius to be bigger than the projection area. So, we make C = bounding sphere center as a first step.

2. When looking to C, discover the Top, Bottom, Leftmost and Rightmost points of the model's projection in the projection plane.

3. Calculate the middle point M of the rectangle defined by that points.
Mx = (Lx+Rx)/2; My = (Ty+By)/2. Not necessarely M is equal to the projection of point C. If model is a cube, they are equal, but will be different if the model is a not a symetrical one.
Example: Lx = -10, Rx = 38, Cx projection = 0. So the projection center x is 14, not 0. This is something different to the code you show us.

4. Rotate the camera such that the projection of C is equal to M. The angle can be computed by reverse calculation, or find by repeating 3 or 4 times the steps 3 and 4, thus refining the exact position of the camera.

5. Now we can zoom in (or out) until left and right (or top and bottom) points are in the view clipping boundaries. You can also add a 5% of left-right (and top-bottom) distance to create a safe viewing area slight bigger than the extreme points. Let's remember that zoom is determined by the distance between the plane and the camera.

Of course, this is true only if we use a conical projection, as stated by nargov. In parallel projection, the distance between projection plane and camera does't matter, so the way is to scaling, as stated by InteractiveMind.


Author Comment

ID: 17199244
@Jose: I am on vacation right now, I will consider your answer when I return.
LVL 11

Expert Comment

ID: 17347045
If you have computed the center of the points and a bounding sphere, then you have a radius outside of which there are no points.

If that's so, then simply:

glOrtho(x-r, x+r, y-r, y+r, z-r, z+r);

and the points will go to the edge. If you want them not to go quite to the edge, increase r by 5 or 10%
LVL 18

Expert Comment

by:Jose Parrot
ID: 17547185

I understand that all participants have contributed with correct comments toward a solution.


Author Comment

ID: 17549974
I already forgat that I posted this here, thanks for reminding. Jose you got the point :)

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