Solved

Zoom to fit function implementation problem

Posted on 2006-06-29
11
2,049 Views
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();
      
      SetContext();
      glLoadIdentity();
      glClear( GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT );

      // 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 );
      
      doc->drawScene(pDC);
      
      SwapGLBuffers();
}

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

      // Change the orthographic viewing volume to
      // reflect the new dimensions of the window
      // and the zoom and position of the viewport.
      SetContext();
      glViewport( 0, 0, cx, cy );
      glMatrixMode( GL_PROJECTION );
      glLoadIdentity();
      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() );
      OnDraw(NULL);
}


I will be glad if anyone helps.
0
Comment
Question by:bitkidoku
  • 4
  • 2
  • 2
  • +2
11 Comments
 

Author Comment

by:bitkidoku
ID: 17031306
I think this is a hard question, lets make it more attractive (raised value to 300)
0
 
LVL 25

Expert Comment

by:InteractiveMind
ID: 17132615
What output are you actually getting?
0
 
LVL 1

Expert Comment

by:nargov
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).
0
Free Tool: Site Down Detector

Helpful to verify reports of your own downtime, or to double check a downed website you are trying to access.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

 

Author Comment

by:bitkidoku
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.
0
 
LVL 25

Expert Comment

by:InteractiveMind
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.
0
 
LVL 18

Accepted Solution

by:
JoseParrot earned 300 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.

Jose
0
 

Author Comment

by:bitkidoku
ID: 17199244
@Jose: I am on vacation right now, I will consider your answer when I return.
0
 
LVL 11

Expert Comment

by:dbkruger
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%
0
 
LVL 18

Expert Comment

by:JoseParrot
ID: 17547185
Hi,

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

Jose
0
 

Author Comment

by:bitkidoku
ID: 17549974
I already forgat that I posted this here, thanks for reminding. Jose you got the point :)
0

Featured Post

Active Directory Webinar

We all know we need to protect and secure our privileges, but where to start? Join Experts Exchange and ManageEngine on Tuesday, April 11, 2017 10:00 AM PDT to learn how to track and secure privileged users in Active Directory.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

As game developers, we quickly learn that Artificial Intelligence (AI) doesn’t need to be so tough.  To reference Space Ghost: “Moltar, I have a giant brain that is able to reduce any complex machine into a simple yes or no answer. (http://www.youtu…
Performance in games development is paramount: every microsecond counts to be able to do everything in less than 33ms (aiming at 16ms). C# foreach statement is one of the worst performance killers, and here I explain why.
Microsoft Active Directory, the widely used IT infrastructure, is known for its high risk of credential theft. The best way to test your Active Directory’s vulnerabilities to pass-the-ticket, pass-the-hash, privilege escalation, and malware attacks …

821 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question