CODEWITHSUNDEEP
iosmatrixopengl-esshader
aledalgrande
aledalgrande

Reputation: 5217

Centering all of the points in iOS OpenGL ES app

I have an OpenGL view that displays a set of 3D points with some basic shaders:

// Fragment Shader

static const char* PointFS = STRINGIFY
(

 void main(void)
{
    gl_FragColor = vec4(0.8, 0.8, 0.8, 1.0);
}

);

// Vertex Shader

static const char* PointVS = STRINGIFY
(
 uniform mediump mat4 uProjectionMatrix;

 attribute mediump vec4 position;

 void main( void )
{
    gl_Position = uProjectionMatrix * position;
    gl_PointSize = 3.0;
}

);

And the MVP matrix is calculated as:

- (void)setMatrices
{
    // ModelView Matrix
    GLKMatrix4 modelViewMatrix = GLKMatrix4Identity;
    modelViewMatrix = GLKMatrix4Scale(modelViewMatrix, 2, 2, 2);

    // Projection Matrix
    const GLfloat aspectRatio = (GLfloat)(self.view.bounds.size.width) / (GLfloat)(self.view.bounds.size.height);
    const GLfloat fieldView = GLKMathDegreesToRadians(90.0f);
    const GLKMatrix4 projectionMatrix = GLKMatrix4MakePerspective(fieldView, aspectRatio, 0.1f, 10.0f);

    glUniformMatrix4fv(self.pointShader.uProjectionMatrix, 1, 0, GLKMatrix4Multiply(projectionMatrix, modelViewMatrix).m);
}

This works fine, but I have a set of 500 points and I see only a few.

How do I scale/translate the MVP matrix to display all of them (they are a dynamic set)? Ideally the "centroid" should be at the origin, and all of the points visible. It should be able to adapt to rotations of the view (gestures are the next step I want to implement).

Upvotes: 0

Views: 72

Answers (1)

Matic Oblak
Matic Oblak

Reputation: 16774

Seeing how you present this you might need quite a lot... I guess best approach might be using "look at", the point you are looking at is (0,0,0) as you stated, camera position should probably be (0,0,Z) and up (0,1,0). So the only issue here is the Z component of camera position.

If you start the Z with for instance -.1 and the iterate through all the points then sin(fieldView*.5f) * (p.z-Z) >= point.y for the point to be visible. So you can compute Z1 = p.z-(point.y/sin(fieldView*.5f)) and if Z1<Z then Z=Z1. This check is only for the positive Y check, you also need the same for negative Y and same for +-X. These evasions are very similar though when checking X you could also take the screen ratio into account.

This procedure should give you the smallest field possible to see all the points (with given limitations such as looking towards (0,0,0)) but is far from the simplest. You also need to consider if the equation will work if p.z<-Z.

Another bit easier approach is to generate the smallest cube around centre which holds all the points: iterate through points and get the coordinate with largest absolute value (any of X,Y or Z). When you have it use it with frustum instead perspective so that all rect parameters (top, bottom, left and right) are generated with this value as +-largest. Then you need to compute the translation which for 90 degrees field is Z = (largest*.5). Z is the zNear for the frustum and then also translate the matrix by -(Z+largest). Again one of the coordinate in frustum must be multiplied by screen ratio.

In any case do watch out what your zFar is, having it only 10.0f might be a bit too short in your case. Until you need the depth buffer you should not worry about that value being too large.

Upvotes: 1

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