Week 5 Tutorial Solutions

Question 1: 3D Affine Transformations

What coordinate frames do the following affine matrices represent?

M1 = [[  1,  0,  0,  0],
      [  0,  1, -1,  0], 
      [  0,  1,  1,  0],
      [  0,  0,  0,  1]]

M2 = [[  2,  0,  0,  2],
      [  0, -1,  0,  1],
      [  0,  0,  1,  0],
      [  0,  0,  0,  1]]

M3 = [[  1,  1,  1,  0],
      [  0,  1,  1,  0],
      [  0,  0,  1,  0]
      [  0,  0,  0,  1]]

Hint: rememeber M = [i, j, k, phi].

M1 = [[ 1, 0, 0, 0], [ 0, 1, -1, 0], [ 0, 1, 1, 0], [ 0, 0, 0, 1]] i = (1,0,0) j = (0,1,1) k = (0,-1,1) phi = (0,0,0) The j/k axes have been rotated 45 degrees about i, and scales to have length sqrt(2). M2 = [[ 2, 0, 0, 2], [ 0, -1, 0, 1], [ 0, 0, 1, 0], [ 0, 0, 0, 1]] i = (2,0,0) j = (0,-1,0) k = (0,0,1) phi = (2,1,0) The origin has been translated to (2,1,0). The i-axis has been scaled by 2. The j-axis has been reflected. M3 = [[ 1, 1, 1, 0], [ 0, 1, 1, 0], [ 0, 0, 1, 0] [ 0, 0, 0, 1]] i = (1,0,0) j = (1,1,0) k = (1,1,1) phi = (0,0,0) The j-axis is sheared towards the i-axis and scaled by sqrt(2). The k-axis is sheared towards both the i-axis and the j-axis and scaled by sqrt(3). The axes are not at right angles to each other. We can check this by using the dot product between the pairs of axes.

Question 2: 3D Rotations

If you rotate a coordinate frame by 90 degrees about x and then rotate the resulting frame by 90 degrees about y, what is the resulting frame? What is the matrix for the combined transformation? What happens if you reverse the order of the rotations?


M = R_x R_y
  =  [1 0  0 0]  [0  0 1 0]
     [0 0 -1 0]  [0  1 0 0]
     [0 1  0 0]  [-1 0 0 0]
     [0 0  0 1]  [0  0 0 1]

  = [0 0 1 0]
    [1 0 0 0]
    [0 1 0 0]
    [0 0 0 1]

i-axis is (0,1,0)
j-axis is (0,0,1)
k-axis is (1,0,0)
i.e. the three axes have been rotated. i->j, j->k, k->i


in the opposite order:

M = R_y R_x
  = [ 0 0 1 0] [1 0  0 0]
    [ 0 1 0 0] [0 0 -1 0]
    [-1 0 0 0] [0 1  0 0]
    [ 0 0 0 1] [0 0  0 1]

  = [ 0 1  0 0]
    [ 0 0 -1 0]
    [-1 0  0 0]
    [ 0 0  0 1]

i-axis is (0,0,-1)
j-axis is (1,0,0)
k-axis is (0,-1,0)

Question 3: 3D Cameras

Consider the following OpenGL code:

gl.glMatrixMode(GL2.GL_MODELVIEW);
gl.glLoadIdentity();
GLU glu = new GLU();
glu.gluLookAt(0, -3, 3, 0, 1, 0, 0, 0, 1);

gl.glMatrixMode(GL2.GL_PROJECTION);
gl.glLoadIdentity();
gl.glOrtho(-4, 4, -3, 3, 1, 4);

What is the camera's local coordinate frame (in world coordinates) after this call?


The origin is (0, -3, 3).

The k-axis points from the target (0, 1, 0) towards the camera at (0, -3, 3).
k  = normalise((0, -3, 3) - (0, 1, 0))
   = (0, -4, 3) / sqrt(4*4 + 3*3)
   = (0, -0.8, 0.6)

The i-axis is the cross of the k axis with the up vector (0,1,0)
i  = normalise(up x k)
   = normalise((0, 0, 1) x (0, -0.8. 0.6))
   = normalise(0.8, 0, 0)
   = (1, 0, 0)

The j-axis is perpendicular to i and k
j = k x i
  = (0, -0.8, 0.6) x (1, 0, 0)
  = (0, 0.6, 0.8)

What is the resulting model-view transform?


The camera-to-world matrix is the combination of the values above

M = [i  j    k   phi]
  = [1  0    0     0]
    [0  0.6 -0.8  -3]
    [0  0.8  0.6   3]
    [0  0    0     1]

The view transform is the inverse of this matrix

V = M-1
  = [1  0   0    0  ]
    [0  0.6 0.8 -0.6]
    [0 -0.8 0.6 -4.2]
    [0  0   0    1  ]

Note: If you do not know how to find the inverse of an arbitrary matrix 

(which you do not need to know how to do for this course) 
you do in the following way:
The gluLookAt performs a translation and then a rotation. You can break up the M matrix into 2 parts.
The matrix M = TR
where
T = [1 0 0  0]
    [0 1 0 -3]
    [0 0 1  3]
    [0 0 0  1]
R = [1    0    0   0]
    [0  0.6 -0.8   0]
    [0  0.8  0.6   0]
    [0  0    0     1]

So to get the inverse matrix we need

V = M-1 = R-1T-1
     
Inverses of purely rotational matrices can be found by transposing them so

R-1 = [1    0     0   0]
      [0   0.6   0.8  0]
      [0  -0.8   0.6  0]
      [0    0     0   1]

and T-1 = [1 0 0   0]
          [0 1 0   3]
          [0 0 1  -3]
          [0 0 0   1]

Matrix multiplication should give the same answer as above.

Sketch the view volume in world coordinates.

The camera is pointing up at angle toward the world Y axis from (0,-3,3) to (0,1,0) (eg like pointing your camera at an angle up towards the ceiling) and the up vector is the z-axis. The view volume is an 8x3x6 rectangular prism centred at (0,-1,1.5) in world co-ordinates (obtained by multiplying matrix for camera to world with the centre in camera coordinates below).

Sketch the view volume in camera coordinates.

The orthographic view volume is an 8x6x3 rectangular prism centred at (0,0,-2.5) in camera coordinates, aligned with the camera's coordinate frame.

What if line 8 were changed to:

gl.glFrustum(-4, 4, -3, 3, 1, 4);

What is the new view volume?

The perspective view volume is a frustum (truncated pyramid). The near plane a(8x6) rectangle centred at (0,0,-1) in camera coordinates. The far plane is a (32x24) rectangle centred at (0,0,-4) in camera coordinates.

Question 4: Perspective Projections

Consider the following code:

        gl.glMatrixMode(GL2.GL_PROJECTION);
        gl.glLoadIdentity();

        GLU glu = new GLU();
        glu.gluPerspective(60, 1, 1, 10);
        
        gl.glMatrixMode(GL2.GL_MODELVIEW);
        gl.glLoadIdentity();

        gl.glPolygonMode(GL2.GL_FRONT_AND_BACK, GL2.GL_LINE);
        gl.glBegin(GL2.GL_QUADS);
        {
            gl.glVertex3d(1, 1, -2);
            gl.glVertex3d(0, 1, -2);
            gl.glVertex3d(0, 0, -2);
            gl.glVertex3d(1, 0, -2);

            gl.glVertex3d(0, 1, -4);
            gl.glVertex3d(-1, 1, -4);
            gl.glVertex3d(-1, 0, -4);
            gl.glVertex3d(0, 0, -4);
        }
        gl.glEnd();

Sketch the view volume.
The view volume is a frustum with a 60 degree fov. The near plane is a square (aspect=1) with side length 2/sqrt(3) centred at (0,0,-1) in camera coordinates. The far plan is a square with side length 20/sqrt(3) cented at (0,0,-10) in camera coordinates
What will be displayed on the screen?

Two squares side by side. The square on the right will be twice as big (both width and height) as the one on the left.

What is the projection transform matrix?
R = [ 2n / (r-l) 0 (r+l)(r-l) 0 ] [ 0 2n/(t-b) (t+b)(t-b) 0 ] [ 0 0 -(f+n)/(f-n) -2fn/(f-n) ] [ 0 0 -1 0 ] = [ sqrt(3) 0 0 0 ] [ 0 sqrt(3) 0 0 ] [ 0 0 -11/9 -20/9 ] [ 0 0 -1 0 ]

What do we get when we multiply the point (1, 1, -2), in camera co-ordinates by this matrix and what will it be equivalent to after perspective division (assuming it survives clipping).
(sqrt(3),sqrt(3),2/9,2) which after perspective division would be (sqrt(3)/2,sqrt(3)/2,1/9)
What happens if the field of view is changed to 90 degrees? 180 degrees? 240 degrees?
As the FOV gets larger the squares shrink towards the middle of the screen. At 180 degress they disappear. A FOV greater than 180 degrees is nonsensical.

Question 5: 3D Modeling

Write a vertex list, normal list and face list for a mesh describing a triangular prism. The front face should be a triangle with its top point in local co-ordinates at (0,1,-1). Assume it lies on the z=-1 plane. The back face should lie on the z=-3 plane at corresponding x and y values. The two faces should be connected by quads.


double vertices[][] = {{0,1,-1},
		       {-0.5,0,-1},
		       {0.5,0,-1},
		       {0,1,-3},
		       {-0.5,0,-3},
		       {0.5,0,-3},
		      };

int faces[][] = {{0,1,2}, //front
	         {3,5,4}, //back
	         {1,4,5,2},//base
	         {0,3,4,1}, //left side
	         {0,2,5,3} //right side
	        };

double normals[][] = { {0,0,1}, //front face
		       {0,0,-1}, //back face
		       {0,-1,0}, //base
                       {-2.0/Math.sqrt(5),1.0/Math.sqrt(5),0}, //left
                       {2.0/Math.sqrt(5),1.0/Math.sqrt(5),0} //right
	             };
	
//Left side normal
//v1 = V3 - V0 = (0,1,-3) - (0,1,-1) = (0,0,-2)
//v2 = V1 - V0 = (-0.5,0,-1) - (0,1,-1) = (-0.5,-1,0)
//v1xv2 = (-2,1,0)
//magnitude = sqrt(5)
//Normalised = (-2/sqrt(5),1/sqrt(5),0);
	
//Right side normal
//v1 = V3 - V0 = (0,0,-2)
//v2 = V2 - V0 = (0.5,0,-1) - (0,1,-1) = (0.5,-1,0)
//v2xv1 = (2,1,0)
//Normalised = (2/sqrt(5),1/sqrt(5),0);

Discuss different ways you could calculate normals for the following triangle mesh. For example, how would you calculate the normal for vertex P when drawing the triangle PDR? You may assume that each triangle is 1 unit in the x direction and 1 unit in the z direction, but has arbitrary y values.
```
     
     |/   |/   |/   |/
...--+----+----+----+--...
    /|   /|   /|   /|           Z
   / |  / |  / |  / |           ^
     | /  | /  | /  | /         |
     |/   |/   |/   |/          |
...--+----P----R----+--...      +-----> X
    /|   /|   /|   /|          
   / |  / |  / |  / |         
     | /  | /  | /  | /      
     |/   |/   |/   |/
...--+----D----+----+--...
    /|   /|   /|   /|
```
There are two main simple ways:

You could calculate the face normal of triangle PDR and use that as the normal for the 3 vertices when drawing that particular face. This would result in flat shading where edges of the triangles are more pronounced.
You could calulate the face normal of each of the 6 triangles that P is a vertex of and take the average. You would do the same thing for D and R so each vertex would have its own individual normal. This would result in a smoother looking surface.
If the normals that you average are not normliased yet, this will result in a weighted average which can give better results.
There are other variations on this approach such as weighting the contribution of each normal by its angle of incidence with the vertex or eliminating normals of faces that meet at sharp angles over a certain threshold etc These are more expensive to compute but may give desired results.

Question 6 (advanced - not examinable):

A rotation about any arbitrary axis can be decomposed into a series of rotations about the X, Y or Z axes. How would you compute such a decomposition? (There are many different ways of doing this.)

See http://en.wikipedia.org/wiki/Euler_angles or Euler Angles in the textbook