Transformations

Rotations in 3D

$(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & \cos θ & - \sin θ & 0 \\ 0 & \sin θ & \cos θ & 0 \\ 0 & 0 & 0 & 1 \end{matrix})$ x-axis rotation
$(\begin{matrix} \cos θ & 0 & \sin θ & 0 \\ 0 & 1 & 0 & 0 \\ - \sin θ & 0 & \cos θ & 0 \\ 0 & 0 & 0 & 1 \end{matrix})$ y-axis rotation
$(\begin{matrix} \cos θ & - \sin θ & 0 & 0 \\ \sin θ & \cos θ & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix})$ z-axis rotation

Rotating about a point in 2D

The homogeneous coordinates seem mystifying and they are. But when you compare it with normal algebra it is easy to see what is going on. Consider the following example of finding a transform that rotates about a given point in 2D by an angle of theta. The standard transformation matrix rotates about the origin and not the point so you need to first shift to the origin, apply the rotation and shift back again.

Let R be the transform matrix,
P be the point being translated,
X₀ be the point being rotated about,
and P' is the transformed point.

$P' = R (P - X_{0}) + X_{0}$

$(\begin{matrix} p_{0}' \\ p_{1}' \end{matrix}) = (\begin{matrix} r_{0} & r_{1} \\ r_{2} & r_{3} \end{matrix}) (\begin{matrix} p_{0} - x_{0} \\ p_{1} - x_{1} \end{matrix}) + (\begin{matrix} x_{0} \\ x_{1} \end{matrix})$
$(\begin{matrix} p_{0}' \\ p_{1}' \end{matrix}) = (\begin{matrix} r_{0} p_{0} - r_{0} x_{0} + r_{1} p_{1} - r_{1} x_{1} + x_{0} \\ r_{2} p_{0} - r_{2} x_{0} + r_{3} p_{1} - r_{3} x_{1} + x_{1} \end{matrix})$

I am going to verify this result with homogeneous coordinates. There are also two similar ways that we can view the computation. Applying the operators to the point by multiplying the transforms in order. Or take the equation and extract the operators order from it.

$P' = X_{0} R (- X_{0}) P$

Comparing this with the normal linear equation you will notice that this equation is ordered from right to left. For example P-X₀ is done first because it is the inner most bracket hence (-X₀)P. Next the multiplication of R comes before addition so the R operator is applied. This gives R(-X₀)P. Finally the addition is done yielding XR(-X₀)P. So essentially the equations are the same with the homogeneous one being exact in the ordering of operations whereas the first equation can be written in many different ways eg P=X₀+R(P-X₀), P=X₀+R(-X₀+P), ...

$(\begin{matrix} 1 & 0 & x_{0} \\ 0 & 1 & x_{1} \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} r_{0} & r_{1} & 0 \\ r_{2} & r_{3} & 0 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} 1 & 0 & - x_{0} \\ 0 & 1 & - x_{1} \\ 0 & 0 & 1 \end{matrix})$

$(\begin{matrix} r_{0} & r_{1} & - r_{0} x_{0} - r_{1} x_{1} + x_{0} \\ r_{2} & r_{3} & - x_{0} r_{2} - x_{1} r_{3} + x_{1} \\ 0 & 0 & 1 \end{matrix})$

Multiply by P on the rhs.

$(\begin{matrix} r_{0} & r_{1} & - r_{0} x_{0} - r_{1} x_{1} + x_{0} \\ r_{2} & r_{3} & - x_{0} r_{2} - x_{1} r_{3} + x_{1} \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} p_{0} \\ p_{1} \\ 1 \end{matrix})$
$(\begin{matrix} p_{0} & r_{0} & + p_{1} r_{1} - r_{0} x_{0} - r_{1} x_{1} + x_{0} \\ p_{0} & r_{2} & + p_{1} r_{3} - x_{0} r_{2} - x_{1} r_{3} + x_{1} \\ 0 & 0 & 1 \end{matrix})$

This is the same as from standard algebra and hence the technique has been verified which gives me a little more confidence.

Affine Transformations

Translations, rotation and sheering, all these transformations preserve parallel lines. Affine transformation are reversible - the translation determinant is non zero and they form a group.

Rigid motions or Euclidean transformations are rotations + translation and preserve distances and angles. They are a subset of affine transformations.

Affine transformation preserve lines. In a sense affine transformations are linear. From a computation point of view transforming a line means transforming the end points and drawing a line between the transformed points. As opposed to transforming every point on the line.

Transition from Vectors to Matrices

A point in space

A point in space is represented as a sum of vectors.

$\underset{~}{p} = \sum a_{i} \underset{~}{v}_{i}$

In 2D.

$\underset{~}{p} = a_{0} \underset{~}{v}_{0} + a_{1} \underset{~}{v}_{1}$

$= (v_{0}, v_{1}) \cdot (\begin{matrix} a_{0} \\ a_{1} \end{matrix})$

The vectors $v_{i}$ have a vertical orientation, realizing this gives.

$(\begin{matrix} (\begin{matrix} v_{00} \\ v_{01} \end{matrix}) & (\begin{matrix} v_{10} \\ v_{11} \end{matrix}) \end{matrix}) (\begin{matrix} a_{0} \\ a_{1} \end{matrix})$

Ok - it looks ugly. The vectors are still being treated as whole units. Someone noticed that the sum can be summed in another order and give the same outcome.

$\sum a_{i} v_{i} = \sum_{i} a_{i} \sum_{k} v_{i k}$
$= \sum_{k} \sum_{i} a_{i} v_{i k}$

Where k is associated with position. This was called matrix multiplication.

The relevance of the above now is clear. The transition matrix comprise of columns of vectors. A transform is a linear combination of these vectors, and is associated with linear algebra. But in a sense is its own coordinate system.

Natural Order of Transformation Operations

Somewhere along the way matrices evolved (I don't know how, its a huge story) so that transforming is multiplying the point with a matrix on the left hand side.

$M X \to X$ or $X' = M X$

For a practical example consider rotating a point anti-clockwise by 45 degrees.

The point is represented as a 2 by 1 matrix - vertical orientation.

$[\begin{matrix} \cos (45) & - \sin (45) \\ \sin (45) & \cos (45) \end{matrix}] = \frac{1}{2^{\frac{1}{2}}} [\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}]$
rotate $(1, 0)$ by 45 degrees.
$\frac{1}{2^{\frac{1}{2}}}$ $[\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}]$ $[\begin{matrix} 1 \\ 0 \end{matrix}]$ $\to \frac{1}{2^{\frac{1}{2}}}$ $[\begin{matrix} 1 \\ 1 \end{matrix}]$
rotate again by 45 degrees.
$\frac{1}{2^{\frac{1}{2}}} [\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}] \frac{1}{2^{\frac{1}{2}}} [\begin{matrix} 1 \\ 1 \end{matrix}] \to [\begin{matrix} 0 \\ 1 \end{matrix}]$

If R is a 45 degree rotation and X is the point then rotating the point 90 degrees is multiplying X by R twice.

$R R X \to X$

Standard algebra can not accept R as a translation. Homogeneous coordinates came into being to address this by alowing R to be a shift operation.

OpenGL and Homogeneous Matrices

OpenGL implements the homogeneous coordinate system. Rather than applying operations directly to the point, the operations are applied to the matrix which ends up multiplying the point(s). In the final stage this matrix is applied to the point.

Let C be the current matrix and M be the new matrix operation.

$C M \to C$

A little bit of algebra

$C_{i + 1} = C_{i} M_{i}$
$C_{N + 1} = C_{0} M_{0} M_{1} M_{2} ... M_{N}$

This shows the final ordering of the transformation matrix, operations operate on the right hand side and not the left as the natural ordering required.

Example

Now consider the order of transforms to move from the large rectangle to the smaller one.

T1: translate or shift to the origin
S: scale the shape
R: rotate the shape
T2: translate it again

The natural ordering can be expressed in algebra.

$T2 R S T1 X \to X$

But OpenGL multiplies matrices on the right hand side, so you will have to reverse the order of transformations.

  glTranslatef(-0.3,0.0,0.0);
  glRotatef(-45.0,0.0,0.0,1.0);
  glScalef(.707,.707,1.0);
  glTranslatef(-0.1,-0.5,0.0);

Expressing this as a multiplication of matricies and the point.

glTranslatef(-0.3,0.0,0.0)*glRotatef(-45.0,0.0,0.0,1.0)*glScalef(.707,.707,1.0)*glTranslatef(-0.1,-0.5,0.0)*X

Now it is easy to see why people say that OpenGL reverses the order of operations because the order that they are written in code is reversed when applied to the point.

glTranslatef(-0.1,-0.5,0.0)
glScalef(.707,.707,1.0)
glRotatef(-45.0,0.0,0.0,1.0)
glTranslatef(-0.3,0.0,0.0)

Note that swapping the scale and rotate operations does not change the outcome because these operations can be done in any order when they are together. Sorry for any confusion.