Transforms

In linear algebra, transform (or transformation matrix) is used to specify the geometrical relationship between two cartesian coordinate system. There are many use cases:

Co-registration between CT and {term}`MRI> volumes
Specify the 3D location of an object using tracking systems,
Tool calibration, e.g. specifying the location of the tip of surgical drill with respect to the tracking sensor, etc.

That is, in an IGI surgical workflow, there are always multiple rigid bodies, i.e. cartesian coordinate system present:

Target anatomy (spine),
Surgical tool (needle),
Live image or video US) and
Geometrical model or representation of anatomy (e.g. surface model derived from CT)

We must keep track of them all the time by understanding how each of these cartesian coordinate system are related with each other.

Translation¶

The simplest transform is moving a object from one location to another, i.e. a translation.

Let the initial position of an object be $p$ , and let $v$ be a translation vector, then the new position of the object $q$ after translation by $v$ is:

q = p + v

(1)

If both $p$ and $v$ are multi-dimensional, then this is equivalant to vector addition. In 3D, we have $$

\begin{align} q_x & = & q_x + v_x\\ q_y & = & q_y + v_y\\ q_z & = & q_z + v_z \end{align}

(2)

$$

In Python, we can simply use the standard addition operation:

import math
import numpy as np 
import numpy.matlib

p = np.array([ 1, 0, 0 ])      # This is a ROW vector
p = p.reshape(-1, 1)           # this is a COLUMN vector
p = np.array([ [0],[1],[0] ])  # alternative

v = np.array( [4, 5, 6] ).reshape(-1,1)
q = p + v

print("The initial position is:\n", p, 
"\nafter translation by:\n", v,
"\nis now located at:\n", q)

The initial position is:
 [[0]
 [1]
 [0]] 
after translation by:
 [[4]
 [5]
 [6]] 
is now located at:
 [[4]
 [6]
 [6]]

For visualization and other implementations, consult the following tutorial.

Rotation¶

In 3D, rotation is specified as $3 \times 3$ matrix $R$ , and using colomn vector, a point $p=(x,y,z)$ is rotated by $R$ into $q=(x',y',z')$ as:

$$

\begin{align} A & = & \begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i \end{bmatrix} \\ q & = & A p\\ & = & \begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} x'\\ y'\\ z' \end{bmatrix} \\ x' & = & a x + b y + c z\\ y' & = & d x + e y + f z \\ z' & = & g x + h y + i z \end{align}

(3)

$$

Rigid Rotation¶

We are particularly interested in rigid rotation, that that simply rotates an object but do not change its geometry (length, volumen, etc.,). For a $3 \times 3$ matrix to be a rigid rotation, the following properties must be satisfied:

Each row and each column is a unit vector,
The dot product between each row with every other is 0.

In other word, a rigid rotation is akin to set up an orthonormal axes: each axis has a unit length one, and each axis is perpendicular with each other.

One consequence of these two properties is:

The determinant of a rigid rotation is 1.

Let’s look at some examples.

Rotation about z-axis¶

Perhaps the most familiar rotation is in fact the rotation about the $z$ -axis. Most of us are used to perform or study rotation in 2D, i.e. the x-y plane, which is in fact equivalent to rotation about the $z$ -axis.

The rotation is performed about an axis, called the axis of rotation, by an angle $\theta$ . Using the RHR, and when looking from the positive towards the negative of the axis of rotation, the CCW rotation is the positive angle.

That is, if the right-thumb points to the positive direction of the axis of rotation, then the curve of the fingers is the positive rotation.

It should be noted that a point on the axis of rotation is invariant under rotation.

Using the RHR, rotation about $z$ -axis is thus defined as:

R_z(\theta) = \begin{bmatrix} \cos{\theta} & -\sin{\theta} & 0\\ \sin{\theta} & \cos{\theta} & 0\\ 0 & 0 & 1 \end{bmatrix}

(4)

Let’s visualize this: suppose a point $p=(1,0,0)$ and $q=(0,1,0)$ is located on the $x$ - and $y$ -axis, respective, and after rotation, we have

\begin{align} p' & = & \begin{bmatrix} \cos{\theta} & -\sin{\theta} & 0\\ \sin{\theta} & \cos{\theta} & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1\\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} \cos{\theta} \\ \sin{\theta} \\ 0 \end{bmatrix} \\ q' & = & \begin{bmatrix} \cos{\theta} & -\sin{\theta} & 0\\ \sin{\theta} & \cos{\theta} & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0\\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -\sin{\theta} \\ \cos{\theta} \\ 0 \end{bmatrix} \end{align}

(5)

Rotation about z-axis — Figure 1:Rotation about $z$ -axis visualized.

And for any point $r=(0,0,z)$ on the $z$ -axis, rotation about the $z$ -axis does not move $r$ :

r' = \begin{bmatrix} \cos{\theta} & -\sin{\theta} & 0\\ \sin{\theta} & \cos{\theta} & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0\\ 0 \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix} = r

(6)

It can be implemented in Python as:

import numpy as np
import math

def rotZ3x3( angle_rad ):
    # return a 3x3 rotation about z-axis, where angle is specified in radian

    c = math.cos( angle_rad )
    s = math.sin( angle_rad )
    # a 3x3 identity matrix
    R = np.identity(3)

    # specify the elements of the rotation
    R[0,0] =  c
    R[0,1] = -s
    R[1,0] =  s
    R[1,1] =  c
    return R

A simple test:

p = np.array([1,0,0]).reshape(-1,1)

# rotation by 90 degree
print( np.around( np.matmul( rotZ3x3( math.pi/2 ) , p) ) )
# print( rotZ3x3( math.pi/2 )@p )  # short cut, the @ sign is matrix multiplication

[[0.]
 [1.]
 [0.]]

Inverse Rotation¶

What happens if, instead rotation by $\theta$ , one rotates by $-\theta$ instead?

\begin{align} R_z(\theta) & = &\begin{bmatrix} \cos{\theta} & -\sin{\theta} & 0\\ \sin{\theta} & \cos{\theta} & 0\\ 0 & 0 & 1 \end{bmatrix}\\ R_z(-\theta) & = & \begin{bmatrix} \cos{-\theta} & -\sin{-\theta} & 0\\ \sin{-\theta} & \cos{-\theta} & 0\\ 0 & 0 & 1 \end{bmatrix}\\ & = & \begin{bmatrix} \cos{\theta} & \sin{\theta} & 0\\ -\sin{\theta} & \cos{\theta} & 0\\ 0 & 0 & 1 \end{bmatrix}\\ \end{align}

(7)

Suppose you rotate an object by an angle $\theta$ , and immediately rotate it by an angle $-\theta$ : nothing changes. Is it true numerically?

R_z(-\theta) \cdot R_z(\theta) = \begin{bmatrix} \cos{\theta} & \sin{\theta} & 0\\ -\sin{\theta} & \cos{\theta} & 0\\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} \cos{\theta} & -\sin{\theta} & 0\\ \sin{\theta} & \cos{\theta} & 0\\ 0 & 0 & 1 \end{bmatrix}

(8)

\begin{align} R_z(-\theta) \cdot R_z(\theta) & = & \begin{bmatrix} \cos{\theta} & \sin{\theta} & 0\\ -\sin{\theta} & \cos{\theta} & 0\\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} \cos{\theta} & -\sin{\theta} & 0\\ \sin{\theta} & \cos{\theta} & 0\\ 0 & 0 & 1 \end{bmatrix} \\ & = & \begin{bmatrix} \cos{\theta} \cos{\theta} + \sin{\theta} \sin{\theta} & -\cos{\theta} \sin{\theta} + \cos{\theta} \sin{\theta} & 0\\ -\cos{\theta} \sin{\theta} + \cos{\theta} \sin{\theta} & \sin{\theta} \sin{\theta} + \cos{\theta} \cos{\theta} & 0\\ 0 & 0 & 1 \end{bmatrix}\\ & = & \begin{bmatrix} \cos{\theta}^2 + \sin{\theta}^2 & 0 & 0\\ 0 & \sin{\theta}^2 + \cos{\theta}^2 & 0\\ 0 & 0 & 1 \end{bmatrix}\\ & = & \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \end{align}

(9)

You should also notice that $R_z(-\theta)=R_z(\theta)^T$ , the transpose.

# rotation by pi/4 followed by rotation by -pi/4
print( np.around( np.matmul( rotZ3x3( -math.pi/4 ) , rotZ3x3( math.pi/4)) ) )

[[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]

Rotation about $x$ -axis¶

To visualize how to perform rotation about the $x$ -axis, position yourself at $+x$ direction looking towards the origin.

Again, a point on the $x$ -axis remains invariant under rotation about the $x$ -axis.

Using this visualization, one can infer that the $3\times 3$ rotation about the $x$ -axis by an angle $\theta$ is:

R_x(\theta) = \begin{bmatrix} 1 & 0 & 0\\ 0 & \cos{\theta} & -\sin{\theta}\\ 0 & \sin{\theta} & \cos{\theta} \end{bmatrix}

(10)

def rotX3x3( angle_rad ):
    # return a 3x3 rotation about z-axis, where angle is specified in radian

    c = math.cos( angle_rad )
    s = math.sin( angle_rad )
    # a 3x3 identity matrix
    R = np.identity(3)

    # specify the elements of the rotation
    R[1,1] =  c
    R[1,2] = -s
    R[2,1] =  s
    R[2,2] =  c
    return R

[[ 1.  0.  0.]
 [ 0.  0. -1.]
 [ 0.  1.  0.]]

Rotation about $y$ -axis¶

To visualize how to perform rotation about the $y$ -axis, position yourself at $+y$ direction looking towards the origin.

Again, a point on the $y$ -axis remains invariant under rotation about the $y$ -axis.

Using this visualization, one can infer that the $3\times 3$ rotation about the $y$ -axis by an angle $\theta$ is:

R_y(\theta) = \begin{bmatrix} \cos{\theta} & 0 & \sin{\theta}\\ 0 & 1 & 0\\ -\sin{\theta} & 0 & \cos{\theta} \end{bmatrix}

(11)

def rotY3x3( angle_rad ):
    # return a 3x3 rotation about z-axis, where angle is specified in radian

    c = math.cos( angle_rad )
    s = math.sin( angle_rad )
    # a 3x3 identity matrix
    R = np.identity(3)

    # specify the elements of the rotation
    R[0,0] =  c
    R[0,2] =  s
    R[2,0] = -s
    R[2,2] =  c
    return R

Successive Rotations¶

Suppose you are interested in rotating an object $p$ about the $x$ -axis by an angle $\alpha$ , and further by rotating the rotated object about the $y$ -axis by an angle $\beta$ , how would this be achieved?

Let $p'$ be the rotated object after the rotation about the $x$ -axis by an angle $\alpha$ :

p'_{3\times 1} = R_x(\alpha)_{3\times 3} \cdot p_{3\times 1}

(12)

and $p''$ be the rotated object after the rotation about the $y$ -axis by an angle $\beta$ :

p''_{3 \times 1} = R_y(\beta)_{3 \times 3} \cdot p'_{3\times 1}

(13)

Thus

\begin{align} p''_{3 \times 1} & = & R_y(\beta)_{3 \times 3} \cdot p'_{3\times 1}\\ & = & R_y(\beta)_{3 \times 3} \cdot R_x(\alpha)_{3\times 3} \cdot p_{3\times 1} \end{align}

(14)

Because matrix multiplication is associative, we can indeed define a combined rotation matrix:

R_{3 \times 3} = R_y(\beta)_{3 \times 3} \cdot R_x(\alpha)_{3\times 3}

(15)

and

p''_{3 \times 1}=R_{3 \times 3} \cdot p_{3\times 1}

(16)

That is, instead of thinking about successive rotations as a series of rotations, it is often more intuitive to think of it as just one rotation about an axis of rotation instead.

For visualization and other implementations, consult the following tutorial.

Rotation About an Unit Vector¶

One omission in the above introduction to rotation, one is often done, is that rotation is performed about a vector. In the previous examples, i.e. rotation about $x$ -, $y$ -, and $z$ -axis, is in fact rotation about the orthonormal basis located at the origin.

That is, rotation about the $x$ -axis should been seen as rotation about the unit $i=(1,0,0)$ originated at $(0,0,0)$

In fact, the rotation can be perform about an arbitrary axis in space. The axis of rotation is specified as a unit orientation vector $v=(v_x,v_y,v_z)$ anchord at a point $p=(p_x,p_y,p_z)$ .

If the axis of rotation $v$ is originated at the origin $(0,0,0)$ , than rotation about the unit vector $v=(v_x,v_y,v_z)$ by an angle $\theta$ can be represented as a $3\times 3$ rotation matrix using the following formula:

R(v,\theta) = \begin{bmatrix} v_x \cdot v_x \cdot (1-\cos{\theta}) + \cos{\theta} & v_x \cdot v_y \cdot (1-\cos{\theta}) - v_z \cdot \sin{\theta} & v_x \cdot v_z \cdot (1-\cos{\theta}) + v_y \cdot \sin{\theta}\\ v_y \cdot v_x \cdot (1-\cos{\theta}) + v_z \cdot \sin{\theta} & v_y \cdot v_y \cdot (1-\cos{\theta}) + \cos{\theta} & v_y \cdot v_z \cdot (1-\cos{\theta}) - v_x \cdot \sin{\theta}\\ v_z \cdot v_x \cdot (1-\cos{\theta}) - v_y \cdot \sin{\theta} & v_z \cdot v_y \cdot (1-\cos{\theta}) + v_x \cdot \sin{\theta} & v_z \cdot v_z \cdot (1-\cos{\theta}) + \cos{\theta}\\ \end{bmatrix}

(17)

def axang2rotm( anang ):
    #
    # Convert the axis-angle rotation into a 3x3 rotation matrix
    #
    # The name is the same as the equivalent function in Matlab (under Navigation Toolbox)
    #
    # Assumes that axang is a row vector of 4:
    #
    # axanx = numpy.array([vx, vy, vz, angle])
    #
    #    where (vx,vy,vz) is axis of rotation, and
    #          angle in radian
    #

    # extract components from the input vector
    x     = anang[0]
    y     = anang[1]
    z     = anang[2]
    angle = anang[3]

    # make sure that and axis of rotation is a unit vector
    l = math.sqrt(x*x + y*y + z*z)
    x = x/l
    y = y/l
    z = z/l
    
    R = np.identity(3)
    
    xx = x * x
    xy = x * y
    xz = x * z

    yy = y * y
    yz = y * z
    
    zz = z * z
    
    ca = math.cos( angle )
    sa = math.sin( angle )
    vers_a = 1 - ca # versine(angle) = 2 * sin^2(angle/2)
    
    R[0,0] = xx * vers_a +     ca
    R[0,1] = xy * vers_a - z * sa
    R[0,2] = xz * vers_a + y * sa
    
    R[1,0] = xy * vers_a + z * sa
    R[1,1] = yy * vers_a +     ca
    R[1,2] = yz * vers_a - x * sa
    
    R[2,0] = xz * vers_a - y * sa
    R[2,1] = yz * vers_a + x * sa
    R[2,2] = zz * vers_a +     ca
    
    return R

You can verify if we can use the above function to reproduce rotation about $x$ -, $y$ -, and $z$ -axis:

print( "rotation about x-axis by 90 degree is:\n", np.around(axang2rotm(np.array([1,0,0,math.pi/2]))))
print( "rotation about y-axis by 90 degree is:\n", np.around(axang2rotm(np.array([0,1,0,math.pi/2]))))
print( "rotation about z-axis by 90 degree is:\n", np.around(axang2rotm(np.array([0,0,1,math.pi/2]))))

rotation about x-axis by 90 degree is:
 [[ 1.  0.  0.]
 [ 0.  0. -1.]
 [ 0.  1.  0.]]
rotation about y-axis by 90 degree is:
 [[ 0.  0.  1.]
 [ 0.  1.  0.]
 [-1.  0.  0.]]
rotation about z-axis by 90 degree is:
 [[ 0. -1.  0.]
 [ 1.  0.  0.]
 [ 0.  0.  1.]]

We can verify our formulation against vtkTransform

import math
import numpy as np

from vtkmodules.vtkCommonTransforms import vtkTransform

myTransform = vtkTransform()
myTransform.PostMultiply()
myTransform.Identity()
myTransform.RotateWXYZ( 33, 1, 2, 3) # angle in degree

print("Our implementation: \n", axang2rotm(np.array([1,2,3,33*math.pi/180])), "\n")
print("VTK implementation: \n", myTransform.GetMatrix())

Our implementation: 
 [[ 0.8501941  -0.41363565  0.3256924 ]
 [ 0.45972978  0.88476469 -0.07641972]
 [-0.25655122  0.21470209  0.94238235]] 

VTK implementation: 
 vtkMatrix4x4 (000002829DD518B0)
  Debug: Off
  Modified Time: 84
  Reference Count: 2
  Registered Events: (none)
  Elements:
    0.850194 -0.413636 0.325692 0 
    0.45973 0.884765 -0.0764197 0 
    -0.256551 0.214702 0.942382 0 
    0 0 0 1

What If the Axis of Rotation does not Originate from the Origin?¶

If the axis of rotation $v$ does not originate from the origin, i.e. $O\ne P + v t$ for any $t$ and a fixed point $P$ on the line, then rotation about $P+v t$ by an angle $\theta$ is performe in 3 steps:

Translate by $-P$ , i.e. so now the rotation is performed about $v$ that is originated from the origin $O=(0,0,0)$ ,
Perform rotation about $v$ by $\theta$ , followed by
Translate by $P$

Homogeneous Transformation¶

Suppower you want to rotate an object (e.g. a point) $a$ about a vector $P+v t$ that does not originate from the origin (i.e. $P\ne (0,0,0)$ ), base on this discussion you would need to perform the following operation in succession:

Translate by $-P$ , i.e.
$a' = -1 * P + a$
(18)
Rotate by $v$ by $\theta$ , i.e.
$a'' = R(v,\theta)_{3\times 3} \cdot a'$
(19)
Translate by $+P$ , i.e.
$a''' = P + a''$
(20)
which involves $1\times 3$ column vectors (translation) and $3\times 3$ square matrices (rotation), with two operators (vectors addition and matrix multiplication). For a longer, successive operations of translations and rotation, both the notation and mathematical operation will become intractable very quickly.

The solution is to work in homogeneous coordinates instead. That is, instead of using $3\times 3$ matrix and $1\times 3$ vector to represent rotation and translation, respectively, one use $4 \times 4$ matrix to represent both.

Translation¶

Traslation by an amount $t=(t_x,t_y,t_z)$ is represented by

T(t) = \begin{bmatrix} 1&0&0&t_x\\ 0&1&0&t_y\\ 0&0&1&t_z\\ 0 & 0 & 0&1 \end{bmatrix}_{4 \times 4}

(21)

Rotation¶

Rotation about a vector $v$ originalted from the origin by an angle $\theta$ is represented by

T(v,\theta) = \begin{bmatrix} R(v,\theta)_{3\times 3} & 0_{1\times 3}\\ 0_{3\times 1} & 1 \end{bmatrix}_{4 \times 4}

(22)

For example, rotation about $x$ -axis by an angle $\theta$ is now represented as:

R_x(\theta) = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & \cos{\theta} & -\sin{\theta} & 0\\ 0 & \sin{\theta} & \cos{\theta} & 0\\ 0&0&0&1 \end{bmatrix}

(23)

Rotation about $y$ -axis by an angle $\theta$ is:

R_y(\theta) = \begin{bmatrix} \cos{\theta} & 0 & \sin{\theta} & 0\\ 0 & 1 & 0 & 0\\ -\sin{\theta} & 0 & \cos{\theta} &0\\ 0&0&0&1 \end{bmatrix}

(24)

Rotation about $z$ -axis by an angle $\theta$ is:

R_z(\theta) = \begin{bmatrix} \cos{\theta} & -\sin{\theta} & 0 & 0\\ \sin{\theta} & \cos{\theta} & 0 & 0\\ 0 & 0 & 1 & 0\\ 0&0&0&1 \end{bmatrix}

(25)

Successive Transforms¶

Notice that both translation and rotation are now represented as $4\times 4$ matrices. Apply either the translation or the rotation operation is now performed as matrix multiplication. To translate a point $P={P_x,P_y,P_z}$ by $t={t_x,t_y,t_z}$ , append a 1 as the $4^{th}$ column:

P' = \begin{bmatrix} P'_x \\ P'_y \\ P'_z \\ 1\end{bmatrix}_{1\times 4} = \begin{bmatrix} 1&0&0&t_x\\ 0&1&0&t_y\\ 0&0&1&t_z\\ 0 & 0 & 0&1 \end{bmatrix}_{4 \times 4} \cdot \begin{bmatrix} P_x \\ P_y \\ P_z \\ 1 \end{bmatrix}_{1 \times 4}

(26)

and after matrix multiplication:

\begin{align} P'_x & = & P_x + t_x \\ P'_y & = & P_y + t_y \\ P'_z & = & P_z + t_z \end{align}

(27)

which is same as before.

Similiarly, rotation by the $x$ -axis by an angle $\theta$ is:

P' = \begin{bmatrix} P'_x \\ P'_y \\ P'_z \\ 1\end{bmatrix}_{1\times 4} = R_x(\theta) = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & \cos{\theta} & -\sin{\theta} & 0\\ 0 & \sin{\theta} & \cos{\theta} & 0\\ 0&0&0&1 \end{bmatrix} \cdot \begin{bmatrix} P_x \\ P_y \\ P_z \\ 1\end{bmatrix}_{1\times 4}

(28)

\begin{align} P'_x & = & P_x \\ P'_y & = & \cos{\theta} P_y - \sin{\theta} P_z \\ P'_z & = & \sin{\theta} P_y + \sin{\theta} P_z \end{align}

(29)

which is same as before.

Successive transformations such as rotation about an arbitrary vector $P + v \cdot t$ is now represented as matrices multiplication:

b = \begin{bmatrix} b_x \\ b_y \\ b_z \\ 1\end{bmatrix}_{1\times 4} = \begin{bmatrix} 1&0&0&P_x\\ 0&1&0&P_y\\ 0&0&1&P_z\\ 0 & 0 & 0&1 \end{bmatrix}_{4 \times 4} \begin{bmatrix} R(v,\theta) & 0_{3\times 1}\\ 0_{1\times 3}&1 \end{bmatrix}_{4\times 4} \begin{bmatrix} 1&0&0&-P_x\\ 0&1&0&-P_y\\ 0&0&1&-P_z\\ 0 & 0 & 0&1 \end{bmatrix}_{4 \times 4} \begin{bmatrix} a_x \\ a_y \\ a_z \\ 1\end{bmatrix}_{1\times 4}

(30)

Moreover, the inverse transform can be trivially found by computing the inverse of the matrices. This left as an exercise for you. You may wish to consult this old but concise document for detail.

For visualization and other implementations, consult the following tutorial.

Change of Persectives¶

Suppose Alice and Bob are sitting on a couch watching TV. The TV is directly in front of Alice, and Bob sits on the right of Alice. Where is the TV from Bob’s perspective?

For the purpose of the discussion, let the COG of the TV denotes the position of the TV. If we assign a cartesian coordinate system to Bob, we can specify a coordinate the location of the TV from Bob’s perspective. For this trivial example, suppose the TV is located at $p=(0,5,0)$ at the coordinate system associated with Bob

Location of TV from Bob's perspective — Figure 4:The position, denoted as a red circle, is specified from Bob’s perspective.

Suppose Alice is sitting sitting at $a=(-3,0,0)$ (from Bob’s perspective). We can also assign a cartesian coordinate system to Alice: for the moment, assume that the orientation of Alice’s and Bob’s cartesian coordinate system have the same orientation. That is, both of their $+x$ -axis point to the same direction, and both their $+y$ -axis point to the same direction.

Location of TV from Alice's perspective — Figure 5:Alice sits next to Bob.

Where is the TV ( $p$ ) from Alice’s perspective?

From Figure 5 you may infer that, because Bob is at $(3,0,0)$ (i.e. inverse of where Alice is from Bob), and the TV is located at $(0,5,0)$ from Bob, the TV is located at $(3,5,0)$ from Alice.

But how do we express this mathematically?

Transformation: Translation¶

Let $^{Bob}T_{Alice}$ denote the geometrical relationship between Alice and Bob from Bob’s perspective:

^{Bob}T_{Alice} = \begin{bmatrix} 1 & 0 & 0 & -3\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}

(31)

We adopt the notation of $^{superscript}$ and $_{subscript}$ to indicate the transformation $_{from}$ one coordinate system to $^{another}$ . For example, the location of Alice, $(0,0,0)$ , in her local coordinate system can be brought from $_{Alice}$ to $^{Bob}$ via:

^{Bob}\begin{bmatrix} -3 \\ 0 \\ 0 \\ 1 \end{bmatrix} = ^{Bob}\begin{bmatrix} 1 & 0 & 0 & -3\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}_{Alice} \cdot ^{Alice}\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}

(32)

Conversely, from Alice’s point of view, Bob is located at $(3,0,0)$ , and

^{Alice}T_{Bob} = \begin{bmatrix} 1 & 0 & 0 & 3\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}

(33)

Notice that $^{Alice}T_{Bob}$ is the inverse of $^{Bob}T_{Alice}$ (You should verify this by computing the inverse of the matrices themselves). Notice that the position of the TV is $p=(0,5,0)$ from Bob’s perspective, we can compute where it is from Alice’s perspective using the transformation:

^{Alice}\begin{bmatrix} 3\\5\\0\\1\end{bmatrix} = ^{Alice}\begin{bmatrix} 1 & 0 & 0 & 3\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}_{Bob} \cdot ^{Bob}\begin{bmatrix} 0\\5\\0\\1\end{bmatrix}

(34)

One Key to help you in writing down this sequence os matrix multiplication is to note that the successive $_{subscript}$ and $^{superscript}$ must match.

To illustrate this point one more time, now we know that the TV, from Alice’s perspective, is located at $(3,5,0)$ . To find out where it is from Bob’s perspective, we use:

^{Bob}\begin{bmatrix} 0 \\ 5 \\ 0 \\ 1\end{bmatrix} = ^{Bob}\begin{bmatrix} 1 & 0 & 0 & -3\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}_{Alice} \cdot ^{Alice}\begin{bmatrix} 3 \\ 5 \\ 0 \\ 1 \end{bmatrix}

(35)

Transformation: Rotation and Translation¶

In the previous discussion, the transform only involves a simple translation (between Alice and Bob). What if Alice defines her local coordinate system differently than Bob (say, she’s left handed and Bob is right handed?).

Alice's rotated perspective — Figure 7:The orientation of the local coordinate system of Alice is different than that of Bob.

You will notice that there is a rotation about the $z$ -axis by $90\degree$ between Alice’s and Bob’s coordinate systems:

^{Bob}T_{Alice} = \begin{bmatrix} 0 & -1 & 0 & -3\\ 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}

(36)

and

^{Alice}T_{Bob} = \begin{bmatrix} 0 & 1 & 0 & 0\\ -1 & 0 & 0 & -3\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}

(37)

And in this new coordinate system of Alice, the location of the TV $p(0,5,0)$ is now:

^{Alice}\begin{bmatrix} 5\\-3\\0\\1\end{bmatrix} = ^{Alice}\begin{bmatrix} 0 & 1 & 0 & 0\\ -1 & 0 & 0 & -3\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}_{Bob} \cdot ^{Bob}\begin{bmatrix} 0\\5\\0\\1\end{bmatrix}

(38)

TV from Alice's perspective — Figure 8:The coordinate of the TV from Alice’s perspective.

But How?¶

To walk you through the process of constructing the transform that links one coordinate system with another, refer to Figure 8. To construct the transformation $^{Bob}T_{Alice}$ that see what Alice sees from Bob’s perspective, consider

How are the base vectors of Alice’s coordinate system mapped to Bob’s, and
How is the origin of Alice’s coordinate system mapped to Bob’s.

We proceed by noticing that

The $+x$ -axis of Alice’s coordinate system, which maps to Bob’s $+y$ -axis ( $1^{st}$ colume of $^{Bob}T_{Alice}$ ),
The $+y$ -axis of Alice’s coordinate system, which maps to Bob’s $-x$ -axis ( $2^{nd}$ colume of $^{Bob}T_{Alice}$ ),
The $+z$ -axis of Alice’s coordinate system, which maps to Bob’s $+z$ -axis ( $3^{rd}$ colume of $^{Bob}T_{Alice}$ , i.e. no change), and
The origin of Alice’s coordinate system is located at $(-3,0,0)$ from Bob’s point of view ( $4^{th}$ column of $^{Bob}T_{Alice}$ ).

That is,

The rotational component of the transformation is to map the base axes from one another, and
The translational component of the transformation is the translation of the origins.

Translation¶

Rotation¶

Rigid Rotation¶

Rotation about z-axis¶

Inverse Rotation¶

Rotation about xxx-axis¶

Rotation about yyy-axis¶

Successive Rotations¶

Rotation About an Unit Vector¶

What If the Axis of Rotation does not Originate from the Origin?¶

Homogeneous Transformation¶

Translation¶

Rotation¶

Successive Transforms¶

Change of Persectives¶

Transformation: Translation¶

Transformation: Rotation and Translation¶

But How?¶

Rotation about $x$ -axis¶

Rotation about $y$ -axis¶