Rotation#

Rotation in $\mathbb{R}^3$ is a linear transformation that rotates a set of points by some angle about one of the $x$, $y$ or $z$-axes. The direction of rotation is assumed to be anti-clockwise when viewed looking down the axis towards the origin (Fig. 37).

../_images/rotation.svg — Fig. 38 Rotation of the vector $\mathbf{u}$ anti-clockwise about the origin.#

Consider the diagram in Fig. 38 that shows the axes from Fig. 37 when viewed looking down the $x$-axis. The point with position vector $\mathbf{u}=(u_x, u_y, u_z)$ is rotated anti-clockwise about the $x$-axis by the angle $\theta$ to $\mathbf{v}=(v_x, v_y, v_z)$. To determine the linear transformation we first consider the rotation of a vector pointing along the $x$-axis which has the same magnitude of $\mathbf{u}$, i.e., $|\mathbf{u}| \mathbf{e}_1$, rotated by the angle $\phi$. Forming a right-angled triangle with the angle $\phi$ and hypotenuse of length $|\mathbf{u}|$ then

(26)#\[\begin{split}\cos(\phi) &= \frac{u_x}{|\mathbf{u}|} & \therefore u_x &= |\mathbf{u}| \cos(\phi), \\ \sin(\phi) &= \frac{u_y}{|\mathbf{u}|} & \therefore u_y &= |\mathbf{u}| \sin(\phi).\end{split}\]

Doing similar for rotating $\mathbf{u}$ by the angle $\theta$ to $\mathbf{v}$ we have

\[\begin{split} \begin{align*} v_x &= |\mathbf{u}| \cos(\phi + \theta), \\ v_y &= |\mathbf{u}| \sin(\phi + \theta), \\ v_z &= u_x \end{align*} \end{split}\]

using the angle sum indentities

\[\begin{split} \begin{align*} \cos(\phi + \theta) &= \cos(\phi) \cos(\theta) - \sin(\phi) \sin(\theta), \\ \sin(\phi + \theta) &= \sin(\phi) \cos(\theta) + \cos(\phi) \sin(\theta), \end{align*} \end{split}\]

then we have

(27)#\[\begin{split}$$ \begin{align*} v_x &= |\mathbf{u}| \cos(\phi) \cos(\theta) - |\mathbf{u}|\sin(\phi) \sin(\theta) \\ v_y &= |\mathbf{u}| \sin(\phi) \cos(\theta) + |\mathbf{u}|\cos(\phi) \sin(\theta). \end{align*} $$\end{split}\]

Substituting equation (26) into equation (27)

\[\begin{split} \begin{align*} v_x &= u_x \cos(\theta) - u_y \sin(\theta), \\ v_y &= u_y \cos(\theta) + u_x \sin(\theta), \\ v_z &= u_z \end{align*} \end{split}\]

which can be written as the matrix equation.

\[\begin{split} \begin{align*} \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} &= \begin{pmatrix} \cos(\theta) & -\sin(\theta) & 0 \\ \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} u_x \\ u_y \\ u_z \end{pmatrix}. \end{align*} \end{split}\]

The square matrix is the transformation matrix for rotation by the angle $\theta$ anti-clockwise about the $z$-axis, i.e., in Fig. 38 we are looking down the $z$-axis. Doing similar we can determine the equivalent matrices for rotating about the $x$ and $y$-axes.

Theorem 8 (Rotation in $\mathbb{R}^3$)

The rotation of a vector in $\mathbb{R}^3$ expressed using homogeneous co-ordinates anti-clockwise by an angle $\theta$ about the $x$, $y$ and $z$ axes is the linear transformation given by the following rotation matrices.

(28)#\[\begin{split}\begin{align*} R_x(\theta) &= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos(\theta) & -\sin(\theta) & 0 \\ 0 & \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \\ R_y(\theta) &= \begin{pmatrix} \cos(\theta) & 0 & \sin(\theta) & 0 \\ 0 & 1 & 0 & 0 \\ -\sin(\theta) & 0 & \cos(\theta) & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \\ R_z(\theta) &= \begin{pmatrix} \cos(\theta) & -\sin(\theta) & 0 & 0 \\ \sin(\theta) & \cos(\theta) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*}\end{split}\]

Since $\cos(-\theta) = \cos(\theta)$ and $\sin(-\theta) = -\sin(\theta)$, the transformation matrices for inverse rotation, rotation in the clockwise direction, are

(29)#\[\begin{split}\begin{align*} R_x^{-1}(\theta) &= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos(\theta) & \sin(\theta) & 0 \\ 0 & -\sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \\ R_y^{-1}(\theta) &= \begin{pmatrix} \cos(\theta) & 0 & -\sin(\theta) & 0 \\ 0 & 1 & 0 & 0 \\ \sin(\theta) & 0 & \cos(\theta) & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \\ R_z^{-1}(\theta) &= \begin{pmatrix} \cos(\theta) & \sin(\theta) & 0 & 0 \\ -\sin(\theta) & \cos(\theta) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*}\end{split}\]

Example 26

A triangular polygon has vertices located at $\mathbf{p}_1 = (4, 1, 2)$, $\mathbf{p}_2 = (6, 1, 2)$ and $\mathbf{p}_3 = (5, 3, 2)$. The triangle is rotated by angle $\theta = \pi/4$ anti-clockwise about the $z$-axis. Calculate the positions of the vertices of the rotated triangle.

MATLAB code#

The following MATLAB code applies the translation from Example 26 and plots the original and rotated polygons.

% Define homogeneous co-ordinate matrix
P = [ 4, 6, 5 ;
      1, 1, 3 ;
      2, 2, 2 ;
      1, 1, 1 ];


% Define rotation matrix
Rz = @(theta) [ cos(theta), -sin(theta), 0, 0 ; 
                sin(theta), cos(theta), 0, 0 ;
                0, 0, 1, 0 ;
                0, 0, 0, 1 ];

% Apply rotation
theta = pi / 4;
P1 = Rz(theta) * P;

% Plot polygons
figure
patch(P(1,:), P(2,:), 'b', FaceAlpha=0.5)
patch(P1(1,:), P1(2,:), 'r', FaceAlpha=0.5)
axis equal
axis([0, 7, 0, 7])
xlabel("$x$", FontSize=12, Interpreter="latex")
ylabel("$y$", FontSize=12, Interpreter="latex")
box on

../_images/rotation_example.png — Fig. 39 Rotating a polygon about the origin.#

Rotation about polygon centre#

We saw in Example 26 that the polygon has shifted position because we rotate about the origin. If we wanted to rotate the polygon about its centre we first need to translate it so that its centre is at the origin, apply the rotation transformation and then translate the centre to the original position (Fig. 40).

../_images/rotate_about_centre.svg — Fig. 40 Steps required to rotate a polygon by its centre.#

If the centre of the polygon is at point $\mathbf{c}=(c_x, c_y, c_z)$ then the first transformation is translation by $-\mathbf{c}$ and the transformation matrix is

\[\begin{split} \begin{align*} T(-\mathbf{c}) = \begin{pmatrix} 1 & 0 & 0 & -c_x \\ 0 & 1 & 0 & -c_y \\ 0 & 0 & 1 & -c_z \\ 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*} \end{split}\]

The second transformation is rotation around the origin using equation (28) and the third transformation is translation by $\mathbf{c}$, so that the centre is back to the original position, using

\[\begin{split} \begin{align*} T(\mathbf{c}) = \begin{pmatrix} 1 & 0 & 0 & c_x \\ 0 & 1 & 0 & c_y \\ 0 & 0 & 1 & c_z \\ 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*} \end{split}\]

The transformation matrix for the composite transformation that achieves rotation of a polygon about its centre is

\[ \begin{align*} A = T(\mathbf{c}) \cdot R (\theta) \cdot T(-\mathbf{c}), \end{align*} \]

where $R(\theta)$ is either $R_x(\theta)$, $R_y(\theta)$ or $R_z(\theta)$.

Example 27

Rotate the polygon from Example 26 by $\theta = \pi/4$ around the $z$-axis about its centre.

Solution

The homogeneous co-ordinate matrix is

\[\begin{split} \begin{align*} P &= \begin{pmatrix} 4 & 6 & 5 \\ 0 & 0 & 0 \\ 1 & 1 & 3 \\ 1 & 1 & 1 \end{pmatrix}, \end{align*} \end{split}\]

and since the polygon is a triangle the co-ordinates of the centre are the average of the vertices

\[\begin{split} \begin{align*} \mathbf{c} &= \frac{1}{3} \left( \begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix} + \begin{pmatrix} 6 \\ 1 \\ 2 \end{pmatrix} + \begin{pmatrix} 5 \\ 3 \\ 2 \end{pmatrix} \right) = \begin{pmatrix} 5 \\ 5/3 \\ 2 \end{pmatrix}. \end{align*} \end{split}\]

The individual transformation matrices are

\[\begin{split} \begin{align*} T(-\mathbf{c}) &= \begin{pmatrix} 1 & 0 & 0 & -5 \\ 0 & 1 & 0 & -5/3 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \\ R_y\left(\frac{\pi}{4}\right) &= \begin{pmatrix} \sqrt{2}/2 & -\sqrt{2}/2 & 0 & 0 \\ \sqrt{2}/2 & \sqrt{2}/2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \\ T(\mathbf{c}) &= \begin{pmatrix} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & 5/3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \end{align*} \end{split}\]

which are multiplied to give the composite transformation matrix

\[\begin{split} \begin{align*} A &= T(\mathbf{c}) \cdot R_y\left(\frac{\pi}{4}\right) \cdot T(-\mathbf{c}) \\ &= \begin{pmatrix} \sqrt{2}/2 & -\sqrt{2}/2 & 0 & 5 - 5\sqrt{2}/3 \\ \sqrt{2}/2 & \sqrt{2}/2 & 0 & 5/3 - 10\sqrt{2}/3 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \end{align*} \end{split}\]

Applying the composite transformation

\[\begin{split} \begin{align*} A \cdot P &= \begin{pmatrix} \sqrt{2}/2 & -\sqrt{2}/2 & 0 & 5 - 5\sqrt{2}/3 \\ \sqrt{2}/2 & \sqrt{2}/2 & 0 & 5/3 - 10\sqrt{2}/3 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 4 & 6 & 5 \\ 1 & 1 & 3 \\ 2 & 2 & 2 \\ 1 & 1 & 1 \end{pmatrix} \\ &= \begin{pmatrix} 5 - \sqrt{2}/6 & 5 + 5\sqrt{2}/6 & 5 - 2\sqrt{2}/3 \\ 5/3 - 5\sqrt{2}/6 & 5/3 + \sqrt{2}/6 & 5/3 + 2 \sqrt{2}/3 \\ 2 & 2 & 2 \\ 1 & 1 & 1 \end{pmatrix} \\ &\approx \begin{pmatrix} 4.7632 & 6.1785 & 4.0572 \\ 0.4882 & 1.9024 & 2.6095 \\ 2 & 2 & 2\\ 1 & 1 & 1 \end{pmatrix}. \end{align*} \end{split}\]

So the vertices of the rotated polygon are $(4.76, 0.49, 2)$, $(6.18, 1.90, 2)$ and $(4.06, 2.61, 2)$. The original polygon and the rotated polygon are plotted below looking along the $z$ axis.

MATLAB code#

The following MATLAB code applies the translation from Example 27 and plots the original and rotated polygons.

% Define homogeneous co-ordinate matrix
P = [ 4, 6, 5 ;
      1, 1, 3 ;
      2, 2, 2 ;
      1, 1, 1 ];

% Calculate polygon centre
c = mean(P, 2);

% Define translation, scaling and rotation matrices
T = @(t) [ 1, 0, 0, t(1) ; 
           0, 1, 0, t(2) ; 
           0, 0, 1, t(3) ; 
           0, 0, 0, 1 ];

S = @(s) [ s(1), 0, 0, 0 ; 
           0, s(2), 0, 0 ; 
           0, 0, s(3), 0 ; 
           0, 0, 0, 1 ];

Rz = @(theta) [ cos(theta), -sin(theta), 0, 0 ;  
                sin(theta), cos(theta), 0, 0 ;
                0, 0, 1, 0 ;
                0, 0, 0, 1 ];

% Apply transformations
theta = pi/4;
P1 = T(c) * Rz(theta) * T(-c) * P;

% plot polygons
figure
patch(P(1,:), P(2,:), 'b', FaceAlpha=0.5)
patch(P1(1,:), P1(2,:), 'r', FaceAlpha=0.5)
axis equal
axis([0, 7, 0, 7])
xlabel("$x$", FontSize=12, Interpreter="latex")
ylabel("$z$", FontSize=12, Interpreter="latex")
box on

../_images/rotation_about_centre_example.png — Fig. 41 Rotating a polygon about its centre.#

Rotation about a line#

Suppose we wish to rotate $\mathbb{R}^3$ by an angle $\theta$ about a general line that passes through the point $\mathbf{p}=(p_x,p_y,p_z)$ with direction $\mathbf{d}$ that does not pass through the origin and $\mathbf{d}=(d_x,d_y,d_z)$ is not parallel to any of the three co-ordinate axes (Fig. 42). To achieve this we first need to apply translation and rotation so that the line lies along one of the co-ordinate axes, preform the rotation about this axis before reversing the rotation and translation operations. Each of the individual transformations can be represented by a matrix so the composite transformation that achieves the rotation about the line is a product of the individual matrices.

../_images/rotate_about_line_1.svg — Fig. 42 Rotation about a line.#

The first transformation is to translate by $-\mathbf{p}$ so that the line passes through the origin so the transformation matrix is

\[\begin{split} \begin{align*} T(-\mathbf{p}) = \begin{pmatrix} 1 & 0 & 0 & -p_x \\ 0 & 1 & 0 & -p_y \\ 0 & 0 & 1 & -p_z \\ 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*} \end{split}\]

../_images/rotate_about_line_2.svg — Fig. 43 The line is rotated about the $z$-axis so that $\mathbf{d}$ is in the $yz$ plane.#

Fig. 43 shows that affect of translating by $-\mathbf{p}$. Now we need to rotate the line so that it lies in a plane the contains two of the three co-ordinate axes. The rotations required will depend on which of the eight directions $\mathbf{d}$ is pointing in. In this case $\mathbf{d}$ points towards the positive direction for all three axes, if we rotate about the $z$ axis anti-clockwise by angle $\phi$ then $\mathbf{d}$ (and therefore the line) will be in the $yz$ plane. The values of $\cos(\phi)$ and $\sin(\phi)$ in $R_z(\phi)$ are

\[ \begin{align*} \cos(\phi) &= \frac{|d_y|}{d_1}, & \sin(\phi) &= \frac{|d_x|}{d_1}, \end{align*} \]

where $d_1 = \sqrt{d_x^2 + d_y^2}$ is the hypotenuse of the right-angled triangle formed by projecting $\mathbf{d}$ onto the $xy$ plane. So the rotation matrix is

\[\begin{split} \begin{align*} R_z(\phi) &= \begin{pmatrix} |d_y|/d_1 & -|d_x|/d_1 & 0 & 0 \\ |d_x|/d_1 & |d_y|/d_1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*} \end{split}\]

../_images/rotate_about_line_3.svg — Fig. 44 The line is rotated about the $x$-axis so that $\mathbf{d}$ points along the $y$ axis.#

Fig. 44 shows that affect of rotating about the $z$-axis. Now we need to rotate clockwise about the $x$ axis so that $\mathbf{d}$ points along the $z$ axis. The values of $\cos(\psi)$ and $\sin(\psi)$ are

\[ \begin{align*} \cos(\psi) &= \frac{d_1}{|\mathbf{d}|}, & \sin(\psi) &= \frac{|d_z|}{|\mathbf{d}|}, \end{align*} \]

so the transformation matrix is (note that the sign of the $\sin(\psi)$ terms are switched because we are rotating in the clockwise direction)

\[\begin{split} \begin{align*} R_x &= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & d_1/|\mathbf{d}| & |d_z|/|\mathbf{d}| & 0 \\ 0 & -|d_z|/|\mathbf{d}| & d_1/|\mathbf{d}| & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*} \end{split}\]

../_images/rotate_about_line_4.svg — Fig. 45 The line is rotated about the $x$-axis so that $\mathbf{d}$ points along the $z$ axis.#

Fig. 45 shows that affect of rotating about the $x$-axis. Now that the line is pointing along the $z$-axis we can perform the rotation about the line using $R_z(\theta)$ and reverse the rotation in the $x$ and $z$ directions before finally reversing the translation. The composite transformation matrix is

\[A = T(\mathbf{p}) \cdot R_z(-\phi) \cdot R_x(-\psi) \cdot R_z(\theta) \cdot R_x(\psi) \cdot R_z(\phi) \cdot T(-\mathbf{p}).\]

Example 28

A flight simulation program is simulating the flight of an aeroplane positioned with its centre of mass at $\mathbf{c} = (10, 5, 50)$ travelling in the direction given by the vector $\mathbf{d} = (2, -1, -3)$. The user performs a roll of the plane by rotating about $\mathbf{d}$ by angle $\theta = \pi/6$ in the anti-clockwise direction. Calculate the composite transformation matrix that performs this action.

Computer Graphics Lecture Notes

Rotation

Contents

Rotation#

MATLAB code#

Rotation about polygon centre#

MATLAB code#

Rotation about a line#