Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

According to my book

Rotations through an angle $\theta$ about the $x$, $y$, and $z$ axes are performed using the following transformation matrices.

For example, $$ R_x(\theta) = \left[ \begin{matrix} 1& 0 & 0 \\\ 0& \cos\theta &-\sin\theta\\\ 0& \sin\theta &\cos\theta \end{matrix}\right] $$

Ry0= | cos0 0 sin0 0|
     | 0 1 0 0|
     | -sin0 0 cos0 0|
     | 0 0 0 1|


Rz0= | cos0 -sin0 0 0 |
     | sin0 cos0 0  0|
     | 0 0 1 0|
     | 0 0 0 1|

And I just need to put the angle in it no matter which rotate about axis? After that, how can I get a $4\times 4$ matrix?

Thank you

share|improve this question

2 Answers 2

up vote 0 down vote accepted

To express ordinary $\mathbb{R}^n \to \mathbb{R}^n$ linear transformation into homogeneous coordinates just add another row and column where every term is equal to $0$ but the diagonal, which should be $1$.

For example if $A$ is your transformation matrix, then the new matrix would be $$A_H = \left[\begin{matrix}A& 0 \\\ 0& 1 \end{matrix}\right].$$ In your example $$A = \left[\begin{matrix}1& 0 & 0 \\\ 0& \cos\theta &-\sin\theta\\\ 0& \sin\theta &\cos\theta \end{matrix}\right],$$ so $$A_H = \left[\begin{matrix}1& 0 & 0 & 0 \\\ 0& \cos\theta &-\sin\theta & 0 \\\ 0& \sin\theta &\cos\theta & 0 \\\ 0 & 0 & 0 & 1 \end{matrix}\right].$$

Hope that helps ;-)

share|improve this answer
    
Thanks , it helps, how about y and z? –  Leo Chan Apr 21 '12 at 10:39
    
Is my try correct ? Thank you........... –  Leo Chan Apr 21 '12 at 10:42
    
Yes, it is, as long you know the difference between $0$ (zero) and $\theta$ (theta). For more information try en.wikipedia.org/wiki/Rotation_matrix and en.wikipedia.org/wiki/Transformation_matrix, actually there is a lot on this on the web, GIYF. –  dtldarek Apr 21 '12 at 10:44
    
Thanks for your help –  Leo Chan Apr 21 '12 at 11:06

Let the rotation axis be a line through a fixed point $(x_0,y_0,z_0,1)^T$ and an infinite point (direction of the line) $(a,b,c,0)^T$, without loss of generality, we assume $a^2+b^2+c^2=1$. Use right-handed rule for rotation, then a general homogeneous rotation with angle $\theta$ can be obtained as:

$$ \boldsymbol{R}^{3D}\left(x_0,y_0,z_0,a,b,c,\theta\right)=\mathscr{C}_1+\left(\sin\theta\cdot\mathscr{A}_2- \left(1-\cos\theta\right)\cdot\mathscr{O}_3\right)\cdot \mathscr{T}_4 $$

where: $$\mathscr{C}_1= \left[\begin{array}{*{20}{c}} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&2-\cos\theta \end{array}\right]$$

$$\mathscr{A}_2 = {\begin{array}{c} \underbrace{ \begin{array}{c} {{\left[ {\begin{array}{*{20}{c}} {\color{blue}0}&{\color{blue}-c}&{\color{blue}b}&0\\ {\color{blue}c}&{\color{blue}0}&{\color{blue}-a}&0\\ {\color{blue}-b}&{\color{blue}a}&{\color{blue}0}&0\\ 0&0&0&0 \end{array}} \right]}} \end{array} } \\ \text{Antisymmetric matrix} \end{array}} $$

$$ \quad\mathscr{O}_3= \begin{array}{c} \underbrace{ {I - {\left[ {\begin{array}{*{20}{c}} a\\ b\\ c\\ 0 \end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}} a&b&c&0 \end{array}} \right]}} }\\ \text{Orthographic parallel projection} \end{array} $$

$$ \mathscr{T}_4=\begin{array}{c} \underbrace{\left[ {\begin{array}{*{20}{c}} 1&0&0&{ - {x_0}}\\ 0&1&0&{ - {y_0}}\\ 0&0&1&{ - {z_0}}\\ 0&0&0&1 \end{array}} \right]}\\ \text{Translation}\end{array}$$

The L-C formulation of homogeneous 3D rotation is similar to Rodrigues' but they are not the same in essence. Details are available here: A submission on homogeneous rotation to arXiv.org

For your case, substitute $x_0=y_0=z_0=0$, $a=b=0$, $c=1$ and $\theta=\dfrac{\pi}{2}$ into the L-C formulation then you can obtain the desired homogeneous rotation matrix.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.