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Spatial Description & Transformation

Descriptions

Note that we use capital form (upper-case) \(X\) or bold form of lower case \(\pmb{x}\) to denote vectors.

Descriptions of a position, an orientation

In a coordinate system \(\{A\}\) (three orthogonal unit vectors), we use \(\hat{X}\) to denote the norm of vector \(X\) equals \(1\), and use subscript \(\hat{X}_A\) to denote \(X\) is the unit vector of frame \(\{A\}\) along \(x\)-axis.

(i) A point \(^AP\) (superscript on the left means reference frame is \(\{A\}\)) is expressed by a vector

\[ ^AP=\left[\begin{array}{c} p_x\\p_y\\p_z \end{array}\right] \]

or a vector expressed by unit vectors

\[ \pmb{r}_{OP}=\left[\begin{array}{ccc}\hat{X}_A&\hat{X}_A&\hat{X}_A\end{array}\right]^AP. \]

(ii) Attach another coordinate system \(\{B\}\), with its unit vectors \(\hat{X}_B\), \(\hat{Y}_B\) and \(\hat{Z}_B\). If they are represented by system \(\{A\}\), as \(^A\hat{X}_B\), \(^A\hat{Y}_B\), and \(^A\hat{Z}_B\). Stack these three vectors together, and we have a Rotation Matrix from \(A\) to \(B\)

\[ ^A_BR=\left[\begin{array}{ccc}^A\hat{X}_B& ^A\hat{Y}_B& ^A\hat{Z}_B\end{array}\right]. \]

Actually, it represents the poze (orientation) transfer from \(\{A\}\) to \(\{B\}\). It is also a method to get a rotation matrix, using expression of \(\{B\}\) basis expressed in \(\{A\}\).

Properties of Rotation Matrix

(i) Calculate elements of rotation matrix.

\[ ^A_BR=\left[\begin{array}{ccc} \langle\hat{X}_B, \hat{X}_A\rangle&\langle\hat{Y}_B, \hat{X}_A\rangle&\langle\hat{Z}_B, \hat{X}_A\rangle\\ \langle\hat{X}_B, \hat{Y}_A\rangle&\langle\hat{Y}_B, \hat{Y}_A\rangle&\langle\hat{Z}_B, \hat{Y}_A\rangle\\ \langle\hat{X}_B, \hat{Z}_A\rangle&\langle\hat{Y}_B, \hat{Z}_A\rangle&\langle\hat{Z}_B, \hat{Z}_A\rangle \end{array}\right] \]

Note that inner product is irrelevant with the choice of frame.

(ii) Show that

\[ ^A_BR=\left[\begin{array}{c} ^B\hat{X}_A^T\\^B\hat{Y}_A^T\\^B\hat{Z}_A^T \end{array}\right], \]

and then

\[ ^B_AR= {^A_B}R^T. \]

(iii) \(^A_BR\) is reversible, and

\[ (^A_BR)^{-1}={^B_A}R^T. \]

If we consider from linear algebra, denoting \(\hat{\pmb{e}}_1,\hat{\pmb{e}}_2,\hat{\pmb{e}}_3\) as basis of \(\{B\}\), \(\pmb{e}_1,\pmb{e}_2,\pmb{e}_3\) as basis of \(\{A\}\), so

\[ \hat{\pmb{e}}_{j}=\left[\begin{array}{ccc}\pmb{e}_1&\pmb{e}_2&\pmb{e}_3\end{array}\right]\left[\begin{array}{c} r_{1j}\\r_{2j}\\r_{3j} \end{array}\right]=\sum_{i=1}^3 r_{ij} \pmb{e}_i,\quad j=1,2,3. \]

From inner product, we have

\[ \langle\pmb{e}_i, \hat{\pmb{e}}_j\rangle=r_{ij}. \]

So a rotation matrix from \(\{A\}\) to \(\{B\}\), we have

\[ ^A_BR=\left[\begin{array}{ccc} r_{11}&r_{12}&r_{13}\\ r_{21}&r_{22}&r_{23}\\ r_{31}&r_{32}&r_{33} \end{array}\right] \]

\(\square\)

from (i) we could know that

\[ ^A\hat{X}_B=\left[\begin{array}{c} \langle\hat{X}_B, \hat{X}_A\rangle\\ \langle\hat{X}_B, \hat{Y}_A\rangle\\ \langle\hat{X}_B, \hat{Z}_A\rangle \end{array}\right] \]

Similarly we have

\[ ^B\hat{X}_A=\left[\begin{array}{c} \langle\hat{X}_B, \hat{X}_A\rangle\\ \langle\hat{Y}_B, \hat{X}_A\rangle\\ \langle\hat{Z}_B, \hat{X}_A\rangle \end{array}\right] \]

which is just the transpose of the first line of \(^A_BR\).

\(\square\)

Since

\[ \begin{align*} ^B_AR\cdot {^A_B}R={^A_B}R^T \cdot{^A_B}R&=\left[\begin{array}{ccc}^A\hat{X}^T_B& ^A\hat{Y}^T_B& ^A\hat{Z}^T_B\end{array}\right]^T\left[\begin{array}{ccc}^A\hat{X}_B& ^A\hat{Y}_B& ^A\hat{Z}_B\end{array}\right]\\ &=\left[\begin{array}{c}^A\hat{X}^T_B\\ ^A\hat{Y}^T_B\\ ^A\hat{Z}^T_B\end{array}\right]\left[\begin{array}{ccc}^A\hat{X}_B& ^A\hat{Y}_B& ^A\hat{Z}_B\end{array}\right]\\ &=I_{3\times 3}. \end{align*} \]

So \((^A_BR)^{-1}={^B_A}R^T\).

\(\square\)

Same Points in different frames

Now we could discuss the description of points within different frames.

Same points within different frames in simple cases

The following process is called analyze points from moving frame into fixed frame.

(i) \(\{B\}\) is a translation of \(\{A\}\). Then

\[ ^AP={^B}P+{^A}O_B. \]

(ii) \(\{B\}\) is a rotation of \(\{A\}\). Then

\[ ^AP={^A_B}R {^B}P. \]

Note the we premultiply rotation matrix to \(^BP\), while we postmultiply rotation matrix to basis vectors in \(\{B\}\) to get basis vectors in \(\{A\}\).

(i) Easy to see.

(ii) Using relationship between basis vectors

\[ \left[\begin{array}{ccc}\hat{X}_B&\hat{X}_B&\hat{X}_B\end{array}\right]=\left[\begin{array}{ccc}\hat{X}_A&\hat{X}_A&\hat{X}_A\end{array}\right]{^A_B}R, \]

we have

\[ \begin{align*} \pmb{r}_{OP}&=\left[\begin{array}{ccc}\hat{X}_B&\hat{X}_B&\hat{X}_B\end{array}\right]{^B}P\\ &=\left[\begin{array}{ccc}\hat{X}_A&\hat{X}_A&\hat{X}_A\end{array}\right]{^A_BR} \cdot{^BP}\\ &:=\left[\begin{array}{ccc}\hat{X}_A&\hat{X}_A&\hat{X}_A\end{array}\right]{^A}P. \end{align*} \]

\(\square\)

Combined from the above two simple cases, we have the following general cases.

Same points within different frames in general cases

Assume \(\{B\}\) is translation and rotation of \(\{A\}\), then

\[ ^AP= {^A_BR} {^BP}+ {^AO_B} \]

Homogeneous Transform Matrix

It is usually good to think of the above formula as a mapping from \(^BP\) to \(^AP\), but if we write

\[ \begin{align} ^AP=\phi({^BP})=T \cdot{^BP},\label{3times3} \end{align} \]

where \(T_{3\times 3}\) is not just the rotation matrix or the translation, which is left out. We are adept to making use of \(^A_BR\) and \(^AO_B\) if we expend one dimension, as the following definition shows.

Homogeneous Transform

Define Homogeneous Transform to be

\[ ^A_BT_{4\times 4}:=\left[\begin{array}{ccc:c} &^A_BR_{3\times 3}& &^AO_B\\ \hdashline 0&0&0& 1 \end{array}\right] \]

such that

\[ \begin{align} \left[\begin{array}{c} ^AP\\1 \end{array}\right]={^A_BT}\left[\begin{array}{c} ^BP\\1 \end{array}\right] \label{4times4} \end{align} \]

Actually it is equivalent to equation \(\ref{3times3}\), because equation \(\ref{4times4}\) demonstrates

\[ \begin{cases} ^AP&= {^A_BR} {^BP}+ {^AO_B}\\ 1&=1. \end{cases} \]

Properties of Homogeneous Transform

Inverse of Homogeneous Transform Matrix.

Assume homogeneous transform matrix \(^A_BT\) from \(\{A\}\) to \(\{B\}\), then it is reversible and

\[ {^B_AT}={^A_BT^{-1}}. \]

We make use of structure inherent of \(^A_BT\).

\[ \begin{align} ^B_AR={^A_BR^T}.\label{Rotation} \end{align} \]
  • For translation vector \(^AO_B\), we want to know \(^BO_A\). This is a little tricky. We apply Same points within different frames in general cases to \(O_B\), in terms of \(\{B\}\). That is, we have the \(O_B\) in frame \(\{B\}\) expressed by frame \(\{A\}\)
\[ \pmb{0}={^BO_B}={^B_AR}{^AO_B}+ {^BO_A} \]

which gives

\[ \begin{align} {^BO_A}=-{^B_AR}{^AO_B}=-{^A_BR^T}{^AO_B}.\label{Translation} \end{align} \]

Conbined with equation \(\ref{Rotation}\) and \(\ref{Translation}\), we have

\[ ^B_AT=\left[\begin{array}{ccc:c} &^A_BR^T& &-{^A_BR^T}{^AO_B}\\ \hdashline 0&0&0& 1 \end{array}\right] \]

So \({^A_BT}\cdot {^B_AT}=I_{4\times 4}\), which is reversible.

\(\square\)

The above proof also gives the method to quickly solve the inverse of \(^A_BT\).

Chain Rule

Chain Rule: Same points with multiple different frames

Assume frame \(\{C\}\) is known relative to frame \(\{B\}\), frame \(\{B\}\) is known relative to frame \(\{A\}\). Their relationship is expressed by \(^B_CT\) and \(^A_BT\), respectively. Then expression of point \(P\) in frame \(A\) could be expressed by transform of its expression in frame \(C\) as

\[ ^AP={^A_BT}{^B_CT}{^CP}. \]

That is, the relationship from \(\{A\}\) to \(\{C\}\) is

\[ {^A_CT}={^A_BT}{^B_CT}. \]

Notice that by Homogeneous Transform

\[ \begin{align*} \left[\begin{array}{c} ^AP\\1 \end{array}\right]&={^A_BT}\left[\begin{array}{c} ^BP\\1 \end{array}\right]\\ &={^A_BT}{^B_CT}\left[\begin{array}{c} ^CP\\1 \end{array}\right]\\ &:={^A_CT}\left[\begin{array}{c} ^CP\\1 \end{array}\right]. \end{align*} \]

\(\square\)

Operators

Operators like translation, rotation could ba accomplished by the same mathematics as mapping the point to a second frame.

Translation Operator

In frame \(\{A\}\), vector \(^AP_1\) is translated by a vector \(^AQ\), the result denoted by

\[ ^AP_2={^AP_1}+{^AQ}. \]

We could use another perspective, i.e. construct a imaginary frame \(\{B\}\) with same orientation of \(\{A\}\) but origins at point \(Q\). The homogeneous transform matrix is

\[ ^A_BT=\left[\begin{array}{ccc:c} &I_{3\times 3}&& {^AO_B}\\ \hdashline 0&0&0&1 \end{array}\right]=\left[\begin{array}{ccc:c} &I_{3\times 3}&& {^AQ}\\ \hdashline 0&0&0&1 \end{array}\right]. \]

so the same point \(P\) (after translation) in frame \(\{A\}\) could be expressed in terms of that in frame \(\{B\}\)

\[ ^AP_2={^A_BT}{^BP_2}={^A_BT}{^AP_1}. \]

Rotation Operator

Similar as we have in translation operator, we shall interpret a rotation to be a transform between two frames, i.e.

\[ ^AP_2={^A_BT}{^AP_1}. \]

where

\[ ^A_BT=\left[\begin{array}{ccc:c} &R_K(\theta)&& \pmb{0}\\ \hdashline 0&0&0&1 \end{array}\right]. \]

Here we discuss a little more about \(R_K(\theta)\) using different perspectives.

Fixed angles

X-Y-Z Fixed angles

Assume frame \(\{B\}\) coincides with a known reference frame \(\{A\}\), then rotate \(\{B\}\) about \(\hat{X}_A\) by an angle \(\gamma\), then rotate about \(\hat{Y}_A\) by an angle \(\beta\), and finally about \(\hat{Z}_A\) by an angle \(\alpha\).

We can interpret these three steps as moving the same vector \(P\) in three steps. Assume initially its expression in frame \(\{A\}\) is \(^AP\), the final point \(P'\) could be expressed by

\[ ^AP'=R_Z(\alpha)\cdot R_Y(\beta)\cdot R_X(\gamma)\cdot{^AP}. \]

On the other hand, expression of \(P'\) in frame \(\{B\}\) is the same as in frame \(\{A\}\), i.e. \({^BP'}={^AP}\). So

\[ ^AP'=R_Z(\alpha)\cdot R_Y(\beta)\cdot R_X(\gamma)\cdot{^BP'}. \]

meaning \(^A_BR=R_Z(\alpha)\cdot R_Y(\beta)\cdot R_X(\gamma)\) by same point in different frames, which is also called X-Y-Z fixed angle, achieved by premultiplying rotation matrices. We define X-Y-Z fixed angle as

\[ ^A_BR_{XYZ}(\gamma, \beta,\alpha)={R_{xyz}(\gamma,\beta, \alpha)}. \]

Euler angles

Z-Y-X Euler angles

Assume frame \(\{B\}\) coincides with a known reference frame \(\{A\}\), then rotate \(\{B\}\) about \(\hat{Z}_B\) by an angle \(\alpha\), then rotate about \(\hat{Y}_B\) by an angle \(\beta\), and finally about \(\hat{X}_B\) by an angle \(\gamma\).

For these three steps, we could construct three frames \(\{C\}\), \(\{D\}\) and \(\{E\}\). Easy to see that

\[ ^A_CR=R_Z(\alpha)=\left[\begin{array}{ccc} 1&0&0\\ 0&\cos \alpha&-\sin \alpha\\ 0 & \sin \alpha& \cos \alpha \end{array}\right]. \]

and

\[ ^C_DR=R_Y(\beta)=\left[\begin{array}{ccc} \cos \beta&0&\sin \beta\\ 0&1&0\\ -\sin \beta&0& \cos \beta \end{array}\right],\quad ^D_ER=R_X(\gamma)=\left[\begin{array}{ccc} \cos \gamma&-\sin \gamma&0\\ \sin \gamma& \cos \gamma&0\\ 0&0&1 \end{array}\right]. \]

By chain rule we have

\[ ^A_BR={^A_ER}={^A_CR}\cdot{^C_DR}\cdot{^D_ER}=R_Z(\alpha)\cdot R_Y(\beta)\cdot R_X(\gamma). \]

meaning the whole process could be achieved by postmultiplying rotation matrices, which is also called Z-Y-X fixed angles. We define Z-Y-X Euler angle as

\[ ^A_BR_{Z'Y'X'}(\beta,\alpha,\gamma)={R_{z'y'x'}(\beta, \alpha,\gamma)}. \]

Notice the above process is the same as we have in fixed angles.

We have 6 symmetric and 6 non-symmetric forms of fixed angles or Euler angles.

Equivalent angle-axis

Definition of Equivalent angle-axis representations

Given a random axis \(^AK={^A\pmb{r}_{OK}}=[\begin{array}{ccc}k_x &k_y&k_z\end{array}]^T\) expressed in frame \(\{A\}\), rotate the original frame \(\{A\}\) about axis \(\pmb{r}_{OK}\) and get another frame \(\{B\}\). How to get the corresponding rotation matrix?

Assume \(\pmb{r}_{OK}\) is a unit vector. We want to find the expression of base vector \(\hat{X}_A\) after it is rotated.

Gimbal Lock

Does every matrix could be expressed by \(Z-Y-X\) angle?

When \(\beta=\frac{\pi}{2}\) in Euler's angle, then we lose a degree of freedom, solution is not unique.