HCR - Lecture 1

Index

• HCR - Reference Frame
• HCR - Pose of a Rigid Body
• HCR - Rotation Matrix
• HCR - Definition of ‘Elementary Rotation’ → HCR - Composition of Elementary Rotation
• HCR - Rotation Matrices Descriptors → HCR - Rotation Matrix Descriptor ‘Euler Angles’ → HCR - Rotation Matrix Descriptor ‘RPY Angles’ → HCR - Rotation Matrix Descriptor ‘Angle and Axis’ → HCR - Rotation Matrix Descriptor ‘Quaternion’
• HCR - Homogeneous Transformation Matrix

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HCR - Reference Frame

The reference frame is a 3 axis system that are usually connected by a body, an object or a system that can be used to acquire information or transmit information. For example a camera can be a reference frame, when you record a video, you are recording from a fixed point of view, the lenses of the camera, which has a mean direction. We can use the object frame (the camera) as reference for a robotic arm that gets information from it, so the camera becomes our reference frame.

Main concepts:

• Reference Frame.
• Position.
• Orientation.

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Reference Frames Examples:

1. Camera Reference Frame
2. Multi joint hand fingertips frame
3. Virtual world reference frame

Everything has a reference frame in our robotic task.

Main task in robotics: Represent a reference frame of an object into another object’s reference frame, that can be the world reference frame, the camera’s, sensor’s, …

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HCR - Pose of a Rigid Body

A rigid body is completely described in a reference frame by position and orientation. The position of point $O{}^{\prime }$ on the rigid body with respect to the coordinate frame $O$ ($x,y,z$) is expressed by a three dimensional vector $<o{}_x^{\prime },o{}_y^{\prime },o{}_z^{\prime }>$.

Where $o{}_x^{\prime },o{}_y^{\prime },o{}_z^{\prime }$ are versors. (??? it doesn't make sense, they should not be versors ???NOT_SURE_ABOUT_THIS )

So the position of $O{}^{\prime }$ with respect to $O$ is defined as:

$$O{}^{\prime }=o{}_x^{\prime }+o{}_y^{\prime }+o{}_z^{\prime }$$

In this case:

• $O$ : is the World reference frame
• $O{}^{\prime }$ : is the Object reference frame

The idea is: I want to know $O{}^{\prime }$ in respect to $O$. This because in my system $O$ is NOT allowed to change (being the world reference frame) so it’s perfect for referencing $O{}^{\prime }$ that can change.

While for the position of $O{}^{\prime }$ with respect to $O$, we can use the position of each vector of the object frame with respect to the reference frame: We obtain a representation of rotation by the Rotation Matrix

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HCR - Rotation Matrix

Given two reference frames like so:

We can write:

$$\{\begin{array}{l}{x}^{\prime }=x_x^{\prime }\cdot x+x_y^{\prime }\cdot y+x_z^{\prime }\cdot z\\ y^{\prime }=y_x^{\prime }\cdot x+y_y^{\prime }\cdot y+y_z^{\prime }\cdot z\\ z^{\prime }=z_x^{\prime }\cdot x+z_y^{\prime }\cdot y+z_z^{\prime }\cdot z\end{array}\to \left[\begin{array}{l}{x}^{\prime }\\ y^{\prime }\\ z^{\prime }\end{array}\right]=\overline{\stackrel{ˉ}{R}}\left[\begin{array}{l}x\\ y\\ z\end{array}\right]$$

Where for example $z_x^{\prime },z_y^{\prime },z_z^{\prime }$ is defined as:

So, $R$ is defined as:

$$R=\left[\begin{array}{lcc}{x}_x^{\prime }& x_y^{\prime }& x_z^{\prime }\\ y_x^{\prime }& y_y^{\prime }& y_z^{\prime }\\ z_x^{\prime }& z_y^{\prime }& z_z^{\prime }\end{array}\right]\triangleq R_{O{}^{\prime }}^O$$

$R_{O{}^{\prime }}^O$ $\triangleq$ Rotation Matrix from reference frame $O$ to $O{}^{\prime }$

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HCR - Definition of 'Elementary Rotation'

Given 2 reference frames we define the elementary rotation of the frame $A$ with respect to frame $B$ as the rotation of only one axis of the frames.

The composition of multiple Elementary Rotation Matrix result in a complete Rotation Matrix. ~Ex.:

~ Example: Rotation along the $x$ axis:

We rotate the frame with an angle of $\theta$ along the $x$ axis, as we can see in the second picture we can rewrite the rotation as:

$$R_{x(\theta )}=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos (\theta )& -\sin (\theta )\\ 0& -\sin (\theta )& \cos (\theta )\end{array}\right]$$

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• ~Example: Other Elementary Rotations:$\begin{array}{l}{R}_{x(\theta )}=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos (\theta )& -\sin (\theta )\\ 0& -\sin (\theta )& \cos (\theta )\end{array}\right]\\ R_{y(\theta )}=\left[\begin{array}{ccc}\cos (\theta )& 0& \sin (\theta )\\ 0& 1& 0\\ -\sin (\theta )& 0& \cos (\theta )\end{array}\right]\\ R_{z(\theta )}=\left[\begin{array}{ccc}\cos (\theta )& -\sin (\theta )& 0\\ \sin (\theta )& \cos (\theta )& 0\\ 0& 0& 1\end{array}\right]\end{array}$

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HCR - Composition of Elementary Rotation

Each orientation could be composed by elementary rotations of the object frame with respect to the reference frame.

• Successive rotations of an object about the object frame is obtained by ==premultiplication== of rotation matrices.
• Successive rotations of an object about the reference frame is obtained by ==postmultiplication== of rotation matrices.

Where premultiplication and postmultiplication only mean when the elementary rotation matrix is multiplied (if at the beginning or at the end).

NOTE: The object frame is referenced to the reference frame

The composition of various elementary rotation matrices gives form to a complete rotation matrix

We need also to define ==how the elementary rotation are composed==, to do this we use the rotation matrix descriptors.

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HCR - Rotation Matrices Descriptors

Using various composition of elementary rotation matrices it can result in a redundant description of the frame orientation.

We can use one of the following descriptors to obtain a ==non-redundant== rotation matrix:

• Euler Angles
• RPY Angles
• Angle and Axis
• Quaternion

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HCR - Rotation Matrix Descriptor 'Euler Angles'

I have to specify beforehand in which axis i will rotate: ~ex.: ZXZ Also all Euler Angles will do 3 elementary rotations: $[\phi ,\theta ,\psi ]$

Where for formality:

• “ROLL” : always means a rotation along the $z$ axis
• “PITCH” : always means a rotation along the $y$ axis
• “YAW” : always means a rotation along the $x$ axis

This are names that sailor uses for rotating the ships

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• ~ Examples of Euler Angles Rotations:
  • ZXZ (Most used for the Euler Angles, ‘Standard for Euler Angles’):$$ R^O_{O \kern 1px ’} = R_z(\varphi) \kern 1px R_x(\theta) \kern 1px R_z(\psi)

$$-\ast \ast ZYX\ast \ast {(}^{\prime }\ast \ast \ast RollPitchandYaw\ast \ast {\ast }^{\prime }o{r}^{\prime }\ast \ast \ast [[HCR-RotationMatrixDescripto{r}^{\prime }RPYAngle{s}^{\prime }|RPYangles]]\ast \ast {\ast }^{\prime }):$$

R^O_{O \kern 1px ’} = R_z(\varphi) \kern 1px R_y(\theta) \kern 1px R_x(\psi)

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HCR - Rotation Matrix Descriptor 'RPY Angles'

Specific Euler Angeles: ZYX (‘Roll Pitch and Yaw’):$R_{O{}^{\prime }}^O=R_{RPY}\triangleq R_z(\phi )R_y(\theta )R_x(\psi )$

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HCR - Rotation Matrix Descriptor 'Angle and Axis'

Formula:$R(\theta ,\overline{v})=R_z(\alpha )R_y(\beta )R_z(\theta )R_y(-\beta )R_z(-\alpha )$Where:

• $R_z$: rotation matrix along the $z$ axis.
• $\theta$ is the angle of rotation along the $\overline{v}$ axis (in the figure this axis/vector it’s represented by the vector $r$ but i wanted to avoid confusion between this axis of rotation and the elements of the rotation matrix $R$ : $r_{i,j}$, so i called it $\vec{v}$).
  The angle of rotation $\theta$ can be found as:$\theta ={\cos }^{-1}\left(\frac{r_{1,1}+r_{2,2}+r_{3,3}-1}{2}\right)$And the axis of rotation $\vec{v}$ as:$\overline{v}=\frac{1}{2\sin (\theta )}\left[\begin{array}{c}{r}_{3,2}-r_{2,3}\\ r_{1,3}-r_{3,1}\\ r_{2,1}-r_{1,2}\end{array}\right]$Where:
  • $r_{i,j}$ is the $i,j$ element of the rotation matrix $R$

So to find the $R$ rotation matrix we can impose $\theta$ and $\overline{v}$ as constraints and find:

$$R=\left[\begin{array}{lcc}{r}_{1,1}& r_{1,2}& r_{1,3}\\ r_{2,1}& r_{2,2}& r_{2,3}\\ r_{3,1}& r_{3,2}& r_{3,3}\end{array}\right]$$

NOTE:IMPORTANTE For $r_{1,1}+r_{2,2}+r_{3,3}=1$ → $\theta =0$ (or $\theta =\pi$) → $\overline{v}=\infty$ (Indeterminate) The Quaternion descriptor is more complicated but can avoid this problem.

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HCR - Rotation Matrix Descriptor 'Quaternion'

NOTE:IMPORTANTE ==There is no possibility that 2 different quaternions give the same final rotation.==

Let’s take an example: in the angle and axis descriptor formulas, we find that:

$$\theta ={\cos }^{-1}\left(\frac{r_{1,1}+r_{2,2}+r_{3,3}-1}{2}\right)$$

Now if we consider $r_{1,1}+r_{2,2}+r_{3,3}=1$ then $\theta$ can be both equal to $0$ or $\pi$, so we have and indetermination on the angle.

Let’s now define $\eta$ and $ϵ$ with respect to $\theta$ and $\vec{v}$ as:\begin{align} &\eta = \cos\left(\frac{\theta}{2}\right) \\[7px] &\varepsilon = \sin\left(\frac{\theta}{2}\right)\cdot \vec v \end{align} Where:

• $\vec{v}$ is a vector ($3\times 1$) and so is $\epsilon$.
• While $\eta$ is a scalar ($1\times 1$).

Notice that:${\eta }^2+{\epsilon }_x^2+{\epsilon }_y^2+{\epsilon }_z^2=1$And:$\begin{array}{c}\eta =\frac{1}{2}\sqrt{r_{1,1}+r_{2,2}+r_{3,3}}\\ \epsilon =\left[\begin{array}{l}\mathrm{sign}(r_{3,2}-r_{2,3})\sqrt{1+r_{1,1}-r_{2,2}-r_{3,3}}\\ \mathrm{sign}(r_{1,3}-r_{3,1})\sqrt{1+r_{2,2}-r_{1,1}-r_{3,3}}\\ \mathrm{sign}(r_{2,1}-r_{1,2})\sqrt{1+r_{3,3}-r_{1,1}-r_{2,2}}\end{array}\right]\end{array}$So by using $\eta$ and $\epsilon$ as constraints we can then find the rotation matrix $R$ :

$$R=\left[\begin{array}{lcc}{r}_{1,1}& r_{1,2}& r_{1,3}\\ r_{2,1}& r_{2,2}& r_{2,3}\\ r_{3,1}& r_{3,2}& r_{3,3}\end{array}\right]$$

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Quaternions re-explained

(As everyone knows them)

• Youtube - ‘Quaternions and 3d rotation, explained interactively’ by ‘3Blue1Brown’

Given a point $\overline{p}$ an axis of rotation $\overline{v}$ and the 2 angles of rotation $\theta$ and $\phi$ we can define:

$$q=\left(\cos \left(\frac{\theta }{2}\right)+\sin \left(\frac{\phi }{2}\right)\right)\cdot (\overline{v}\cdot [i,j,k])$$

Where:

• $i^2=j^2=k^2=ijk=-1$

So once we have defined $q$ the new point ${\overline{p}}^{\prime }$ location after the rotation is given by:

$${\overline{p}}^{\prime }=q\cdot p\cdot q^{-1}$$

Where:

$$\begin{array}{lcc}\text{if:}& q& =w_0+x_{0}i+y_{0}j+z_{0}k\\ \text{then:}& q^{-1}& =w_0-x_{0}i-y_{0}j-z_{0}k\end{array}$$Link to original

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HCR - Homogeneous Transformation Matrix

The entire pose of an object, given one reference frame could be represented in another frame, just by multiplying the first one with a matrix called homogeneous transformation matrix.

Contrary to the rotation matrix, where we tackle only the problem of orientation, now we want to resolve also the problem of position.

After all ==a pose is just a collection of position and orientation==

• So, first I resolve the problem of position (translation) then the problem of orientation (rotation).
  1. Solve the difference of $O$ and $O{}^{\prime }$.
  2. Considering the $O$ and $O{}^{\prime }$ in the same center i calculate the rotation matrix.

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Definition of ‘Transformation Matrix’

The Transformation Matrix that change one pose to another is denoted as $T$.

\end{array} & \kern5px 1 \end{array} \right]_{4\times4} $$ Where: - $R^O_{O \kern 1px '}$ : [[HCR - Rotation Matrix|rotation matrix]]. - $O \kern 1px '$ : distance $\overline{O \kern 2px O \kern 1px '}$ Then we can write the following transformation given a point $p^{O}$ in the reference frame $O$ we can find the point $p^{O \kern 1px '}$ in the reference frame $O \kern 1px '$:$$p^{O \kern 1px '} = T_{O \kern 1px '}^O \cdot p^{O}$$ > **NOTE**: > The 3 zeros elements in the last row are usually unchanged but sometimes they do change, when they do it's because we want to impose the **concept of prospective**, for example when using a camera some dimensions are forcefully distorted (closer objects appear bigger). > **NOTE**: > The last element $(4,4)$ is often $1$, but it can be changed to reference the **scale of my robot or environment** ($0.1$, $0.01$, $0.001$, ... when working with small robots, while we'll change it to $10$, $100$, $1000$, $10000$ for big robots) ---- The **inverse** of a **tranformation matrix** $T$ is also a transformation matrix, and can be computed as:$$\begin{array}{l} T^{-1} &=& \left[\begin{array}{c} R^O_{O \kern1px '} & O' \kern2px \\[5px] \begin{array}{c} 0 & 0 & 0 \end{array} & 1 \end{array}\right]^{-1} \\[-5px]&\kern3px\downarrow&\\[-5px] &=& \left[\begin{array}{c} \left(R^O_{O \kern1px '}\right)^{\tiny T} & -\left(R^O_{O \kern1px '}\right)^{\tiny T} \cdot O' \kern2px \\[5px] \begin{array}{c} 0 & 0 & 0 \end{array} & 1 \end{array}\right] \end{array}$$***Source***: [Mecharithm (reading)](https://mecharithm.com/learning/lesson/homogenous-transformation-matrices-configurations-in-robotics-12#:~:text=The%20inverse%20of%20a%20transformation%20matrix%20T%20%E2%88%88%20SE(3)%20is%20also%20a%20transformation%20matrix%20and%20can%20be%20computed%20as) ^inverse-of-a-transformation-matrixLink to original