Useful Notes and Equations

For a comprehensive and thorough summary of the theory, check MuChenSun’s wonderful note here.

My own notes:

Useful Equations:

For \(R = \text{Rot}(\hat{v}_{i},-90^{\circ})\), if:

1. \(R^{\prime}_{ab} = RR_{ab}\), then the rotation is about \(\hat{v}_{a}\) (in frame {a});
2. \(R^{\prime}_{ab} = R_{ab}R\), then the rotation is about \(\hat{v}_{b}\) (in frame {b}).

Adjacent Common Subscripts Cancellation Rule: if two subscripts are adjacent and common, then they can be cancelled. This rule applies to the product of two rotation matrix and the product of one rotation matrix with a vector.

Matrix Exponentials \(\hat{\omega}\theta = \vec{v} = \frac{1}{\lvert v \rvert} \begin{bmatrix} \frac{v_{1}}{\lvert v \rvert} \\ \frac{v_{2}}{\lvert v \rvert} \\ \frac{v_{3}}{\lvert v \rvert} \\ \end{bmatrix}\)

Adjoint representation formula: \(\left[\text{Ad}_{T}\right] = \begin{bmatrix} R & 0 \\ [p]R & 0 \\ \end{bmatrix} \in \mathbb{R}^{6\times6}\)

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Textbook Exercises Attempts

Exercise 3.1 In terms of the \(\hat{x}_{s}\), \(\hat{y}_{s}\), \(\hat{z}_{s}\) coordinates of a fixed space frame {s}, the frame {a} has its \(\hat{x}_{a}\)-axis pointing in the direction (0,0,1) and its \(\hat{y}_{a}\)-axis pointing in the direction (−1,0,0), and the frame {b} has its \(\hat{x}_{b}\)-axis pointing in the direction (1, 0, 0) and its \(\hat{y}_{b}\)-axis pointing in the direction (0, 0, −1).

• (a) Draw by hand the three frames, at different locations so that they are easy to see.
• (b) Write down the rotation matrices \(R_{sa}\) and \(R_{sb}\).
• (c) Given \(R_{sb}\), how do you calculate \(R^{-1}_{sb}\) without using a matrix inverse? Write down \(R_{sb}^{-1}\) and verify its correctness using your drawing.
• (d) Given \(R_{sa}\) and \(R_{sb}\), how do you calculate \(R_{ab}\) (again without using matrix inverses)?Compute the answer and verify its correctness using your drawing.
• (e) Let R = \(R_{sb}\) be considered as a transformation operator consisting of a rotation about \(\hat{x}\) by \(−90^{\circ}\). Calculate \(R_{1} = R_{sa}R\), and think of \(R_{sa}\) as a representation of an orientation, R as a rotation of \(R_{sa}\), and \(R_{1}\) as the new orientation after the rotation has been performed. Does the new orientation \(R_{1}\) correspond to a rotation of \(R_{sa}\) by \(−90^{\circ}\) about the world-fixed \(\hat{x_{s}}\)-axis or about the body-fixed \(\hat{x}_{a}\)-axis? Now calculate \(R_{2} = RR_{sa}\). Does the new orientation \(R_{2}\) correspond to a rotation of \(R_{sa}\) by \(−90^{\circ}\) about the world-fixed \(\hat{x}_{s}\)-axis or about the body-fixed \(\hat{x}_a\)-axis?
• (f) Use \(R_{sb}\) to change the representation of the point \(p_{b} = (1,2,3)\) (which is in {b} coordinates) to {s} coordinates.
• (g) Choose a point p represented by \(p_{s} = (1, 2, 3)\) in {s} coordinates. Calculate \(p^{\prime} = R_{sb}p_{s}\) and \(p^{\prime\prime} = R^{T}_{sb}p_{s}\). For each operation, should the result be interpreted as changing coordinates (from the {s} frame to {b}) without moving the point p or as moving the location of the point without changing the reference frame of the representation?
• (h) An angular velocity w is represented in {s} as ωs = (3, 2, 1). What is its representation ωa in {a}?
• (i) By hand, calculate the matrix logarithm \([\hat{\omega}]\theta\) of \(R_{sa}\). (You may verify your answer with software.) Extract the unit angular velocity \(\hat{\omega}\) and rotation amount \(\theta\). Redraw the fixed frame {s} and in it draw \(\hat{\omega}\).
• (j) Calculate the matrix exponential corresponding to the exponential coordinates of rotation \(\hat{\omega}\theta = (1, 2, 0)\). Draw the corresponding frame relative to {s}, as well as the rotation axis \(\hat{\omega}\).

• (a) As shown below:

• (b) By observation we have \(R_{sa}= \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 1 & 0 & 0 \\ \end{bmatrix}, R_{sb}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{bmatrix}\)

• (c) \(\begin{align} \begin{split} R_{sb}^{-1} &= R_{sb}^{T}\\ &= R_{bs}\\ &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \\ \end{bmatrix} \end{split} \end{align}\)

• (d) \(\begin{align} \begin{split} R_{ab} &= R_{as} R_{sb}\\ &= R_{sa}^{T} R_{sb}\\ &= \begin{bmatrix} 0 & 0 & 1 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{bmatrix} \\ &= \begin{bmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \\ \end{bmatrix} \end{split} \end{align}\)

• (e) N.B. if \(R=\text{Rot}(\hat{u}_{i},\theta)\) and R is pre-multiplied, that is, \(R^{\prime}=RR_{ab}\), then \(\hat{u}_{i}\) is \(\hat{u}_{a}\); if R is post-multiplied, that is, \(R^{\prime\prime}=R_{ab}R\), then \(\hat{u}_{i}\) is \(\hat{u}_{b}\). In this case, the rotation matrix is defined as, \(R = R_{sb}=\text{Rot}(\hat{x}_{i},-90^{\circ})\) \(\begin{align} \begin{split} R_{1} &= R_{sa} R \\ &= R_{sa} \text{Rot}(\hat{x}_{a},-90^{\circ}) \\ &= \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 1 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{bmatrix} \\ &= \begin{bmatrix} 0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{bmatrix} \end{split} \end{align}\) Clearly, \(R_{1}\) is \(R_{sa}\) rotated \(-90^{\circ}\) about \(\hat{x}_{a}\). \(\begin{align} \begin{split} R_{2} &= R R_{sa} \\ &= \text{Rot}(\hat{x}_{s},-90^{\circ})R_{sa}\\ &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \end{split} \end{align}\) Clearly, \(R_{2}\) is \(R_{sa}\) rotated \(-90^{\circ}\) about \(\hat{x}_{s}\).

• (f) \(\begin{align} \begin{split} p_{s} &= R_{sb} p_{b} \\ &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3 \\ \end{bmatrix} \\ &= \begin{bmatrix} 1 \\ 3 \\ -2 \\ \end{bmatrix} \end{split} \end{align}\)

• （g） \(\begin{align} \begin{split} p^{\prime} &= R_{sb}p_{s}\\ &= \begin{bmatrix} 1 \\ 3 \\ -2 \\ \end{bmatrix} \text{(from (f))}\\ p^{\prime\prime} &= R^{T}_{sb}p_{s}\\ &= R_{bs}p_{s} \\ &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3 \\ \end{bmatrix} \\ &= \begin{bmatrix} 1 \\ -3 \\ 2 \\ \end{bmatrix} \end{split} \end{align}\)

\(p^{\prime}\) should be interpreted as moving the location of the point without changing the reference frame. The reason is that the two adjacent subscripts are \(b\) and \(s\), which aren’t common and therefore cannot be canceled. \(p^{\prime\prime}\) should be interpreted as changing coordinates from {s} to {b} frame without moving the point \(p\). The reason is that the two adjacent subscripts are common and can therefore be canceled. This cancellation indicates coordinate change.

• (h) \(\begin{align} \begin{split} \omega_{a} &= R_{as}\omega_{s} = R_{sa}^{T}\omega_{s}\\ &= \begin{bmatrix} 0 & 0 & 1 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \\ \end{bmatrix} \begin{bmatrix} 3 \\ 2 \\ 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} 1 \\ -3 \\ -2 \\ \end{bmatrix} \end{split} \end{align}\)

• (i) Because \(\text{tr}(R_{sa}) = 0 + 0 + 0 = 0\), we should use the third general solution (see my own scribbling note). By the general formula, we have \(\begin{align} \begin{split} \theta &= \cos^{-1}\left( \frac{\text{tr}(R)-1}{2} \right) = \cos^{-1}\left(-\frac{1}{2}\right) \\ &= \frac{2\pi}{3} \in [0,\pi) \\ \end{split} \end{align}\) With \(\theta\), \([\hat{\omega}]\) can be obtained using formula as well, \(\begin{align} \begin{split} [\hat{\omega}] &= \frac{1}{2\sin(\frac{2\pi}{3})}(R-R^{T}) \\ &= \frac{\sqrt{3}}{3}(R_{sa}-R_{as}) \\ &= \frac{\sqrt{3}}{3}\left( \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 1 & 0 & 0 \\ \end{bmatrix} - \begin{bmatrix} 0 & 0 & 1 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \\ \end{bmatrix} \right) \\ &= \frac{\sqrt{3}}{3}\begin{bmatrix} 0 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & 1 & 0 \\ \end{bmatrix}\\ \end{split} \end{align}\) By utilizing the equation \(R=e^{[\hat{\omega}]\theta}\), we can find, \(\begin{align} \begin{split} \omega_{1} &= \frac{r_{32}-r_{23}}{2s_{\theta}} \\ &= \frac{1-(-1)}{2s_{120^{\circ}}} = \frac{\sqrt{3}}{3}\cdot(2)\\ \omega_{2} &= \frac{r_{13}-r_{31}}{2s_{\theta}} \\ &= \frac{-1-1}{2s_{120^{\circ}}} = \frac{\sqrt{3}}{3}\cdot(-2)\\ \omega_{3} &= \frac{r_{21}-r_{12}}{2s_{\theta}} \\ &= \frac{1-(-1)}{2s_{120^{\circ}}} = \frac{\sqrt{3}}{3}\cdot(2)\\ \end{split} \end{align}\) To obtain the unit vector \(\hat{\omega}\), \(\begin{align} \begin{split} \hat{\omega} &= \frac{1}{\lvert \omega \rvert} \begin{bmatrix} \omega_{1} \\ \omega_{2} \\ \omega_{3} \\ \end{bmatrix} \\ &= \frac{1}{2} \cdot \left( \frac{\sqrt{3}}{3} \begin{bmatrix} 2 \\ -2 \\ 2 \\ \end{bmatrix} \right) = \frac{\sqrt{3}}{3} \begin{bmatrix} 1 \\ -1 \\ 1 \\ \end{bmatrix}\\ \end{split} \end{align}\)

• (j) Since \(\hat{\omega}\theta = (1, 2, 0)\), we can find them separately by obtaining the unit vector: \(\hat{\omega}\theta = (1,2,0)^{T} = \sqrt{5} \begin{bmatrix} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \\ 0 \\ \end{bmatrix}\) Then it’s obvious that we have \(\theta=\sqrt{5}\) and \(\hat{\omega}=(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}},0)^{T}\). Ipso facto, the skew symmetric form of \(\hat{\omega}\) is, \([\hat{\omega}] = \begin{bmatrix} 0 & 0 & \frac{2}{\sqrt{5}} \\ 0 & 0 & -\frac{1}{\sqrt{5}} \\ -\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} & 0 \\ \end{bmatrix}\) The rotation is obtained as \(\begin{align} \begin{split} R &= e^{[\hat{\omega}]\theta} \\ &= I + \sin(\theta)[\hat{\omega}] + (1-\cos{\theta})[\hat{\omega}]^{2}\\ &= \begin{bmatrix} -0.29 & 0.65 & 0.70 \\ 0.65 & 0.68 & -0.35 \\ -0.70 & 0.35 & -0.62 \\ \end{bmatrix}\\ \end{split} \end{align}\)

Exercise3.2 Let \(p\) be a point whose coordinates are \(p=\left( \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{2}}\right)\) with respect to the fixed frame \(\hat{x}-\hat{y}-\hat{z}\). Suppose that \(p\) is rotated about the fixed-frame \(\hat{x}\)-axis by 30 degrees, then about the fixed-frame \(\hat{y}\)-axis by 135 degrees, and finally about the fixed-frame \(\hat{z}\)-axis by −120 degrees. Denote the coordinates of this newly rotated point by \(p^{\prime}\).

• (a) What are the coordinates \(p^{\prime}\)?
• (b) Find the rotation matrix \(R\) such that \(p^{\prime} = Rp\) for the \(p^{\prime}\) you obtained in (a).

• (a) Just use the formulae: \(p^{\prime} = \text{Rot}(\hat{z},-135^{\circ})\text{Rot}(\hat{y},135^{\circ})\text{Rot}(\hat{x},30^{\circ})p\). The calculation and result are skipped here.

• (b) \(R = \text{Rot}(\hat{z},-135^{\circ})\text{Rot}(\hat{y},135^{\circ})\text{Rot}(\hat{x},30^{\circ})\).

Exercise 3.3 Suppose that \(p_{i} \in \mathbb{R}^{3}\) and \(p^{\prime}_{i} \in \mathbb{R}^{3}\) are related by \(p^{\prime}_{i} = Rp_{i},\text{where} i = 1, 2, 3\), for some unknown rotation matrix \(R\). Find, if it exists, the rotation \(R\) for the three input–output pairs \(p_{i} \Rightarrow p^{\prime}_{i}\), where: \(\begin{align} \begin{split} p_{1} &= (\sqrt{2}, 0, 2) \Rightarrow p^{\prime}_{1} = (0, 2, \sqrt{2}), \\ p_{2} &= (1,1,-1) \Rightarrow p^{\prime}_{2} = (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},-\sqrt{2}), \\ p_{3} &= (0, 2\sqrt{2}, 0) \Rightarrow p^{\prime}_{3} = (-\sqrt{2},\sqrt{2},-2) \\ \end{split} \end{align}\)

Easy question. Since \(R\left[p_{1}^{T}, p_{2}^{T}, p_{3}^{T}\right] = \left[(p^{\prime}_{1})^{T}, (p^{\prime}_{2})^{T}, (p^{\prime}_{3})^{T}\right]\), then \(R\) can be found if the inverse of the term exists: \(R=\left[(p^{\prime}_{1})^{T}, (p^{\prime}_{2})^{T}, (p^{\prime}_{3})^{T}\right]\left[p_{1}^{T}, p_{2}^{T}, p_{3}^{T}\right]^{-1}\).

Exercise 3.5 Find the exponential coordinates \(\hat{\omega}\theta \in \mathbb{R}^{3}\) for the SO(3) matrix \(\begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 1 & 0 & 0 \\ \end{bmatrix}\)

Find \(\theta\) first by \(\begin{align} \begin{split} 1+2c_{\theta} &= \text{tr}(R)=0 \\ \theta&= \frac{2\pi}{3} \\ \end{split} \end{align}\)

Then find \(\hat{\omega}\) by \(\begin{align} \begin{split} 2s_{\theta} \begin{bmatrix} \omega_{1} \\ \omega_{2} \\ \omega_{3} \\ \end{bmatrix} &= \begin{bmatrix} r_{32} - r_{23} \\ r_{13} - r_{31} \\ r_{21} - r_{12} \\ \end{bmatrix} = \begin{bmatrix} 0-(-1) \\ 0-1 \\ 0-(-1) \\ \end{bmatrix} \\ \hat{\omega} &= \frac{1}{ 2s_{\frac{2\pi}{3}} } \begin{bmatrix} 1 \\ -1 \\ 1 \\ \end{bmatrix} = \frac{1}{\sqrt{3}}\begin{bmatrix} 1 \\ -1 \\ 1 \\ \end{bmatrix} \\ \end{split} \end{align}\)

The exponential coordinate \(\hat{\omega}\theta\) can therefore be found as \(\begin{align} \begin{split} \hat{\omega}\theta &= \left( \frac{2\pi}{3} \right) \left( \frac{\sqrt{3}}{3} \begin{bmatrix} 1 \\ -1 \\ 1 \\ \end{bmatrix}\right)\\ \end{split} \end{align}\)

Exercise 3.16 In terms of the \(\hat{x}_{s}, \hat{y}_{s}, \hat{z}_{s}\) coordinates of a fixed space frame {s}, frame {a} has its \(\hat{x}_{a}\)-axis pointing in the direction \((0,0,1)\) and its \(\hat{y}_{a}\)-axis pointing in the direction \((−1, 0, 0)\), and frame {b} has its \(\hat{x}_{b}\)-axis pointing in the direction \((1, 0, 0)\) and its \(\hat{y}_{b}\)-axis pointing in the direction \((0, 0, −1)\). The origin of {a} is at \((3,0,0)\) in {s} and the origin of {b} is at \((0,2,0)\) in {s}.

• (a) Draw by hand a diagram showing {a} and {b} relative to {s}.
• (b) Write down the rotation matrices \(R_{sa}\) and \(R_{sb}\) and the transformation matrices \(T_{sa}\) and \(T_{sb}\).
• (c) Given \(T_{sb}\), how do you calculate \(T^{−1}_{sb}\) without using a matrix inverse? Write \(T_{sb}^{-1}\) and verify its correctness using your drawing.
• (d) Given \(T_{sa}\) and \(T_{sb}\), how do you calculate Tab (again without using matrix inverses)? Compute the answer and verify its correctness using your drawing.
• (e) Let T = \(T_{sb}\) be considered as a transformation operator consisting of a rotation about \(\hat{x}\) by \(−90^{\circ}\) and a translation along \(\hat{y}\) by 2 units. Calculate \(T_{1} = T_{sa}T\). Does \(T_{1}\) correspond to a rotation and translation about \(\hat{x}_{s}\) and \(\hat{y}_{s}\), respectively (a world-fixed transformation of \(T_{sa}\)), or a rotation and translation about \(\hat{x}_{a}\) and \(\hat{y}_{a}\), respectively (a body-fixed transformation of \(T_{sa}\))? Now calculate \(T_{2} = TT_{sa}\). Does \(T_{2}\) correspond to a body-fixed or world-fixed transformation of \(T_{sa}\)?
• (f) Use \(T_{sb}\) to change the representation of the point \(p_{b} = (1,2,3)\) in {b} coordinates to {s} coordinates.
• (g) Choose a point \(p\) represented by \(p_{s} = (1, 2, 3)\) in {s} coordinates. Calculate \(p^{\prime} = T_{sb}p_{s}\) and \(p^{\prime\prime} = T^{-1}_{sb}p_{s}\). For each operation, should the result be interpreted as changing coordinates (from the {s} frame to {b}) without moving the point p, or as moving the location of the point without changing the reference frame of the representation?
• (h) A twist \(\mathscr{V}\) is represented in {s} as \(\mathscr{V}_{s} = (3, 2, 1, −1, −2, −3)\). What is its representation \(\mathscr{V}_{a}\) in frame {a}?
• (i) By hand, calculate the matrix logarithm \([\mathscr{S}]\theta\) of \(T_{sa}\). (You may verify your answer with software.) Extract the normalized screw axis \(\mathscr{S}\) and rotation amount \(\theta\). Find a \(\{q,\hat{s},h\}\) representation of the screw axis. Redraw the fixed frame {s} and in it draw \(\mathscr{S}\).
• (j) Calculate the matrix exponential corresponding to the exponential coordinates of rigid-body motion \(\mathscr{S}\theta = (0, 1, 2, 3, 0, 0)\). Draw the corresponding frame relative to {s}, as well as the screw axis \(\mathscr{S}\).

• (a)

• (b) \(R_{sa} = \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 1 & 0 & 0 \\ \end{bmatrix}, R_{sb} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{bmatrix}, T_{sa} = \begin{bmatrix} 0 & -1 & 0 & 3 \\ 0 & 0 & -1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}, T_{sb} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}\)

• (c) \(\begin{align} \begin{split} T_{sb}^{-1} &= \begin{bmatrix} R_{sb} & p_{b} \\ 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} R_{sb}^{T} & -R_{sb}^{T}p_{b} \\ 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \\ \end{split} \end{align}\)

• (d) \(\begin{align} \begin{split} T_{ab} &= T_{as}T_{sb} \\ &= T_{sa}^{-1}T_{sb}\\ &= \begin{bmatrix} R_{sa}^{T} & -R_{sa}^{T}p_{a} \\ 0 & 1 \\ \end{bmatrix} T_{sb}\\ &= \begin{bmatrix} 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & -3 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}\\ \end{split} \end{align}\)

• (e) \(\begin{align} \begin{split} T_{1} &= T_{sa}T\\ &=\begin{bmatrix} 0 & 0 & -1 & 1 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}\\ \end{split} \end{align}\)

By the drawing above, \(T_{1}\) corresponds to a rotation about \(\hat{x}_{a}\) of \(-90^{\circ}\) plus a translation about \(\hat{y}_{a}\) of \(2\) units. This behaviour is predicted by the adjacent subscripts cancellation rule.

\[\begin{align} \begin{split} T_{2} &= TT_{sa}\\ &=\begin{bmatrix} 0 & -1 & 0 & 3 \\ 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}\\ \end{split} \end{align}\]

By the drawing above, \(T_{2}\) corresponds to a rotation about \(\hat{x}_{s}\) of \(-90^{\circ}\) plus a translation about \(\hat{y}_{s}\) of \(2\) units. This behaviour is predicted by the adjacent subscripts cancellation rule.

• (f) \(p_{s} = T_{sb}p_{b}= \begin{bmatrix} 1 \\ 5 \\ -2 \\ 1 \\ \end{bmatrix}\)

Truncate the meaningless last row of the vector, we have

\[p_{s} = \begin{bmatrix} 1 \\ 5 \\ -2 \\ \end{bmatrix}\]

• (g) From part (f) we have, \(p^{\prime} = \begin{bmatrix} 1 \\ 5 \\ -2 \\ \end{bmatrix}\)

\(p^{\prime}\) should be interpreted as moving the location of point \(p_{s}\) without changing the reference frame, because the two adjacent subscripts are not common and cannot be cancelled.

\[\begin{align} \begin{split} p^{\prime\prime} &= T_{sb}^{-1}p_{s}\\ &= T_{bs}p_{s}\\ &= \begin{bmatrix} 1 \\ -3 \\ 0 \\ \end{bmatrix}\text{last row truncated.}\\ \end{split} \end{align}\]

\(p^{\prime\prime}\) should be interpreted as changing the reference frame from {s} to {a} of point \(p_{s}\) without moving its location, because the two adjacent subscripts are common and can therefore be cancelled.

• h) We know that \(\mathscr{V}_{a}=\left[\text{Ad}_{T_{as}}\right]\mathscr{V}_{s}\). To find \(\text{Ad}_{T_{as}}\), we need to find \(T_{as}\) first, \(T_{as} = T_{sa}^{-1} = \begin{bmatrix} 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 3 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}\)

Then, using the adjoint representation formula, we have \(\left[\text{Ad}_{T_{sa}}\right] = \begin{bmatrix} 0 & 0 & 1 & \; & 0 & 0 & 0 \\ -1 & 0 & 0 & \; & 0 & 0 & 0 \\ 0 &-1 & 0 & \; & 0 & 0 & 0 \\ \,&\, &\, & \, &\, &\, &\, \\ 0 &-3 & 0 & \; & 0 & 0 & 1 \\ 0 & 0 & 0 & \; &-1 & 0 & 0 \\ 0 & 0 &-3 & \; & 0 &-1 & 0 \\ \end{bmatrix}\) Finally, we have \(\mathscr{V}_{a} = \begin{bmatrix} 1 \\ -3 \\ -2 \\ -9 \\ 1 \\ -1 \\ \end{bmatrix}\)

• (i) This is a big chunk. Get Ready! The calculation is done on MATLAB. We have found \(R_{sa}\) in previous parts, now we need to find \(\theta\) and \([\hat{\omega}]\). Because \(\text{tr}(R_{sa})=0\), the third choice of the algorithm should be used. Therefore, we have

\[\begin{align} \begin{split} \theta &= \cos^{-1}\left(\frac{\text{tr}(R_{sa})-1}{2}\right)=\frac{2\pi}{3}, \\ [\hat{\omega}] &= \frac{1}{2\sin(\theta)}(R_{sa}-R_{sa}^{T}) = \frac{\sqrt{3}}{3} \begin{bmatrix} 0 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & 1 & 0 \\ \end{bmatrix} \\ \end{split} \end{align}\]

%% Exercise 3.16
%% (i)
R_sa = [0 -1  0;
        0  0 -1;
        1  0  0];
theta = acos(-1/2);
skew_omega = 1/(2*sin(theta)) .* (R_sa - transpose(R_sa));

Now we need to find \(v\). From the formula we know that \(v=G^{-1}(\theta)p\). Therefore, we need to first find \(G^{-1}(\theta)\) using, \(\begin{align} \begin{split} G^{-1}(\theta) &= \frac{1}{\theta}I-\frac{1}{2}[\theta]+\left( \frac{1}{\theta} - \frac{1}{2}\cot\frac{\theta}{2} \right)[\omega]^{2} \\ &= \begin{bmatrix} 0.3516 & 0.2257 & 0.3516 \\ -0.3516 & 0.3516 & 0.2257 \\ -0.2257 & -0.3516 & 0.3516 \\ \end{bmatrix} \\ v &= G^{-1}(\theta)p \\ &= \begin{bmatrix} 1.0548 \\ -1.0548 \\ -0.6772 \\ \end{bmatrix} \end{split} \end{align}\)

Since \(\omega\) does not equal to 0, we know that \(\mathscr{S}\) is just the normalized \(\mathscr{V}\). Therefore we have \(\mathscr{S} = (0.5774, -0.5774, 0.5774, 1.0548, -1.0548, -0.6772) ^{T}\) The unit screw axis \(\hat{s}\) is simply \([0.5774, -0.5774, 0.5774]^{T}\). Then find \(h=\frac{\hat{\omega}^{T}v}{\dot{\theta}}=0.8271\).

I = eye(3);     
inv_G = 1/theta .* I - 1/2 .* skew_omega + ( 1/theta - 1/2 * cot(theta/2) ) .* skew_omega^2;
p = [3; 0; 0];
v = inv_G * p;
skew_S = [skew_omega v;
          0  0   0   0];
s_hat = [0.5774; -0.5774; 0.5774]; % by hand
omega_hat = s_hat;
h = transpose(omega_hat) * v;

Because in \(v=-\omega\times q + h \omega\) there is a cross product, we need to solve a system of equations. Rearrange the aforesaid equation as \(v-h\omega = -\omega\times q\). Let \(q=(a,b,c)^{T}\). Then we have, \(\begin{align} \begin{split} \begin{bmatrix} 0.5773 \\ -0.5773 \\ -1.1548 \\ \end{bmatrix} &= -\left\lvert \begin{array}{ccc} i & j & k \\ 0.5773 & -0.5773 & 0.5773 \\ a & b & c \\ \end{array}\right\rvert \\ \end{split} \end{align}\)

\[\left[ \begin{array}{ccc|c} 0 & 0.5773 & 0.5773 & 0.5773 \\ -0.5773 & 0 & 0.5773 & -0.5773 \\ -0.5773 & -0.5773 & 0 & -1.1548 \\ \end{array} \right] \Longrightarrow q = \begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix}\]

At last, the screw axis represented in \(\{q,\hat{s},h\}\) form is

\[\{q,\hat{s},h\} = \left\{\begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix}, \begin{bmatrix} 0.5774 \\ -0.5774 \\ 0.5774 \\ \end{bmatrix}, 0.8271 \right\}\]

• (j) The calculation steps are:
  1. Because \(\mathscr{S}\theta=[\hat{\omega}\theta,\,v\theta]^{T}\), we can find \(\hat{\omega}\) and \(\theta\).
  2. Since \(e^{[\mathscr{S}]\theta}=\begin{bmatrix} e^{[\hat{\omega}]\theta} & G(\theta)v \\ 0 & 1 \\ \end{bmatrix}\), we need to first find each term first.
  3. Done!

Here is the actual MATLAB code:

%% (j)
s_theta = [0; 1; 2; 3; 0; 0];
omega_theta = [0; 1; 2];
v = [3; 0; 0];
theta = norm(omega_theta);
hat_omega = omega_theta ./ theta;
bracket_omega = [      0       -hat_omega(3)  hat_omega(2);
                  hat_omega(3)      0             0;
                 -hat_omega(2)      0             0;       ];
exp_omega_theta = eye(3) + sin(theta).* bracket_omega + (1-cos(theta)) ...
                  .* bracket_omega^2;
G = eye(3) .* theta + (1-cos(theta)).*bracket_omega + (theta-sin(theta))...
    .* bracket_omega^2;
exp_S_theta = [exp_omega_theta   G*v;
                0     0    0      1 ];

The final result is, \(e^{[\mathscr{S}]\theta} = \begin{bmatrix} -0.6173 & -0.7037 & 0.3518 & 2.3602 \\ 0.7037 & -0.2938 & 0.6469 & 4.3396 \\ -0.3518 & 0.6469 & 0.6765 & -2.1698 \\ 0 & 0 & 0 & 1.0000 \\ \end{bmatrix}\)

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