HW1.2 - Coordinate Transformations

87/95, lowest

Q2b [-2] Well done! Your approach is correct, however you should have the final result calculated in decimal form.

Q2c [-2] Well done! Your approach is correct, however you should have the final result calculated in decimal form.

Q2d [-3] Formulas are correct but we were looking for you to actually plug in the numbers and solve for the degrees of the angles.

Q3b [-1] need calculation

Q1

$$F_b=H_{rot}(Y,90)H_{trans}(0,1,0)H_{rot}(Z,45)H_{rot}(Y,30)H_{rot}(X,90)H_{trans}(0,0,1)F_0$$

a

$$\begin{aligned} \vec p^0 &=F_b \vec p^b\\ &=H_{rot}(Y,90)H_{trans}(0,1,0)H_{rot}(Z,45)H_{rot}(Y,30)H_{rot}(X,90)H_{trans}(0,0,1) \vec p^b \end{aligned}$$

b

$$\begin{aligned} \vec v^b &=F_b^{-1} \vec p^0\\ &=(H_{rot}(Y,90)H_{trans}(0,1,0)H_{rot}(Z,45)H_{rot}(Y,30)H_{rot}(X,90)H_{trans}(0,0,1))^{-1}\vec p^b\\ &=(H_{trans}(0,0,1))^{-1}(H_{rot}(X,90))^{-1}(H_{rot}(Y,30))^{-1}(H_{rot}(Z,45))^{-1}(H_{trans}(0,1,0))^{-1}(H_{rot}(Y,90))^{-1} \vec p^b\\ &=H_{trans}(0,0,-1)H_{rot}(X,-90)H_{rot}(Y,-30)H_{rot}(Z,-45)H_{trans}(0,-1,0)H_{rot}(Y,-90) \vec p^b \end{aligned}$$

Q2

a

$$\begin{aligned} R_{ZYX}(90,30,45) &=R_Z(45)R_Y(30)R_X(90)\\ &= \begin{pmatrix} \frac{\sqrt2}{2}&-\frac{\sqrt2}{2}&0\\ \frac{\sqrt2}{2}&\frac{\sqrt2}{2}&0\\ 0&0&1 \end{pmatrix} \begin{pmatrix} \frac{\sqrt3}{2}&0&\frac{1}{2}\\ 0&1&0\\ -\frac{1}{2}&0&\frac{\sqrt3}{2} \end{pmatrix} \begin{pmatrix} 1&0&0\\ 0&0&-1\\ 0&1&0 \end{pmatrix}\\ &=\begin{pmatrix} \frac{\sqrt6}{4}&\frac{\sqrt2}{4}&\frac{\sqrt2}{2}\\ \frac{\sqrt6}{4}&\frac{\sqrt2}{4}&-\frac{\sqrt2}{2}\\ -\frac{1}{2}&\frac{\sqrt3}{2}&0 \end{pmatrix} \end{aligned}$$

b

-2, final result calculated in decimal form.

0.701, 0.5618, 0.4310, 0.0925

$$s=\frac{\sqrt{1+r_{11}+r_{22}+r_{33}}}{2}\\ x=\frac{r_{32}-r_{23}}{4w}\\ y=\frac{r_{13}-r_{31}}{4w}\\ z=\frac{r_{21}-r_{12}}{4w}$$

c

-2, final result calculated in decimal form.

$\theta=2\arccos(0.701)=90.99\degree$

$\hat u=\frac{1}{\|v\|}[x,y,z]^T=[0.7867,0.6036,0.1295]^T$

$$\begin{align} \hat u &=\frac{\vec v}{||v||}\\ &=\frac{\vec v}{\sin\frac{\theta}{2}}\\ \theta &=2\cdot\cos^{-1}(s) \end{align}$$

d

-3, final result calculated in decimal form.

$R_{ZYX}=\begin{bmatrix}0.6124&0.3535&0.707\\0.6124&0.3535&-0.707\\-0.5&0.866&0\end{bmatrix}$

30, 45, 90

Because

$$\begin{aligned} R_E &=R_{ZYX}(\phi,\theta,\psi)\\ &=R_z(\psi)R_Y(\theta)R_X(\phi)\\ &=\begin{pmatrix} \cos\psi & -\sin\psi & 0 \\ \sin\psi & \cos\psi & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} \cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\ -\sin\theta & 0 & \cos\theta \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\phi & -\sin\phi \\ 0 & \sin\phi & \cos\phi \end{pmatrix}\\ &=\begin{pmatrix} \cos\theta\cos\psi & \sin\phi\sin\theta\cos\psi-\cos\phi\sin\psi & \sin\phi\sin\psi+\cos\phi\sin\theta\cos\psi \\ \cos\theta\sin\psi & \sin\phi\sin\theta\sin\psi+\cos\phi\cos\psi & \sin\psi\sin\theta\cos\phi-\cos\psi\sin\phi \\ -\sin\theta & \cos\theta\sin\phi & \cos\theta\cos\phi \end{pmatrix} \end{aligned}$$

Therefore,

$$\phi_y=-\arcsin r_{31} \\ \phi_x = \arctan\frac{r_{32}}{r_{33}}\\ \phi_z = \arctan\frac{r_{21}}{r_{11}}$$

Q3

a

$$F=H_{trans}(1,1,1)H_{rot}(z,45)H_{rot}(y,30)H_{rot}(x,90)$$

b

-1, final result calculated in decimal form.

$$\begin{aligned} \because & [\phi^L,\theta^L,\psi^L] \Leftrightarrow [-\phi^R,-\theta^R,\psi^R]\\ &x^R=x^L, y^R=y^L, z^R=-z^L\\ \therefore &F^L=H_{trans}(1,1,-1)H_{rot}(z,45)H_{rot}(y,-30)H_{rot}(x,-90) \end{aligned}$$