HW1.1 - Coordinate Transformations

80/100, mean

Q1 [-5] The columns of the matrices correspond to the axes. Your sin should be negative in your y component but in the i hat direction not in your x component.

Q3a [-1] Should explain how the answer was derived.

Q3b [-1] Should explain how the answer was derived.

Q4c [-10] no submission for 4c

Q4d [-3] incorrect order

1

Given a rotation angle about the x-axis, I showed in class how to derive the associated rotation matrix

$$R_1^0(x,\phi)=R_X=\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\phi & -\sin\phi \\ 0 & \sin\phi & \cos\phi \end{pmatrix}$$

representing how vectors specified in the rotated (1) coordinate system can be represented in the unrotated (0) coordinate system. Using a similar approach, show that:

$$R_1^0(z,\psi)=R_Z = \begin{pmatrix} \cos\psi & -\sin\psi & 0 \\ \sin\psi & \cos\psi & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

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Answer:

20-5, The columns of the matrices correspond to the axes. Your sin should be negative in your y component but in the i hat direction not in your x component.

Let $\vec p$ represent the vector

$$\vec p=p^0_x\hat i_0+p^0_y\hat j_0+p^0_z\hat k_0\\ \vec p=p^1_x\hat i_1+p^1_y\hat j_1+p^1_z\hat k_1$$

We want to find the equivalent vector $\vec p$ represnted in the System 1

$$\begin{equation} \begin{aligned} p_x^0\hat i_0&=p_x^0\cos\psi\hat i_1-p_x^0\sin\psi\hat j_1&\\ p_y^0\hat j_0&=p_y^0\sin\psi\hat i_1+p_y^0\cos\psi\hat j_1\\ p_z^0\hat k_0&=p_z^0\hat k_1 \end{aligned}\\ \Rightarrow \begin{pmatrix} \hat i_0\\\hat j_0\\\hat k_0 \end{pmatrix}=\begin{pmatrix} \cos\psi & -\sin\psi & 0 \\ \sin\psi & \cos\psi & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \hat i_1\\\hat j_1\\\hat k_1 \end{pmatrix} \end{equation}$$

So we can get the result:

$$\vec p^0=R_1^0(z,\psi)\vec p^1\\ R_1^0(z,\psi)=R_Z = \begin{pmatrix} \cos\psi & -\sin\psi & 0 \\ \sin\psi & \cos\psi & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

2

a) (10pts) Derive the symbolic form of the Euler angle rotation matrix $R_{ZYX}$ that corresponds to a sequence of three rotations where the first rotation ($\psi$) is about the local z-axis, the second rotation ($\theta$) is about the local y-axis, and the third rotation ($\phi$) is about the local x-axis. That is, the order of rotation corresponds to ZYX.

b) (5pts) Does the rotation matrix $R_{ZYX}$ you have derived transform vectors from the unrotated frame to the rotated frame, or from the rotated frame to the unrotated frame? How do you know?

c) (10pts) Given $R_z, R_y, R_x$ show how you would compute $(R_{zyx}^{-1})$

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a

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

b

The matrix $R_{ZYX}$ transforms vectors from the rotated frame to the unrotated frame. $R_{ZYX}=R_ZR_YR_X$ Because $R_X$ is rotated frame to the unrotated frame, $R_Y$ is rotated frame to the unrotated, and $R_Z$ is rotated frame to the unrotated. So do $R_{ZYX}$ transforms vectors from the rotated frame to the unrotated frame.

c

$$\begin{aligned} (R_{ZYX})^{-1} &=(R_ZR_YR_X)^{-1}\\ &=(R_X)^{-1}(R_Y)^{-1}(R_Z)^{-1}\\ &=(R_X)^{T}(R_Y)^{T}(R_Z)^{T} \end{aligned}$$

3

a

-1, Should explain how the answer was derived.

$$H_{trans}^{-1}=\begin{pmatrix} I & -d \\ 0 & 1 \end{pmatrix}$$

b

-1, Should explain how the answer was derived.

$$H_{rot}^{-1}=\begin{pmatrix} R^T & 0 \\ 0 & 1 \end{pmatrix}$$

c

$$\begin{aligned} (H)^{-1} &=(H_{trans}H_{rot})^{-1}\\ &=(H_{rot})^{-1}(H_{trans})^{-1}\\ &=\begin{pmatrix} R^T & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} I & -d \\ 0 & 1 \end{pmatrix}\\ &=\begin{pmatrix} R^T & -R^Td \\ 0 & 1 \end{pmatrix} \end{aligned}$$

4

a

$$\begin{aligned} F_1^0&=\begin{pmatrix} 1 & 0 & 0 & 3\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\\ F_2^0&=\begin{pmatrix} 1 & 0 & 0 & 8\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} \frac{\sqrt2}{2} & -\frac{\sqrt2}{2} & 0 & 0\\ \frac{\sqrt2}{2} & \frac{\sqrt2}{2} & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\\ &= \begin{pmatrix} \frac{\sqrt2}{2} & -\frac{\sqrt2}{2} & 0 & 8\\ \frac{\sqrt2}{2} & \frac{\sqrt2}{2} & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\\ F_3^0&= \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 3\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 4\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & -1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1 & 0 & 4\\ 1 & 0 & 0 & 3\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \end{aligned}$$

b

$$|d_3^0|=|[4, 3, 0]^T|=5$$

c

$$|d_3^2|=|d_3^0-d_2^0|=|[-4,3,0]^T|=5$$

-10, no submission for 4c

$$F_3^2=(F_2^0)^{-1}F_3^0=\begin{bmatrix} \frac{\sqrt2}{2}&\frac{\sqrt2}{2}&0&-\frac{\sqrt2}{2}\\ -\frac{\sqrt2}{2}&\frac{\sqrt2}{2}&0&\frac{7\sqrt2}{2}\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix}\\ d_3^2=[-\frac{\sqrt2}{2},\frac{7\sqrt2}{2},0]^T$$

d

-3, incorrect order

$$\begin{aligned} H&=H_{trans}(-4,3,0)H_{rot}(Z^2,-45)\\ &=\begin{pmatrix} 1&0&0&-4\\ 0&1&0&3\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix}\begin{pmatrix} \frac{\sqrt2}{2}&\frac{\sqrt2}{2}&0&0\\ -\frac{\sqrt2}{2}&\frac{\sqrt2}{2}&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix}\\ &=\begin{pmatrix} \frac{\sqrt2}{2}&\frac{\sqrt2}{2}&0&-4\\ -\frac{\sqrt2}{2}&\frac{\sqrt2}{2}&0&3\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix} \end{aligned}$$

❌

$$F_4^0=HF_2^0\Rightarrow H=F_4^0(F_2^0)^{-1}$$