# Forward Kinematics of Cylindrical Arm (Figure 2) Let's solve the forward kinematics for the cylindrical robot arm shown. ## **Step 1: Assign D-H Parameters** The cylindrical arm has: - 1 revolute joint (\(\theta_1\)) - 2 prismatic joints (\(d_2, d_3\)) Let’s define the D-H parameters for each joint. | Link | \(\theta_i\) | \(d_i\) | \(a_i\) | \(\alpha_i\) | |------|:-----------:|:-------:|:-------:|:------------:| | 1 | \(\theta_1\)| \(d_1\) | | \(^\circ\) | | 2 | \(\) | \(d_2\) | | \(90^\circ\) | | 3 | \(\) | \(d_3\) | | \(^\circ\) | Note: - \(d_1\): height along \(z_\) - \(\theta_1\): rotation about \(z_\) - \(d_2\): translation along \(z_1\) - \(d_3\): translation along \(x_2\) (see figure) ## **Step 2: Write Each Transformation Matrix** The standard DH transformation: \[ A_i = \begin{bmatrix} \cos\theta_i & -\sin\theta_i\cos\alpha_i & \sin\theta_i\sin\alpha_i & a_i\cos\theta_i \\ \sin\theta_i & \cos\theta_i\cos\alpha_i & -\cos\theta_i\sin\alpha_i & a_i\sin\theta_i \\ & \sin\alpha_i & \cos\alpha_i & d_i \\ & & & 1 \end{bmatrix} \] ### **Link 1:** \[ A_1 = \begin{bmatrix} \cos\theta_1 & -\sin\theta_1 & & \\ \sin\theta_1 & \cos\theta_1 & & \\ & & 1 & d_1 \\ & & & 1 \end{bmatrix} \] ### **Link 2:** \(\theta_2 = , \alpha_2 = 90^\circ\) \(\cos 90^\circ = , \sin 90^\circ = 1\) \[ A_2 = \begin{bmatrix} 1 & & & \\ & & -1 & \\ & 1 & & d_2 \\ & & & 1 \end{bmatrix} \] ### **Link 3:** \(\theta_3 = , \alpha_3 = ^\circ\) \[ A_3 = \begin{bmatrix} 1 & & & d_3 \\ & 1 & & \\ & & 1 & \\ & & & 1 \end{bmatrix} \] --- ## **Step 3: Compute the Overall Transformation** \[ T = A_1 A_2 A_3 \] Let’s multiply step by step. ### **First: \(A_1 A_2\)** \[ A_1 A_2 = \begin{bmatrix} \cos\theta_1 & -\sin\theta_1 & & \\ \sin\theta_1 & \cos\theta_1 & & \\ & & 1 & d_1 \\ & & & 1 \end{bmatrix} \begin{bmatrix} 1 & & & \\ & & -1 & \\ & 1 & & d_2 \\ & & & 1 \end{bmatrix} \] Calculate: - Row 1: \( [\cos\theta_1, -\sin\theta_1, ] \) - Dot with col 1: \( \cos\theta_1 \) - Dot with col 2: \( -\sin\theta_1 \cdot + = \) - Dot with col 3: \( -\sin\theta_1 \cdot -1 = \sin\theta_1 \) - Dot with last col: \( -\sin\theta_1 \cdot d_2 \) - Row 2: \( [\sin\theta_1, \cos\theta_1, ] \) - Dot with col 1: \( \sin\theta_1 \) - Dot with col 2: \( \cos\theta_1 \cdot + = \) - Dot with col 3: \( \cos\theta_1 \cdot -1 = -\cos\theta_1 \) - Dot with last col: \( \cos\theta_1 \cdot d_2 \) - Row 3: \( [, , 1] \) - Dot with col 1: \( \) - Dot with col 2: \( 1 \) - Dot with col 3: \( \) - Dot with last col: \( d_1 \) So, \[ A_1 A_2 = \begin{bmatrix} \cos\theta_1 & & \sin\theta_1 & -\sin\theta_1 d_2 \\ \sin\theta_1 & & -\cos\theta_1 & \cos\theta_1 d_2 \\ & 1 & & d_1 \\ & & & 1 \end{bmatrix} \] --- ### **Then: \((A_1 A_2) A_3\)** Multiply by \(A_3\): \[ \begin{bmatrix} \cos\theta_1 & & \sin\theta_1 & -\sin\theta_1 d_2 \\ \sin\theta_1 & & -\cos\theta_1 & \cos\theta_1 d_2 \\ & 1 & & d_1 \\ & & & 1 \end{bmatrix} \begin{bmatrix} 1 & & & d_3 \\ & 1 & & \\ & & 1 & \\ & & & 1 \end{bmatrix} \] - The rotation part (upper \(3\times3\)) remains the same. - The position column: \[ \mathbf{p} = \begin{bmatrix} -\sin\theta_1 d_2 + \cos\theta_1 d_3 \\ \cos\theta_1 d_2 + \sin\theta_1 d_3 \\ d_1 \end{bmatrix} \] --- ## **Step 4: Final Transformation Matrix** \[ \boxed{ T = \begin{bmatrix} \cos\theta_1 & & \sin\theta_1 & -\sin\theta_1 d_2 + \cos\theta_1 d_3 \\ \sin\theta_1 & & -\cos\theta_1 & \cos\theta_1 d_2 + \sin\theta_1 d_3 \\ & 1 & & d_1 \\ & & & 1 \end{bmatrix} } \] --- ## **Summary** - The forward kinematics of the cylindrical arm is given by the transformation above. - The position of the end-effector (origin of frame 3) in the base frame is: - \(x = -\sin\theta_1 d_2 + \cos\theta_1 d_3\) - \(y = \cos\theta_1 d_2 + \sin\theta_1 d_3\) - \(z = d_1\) **This matrix gives the pose of the end-effector with respect to the base frame for any given joint variables \((\theta_1, d_1, d_2, d_3)\).**

# Forward Kinematics of Cylindrical Arm Let's solve the forward kinematics for the cylindrical robot arm. ## Step 1: Assign D-H Parameters The cylindrical arm has: - 1 revolute joint (\(\theta_1\)) - 2 prismatic joints (\(d_2, d_3\)) ### D-H Parameter Table | Link | \(\theta_i\) | \(d_i\) | \(a_i\) | \(\alpha_i\) | |------|:--------------:|:-------:|:-------:|:------------:| | 1 | \(\theta_1\) | \(d_1\) | 0 | 0 | | 2 | 0 | \(d_2\) | 0 | \(90^\circ\) | | 3 | 0 | \(d_3\) | 0 | 0 | ## Step 2: Write Each Transformation Matrix The standard DH transformation is given by: \[ A_i = \begin{bmatrix} \cos\theta_i & -\sin\theta_i \cos\alpha_i & \sin\theta_i \sin\alpha_i & a_i \cos\theta_i \\ \sin\theta_i & \cos\theta_i \cos\alpha_i & -\cos\theta_i \sin\alpha_i & a_i \sin\theta_i \\ 0 & \sin\alpha_i & \cos\alpha_i & d_i \\ 0 & 0 & 0 & 1 \end{bmatrix} \] ### Link 1 Transformation Matrix \(A_1\) \[ A_1 = \begin{bmatrix} \cos\theta_1 & -\sin\theta_1 & 0 & 0 \\ \sin\theta_1 & \cos\theta_1 & 0 & 0 \\ 0 & 0 & 1 & d_1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] ### Link 2 Transformation Matrix \(A_2\) \(\alpha_2 = 90^\circ\) \[ A_2 = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & d_2 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] ### Link 3 Transformation Matrix \(A_3\) \[ A_3 = \begin{bmatrix} 1 & 0 & 0 & d_3 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] ## Step 3: Compute the Overall Transformation The overall transformation \(T\) is calculated as: \[ T = A_1 A_2 A_3 \] ### Multiply \(A_1\) and \(A_2\) \[ A_1 A_2 = \begin{bmatrix} \cos\theta_1 & -\sin\theta_1 & 0 & 0 \\ \sin\theta_1 & \cos\theta_1 & 0 & 0 \\ 0 & 0 & 1 & d_1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & d_2 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] The resulting matrix is: \[ A_1 A_2 = \begin{bmatrix} \cos\theta_1 & \sin\theta_1 & 0 & -\sin\theta_1 d_2 \\ \sin\theta_1 & -\cos\theta_1 & 0 & \cos\theta_1 d_2 \\ 0 & 0 & 1 & d_1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] ### Multiply \((A_1 A_2)\) and \(A_3\) \[ T = (A_1 A_2) A_3 = \begin{bmatrix} \cos\theta_1 & \sin\theta_1 & 0 & -\sin\theta_1 d_2 \\ \sin\theta_1 & -\cos\theta_1 & 0 & \cos\theta_1 d_2 \\ 0 & 0 & 1 & d_1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & d_3 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] The final transformation matrix \(T\) is: \[ T = \begin{bmatrix} \cos\theta_1 & \sin\theta_1 & 0 & -\sin\theta_1 d_2 + \cos\theta_1 d_3 \\ \sin\theta_1 & -\cos\theta_1 & 0 & \cos\theta_1 d_2 + \sin\theta_1 d_3 \\ 0 & 0 & 1 & d_1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] ## Step 4: Final Transformation Matrix \[ \boxed{ T = \begin{bmatrix} \cos\theta_1 & \sin\theta_1 & 0 & -\sin\theta_1 d_2 + \cos\theta_1 d_3 \\ \sin\theta_1 & -\cos\theta_1 & 0 & \cos\theta_1 d_2 + \sin\theta_1 d_3 \\ 0 & 0 & 1 & d_1 \\ 0 & 0 & 0 & 1 \end{bmatrix} } \] ## Summary - The forward kinematics of the cylindrical arm provides the pose of the end-effector with respect to the base frame. - The position of the end-effector (origin of frame 3) in the base frame is: - \(x = -\sin\theta_1 d_2 + \cos\theta_1 d_3\) - \(y = \cos\theta_1 d_2 + \sin\theta_1 d_3\) - \(z = d_1\)