1. **Problem Statement:** We need to assign frames to the manipulator using Denavit-Hartenberg (DH) convention and derive the forward kinematics to find the position of the wrist point $W$ in terms of joint angles $\theta_1, \theta_2, \theta_3$ and link lengths $L_1, L_2, L_3$. 2. **DH Parameter Assignment:** For a planar 3R manipulator, the DH parameters are: - $\alpha_i = 0$ (all axes are parallel in the plane) - $a_i = L_i$ (link lengths) - $d_i = 0$ (revolute joints, no offset along $z$) - $\theta_i$ are the joint angles The DH table is: | Link $i$ | $\theta_i$ | $d_i$ | $a_i$ | $\alpha_i$ | |----------|------------|-------|-------|------------| | 1 | $\theta_1$ | 0 | $L_1$ | 0 | | 2 | $\theta_2$ | 0 | $L_2$ | 0 | | 3 | $\theta_3$ | 0 | $L_3$ | 0 | 3. **Transformation Matrices:** Each link transformation matrix $T_i^{i-1}$ is given by: $$ T_i^{i-1} = \begin{bmatrix} \cos\theta_i & -\sin\theta_i & 0 & a_i \\ \sin\theta_i & \cos\theta_i & 0 & 0 \\ 0 & 0 & 1 & d_i \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ Since $d_i=0$ and $\alpha_i=0$, this simplifies to: $$ T_i^{i-1} = \begin{bmatrix} \cos\theta_i & -\sin\theta_i & 0 & L_i \\ \sin\theta_i & \cos\theta_i & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ 4. **Forward Kinematics:** The overall transformation from base to wrist $W$ is: $$ T = T_1^0 T_2^1 T_3^2 $$ Calculate stepwise: $$ T_1^0 = \begin{bmatrix} \cos\theta_1 & -\sin\theta_1 & 0 & L_1 \\ \sin\theta_1 & \cos\theta_1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, \quad T_2^1 = \begin{bmatrix} \cos\theta_2 & -\sin\theta_2 & 0 & L_2 \\ \sin\theta_2 & \cos\theta_2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, \quad T_3^2 = \begin{bmatrix} \cos\theta_3 & -\sin\theta_3 & 0 & L_3 \\ \sin\theta_3 & \cos\theta_3 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ Multiply $T_1^0$ and $T_2^1$: $$ T_1^0 T_2^1 = \begin{bmatrix} \cos\theta_1 & -\sin\theta_1 & 0 & L_1 \\ \sin\theta_1 & \cos\theta_1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos\theta_2 & -\sin\theta_2 & 0 & L_2 \\ \sin\theta_2 & \cos\theta_2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ Calculate the product: $$ = \begin{bmatrix} \cos\theta_1 \cos\theta_2 - \sin\theta_1 \sin\theta_2 & -\cos\theta_1 \sin\theta_2 - \sin\theta_1 \cos\theta_2 & 0 & L_1 + L_2 \cos(\theta_1 + \theta_2) \\ \sin\theta_1 \cos\theta_2 + \cos\theta_1 \sin\theta_2 & -\sin\theta_1 \sin\theta_2 + \cos\theta_1 \cos\theta_2 & 0 & L_2 \sin(\theta_1 + \theta_2) \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ Using angle sum identities: $$ \cos(\theta_1 + \theta_2) = \cos\theta_1 \cos\theta_2 - \sin\theta_1 \sin\theta_2 $$ $$ \sin(\theta_1 + \theta_2) = \sin\theta_1 \cos\theta_2 + \cos\theta_1 \sin\theta_2 $$ So the matrix simplifies to: $$ = \begin{bmatrix} \cos(\theta_1 + \theta_2) & -\sin(\theta_1 + \theta_2) & 0 & L_1 + L_2 \cos(\theta_1 + \theta_2) \\ \sin(\theta_1 + \theta_2) & \cos(\theta_1 + \theta_2) & 0 & L_2 \sin(\theta_1 + \theta_2) \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ Now multiply by $T_3^2$: $$ T = (T_1^0 T_2^1) T_3^2 = \begin{bmatrix} \cos(\theta_1 + \theta_2) & -\sin(\theta_1 + \theta_2) & 0 & L_1 + L_2 \cos(\theta_1 + \theta_2) \\ \sin(\theta_1 + \theta_2) & \cos(\theta_1 + \theta_2) & 0 & L_2 \sin(\theta_1 + \theta_2) \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos\theta_3 & -\sin\theta_3 & 0 & L_3 \\ \sin\theta_3 & \cos\theta_3 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ Calculate the product: $$ = \begin{bmatrix} \cos(\theta_1 + \theta_2) \cos\theta_3 - \sin(\theta_1 + \theta_2) \sin\theta_3 & -\cos(\theta_1 + \theta_2) \sin\theta_3 - \sin(\theta_1 + \theta_2) \cos\theta_3 & 0 & L_1 + L_2 \cos(\theta_1 + \theta_2) + L_3 \cos(\theta_1 + \theta_2 + \theta_3) \\ \sin(\theta_1 + \theta_2) \cos\theta_3 + \cos(\theta_1 + \theta_2) \sin\theta_3 & -\sin(\theta_1 + \theta_2) \sin\theta_3 + \cos(\theta_1 + \theta_2) \cos\theta_3 & 0 & L_2 \sin(\theta_1 + \theta_2) + L_3 \sin(\theta_1 + \theta_2 + \theta_3) \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ Using angle sum identities again: $$ \cos(\theta_1 + \theta_2 + \theta_3) = \cos(\theta_1 + \theta_2) \cos\theta_3 - \sin(\theta_1 + \theta_2) \sin\theta_3 $$ $$ \sin(\theta_1 + \theta_2 + \theta_3) = \sin(\theta_1 + \theta_2) \cos\theta_3 + \cos(\theta_1 + \theta_2) \sin\theta_3 $$ So the final transformation matrix is: $$ T = \begin{bmatrix} \cos(\theta_1 + \theta_2 + \theta_3) & -\sin(\theta_1 + \theta_2 + \theta_3) & 0 & L_1 + L_2 \cos(\theta_1 + \theta_2) + L_3 \cos(\theta_1 + \theta_2 + \theta_3) \\ \sin(\theta_1 + \theta_2 + \theta_3) & \cos(\theta_1 + \theta_2 + \theta_3) & 0 & L_2 \sin(\theta_1 + \theta_2) + L_3 \sin(\theta_1 + \theta_2 + \theta_3) \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ 5. **Wrist Point Position:** The position of the wrist point $W$ in the base frame is the translation part of $T$: $$ x_W = L_1 \cos\theta_1 + L_2 \cos(\theta_1 + \theta_2) + L_3 \cos(\theta_1 + \theta_2 + \theta_3) $$ $$ y_W = L_1 \sin\theta_1 + L_2 \sin(\theta_1 + \theta_2) + L_3 \sin(\theta_1 + \theta_2 + \theta_3) $$ This gives the forward kinematics of the wrist point $W$. --- **Final answer:** $$ \boxed{ \begin{cases} x_W = L_1 \cos\theta_1 + L_2 \cos(\theta_1 + \theta_2) + L_3 \cos(\theta_1 + \theta_2 + \theta_3) \\ y_W = L_1 \sin\theta_1 + L_2 \sin(\theta_1 + \theta_2) + L_3 \sin(\theta_1 + \theta_2 + \theta_3) \end{cases} } $$