A planar elbow manipulator is shown. Write the D-H parameters and hence locate the position and orientation of the end E of the manipulator. Given \(L_1=L_2=250\) mm and \(\theta_1=\theta_2=30^\circ\). (asked in MU June 2025 KT)Figure (described in w…

Answer: E is at \((341.5,\ 341.5)\) mm and the end link is oriented at \(\theta_1+\theta_2=60^\circ\) to \(X_0\).

D-H parameter table (both joints revolute, all axes \(Z_i\) perpendicular to the plane, so \(d_i=0\), \(\alpha_i=0\)):

Transformation for each link (with \(\alpha=0\), \(d=0\)):

\[{}^{i-1}T_i=\begin{bmatrix}\cos\theta_i&-\sin\theta_i&0&a_i\cos\theta_i\\ \sin\theta_i&\cos\theta_i&0&a_i\sin\theta_i\\0&0&1&0\\0&0&0&1\end{bmatrix}\]

\[{}^0T_1={}^1T_2=\begin{bmatrix}0.866&-0.5&0&216.5\\0.5&0.866&0&125\\0&0&1&0\\0&0&0&1\end{bmatrix}\]

Overall: \({}^0T_2={}^0T_1\,{}^1T_2\):

\[{}^0T_2=\begin{bmatrix}\cos60^\circ&-\sin60^\circ&0&L_1\cos30^\circ+L_2\cos60^\circ\\ \sin60^\circ&\cos60^\circ&0&L_1\sin30^\circ+L_2\sin60^\circ\\0&0&1&0\\0&0&0&1\end{bmatrix}=\begin{bmatrix}0.5&-0.866&0&341.5\\0.866&0.5&0&341.5\\0&0&1&0\\0&0&0&1\end{bmatrix}\]

\(x_E=216.5+125=341.5\) mm, \(y_E=125+216.5=341.5\) mm. The rotation part shows the end frame is turned by 60° about \(Z_0\), so link 2 (and the end E) points at 60° to \(X_0\).