A three-link planar (3R) robot arm has link lengths \(L_1=0.5\) m, \(L_2=0.4\) m and \(L_3=0.2\) m. At an instant the joint angles are \(\theta_1=30^\circ\) (link 1 measured anticlockwise from the base \(X_0\) axis), \(\theta_2=45^\circ\) and \(\thet…

Answer: End effector at \((0.678,\ 0.778)\) m, oriented at \(45^\circ\) to \(X_0\).

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

For each link \({}^{i-1}T_i=\begin{bmatrix}c_i&-s_i&0&a_ic_i\\s_i&c_i&0&a_is_i\\0&0&1&0\\0&0&0&1\end{bmatrix}\) (\(c_i=\cos\theta_i\), \(s_i=\sin\theta_i\)).

\[{}^0T_1=\begin{bmatrix}0.866&-0.5&0&0.433\\0.5&0.866&0&0.25\\0&0&1&0\\0&0&0&1\end{bmatrix},\ {}^1T_2=\begin{bmatrix}0.7071&-0.7071&0&0.2828\\0.7071&0.7071&0&0.2828\\0&0&1&0\\0&0&0&1\end{bmatrix},\ {}^2T_3=\begin{bmatrix}0.866&0.5&0&0.1732\\-0.5&0.866&0&-0.1\\0&0&1&0\\0&0&0&1\end{bmatrix}\]

Step 1: \({}^0T_2={}^0T_1\,{}^1T_2\). The angles add: \(\theta_1+\theta_2=75^\circ\).

\[{}^0T_2=\begin{bmatrix}0.2588&-0.9659&0&0.5365\\0.9659&0.2588&0&0.6364\\0&0&1&0\\0&0&0&1\end{bmatrix}\]

(\(0.5\cos30^\circ+0.4\cos75^\circ=0.433+0.1035=0.5365\); \(0.5\sin30^\circ+0.4\sin75^\circ=0.25+0.3864=0.6364\).)

Step 2: \({}^0T_3={}^0T_2\,{}^2T_3\), with \(\phi=\theta_1+\theta_2+\theta_3=45^\circ\):

\[{}^0T_3=\begin{bmatrix}0.7071&-0.7071&0&0.678\\0.7071&0.7071&0&0.778\\0&0&1&0\\0&0&0&1\end{bmatrix}\]

\(x=0.5365+0.2\cos45^\circ=0.5365+0.1414=0.678\) m; \(y=0.6364+0.2\sin45^\circ=0.6364+0.1414=0.778\) m.

So the end effector is at \((0.678,\,0.778)\) m and its axis points at \(45^\circ\) to \(X_0\).