I don't know if I could upload these things here, but I want solve the following exercise:.

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I undestand that I should use the following formula:

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But, I have to solve every joint torque, so how could I get those values?


  • $\begingroup$ "I don't know if I could upload these things here" of course you can. However, using this website does not mean we will do your homework for you. All the necessary details must be provided and you must demonstrate your effort. For example, you provided $\tau = J^T(\theta) f$, which is the correct first step, but you failed to show what is $J(\theta)$ and in which frame it is expressed? Calculate the Jacobian matrix according to your textbook first and if you still have trouble, please come back. $\endgroup$
    – CroCo
    Commented Mar 18, 2023 at 18:08

1 Answer 1


Here below are the expressions you ought to recruit:

$$ \mathbf{J} \left( \mathbf{\theta} \right) = \left( \begin{matrix} \frac{\partial x_e}{\partial \theta_1} & \frac{\partial x_e}{\partial \theta_2} & \frac{\partial x_e}{\partial \theta_3} \\ \frac{\partial y_e}{\partial \theta_1} & \frac{\partial y_e}{\partial \theta_2} & \frac{\partial y_e}{\partial \theta_3} \\ \frac{\partial \phi_e}{\partial \theta_1} & \frac{\partial \phi_e}{\partial \theta_2} & \frac{\partial \phi_e}{\partial \theta_3} \end{matrix} \right), $$

being $\left( x_e, y_e, \phi_e \right)^T$ the coordinates of the tip,

$$ \mathbf{f}_{tip} = \left( \begin{matrix} 2 \\ 0 \\ 0 \end{matrix} \right), $$

$$ x_e = \cos\left(\theta_1\right) + \cos\left(\theta_1+\theta_2\right) + \cos\left(\theta_1+\theta_2+\theta_3\right), $$

$$ y_e = \sin\left(\theta_1\right) + \sin\left(\theta_1+\theta_2\right) + \sin\left(\theta_1+\theta_2+\theta_3\right), $$

$$ \phi_e = \theta_1 + \theta_2 + \theta_3, $$

$$ \theta_1 = \theta_3 = 0; \; \theta_2 = \pi/4, $$

$$ \mathbf{\tau} = \left( \begin{matrix} \tau_1 \\ \tau_2 \\ \tau_3 \end{matrix} \right) = \mathbf{J}^T \cdot \mathbf{f}_{tip} = \left( \begin{matrix} 2\frac{\partial x_e}{\partial \theta_1} \\ 2\frac{\partial x_e}{\partial \theta_2} \\ 2\frac{\partial x_e}{\partial \theta_3} \end{matrix} \right) = -\left( \begin{matrix} 2\sqrt{2} \\ 2\sqrt{2} \\ \sqrt{2} \end{matrix} \right). $$

  • $\begingroup$ But, how i solve that multiplication? How I know the Jacobian matrix? $\endgroup$ Commented Mar 16, 2023 at 16:06
  • $\begingroup$ The expression of $\mathbf{J}$ is the standard definition of the Jacobian. I've provided $x_e$, so you just need to compute the partial derivatives of $x_e$ with respect to the joint angles $\theta_i$. $\endgroup$ Commented Mar 17, 2023 at 10:05
  • $\begingroup$ (upvoted +1). This answer appears to be correct. However, it seems that this question is taken from the book Modern Robotics, which uses screw theory to represent rigid motion mathematically. In addition, the Jacobian of a fixed frame $J_s(q)$ must be indented according to the details given. A student who wants to submit this answer must be careful not to do so if the teacher indents to use a screw theory formula. $\endgroup$
    – CroCo
    Commented Mar 18, 2023 at 18:01
  • $\begingroup$ "The expression of $J$ is the standard definition of the Jacobian.", This is something I disagree with. The form of the Jacobian matrix depends on the task and the mathematical formalism being used. $\endgroup$
    – CroCo
    Commented Mar 18, 2023 at 18:20
  • $\begingroup$ The term standard refers to the fact that once the task is defined (here the task has to do with the tip of the manipulator), one can derive the analytic Jacobian according to a standard procedure, pretty much as detailed at en.wikipedia.org/wiki/Jacobian_matrix_and_determinant. Then, as you said, we don't solve homework. Mine was just a possible approach. $\endgroup$ Commented Mar 19, 2023 at 9:45

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