Model of two link rigid manipulator

Question

I am working on parameter estimation of a rigid two-link manipulator as shown in the picture below.

Unlike the cylindrical links shown in the above image, my robot has links of oval shape.

So the centre of mass of links is not the midpoint of the link. I tried to search for the model equation but each one assumed the cylindrical shape of the links and the centre of mass as the centre of the link.

What I know till now is that we write Langragian equation $$L=K E-P E$$ And then derive the torque equation using $$ \frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}}\right)-\frac{\partial L}{\partial q}=\tau $$ I referenced these papers 2 Dof Manipulator Dynamics, Parameter Estimation of Manipulator Robot Dynamics.

Even though both have considered the same robot structure with the same joint type and link shape, they both have different model equation. The former (page 6 to 9) has considered the length of the links while the latter (page 22) has considered the centre of mass of the links.

Which one is correct for my robot? How should I derive the model for a rigid two-link manipulator with oval-shaped links?

Answer 1

Those two papers have a different equation for the same robot because they have taken different assumption. The first paper has considered the centre of mass to be at the end of the link, and the other paper has considered the centre of mass at the middle of the link.

Now the dynamics of any two-link manipulator irrespective of the link shape is $$ \tau = H(q)\ddot q + C(q,\dot q)\dot q + G(q) $$ $$ \begin{bmatrix} \tau_1 \\ \tau_2 \\ \end{bmatrix} = \begin{bmatrix} I_1 + I_2 + m_2l_1^2 + 2m_2l_1l_{c2}cos(q_2) & I_2 + m_2l_1l_{c2}cos(q) \\ I_2 + m_2l_1l_{c2}cos(q) & I_2 \\ \end{bmatrix} \begin{bmatrix} \ddot{q}_1\\ \ddot{q}_2 \\ \end{bmatrix} \\ + \begin{bmatrix} -m_2l_1l_{c2}\dot{q}_2sin(q_2) & -m_2l_1l_{c2}(\dot{q}_1 + \dot{q}_2)sin(q_2)\\ m_2\dot{q}_1l_1l_{c2}sin(q_2) & 0 \end{bmatrix} \begin{bmatrix} \dot{q}_1\\ \dot{q}_2 \\ \end{bmatrix} \\ + \begin{bmatrix} m_1gl_{c1}sin(q_1) + m_2g(l_1sin(q_1) + l_{c2}sin(q_1 + q_2))\\ m_2gl_{c2}sin(q_1 + q_2) \\ \end{bmatrix} $$

Here $I_1$ and $I_2$ are the moment of inertia about the rotating axis of link-1 and link-2 respectively. $l_{c1}$ and $l_{c2}$ are the centre of mass of the two links. Moment of inertia can be calculated by any CAD software or separating the shape of link into basic shape and finding the moment of inertia by the composite method as mentioned in the other answer.

Now, the derivation of this dynamic is simple. Proceed with the Lagrangian equation and differential equation as written in the question. The Kinetic and Potential energy should be about the centre of mass of the links. For more information read this book Introduction to Robotics chapter 7.

Answer 2

the model would be different because of the location of the center of mass of the links, so maybe in one model is just the total length d while in others might only be d/2.

To calculate the center of mass of your oval kind shape link you have two options: separate the shape of your link into the basic shapes like a trapezoid and two half circles https://www.youtube.com/watch?v=8jo8gVUmTEw

or take the data from a CAD program like solidworks/solid edge/catia etc.

and use that vector to get the kinetic and potential energy of your links and derived the model.