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I am a software developer running this simulation and have some trouble understanding the underlying physical problem. My description of the problem is therefore probably not perfect, but I hope understandable. I am working on a simulation that consists of an inverted pendulum actuated through 2 linear actuators to the left and right. The inverted pendulum with linear actuators 1 and 2

Rotational joint stiffness is defined as $K_j=\frac{d\tau}{d\theta}$ with $\tau$ being the torque acting on the joint and $\theta$ the joint angular position.

The high level goal of my simulation is to control the joint stiffness only through the force of the 2 actuators.

Now to the part about my intuition and lack of formalization: Stiffness is used to describe the amount the pendulum displaces for a change in applied torque. That is especially useful when looking at what happens if an external force is applied. How much does it move if i tap it with a certain force? Intuitively I would assume, that there are infinitely many solutions of balancing the pendulum at an angle $\theta$ with different stiffness - depending on the actuator activation. The harder each side pulls, the stiffer the joint becomes.

However, thinking about forces that doesn't make sense to me. If we balance the pendulum at a fixed $\theta$ and it is not moving, the net torque on the joint has to equal $\tau = 0$ If I apply an external force now, in the moment of application the net force on the joint is therefore just that external force. So what difference does different actuator activation actually make? What happens over time after that external force is applied and how do the actuators influence that?

In fact in the tests I ran in the simulator, if I balance the pendulum in an upright position ($\theta = 90°$) with low force and apply a push to the right, the pendulum falls slower. If I conduct the same experiment but both actuators pull hard, the pendulum falls faster.

Why is that happening?

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  • $\begingroup$ Your first and third paragraph are at odds, which might be something to tease apart. In the first paragraph you define stiffness with $K_j=\frac{d\tau}{d\theta}$, but in the third paragraph you define stiffness as "the amount the pendulum displaces for a change in applied torque," which is the inverse of what you had earlier. $\endgroup$
    – Cort Ammon
    Mar 14 at 14:12
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Mar 14 at 14:15
  • $\begingroup$ I take it you are starting with a good bearing, and want to make it act like a stiff joint by adding actuators. How do you want your stiff joint to act? Some joints don't move until you apply a force above a threshold. Then they break free. It sounds like you are thinking of a joints like an open safety clip. It resists being pushed away from its equilibrium position. $\endgroup$
    – mmesser314
    Mar 14 at 14:17
  • $\begingroup$ To make a joint that breaks free, you need to sense the force. Can your actuators do this? $\endgroup$
    – mmesser314
    Mar 14 at 14:19
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    $\begingroup$ Why do you need 2 actuators? Are they asymmetric such that they only provide tension or only compression? $\endgroup$ Mar 14 at 17:03

2 Answers 2

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The torque applied is $\tau=F\cos\phi$, where $\phi$ is the angle between the rod and the actuator. There will be a $\phi_1$ and $F_1$ for one actuator, and a $\phi_2$ and $F_2$ for the other. The relationship between $\phi_1$ and $\phi_2$ will be defined by your particular configuration of joints and actuators. The "set point" for the whole system will be the point where the angles and forces are such that the torques balance out (if the configuration is stable. It wont have a set point if it is not stable). I believe it is the $\cos\phi$ term that is missing from your prose. When you apply the chain rule to your stiffness equation, you will see that there will be non-zero terms involving those forces and phis (We don't have enough information to calculate them from the picture given, but you should have all you need).

Intuitively, for any set point $\theta$ (and thus a calculated $\phi_1$ and $\phi_2$), there will be some constant of proportionality $a$ such that $F_2=aF_1$ balances the torques. As you say, any one of an infinite number of force-pairs will result in a balance, but the stiffness, $K_j$ will depend on $F_1$.

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  • $\begingroup$ Thanks for the answer, how do I know if my system is stable? Right now as I described the rod always falls down if i apply an external force, no matter how hard the actuators pull. Actually, if they pull harder, the rod falls faster. I would have thought the harder they pull the stiffer the joint must be so the slower it should fall. Is that would you mean that my system is not stable? $\endgroup$
    – mavex857
    Mar 15 at 9:54
  • $\begingroup$ Stability is a whole topic, but the first step is negative feedback. If your arm is rotated by $d\theta$, the response of your system needs to apply a $d\tau$ in the opposite direction, driving the arm back to the set-point. If the the system applies a $d\tau$ in the same direction as $d\theta$, then the system will be unstable, fleeing the set point. You will also most likely need some form of damping, like a friction force. An undamped system can oscilate forever (like how the moon orbits the Earth "forever") $\endgroup$
    – Cort Ammon
    Mar 15 at 14:31
  • $\begingroup$ If you're comfortable with what I did with $d\theta$ and $d\tau$ being on their own rather than part of a derivative operation, I'd recommend that. It's called a linearized model (or a small-signal model, depending on your discipline) and its very useful for exploring things like stability. In these cases, you pick a configuration you are interested in (such as when the arm is near the set-point), and then "linearize" around that point by considering the infintessimal displacement between $\theta_{set}$ and your actual angle. This turns all of the derivatives into simple algebra which is... $\endgroup$
    – Cort Ammon
    Mar 15 at 14:37
  • $\begingroup$ easier to reason about. You may find that your particular arrangement is capable of providing negative feedback for some $\theta_{set}$ but not others. $\endgroup$
    – Cort Ammon
    Mar 15 at 14:40
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Statics can provide you with your answer.

You are looking for $$ \tau = - k_{\rm rot} \theta \tag{1}$$

and you have a situation where the actuators can provide some axial tension $\vec{F_1}$ and $\vec{F_2}$ on the rod

fig1

then torque about the pivot is

$$ \vec{\tau} = \vec{r}_A \times (\vec{F}_1 + \vec{F}_2) \tag{2}$$

where $\vec{r}_A = \pmatrix{\ell \cos \theta \\ \ell \sin \theta} $

So you just work out the trigonometry needed to do the cross product my component to relate the torque magnitude $\tau$ to the actuator force magnitudes $F_1$ and $F_2$.

For example, with the following simple distances defined

fig2

you have

$$ \tau = \frac{F_1 \ell d_1 \sin \theta}{\sqrt{d_1^2+\ell^2-2 \ell d_1 \cos \theta}} - \frac{F_2 \ell d_2 \sin \theta}{\sqrt{d_2^2+\ell^2-2 \ell d_2 \cos \theta}} \equiv - k_{\rm rot} \theta \tag{3}$$

this can be achieved with $F_1=0$ and

$$ F_2 = k_{\rm rot} \theta \frac{\sqrt{d_2^2+\ell^2-2 \ell d_2 \cos \theta}}{\ell d_2 \sin \theta} \tag{4} $$

but there are other solutions where you can define one or both actuator forces to give you the desired torque.


PS. You would be better off defining the angle $\theta$ from the neutral position, like vertical, instead of horizontal so you can do small angle approximations if needed.

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  • $\begingroup$ Thanks for the long answer and illustration :) What you described is how I have to set my actuator force to produce a certain joint torque correct? But, what I need is the question: "How do I have to set my actuator force to produce a certain joint stiffness?" Is that question understandable and solvable? $\endgroup$
    – mavex857
    Mar 15 at 10:16
  • $\begingroup$ It is unclear what is given and what is required. If you have a set performance characteristic for the actuators and are looking for the geometry to get some required rot. stiffness, then note that this can be done for one angle only, as the rot. stiffness will vary with actuation due to geometric non-linearities. $\endgroup$ Mar 15 at 14:30
  • $\begingroup$ Okay, let me try to clarify: I have given: The configuration/geometry of the pendulum setup, so the distances that you used in your inital answer. Some angle $\theta$. Some joint stiffness $k_j$. What I want: The needed forces $F_1$ and $F_2$ of the actuators to achieve the given angle and stiffness. How I understood your previous answer that is only possible for one specific angle $\theta$ and not in general for any angle and stiffness? $\endgroup$
    – mavex857
    Mar 16 at 11:46
  • $\begingroup$ If you want the force to be one constant value, it will correspond to one angle configuration for the specified stiffness. If you can control the force and make it a function of angle, then you can achieve your goal. Use $$F_1 = k_{\rm j} \theta \frac{\sqrt{d_1^2+\ell^2-2 \ell d_1 \cos \theta}}{\ell d_1 \sin \theta}$$ or $$F_2 = k_{\rm j} \theta \frac{\sqrt{d_2^2+\ell^2-2 \ell d_2 \cos \theta}}{\ell d_2 \sin \theta}$$ depending on which actuator is active (depends on the sign of $\dot{\theta}$) $\endgroup$ Mar 17 at 19:09

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