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13

Torque is analogous to force for rotating systems, in that: $$ F = m a \\ \tau = I \alpha \\ $$ Where $\alpha$ is angular acceleration and $I$ is moment of inertia. $m$ and $a$ are mass and linear acceleration, respectively. So, in a way, a position controller, a velocity controller, and an acceleration (torque) controller are all different ...


11

I'm going to take a slightly different tack to Chuck. What is Torque Control? For me, Torque Control is about performing a move with an explicitly defined torque, rather considering torque just the means to the end of Position or Velocity control. Normally when you move a robot, you specify position and speed, with the robot allowed to use any and all ...


9

If you're only using proportional force, then at some point it will be balanced by the force of gravity -- your error will converge on that balance, not zero. To compensate for the mass of the arm, you'll need to add an integral force term. This will increase over time to counterbalance the constant force of gravity. See also: this answer on the integral ...


9

The two views are not contradictory; they apply to two different situations, which you are treating as a single one. Your personal experience about having a low center of mass applies to situations where there is a stable position -- a local minimum height for the center of mass. For example, if you have a vase on a shelf, the height of the vase's center ...


8

Disclaimer: I have never done this myself, but only have seen a description of it being done through Georgia Tech's "Control of Mobile Robotics" on Coursera. My knowledge of controls is spotty, too. Thus... take this with a grain of salt. :) To keep the robot upright (and still), you're trying to stabilize (send to $0$) the state $x$, where: $$x=\left[\...


8

For a nonholonomic system, you can at best determine a differential relationship between state and inputs. You cannot determine a closed-form geometric relationship. This means that the history of states is needed in order to determine the current state. Vehicles are a good example because you can intuitively see that turning the right wheel 100 ...


8

Yes, a state matrix with zero rows and/or columns makes sense and is viable. It typically signify pure integrators in the system. In the example you give, $$ \dot{v} = -\frac{b}{m} v +\frac{1}{m} u $$ where $v$ is the speed, $u$ is the externally applied force, and $bv$ is some viscous damping force. Now if the viscous damping coefficient is zero (no ...


7

This is video is from an Edx course on Autonomous Quadcopters...It is 10 minutes and goes over the "Flying Principle" of quadrotors. But basically what you are missing is the reactions. The famous Newtown's third law says (paraphrasing) "for every action there is an equal and opposite reaction". So by spinning the clockwise blades faster, a counter-...


7

I would like to use P (proportional) controller for now. Just a proportional controller will never make your error stay at 0. Your system is not damped and a proportional controller acts like an undamped spring. Look at the controller equation that you wrote: τ=−K(θ−θd) and compare it to a spring equation: F=Kx or F=K(x1-x2) Your controller is acting ...


5

This is a sweeping generalization that I'd be very cautious about. Engineering is about tradeoffs. But there are two things that I'd be comfortable generalizing, from my own personal biases. I'm 99% serious when I say robots become more effective as the amount of money you are willing to invest becomes higher. This is because you can eventually custom-make ...


5

Series elastic actuators tend to have more stable force control because the spring filters out the high-frequency motion of the mechanism. A low frequency in the system dynamics means that you can use slower control techniques, which is important when using digital controllers with naive control implementations, and sensors with significant abbe error and ...


5

A force balance equation is typically written as: $$ m\ddot{x} + b\dot{x} + k{x} = F \\ $$ where $F$ is an applied force, $x$ is position, $\dot{x}$ is velocity (first derivative of position), and $\ddot{x}$ is acceleration (second derivative of position). $m$ is mass, $k$ is a spring constant, and $b$ is a viscous damping term. This force balance is one ...


5

the Simulink diagram is straightforward. It is a matter of connecting blocks. For the differential equations provided in your post, the simulink is For $u_1$ and $u_2$, I've chosen the unit step. You can change that of course. For $L$, I've set it to 0.5 since you didn't provide the actual value. The result of the position of the vehicle is shown below:


5

In short answer: yes Kalman filter is a special case of an $H_2$ observer Yes Yes ... LQG is just Kalman filter + LQR controller, which are both special cases of $H_2$ Depends on use case. $H_2$ minimizes maximum error while $H_{\infty}$ minimizes error function 2-norm Very complicated The somewhat longer answer: $H_2$ and $H_{\infty}$ control are both ...


5

Short answer Robot Dynamics and Control by Spong et al. (especially Chapter 5) can definitely help you on this matter. Long answer First of all, you are partially correct about a Jacobian. It is indeed a function of joint values (say $q \in \mathbf{R}^n$). However, as a Jacobian maps a joint velocity to an end-effector velocity, its dimension is not ...


4

http://en.wikipedia.org/wiki/Square-cube_law The square-cube law essentially states that larger robots are more fragile. "You can drop a mouse down a thousand-yard mine shaft and, on arriving at the bottom, it gets a slight shock and walks away. A rat is killed, a man is broken, a horse splashes." — J.B.S. Haldane, biologist Note that this only ...


4

Controllers type A more mathematical approach to the error. Suppose you have a close loop system like above. The equation is: $\hspace{2.5em}$ $Y(s) = \frac{G(s)C(s)}{1+G(s)C(s)} R(s)$ The error equation is: $\hspace{2.5em}$ $E(s) = R(s) - Y(s)$ $\hspace{2.5em}$ $E(s) = \frac{1}{1+G(s)C(s)} R(s)$ $\hspace{2.5em}[1]$ The final value theorem states ...


4

I will try to make it as simple as possible. Imagine you have a SCREW, when you WRENCH it, it TWIST forward or backward. From your wiki link The components of the screw define the Plücker coordinates of a line in space and the magnitudes of the vector along the line and moment about this line. It means that any system can be described as those ...


4

You can use the 'simplify' command , e.g : a=simplify(A*B) .This command is calculating some trigonometric functions thus making your expression shorter .


4

In that context, SE means "Special Euclidean" group, e.g. SE(3) which is shorthand for "the special Euclidean group of rigid body displacements in three-dimensions". This paper may explain more.


4

The author appears to be writting in a self promotional style, where the details listed about other work are simplified or ignored so that their claim, that their approach is better, can be established. I suspect the author is making the statement 'for good reasons' because they don't really know. I would ignore their claims and go read their calculations ...


4

About why screw axes: According to Kevin Lynch in his video of Twists, "just like the time-derivative of a rotation matrix is not equivalent to the angular velocity, the time-derivative of a transformation matrix is not equivalent to the rigid-body velocity" (linear and angular). Also he mentions that, instead, "any rigid-body velocity is equivalent to the ...


4

I will use Mr. Richard Feynman's quote to answer your first question: Finally, we make some remarks on why linear systems are so important. The answer is simple: because we can solve them! Yes. As simple as that. Dealing with linear systems is well-understood. We can generalize any notion about linear systems, however, this is not the case with ...


3

It really doesn't sound very unexpected for your equations to become this long. Something that might help a little, is using assumptions for defining symbolic variables. x = sym('x','real'); y = sym('y','positive'); z = sym('z','integer'); t = sym('t','rational');


3

Since both $\theta_d\left(t\right)$ and $\dot\theta_d\left(t\right)$ are references at your disposal, i.e. you have to provide them in some way, why don't you simply play with $\dot\theta_d\left(t\right)$ and then compute $\theta_d\left(t\right)$ accordingly by means of integration? As you might know, integration is a well posed operation compared with the ...


3

I think you are confusing 2 issues. Inverse dynamics is the process of mapping end effector position, velocity, and acceleration to joint torques. as described in this book, page 298: http://books.google.com/books?id=jPCAFmE-logC&lpg=PR2&pg=PA298#v=onepage&q=inverse%20dynamics&f=false But the paper you posted is simply modeling and ...


3

This is just basic trigonometry; you'll covert your world-relative calculations of roll and pitch ($\phi$ and $\theta$) into vehicle-relative values, based on yaw ($\psi$). Just so we're on the same page, I'm assuming measurements like the following, with roll, pitch, and yaw being zero when levelly flying North: $$\phi_{vehicle} = \phi_{world}\cos(\psi) - ...


3

Aside from the square-cube law for measuring the strength of the bot mentioned by user65, you have a few more effects. I don't know of any theorem for this, though. Firstly, note that "bigger batteries" doesn't mean "bigger motors". And "bigger motors" doesn't mean "stronger motors" or "more powerful motors". If we're talking about the same type of battery, ...


3

Actuators Forces Do I get this right: you have a theoretical model of a rigid multibody system and would like to perform rigid body dynamics computations. You have implemented the model and now would like to compute how the model behaves when driven by an actuator. However what is an actuator for you? Is it simply a force acting at that joint? Is it a DC ...


3

If your computed torques are out of the range for your robot, then you are no longer modelling your robot but a different one. You can ignore the torque limits as good as the joint angle limits of your robot (i.e. not at all). If the required torque is to high, your robot will either be much slower (if you are lucky) or won't be able to reach its goal ...


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