5

A hint towards what the answer is given in the paper. Namely, one can use the generalized eigenvalue decomposition, which in this case can be formulated as finding eigenvalues $\lambda \in \mathbb{R}$ and eigenvectors $v \in \mathbb{R}^n$ such that $$ (\lambda\,\Lambda - K_d)\,v = 0. \tag{1} $$ Consider two distinct solutions $(\lambda_i,v_i)$ and $(\...


5

The workspace of a manipulator is strictly determined by its kinematics. Since kinematics only consider the geometry of motion, without regard to forces and torques needed to accomplish tasks, you need the dynamics (and controls) to determine what motion profiles are achievable within the workspace. But those dynamics do nothing to determine the workspace ...


5

You essentially want to find the time derivative of a linear interpolation between two rotations. The easiest way to obtain this would probably to convert the rotation matrix between the two orientations to a axis-angle representation and the angular velocity would simply be the axis times the angle divided by $T$.


4

The term "singularity" characterizes those configurations in the joint space where the Jacobian matrix loses rank and thus it is not directly invertible. The Jacobian, in turn, is used to remap a trajectory from the Cartesian space to the joint space. Therefore, if you plan the trajectory straight in the joint space, then you are not going to use the ...


4

So it sounds like you have an external camera system to track the end effector position and orientation? In that case I would use QR codes or april tags. Stick a couple of these on your end effector, and make sure that you put enough of them that 1 is always visible from your camera. The apriltags library provides the ability to extract position and ...


4

The original manipulators referred to in that article were through-the-wall pantographs which moved radioactive materials without the human operator having direct contact with those materials. The end effectors of those manipulators did have direct contact with the materials - just not the human. Although the article is not specific about this, it seems ...


3

Your problem statement is: using a bang-coast-bang acceleration profile with symmetric acceleration and deceleration phases, each of duration $T_s=\frac{T}{4}$. So, right off, you know the time spent in acceleration is $T/4$. The peak velocity is not $T/4$. I'll agree that your professor's notation is hard to read (for me at least), but it looks like ...


3

There is not one correct DH Frame assignment, but there are many. Even if you do not get the same results, you might still have a correct frame assignment. The first, base frame at the root of the robot, sometimes called robroot frame is the reference frame, relative to this are the TCP coordinates defined, this is why it is needed. A this is a frame ...


3

When you apply torque to the first joint, it makes the first link rotate. This means that the far end of the first link translates. Because the second joint pins the near end of the second link to the far end of the first link, this means that the near end of the second link must also move. In order for the second link to move, the pin in the second joint ...


3

As you already have the Jacobian vector, I assume you also solved the inverse kinematics problem. I will refer to the IK problem as $f^{-1}$ In order to get the joint space equivalent of the Cartesian pose A (position and orientation, denoted as $X_A$) you can use the inverse kinematics: $$Q_A = f^{-1}(X_A) $$ similarly $$Q_B = f^{-1}(X_B) $$ (Please note ...


3

The interpretation of error for a robot manipulator pose is subjective in terms of what is the error being used for. A robot manipulator pose is combination of both position (x, y, z) and the orientation (quaternions or euler angles) and thus, designing the error could partially or completely include all terms. Your representation of error is essentially ...


2

In general i would say there is not enough information on this drawing, so we all ASSUME. If you look at this and assume it is a multicopter with 2 counter rotating propellers (can not fly in a normal fashion without a pivoting mechanism, like a Chinook) you can only have yaw, roll and up down control which makes for 3-DoF. For the arm it give you 2 DoF (...


2

This ruler has a resolution of 1 inch: It has a resolution of 1 inch because the only readings on the ruler are 0, 1, 2, or 3 inches. It can't measure anything smaller than an inch; it only measures inches. The ruler below has a resolution of a half-inch (0.5 inches). This ruler now has the capacity to measure 0, 0.5, 1.0, 1.5, ..., 3.5 inches. For ...


2

There are 2 main rules in assigning frames following DH convention: source: Robot Modeling and Control, Spong et. al $x_i \perp z_{i-1}$ $x_i$ intersects $z_{i-1}$ In your attempts, Your first attempt is incorrect as your $z_2$ axis is not in the direction of actuation of the prismatic joint. I'm not clear on your second attempt as $x_1$ appears to be ...


2

The Jacobian is a 6x6 matrix and there are multiple ways it can be derived. The concept of the Jacobian has its roots in Mathematics, where it is not stricly tied to any problem, it is just a matrix of partial derivates of an equation system. As such it is obtained as: $$ J = \frac{\partial f}{\partial \theta} = \begin{bmatrix} \frac{\partial f_{\text{1}...


2

Yes, it is the same, $F$ should be the same as $F_{ext}$. And $F$ comes from a sensor or an estimation, in general from a force sensor. The diagram seems correct and you may have $F$ and $F_{ext}$ coming from the same source. The impedance controller is an indirect force control, you directly do not control the force exerted to the environment but rather ...


1

If you should consider not just points, but poses (position and orientations). If you can write these as a table, where the columns are the coordinates +-------+--------+--------+-------+---------+--------+--------+ | Point | X [mm] | Y [mm] | Z [mm]| A [deg] | B[deg] | C[deg] | +-------+--------+--------+-------+---------+--------+--------+ | P1 | ...


1

If I understand correctly, your end effector is at pose A, you want to move it to pose B, and you have Jacobian pseudo-inverse control set up so that you can specify a pose velocity $v$ and get a corresponding joint velocity $\omega$ that will produce that pose velocity. There are several ways to produce the $v$ vector. All of them start by defining a time ...


1

Derivative of $\sin$ is $\cos$, and the derivative of $\cos$ is $-sin$. Given a quaternion definition of: $q = \cos{a} + \mathbf{r}\sin{a}$ $\mathbf{r}^2 = -1$ I would expect to see what is effectively a phase shift at every derivative level, and that's what I'm seeing in your curves. I noticed your magnitude seems to be growing, but that may be because ...


1

maybe need some transformation from centers of mass to the joint frame? Isn't that what $A_i$ is? I don't have the book with me, but from your excerpt: Let $A_i$ be the screw axis of joint $i$ defined in $\{i\}$ where it says at the top, frames $\{1\}$ to $\{N\}$ [are attached] to the centers of mass of links $\{1\}$ to $\{N\}$


1

Deriving analytic IK for higher DOFs is tricky. Numerical IK solvers are a good alternative usually, unless you absolutely must use analytic IK. In various robotics textbooks, there are examples on how to derive analytical IK for 6R robots. The way it's done is by considering position IK and rotation IK independently (note that this is possible only if the ...


1

I answered a similar question DH parameters for a PPP arm, take a look for a detailed answer. If your first joint is the origin, you start the parameters by looking to the next joint. In this case, the last line of the table will give you the parameters of the transformation between the last joint and the end effector. For your RRP arm, the DH parameters ...


1

A rotation matrix is also called a director cosine matrix. The elements of the rotation matrix are the cosines of the unit vectors of two coordinate systems involved. You can find a more generic explanation here. Let $\angle (e_{2,i}, e_{e,j})$ denote the angle between the angle between unit vector on the i axis of the fixed reference frame and the unit ...


1

It looks like both frames are attached to the end effector at point P, and are offset by a constant angle $\beta$. So you don't really have to consider the D-H parameters to answer this question - just do a simple rotation matrix evaluation. So how are the frames, $F_2$ = $[\hat{x_2} \; \hat{y_2} \; \hat{z_2}]^T$ and $F_e$ = $[\hat{x_e} \; \hat{y_e} \; \hat{...


1

First of all, I have to share with you this link: https://www.youtube.com/watch?v=rA9tm0gTln8 I like to think of a few tricks with D-H params. First, the Z-axis is the axis of rotation/translation. Then, theta rotates along the Z-axis to align with the start of the next D-H frame. You then translate along the Z axis such that the point is tangent to the ...


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 ...


1

Kinematics is based on geometry and joint constraints, so no change there at all. Dynamic motion results from the sum of forces, so the only change to a dynamic analysis is that you don't include gravity, but all other forces would remain.


1

No, unfortunately there is no automated way to convert between different DH parameter conventions. See my similar question here: How to convert between classic and modified DH parameters? Also, you might be interested in this paper: "Lipkin 2005: A Note on Denavit-Hartenberg Notation in Robotics". It explains the 3 main DH parameter conventions and how ...


1

Seems that the book has an error. You are right, they should correct that. In the drawing it can be seen that those two z-axis are indeed parallel. I recommend that you considered them parallel and try to draw them to see that it fits with the image.


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