9
You are right he made a mistake there.
This is probably one of many typos in this preprint of Mark Spong. You should rather turn to other good books, such as the mathematically more elegant book of Richard Murray,Zexiang Li and Sankar Sastry, A Mathematical Introduction to Robotics Manipulation (MLS94). The mathematics they use is consistent in other of ...
9
To answer your first question: if you really want to find the true kinematic equations for differential drive, I wouldn't start approximating by assuming that each wheel has moved in a straight line. Instead, find the turning radius, calculate the center point of the arc, and then calculate the robot's next point. The turning radius would be infinite if the ...
8
Craig uses the modified DH parameters, while Spong uses the classic DH parameters.
The difference between them are the locations of the coordinates system attachment to the links: in the modified DH, the coordinates of frame $O_{i}$ is put on axis $i$, while in the classic DH convention the coordinates of frame $O_{i}$ is put on axis $i+1$.
Update (2014....
8
Let me give you a mathematician's perspective on the difference between the two kinds problems.
Forward kinematics asks the question: given a certain input (i.e. control command), what will be the output (i.e. robot configuration, pose, etc.). Inverse kinematics asks the reverse question: given a certain desired output, what is the necessary input.
...
8
Forward kinematics uses joint angles (with known link lengths) to compute the tool position and orientation. Inverse kinematics uses tool position and orientation, to compute joint angles. Note: if your device has prismatic links (the length changes) then those are used just like joint angles in the above.
To compute the Jacobian, start with the forward ...
7
Start with coordinate systems. I've drawn one example.
In my analysis, if all $q_i = 0$ then the manipulator would point straight up. You can choose other coordinate frames to get the same result. Build your rotation matrices from the coordinate systems you set up.
The rotation matrix from coordinate system $0$ to coordinate system $1$ is
$$_0^1R = \...
7
Your professor has made an error, but he or she is only human.
The upper-left 3x3 matrix must be an orthonormal rotation matrix. Every column of that must have a unit norm. The second column $[0, 1, -1]^T$ has a norm of $\sqrt{2}$ which makes the rotation matrix invalid.
6
Yes, the Jacobian relates the joint velocities to end-effector velocity through this equation:
$$
\mathbf{v}_e = \mathbf{J}(\mathbf{q}) \dot{\mathbf{q}}
$$
Where $\mathbf{q}$ is the joint angles, $\dot{\mathbf{q}}$ is the joint velocities, and $\mathbf{v}_e$ is the end-effector velocity. As you can see, the Jacobian, $\mathbf{J}$, is configuration ...
6
In order to answer my own question Configuration space and Joint space must be defined.
Configuration space of a rigid body is a minimum set of parameters that can determine position of each point in that body or Configuration space is set of all possible configurations of that body.
Configuration space of the end-effector is set of all possible positions ...
6
The DH Matrix section of the DH page on wikipedia has the details.
Basically you want to use the information in your table to create a set of homogeneous transformation matrices. We do so because homogeneous transformations can be multiplied to find the relation between frames seperated by one or more others. For example, $^0T_1$ represents the ...
6
I agree with SteveO that there is nothing wrong with reinventing the wheel if you want to learn about wheels. And for a single application, 4 DoF arm, the IK is probably not too hard.
But I feel like I should mention that most of the kinematics libraries out there are mostly targeted towards Linux. And as such, probably not too hard to compile from ...
5
In robotics the configuration space is exactly the joint space of the manipulator.
Differently, to indicate the space where the forward kinematic law maps the joints configuration into, we use the terms task space and operational space, equivalently.
5
It is rather straightforward to implement inverse kinematics for a particular manipulator in C++. Of course, you need to begin with the inverse kinematic equations themselves. Putting those into code will only involve a few trigonometric functions such as acos, asin, and atan2 (use atan2 instead of atan), and probably a couple of square and square root ...
5
EDIT: Improved based on the comments below.
If you have a CAD assembled, that means that you have one valid configuration given. You move the TCP (Tool Center Point) only a small amount, since your mouse (which drags the mechanism) travels only a small amount between two update periods, that means that the two solutions are close to each other. Each refresh ...
5
The updated image solves the problem. You did not consider the end-effector coordinate frame earlier.
Also, the crosses (going into) in the diagrams should be replaced by dots(coming out), because the crosses don't hold the right hand rule in case you are using a right hand coordinate system.
5
In general you need 6 parameters to describe the position and orientation of any joint with respect to a link coordinate frame. The DH parameterisation includes 2 constraints so only 4 parameters are required. The constraints are:
axis $x_j$ intersects axis $z_{j-1}$
axis $x_j$ is perpendicular to axis $z_{j-1}$
(see Robotics, Vision & Control, ...
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 ...
4
Actually, the 6x1 vector is sometimes better referred to as the coordinates of the twist. The twist itself is a 4x4 matrix, element of $SE(3)$, found by
$$
\begin{align}
A &= \begin{bmatrix} \widehat{\omega} & v \\ 0 & 0 \end{bmatrix}
\\
v &\triangleq -\omega \times q
\end{align}
$$
where $\omega$ is the unit vector pointing along the axis of ...
4
In your forward kinematics transformation matrix (4x4, incl. also translation, of just 3x3) the orientation of the end-effector is expressed relative to the base (or world) coordinate system.
SteveO described very well how to obtain the 3x3 rotation matrix, similarly you can obtain also the 4x4 transformation matrix
If you have the matrix you have to ...
4
The best way to understand forward and inverse kinematics is to write a library for its own. A good starting point is to implement Cyclic Coordinate Descent.
So called "ready-to-run" librarys like OpenRave or OMPL have an extensive C++ tutorial section too but are notorius difficult to install. Most of them only run under linux, and are not even part of ...
4
If you can write the forward kinematics equations of a parallel robot in an explicit form, you can derivate those equations and you get the formula for the velocities. This is generally valid approach for all robots, but the formulae obtained are only valid for the specific structure.
If the jacobi matrix is diagonal it means that the motions of the robot ...
4
(EDITED TO CLARIFY PARENTHETICAL ABOUT CARTESIAN MANIPULATORS)
Your equation is true in general only for those manipulators in which $J_a$ is independent of $\theta$ (such as with Cartesian manipulators). Otherwise, the expression is only true in the small (the region of $\theta$ close to $\theta_{t=0}$.
The equation is not true in general because ...
4
Add a coordinate system that matches the previous coordinate system exactly. The last rotary joint will be the parameter for the next-to-last coordinate system, and the link length will be the parameter for the prismatic joint.
Note: a simpler approach is possible, but this gets the job done easily.
4
When working with rigid-body transformations, it is crucial to understand which coordinate frame the transformation is defined in. Further, there are different notations for this, so it is important to know which is in use. Let's assume that $T_b^a$ describes the coordinate axes of frame $b$ with respect to the coordinate axes of frame $a$. (Note that this ...
3
In general, Euler angles (or specifically roll-pitch-yaw angles) can be extracted from any rotation matrix, regardless of how many rotations were used to generate it. For a typical x-y-z rotation sequence, you end up with this rotation matrix where $\phi$ is roll, $\theta$ is pitch, and $\psi$ is yaw:
$R = \begin{bmatrix} c_\psi c_\theta & c_\psi s_\...
3
In DH, the Z axis always goes along the direction of variability. For a rotational (revolute) joint, that means Z is the axis of rotation. For a translational joint (prismatic) the Z axis is in the direction that the joint can translate. For each direction the robot can be actuated, a coordinate system is needed. So your example of a gantry crane, if that ...
3
You should read this paper: "Lipkin 2005: A Note on Denavit-Hartenberg Notation in Robotics". It explains the 3 main DH parameter conventions and how they differ.
3
Actually I wish to implement my own algorithm (like some variation of
RRT) without MoveIt!/OMPL hence it is important for me to know all the
details.
I am really confused about this. Any explanations or links where I can
find the details and understand them would be really helpful.
OMPL and MoveIt have a ton of features that are already ...
2
For a repeated calculation, it doesn't matter whether you find $\Delta\theta$ before or after you apply $\theta$ to the $\Delta{x}, \Delta{y}$ calculation. You will always be alternating between a position and an orientation calculation.
In a practical sense, it might be better to calculate $\Delta\theta$ after you calculate $\Delta{x}, \Delta{y}$, since ...
2
Seems like you are using the Proximal convention. It's always good practice to have an extra frame called {X0,Y0,Z0} at the base. It makes visualizing so much easier.
Anyways, from the image below, I know, it's not good, but hey common, I made it using a mouse. Alright, so I'll Z0 is the base frame and Z1 is parallel to it because, its the rotating Joint1 ...
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