# Tag Info

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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. ...

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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 ...

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Let's start from the forward kinematics equation $$x = f(q),$$ where $x \in \mathbf{R}^6$ is the end-effector position, $q$ is the joint angles, and $f$ is a (usually highly nonlinear) forward kinematics mapping. Due to the nature of $f$, computing the image of $q$ under $f$ (i.e. $f(q)$) is not difficult but computing the preimage of $x$ under $f$ (i.e. $f^{... 6 Writing the equations by hand and deriving them is certainly the best way to understand what is happening "in the background". Generating the equations and deriving them using a syombolics engine, like @SteveO suggested is essentially the same process but someone else, in this case a symbolic engine, is doing the work for you. There are however different ... 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 ... 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 ... 5 The geometric Jacobian provides all the information you need for singularity or manipulability analysis. Linearly dependent columns correspond to joints with parallel axes. More information about Jacobians for under-actuated manipulators (as is your case) can be found in my book Robotics, Vision & Control" section 8.4.1. For information about ... 5 Write the forward kinematic equations $$\vec(x) = F\vec(\theta)$$ Taking the partial derivatives of each$\vec (x)$term with respect to each joint variable$\vec(\theta)$will give you$J$. 4 Nowadays we no longer employ exact solutions for the IK problem, simply because the number of degrees of freedom so as the number of constraints the final configuration needs to comply with make the so called geometric approach intractable. By contrast, iterative methods are used to converge to the most plausible solution. Therefore, the reachability is ... 4 The math required to deal with the orientation error (in terms of error with respect to both desired position and desired velocity) is well described in the book of Sciavicco-Siciliano. See section 3.7.3 starting from page 137. I'm personally employing the axis-angle notation, thus I coded the formula (3.89) given at page 139 with correct results. In short,... 4 The Jacobian in that equation is from the joint velocity to the "spatial velocity" of the end effector. The spatial velocity of an object is a somewhat unintuitive concept: it is the velocity of a frame rigidly attached to the end effector but currently coincident with the origin frame. It may help to think of the rigid body as extending to cover the whole ... 4 There are more approaches to solve the inverse kinematics equations. If you want to continue to use the pseudo-inverse based approach and still obtain more then 1 solution you can flip the sign of joint angle 1 for example (if it is a puma type robot arm) in the initial guess and run the iterative solver again. Same for joint angle 3. You can have ... 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 Condition number and manipulability are measured at a specific joint configuration, not end-effector location. You already understand it correctly that the values change according to the robot configuration as they are computed from the Jacobian. You may also want to check out Yoshikawa's original paper on manipulability. 3 You have to analytically compute all IK solutions. This is basically done with straightforward geometry. Most robotics textbooks with a section on manipulation will have a detailed explanation of this. But you might also want to check out the ikfast library. I am not sure if your gradient decent solution is guaranteed to have any special properties. But ... 3 Well, although, I didn't well understand what have wrote as a solution to the first part (upper) half of the Jacobian, but AFAIK, the manipulator Jacobian is a$6\times n$matrix, for that let's say$ J$is the Jacobian thus: $$J = \begin{bmatrix} J_v \\ J_\omega \end{bmatrix} \\$$ where both the upper part$ J_v$and the lower one$ J_\omega$are$ 3 \...

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Ugo's answer refers to "Sciavicco-Siciliano" which is a good book I'll quote as well. Chapter 3.6 introduces the so-called analytical Jacobian which is not the same as the so called geometrical Jacobian as it shows up in: $\omega=J_{geom} \cdot \dot{q},$ but has to be obtained from $J_{geom}$ with the help of the transformation matrix $T_A$: $J_{... 3 Yes, 6x9. Since $$\dot{x} = J \dot{\theta}$$ each column of the Jacobian represents the differential change in one of the six$x$coordinates with respect to each of the joints. 3 There are a lot of definitional problems and inconsistencies in this area. Geometric Jacobian. I'm not sure this has a precise and agreed upon meaning. But across the more classical robotics books (Siciliano etal., Spong etal., Corke) it relates joint velocities to end-effector velocity (translational and rotational) expressed in either the world or end-... 2 You don't need the positional IK to solve this problem. All that is required is, that your Jacobian is invertible, i.e. keep away from sinuglar joint configurations. Consider, how the end effector's velocity is formed: $$\begin{bmatrix} v_{trans} \\ v_{rot} \end{bmatrix} = J(\boldsymbol{q})\cdot \boldsymbol{\dot{q}}$$ So, if I understand correctly, you ... 2 The math involved in determining orientation error is described in this thread: Jacobian-based trajectory following 2 It depends on how theroetical/practical solution you are looking for. If you are considering a theoretical workspace, with no angular limits of your joints (e.g. due to mechanical constraints) then the calculating the inverse kinematics for a Cartesian pose which is out of the workspace would result in complex joint angles (at least one of the 7 joint ... 2 Assuming it is a robot with a serial structure: You will always be able to move every joint of the robot in any pose. However the singularitites still exist, and you can end up in a singular pose, but if you do not have any cartesian space contrains (and do not even calculate inverse kinematics/inv. jacobi), then from a motion point of view these are ... 2 I guess you want to find a cubic polynomial for the end effector. You have 3 coordinates for your points A and B, from your question is not clear if they are$x,y,z$or$x,y,\theta$. Anyway, I'll show here the procedure for$x$, and you can repeat it for the other two coordinates. Given the cubic parametric form$x = a_0+a_1 t + a_2 t^2 + a_3 t^3$($*$), ... 2 Worked example$\hspace{2.5em}\vec{q}$=$[q_{1}\hspace{1em}q_{2}]^{T}\hspace{1.5em}$[Generalized coordinate]$\hspace{2.5em}\vec{J}$=$\frac{\partial \vec{r}_{OA}(\vec{q})}{\partial\vec{q}}$=$\begin{bmatrix} \frac{\partial \vec{r}_{1}}{\partial\vec{q}_{1}} & ... & \frac{\partial \vec{r}_{1}}{\partial\vec{q}_{n}} \\ ... & & .....

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Look at example 283 and its derivation here. You just take the time derivative of each element of J, paying particular attention to the chain rule.

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"Why the end effector?" Because a robot working on a task uses its end effector to interact with the objects that make up the task. For example, it is common to align the jaws ("fingers") of a parallel-jaw gripper with the sides of an object to be picked up. The simplest way to describe that object is by using a fixed coordinate system in which the ...

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You can use Matlab to compute $\dot{J}$. For example, to compute $\dot{J}_{11}$, we can use the following code clear all clc syms T1 T2 T3 l1 l2 l3 t T1(t) = symfun(sym('T1(t)'), t); T2(t) = symfun(sym('T2(t)'), t); T3(t) = symfun(sym('T3(t)'), t); J11 = -l2*sin(T1(t)+T2(t)) - l1*sin(T1(t)) -l3*sin(T1(t)+T2(t)+T3(t)); dJ11dt = diff(J11,t) which yields ...

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Torque is pretty easy to calculate for a single static arm configuration. Torque is just the length of the moment arm * the perpendicular force. And it is easy to decompose the problem into X and Y components, then sum. So in your example you would have the torque about joint 1 to be: $$T_1 = a * F_y + b * F_x$$ (Assuming $F_x$ and $F_y$ are in the ...

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