I am currently solving a kinematics example that asks for the Jacobian. However, when I solve it using the Screw method I get different results from the analytical method, The example is kinda hard to write here but I will demonstrate on a 2D RR robot which shows the same behavior.

Here is the simple problem:

enter image description here

Now lets find the Jacobian when $θ_1=pi/4 $ while $θ_2=0$, $L1=L2=1$

Here is the Jacobian using Screw Method:

$J = \begin{bmatrix} 0 & 0\\ 0 & 0 \\ 1&1 \\ 0& 0.7\\ 0& -0.7\\ 0&0 \end{bmatrix} $

Here is the Jacobian using Analytical Method:

$J = \begin{bmatrix} -L_1 sin(θ_1) - L_2 sin(θ_1+θ_2) & -L_2 sin(θ_1+θ_2)\\ L_1 cos(θ_1) + L_2 cos(θ_1+θ_2) & L_2 cos(θ_1+θ_2) \\ \end{bmatrix}, $

$J = \begin{bmatrix} -\sqrt{{2}} & -\sqrt{{2}} / 2 \\ \sqrt{{2}} & \sqrt{{2}} / 2 \\ \end{bmatrix} $

Notice that even singularity of both methods do not match. Where did I make the mistake?

Also, why the screw method does not take into account the length of last link? I mean if the length of last link is 0 or 1000, the linear velocity W x R should change.

Edit1: Here is the method I used to calculate the first jacobian. enter image description here

  • $\begingroup$ Can you provide sources of formulae you were using? It seems to me that they are Jacobian of different equations. The first one maybe maps joint velocities to a screw. The second one maps joint velocities to tool velocities. $\endgroup$ Nov 18 '19 at 13:26
  • $\begingroup$ @PetchPuttichai I have updated the question with the method used to calculate the first jacobian. The second one is simply FwD kinematics then differentiation. $\endgroup$ Nov 18 '19 at 18:31
  • $\begingroup$ I haven't had time to get back to this yet. But I guess both of them are representing the same thing. Maybe if you try writing the 6D tool twist (from the equation with screw method Jacobian) in terms of just $\dot{x}$ and $\dot{y}$, you might get the tool x-y velocity similar to what you'd get from using the equation with analytical method. $\endgroup$ Nov 25 '19 at 11:34
  • $\begingroup$ The first one, based on screw theory does not seem right. Could you complete the question with an indication how was it derived? $\endgroup$
    – 50k4
    Jun 4 '20 at 19:02

There are in fact two types of Jacobians, a geometric Jacobian and an analytical Jacobian. The intro to chapter 3 in the book: Robotics: Modelling, Planning and Control by Bruno Siciliano, Lorenzo Sciavicco, Luigi Villani, Giuseppe Oriolo says it well:

...differential kinematics is the relationship between the joint velocities and the corresponding end-effector linear and angular velocity. This mapping is described by a matrix, termed geometric Jacobian, which depends on the manipulator configuration. Alternatively, if the end-effector pose is expressed with reference to a minimal representation in the operational space, it is possible to compute the Jacobian matrix via differentiation of the direct kinematics function with respect to the joint variables. The resulting Jacobian, termed analytical Jacobian, in general differs from the geometric one.


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