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:
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.