# Converting product of exponentials from base frame to end-effector frame

This is a homework question from edx course Robot Mechanics and Control, Part II

Given the following and expressing its forward kinematics as $T = e^{[S_1]\theta_1} ... e^{[S_6]\theta_6}M$

It is can be found (and also shown in the answer) that

$$[S_2] = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & -2L \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$$ and $$M = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 3L \\ 0 & 0 & -1 & -2L \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$$

Part 4b) requires expressing the forward kinematics in the form of $T = Me^{[B_1]\theta_1} ... e^{[B_n]\theta_n}$ and finding $[B_2]$

I wanted to try deriving the answer using the following property (as found in the lecture notes page 20 and in the lecture around 3:00, basically using property $Pe^A = e^{PAP^{-1}}P$ for any invertible $P$):

$$e^{[S_1]\theta_1} ... e^{[S_n]\theta_n}M = Me^{[B_1]\theta_1} ... e^{[B_n]\theta_n}$$ where $$[B_i] = M^{-1}[S_i]M$$

I get $$M^{-1} = \begin{bmatrix} 0 & 1 & 0 & -3L \\ 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & -2L \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$$

Using the property $$[B_i] = M^{-1} [S_i] M$$ to calculate $B_2$ and I get $$[B_2] = \begin{bmatrix} 0 & 0 & 1 & -3L \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & -5L \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$$

which is obviously wrong. What am I doing incorrectly?

$$[B_2] = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & -3L \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$$

Full question with answers below: • Welcome to Robotics, Woofas. The last section of your image post gives an explanation of how/why the $B_2$ matrix was calculated using their method. You posted a procedure - $[B_i]=M^{−1}[S_i]M$ - could you please link to a derivation of that property? I suspect that's where your problem is. I would imagine that there's another step in the procedure, like a summation, or a multiplication, etc., but I can't personally find any details on the step you posted. Please edit your question to include the link to the procedure.
– Chuck
Aug 18, 2017 at 14:05
• Hi, thanks for helping, I've attached the source (lecture notes and lecture) for said equation. Hope it helps you help me! Thanks a lot!
– user16060
Aug 21, 2017 at 3:56
• @Chuck Sorry it's been a while, but would I be able to provide any more information that may help you help me? Thanks!
– user16060
Sep 18, 2017 at 8:27

$[S_2] = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & -2L \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$ should be $[S_2] = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & -2L \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$
$[B_2] = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & -3L \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$ should be $[B_2] = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & -3L \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$