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

Given the following Figure 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?

Thanks in advance

The correct answer is

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


Full question with answers below: Full question

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  • $\begingroup$ 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. $\endgroup$ – Chuck Aug 18 '17 at 14:05
  • $\begingroup$ 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! $\endgroup$ – user16060 Aug 21 '17 at 3:56
  • $\begingroup$ @Chuck Sorry it's been a while, but would I be able to provide any more information that may help you help me? Thanks! $\endgroup$ – user16060 Sep 18 '17 at 8:27
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So I finally found my mistake...

The mistake was in the matrix form of the screw

$ [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} $

and I misinterpreted the answer...

$ [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} $

What a fool...

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