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The equation (14) in this paper, $$ P_{Cj}=RP_{Pj}+T+e_j $$ where $P_{Cj}$ and $P_{Pj}$ are the observed positions of feature $j$ prior to and after the current robot motion, $R$(otation) and $T$(ranslation)

If ignoring $e_j$ and assuming there is no translation,

I don't understand why applying $R$ and not $R^{-1}$.

For example,

the previous robot orientation is GREEN and the blue dot in GREEN coord is $(0,3)$.

Then the robot rotates clockwise $\frac{\pi}{2}$ degree to RED coord and the blue dot in RED coord is (-3,0)

If clockwise rotation is negative,

$R$ in this case is:

\begin{bmatrix}\cos(\frac{-\pi}{2})&-sin(\frac{-\pi}{2})\\sin(\frac{-\pi}{2})&cos(\frac{-\pi}{2})\end{bmatrix}

equal to

\begin{bmatrix}0&1\\-1&0\end{bmatrix}

So to get $(-3,0)$, $R^{-1}$ seems to be the correct one.

Do I misunderstand anything?

Thank you for precious time on my question.

enter image description here

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It's been years since I've used any of this math, so I'll appeal to Wikipedia and hope I'm not off-base.

The examples in this article apply to active rotations of vectors counterclockwise in a right-handed coordinate system by pre-multiplication. If any one of these is changed (such as rotating axes instead of vectors, a passive transformation), then the inverse of the example matrix should be used, which coincides with its transpose.

So it's the rotation of the coordinate system GREEN into RED that uses the inverse $$\color{red}{C} = R^{-1}\color{lime}{C}$$ and determining the location of a fixed point under the new coordinates uses $R$ itself.

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    $\begingroup$ Thank you for the quick response, ShapeOfMatter. Your answer inspires me to look at Active and passive transformation and Rotation of axex. Based on my understanding after reading these articles, I incorrectly use "active transformation" R in my original question. The correct approach is using "passive transformation" R in both coordinate axes and the observed points. Example 2 in Rotation of axes provides a good example. $\endgroup$ Apr 26, 2019 at 5:43

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