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I have a pose $\boldsymbol{x}_0 = (x, y, \theta)$ in an absolute frame, right-handed and centered in (0, 0, 0). I obtain a measurement $\boldsymbol{m} = (\Delta x, \Delta y, \Delta \theta)$ in the previous pose coordinate system.

What is the new pose $\boldsymbol{x}_1$ in the absolute frame? If the previous pose had a covariance $\Sigma_{x_0}$, and the measurement has a covariance $\Sigma_{m}$, what is the covariance of the new pose $\Sigma_{x_1}$?

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For your first question, it's just a a change in coordinate frames, you have the absolute coordinate frame and a coordinate frame "local" which is centered in x0 and the pose is in the local coordinate frame.

The equation would look like that:

$ \bigg( \begin{matrix} x'\\ y' \\ \theta' \end{matrix}\bigg) = \bigg( \begin{matrix} cos(\theta) & -sin(\theta) & x0 \\ sin(\theta) & cos(\theta) & y0 \\ 0 & 0 & 1 \end{matrix}\bigg) \bigg( \begin{matrix} x \\ y \\ \theta \end{matrix}\bigg)$

For the second question I think it's ok to also take into account the covariance matrices in the previous eq. $ \bigg( \begin{matrix} x'\\ y' \\ \theta' \end{matrix}\bigg) = \bigg( \begin{matrix} cos(\theta) & -sin(\theta) & x0 \\ sin(\theta) & cos(\theta) & y0 \\ 0 & 0 & 1 \end{matrix}\bigg) Q_{m} \bigg( \begin{matrix} x \\ y \\ \theta \end{matrix}\bigg)Q_{x1}$

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