2

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(\...


1

You can use the rotation matrix. linear_v= R*sensor_linear_v + p X sensor_angular_v p is the translation vector and the X is the cross operator angular_v=R*sensor_angular_v


1

Do I understand correctly, that the problem is with the matrix equation you have posted and how to calculate $T_{sb}$ from that equation? You can solve the matrix equation by multiplying from the right with the inverse of the last matrix on the rhs, step by step. $$\require{cancel} $$ $$T_{se} = T_{sb} \times T_{b0} \times T_{0e}$$ $$T_{se} \times T_{0e}^{-1}...


1

Assuming frame $0$ is the 'absolute frame', if we let $^j P_i$ be the $i^{th}$ position/orientation expressed in the $j^{th}$ coordinate frame, then what you're asking for is the sequence $$\{(^0P_i)_{i=1 ... N}\},$$ correct? Using your sequence of measurements $\{(x_i, y_i, \theta_i)_{i=1 ... N}\}$, it's straightforward to compute the $i^{th}$ 2-D ...


1

Each step can be represented by its transformation matrix, $$ \begin{bmatrix} \cos{\theta'_{i}} & -\sin{\theta'_{i}} & x'_{i}\\ \sin{\theta'_{i}} & \phantom{-}\cos{\theta'_{i}} & y'_{i} \\ 0 & 0 & 1 \end{bmatrix}. $$ The world position at each step $i$ is the rightward-propagating product of the transforms up to that point, $$ \...


1

1) There are many ways to express velocities. All of them are mathematical constructs to describe the same motion. They can have some minor advantages/disadvantages depending on the applications. The one main disadvantage of an Euler angle based approach is the gimbal lock problem. An advantage of Euler angle based approaches is, that they can be more ...


1

Here's a graphical answer. Given some vector $r$ in Frame 1: You can rotate to a new frame, F2: All you do is project to the new axes. Counting along the x-axis in $F_2$, you can see that you go right (positive) in an amount $x_a \cos{\theta}$, then you go right (positive) in the amount $y_a \sin{\theta}$, so the total x-axis location is $$ x_{F_2} = +\...


Only top voted, non community-wiki answers of a minimum length are eligible