Transform this matrix of distances into a more general one

Question

I need to make the navigation and guidance of a vehicle (a quadcopter) in a platform. This platform can be seen like this:

where the blue dots are the center of each square, and the $x$ distances are all the same, and the $y$ distances are all the same.

I need the distance between each blue dot to the center (the blue dot of the $(2;2)$), but that distance depends on the $yaw$ angle. For example, if $yaw=0^\circ$, the situation is like this:

and the distances are:

$$d_{1;1} = (-d_x; -d_y)$$ $$d_{1;2} = (-d_x; 0)$$ $$d_{1;3} = (-d_x; d_y)$$

$$d_{2;1} = (0; -d_y)$$ $$d_{2;2} = (0; 0)$$ $$d_{2;3} = (0; d_y)$$

$$d_{3;1} = (d_x; -d_y)$$ $$d_{3;2} = (d_x; 0)$$ $$d_{3;3} = (d_x; d_y)$$

If the situation is with $yaw=180^\circ$:

the distances are the same but with the opposite sign, i.e,

$$d_{1;1} = (d_x; d_y)$$ $$d_{1;2} = (d_x; 0)$$ $$d_{1;3} = (d_x; -d_y)$$

$$d_{2;1} = (0; d_y)$$ $$d_{2;2} = (0; 0)$$ $$d_{2;3} = (0; -d_y)$$

$$d_{3;1} = (-d_x; d_y)$$ $$d_{3;2} = (-d_x; 0)$$ $$d_{3;3} = (-d_x; -d_y)$$

If $yaw=90^\circ$, the situation is like this:

and the distances (see the difference between $d_x$ and $d_y$) would be:

$$d_{1;1} = (-d_y; d_x)$$ $$d_{1;2} = (-d_y; 0)$$ $$d_{1;3} = (-d_y; d_x)$$

$$d_{2;1} = (0; -d_x)$$ $$d_{2;2} = (0; 0)$$ $$d_{2;3} = (0; d_x)$$

$$d_{3;1} = (d_y; -d_x)$$ $$d_{3;2} = (d_y; 0)$$ $$d_{3;3} = (d_y; d_x)$$

If $yaw = -90^\circ$:

the distances would be:

$$d_{1;1} = (d_y; d_x)$$ $$d_{1;2} = (d_y; 0)$$ $$d_{1;3} = (d_y; -d_x)$$

$$d_{2;1} = (0; d_x)$$ $$d_{2;2} = (0; 0)$$ $$d_{2;3} = (0; -d_x)$$

$$d_{3;1} = (-d_y; d_x)$$ $$d_{3;2} = (-d_y; 0)$$ $$d_{3;3} = (-d_y; -d_x)$$

I need to write a matrix that uses the information of the $yaw$ angle and returns the distances from each angle (not just 0, 90, -90 and 180, but also 1, 2, 3, ...)

I tried to write it but I couldn't find the solution.

Thank you very much. I really need this help

Edit: please note that the coordinate frame moves with the quadcopter, like in this image:

Edit 2: for example, if $yaw=45^\circ$, then the distance from $(3;3)$ to $(2;2)$ is $\sqrt{d_x^2+d_y^2}$ in $x$ and $0$ in $y$.

Answer 1

Place a fixed coordinate system at the upper left corner of your grid, with the x axis pointing down and y to the right. Call this the 0 coordinate system.

Imagine the origin of the moving coordinate system to be located at the origin of the 0 coordinate system. Unfortunately you have defined that coordinate system to have the x axis pointing up, and the y axis pointing to the left. We’ll handle that, but the math would be a bit simpler if the coordinate systems were aligned when $yaw = 0$. Call the moving coordinate system the 1 system.

Define a rotation matrix $R_{01}$, where $$ \vec {p_1} = R_{01} \vec {p_0}$$

The above equation, then, can be used to map vectors from the 0 coordinate system (your grid) to vectors in the rotating coordinate system.

Beginning kinematics can be used to find that

$$ R_{01} = \begin{bmatrix} - \cos(yaw) & \sin(yaw) \\ - \sin(yaw) & - \cos(yaw) \\ \end{bmatrix} $$

Note that each element’s sign would be the opposite of what it is above had the two coordinate systems aligned when $yaw = 0$.

Vector math gives:

$$x_1 = - \cos(yaw) x_0 + \sin(yaw) y_0$$ $$y_1 = - \sin(yaw) x_0 - \cos(yaw) y_0$$

where $x_0$ is what you call $d_x$, and $y_0$ is your $d_y$.