I have quaternion representing a camera pose in a coordinate space called "World Frame 2":
- x+ = right
- y+ = up
- z+ = out
...and I want to convert it to another space called "World Frame 1":
- x+ = right
- y+ = down
- z+ = in
(So, y and z are each inverted)
How can I do this conversion?
In another thread, I reasoned it should work like so: (Call this "Approach 1")
When changing the camera coordinate system, we alter the transformation of a 3D point from the world frame to the camera frame. The original transformation from the world to the original camera is:
$$P_{oc} = R \cdot P w + T$$
The transformation from the original camera to Habitat's camera is:
$$P_{hc} = \begin{bmatrix} 1&0& 0 \\ 0& −1& 0 \\ 0 &0& −1 \end{bmatrix} P_{oc} = R_x ( \pi) P_{o c} $$
By combining the above transformations, the translation \$( t_x, t_y, t_z )\$ will change to \$( t_x, − t_y, − t_z )\$, the rotation R will change to \$R_x ( 180 ° ) R\$, and in quaternion form: \$ i \cdot ( w + x i + y j + z k ) = − x + w i − z j + y k \$. Here, i represents a 180-degree rotation around the x-axis.
Another user proposed this: ("Approach 2")
I agree that it's a 180-degree rotation around the x-axis. So the angle axis is:
$$ (\text{angle, axis}) = (\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \pi) $$
If we convert it to quaternion, it should be
$$ q = \begin{pmatrix} \cos(\pi / 2) \\ \sin(\pi / 2) \\ \cos(\pi / 2) \\ \cos(\pi / 2) \end{pmatrix} = \begin{pmatrix}0 \\ 1 \\ 0 \\ 0\end{pmatrix} $$
So the result will be $$q p q^{-1} = [0 1 0 0] [w x y z] [0 1 0 0]^{-1} = [w x -y -z]$$
I asked another user and they linked to this thread ("Approach 3"), which says the resulting quaternion should be:
$$q = w + ix - jy - kz$$
Which one is correct, and why are the others wrong?
