I’ve quaternion representing a digital camera pose in a coordinate area known as “World Body 2”:
- x+ = proper
- y+ = up
- z+ = out
…and I need to convert it to a different area known as “World Body 1”:
- x+ = proper
- y+ = down
- z+ = in
(So, y and z are every inverted)
How can I do that conversion?
In one other thread, I reasoned it ought to work like so: (Name this “Method 1”)
When altering the digital camera coordinate system, we alter the transformation of a 3D level from the world body to the digital camera body.
The unique transformation from the world to the unique digital camera is:$$P_{oc} = R cdot P w + T$$
The transformation from the unique digital camera to Habitat’s digital camera is:
$$P_{hc} = start{bmatrix} 1&0& 0
0& −1& 0
0 &0& −1 finish{bmatrix} P_{oc} = R_x ( pi) P_{o c} $$By combining the above transformations, the interpretation $( t_x, t_y, t_z )$ will change to $( t_x, − t_y, − t_z )$, the rotation R will change to $R_x ( 180 ° ) R$, and in quaternion kind: $ i cdot ( w + x i + y j + z ok ) = − x + w i − z j + y ok $. Right here, i represents a 180-degree rotation across the x-axis.
One other consumer proposed this: (“Method 2”)
I agree that it is a 180-degree rotation across the x-axis. So the angle axis is:
$$
(textual content{angle, axis}) = (start{pmatrix}
1
0
0
finish{pmatrix}, pi)
$$If we convert it to quaternion, it needs to be
$$
q = start{pmatrix}
cos(pi / 2)
sin(pi / 2)
cos(pi / 2)
cos(pi / 2)
finish{pmatrix}
= start{pmatrix}0 1 0 0end{pmatrix}
$$So the end result might be $$q p q^{-1} = [0 1 0 0] [w x y z] [0 1 0 0]^{-1} = [w x -y -z]$$
I requested one other consumer and they linked to this thread (“Method 3”), which says the ensuing quaternion needs to be:
$$q = w + ix – jy – kw$$
Which one is appropriate, and why are the others flawed?
