Given an extrinsic matrix, it is granted that it transforms a point in world coordinate to a point in camera coordinate.
Let E be the extrinsic matrix:
R11 R12 R13 t1
R21 R22 R23 t2
R31 R32 R33 t3
0 0 0 1
Technically, extrinsic matrix gives you the orientation of world coordinate frame with respect to camera coordinate frame. It also gives you the translation of world coordinate frame with respect to camera coordinate frame. Thus, t1 should be distance from camera coordinate origin to world coordinate origin in x direction and t2 in y direction and t3 in z direction. This indirectly means that camera coordinate origin is located -t1 in x direction, -t2 in y direction and -t3 in z direction from the world coordinate origin. That is, (-t1, -t2, -t3) is the camera coordinate origin in world coordinate frame. Camera coordinate origin should correspond to camera position and thus (-t1, -t2, -t3) should correspond to camera position in world coordinate frame. However, it isn't so.
Let Xw be the camera position in world coordinate frame and Xc be the camera position in camera coordinate frame. We need to find Xw.
Then:
E * Xw = Xc
Xc is (0,0,0) and hence:
Xw = -1 * (R)' * (t), where R (3*3) and t (3*1) are the rotation and translation components of E respectively.
Why is Xw not equal to (-t1, -t2, -t3)?
Hi Dennis,
You have answered your own question by deriving Xw...
If you are looking for the intuition of why this is the case, think about it this way: the extrinsic matrix transforms a point from the world coordinates into the camera's coordinates. it applies rotation first, and then translation:
Pc = R * Pw + t
The order matters. To find camera's location in the world coordinates, we need to reverse the transformation: apply the negative of the translation, followed by the inverse of rotation. That is why Xw = R' * [-t1, -t2, -t3], and not just [-t1, -t2, -t3].
This is also why the camera matrix is sometimes given on the form P = [R -R*t]. For this P the camera is centred at t.
Thanks for your answers. I understand the fact that you have to apply the reverse of transformation. But even then, rotation doesn't affect the position of an object, right? It just changes the orientation. That's why I thought the position depends entirely on the translation vector of extrinsic matrix
The rotation is about the origin. So it can affect the position quite dramatically.
Yes, but the position of the origin (which is what we are concerned with), does not change by rotation in X,Y or Z axes.
Oh, the rotation is about the origin of the world coordinate system, not the camera's coordinate system. So the camera's origin moves quite a bit.
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