View Full Version : Eye position in object space?
mattjakob
11-06-2009, 03:50 AM
I know it's very dumb but I cant get around this.
I need to calculate the position of the camera (eye) in object space, given a transformation matrix M.
I think in theory is should be:
eye_obj = col * (V.transpose).inverse * M.inverse
Where:
col = {0,0,0,1}
V = 4x4 camera transform matrix retrieved via glGetFloatv(GL_MODELVIEW_MATRIX, V)
M = 4x4 transform matrix defining the center of the object
DmitryM
11-06-2009, 06:30 AM
I guess you need:
eye_obj = col * V * M.inverse = V.pos * M.inverse
Dark Photon
11-06-2009, 07:06 AM
I need to calculate the position of the camera (eye) in object space, given a transformation matrix M.
I think in theory is should be:
eye_obj = col * (V.transpose).inverse * M.inverse
Dmitry's got you covered, but it looks like you've got it, except for that transpose in there.
MODELVIEW = Object-to-eye-space transform
so...:
v_eye = MODELVIEW * v_obj
But you want the opposite, so multiply both sides by INV_MODELVIEW:
INV_MODELVIEW * v_eye = INV_MODELVIEW * MODELVIEW * v_obj
INV_MODELVIEW * v_eye = v_obj
inverse(M) * inverse(V) * v_eye = v_obj
So now just plug-n-chug. The eye-space vertex position (v_eye) you've got is (0,0,0,1) -- the eyepoint. So plugging it in, this will effectively just extract out the translation component in INV_MODELVIEW.
You can multiply V*M and invert the result to get INV_MODELVIEW, or you can invert M and V separately, and multiply them in the opposite order to get INV_MODELVIEW. Then multiply that by the vector (v_eye) of (0,0,0,1).
Note that my text above uses OpenGL transform ordering (operator-on-the-left -- i.e. Ax=y). It looks like you're using the opposite in your notation (what you'd expect for C++), so just flip everything left-to-right. That said, looks like you already have the right answer, except for that transpose.
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