Camera Vector
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Have a picture, how I can get the location of the camera view direction?
Say, I know that somewhere in the sample A, with coordinates (xA, yA, ZA), which is in position (XCA, YCA) on the screen of the camera, and some more points B, C, D as well. (Requires point as the point of sampling, which can measure and its coordinate position on the screen of the camera. But considering the cost of measuring, take the sample of 4 points lower if possible) How I can get the camera position (x, y, z) and its normal vector (direction of sight)? Or a much simpler problem, I know the camera position (x, y, z), how I can calculate the effective leadership of the view (as vector)? Sorry for my poor English: (. Thanks who helps me solve this fundamental problem. To Jay: Thanks for answering. I have discovered what is solvable, and how many points. I can even write some equations for this, the problem is, those that are not linear and I can not solve. I'm reading the link provided (thanks again for that: P) in the hope of obtaining a neat solution.
I'm not going to be very helpful … I have not studied this area much but I'm curious. I'm not sure that the first problem is solvable with a real camera. Assuming you are projecting 3 or fewer unique identification points (a triangle in 3D space) on a plane, does not seem to be at 3d least two possible orientations of at least two of the points in any projection. Add another point would help with that. But then all points should be visible from 4 points. Knowing the distance between the camera and 3 points would probably work well. If you know the camera position, camera becomes the point 4 and it seems reasonable that if the other (unique identification three) points are visible, you should at least be able to view address sight. may not even need a point of 3. This is not a source, but it might be useful? I do not really understand everything. Http: / / people.csail.mit.edu / bkph / articles / Harmful.pdf Other places to look would be to consider googling projective geometry, Computer Vision, etc.
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