Imroi and all the im* functions in MATLAB use so much processing power that my MacBook’s fan is on full! Haven’t found an explanation for the code inefficiency though :/ ]]>

This was such a long time ago that I forgot what I used it for. I think order(VECTOR) would have also served my purposes, and its simpler. Thx for the feedback :)

]]>This is how the book answers your question: “You can envision the chessboard’s location as being expressed by (1) “creating” a chessboard at the origin of your camera coordinates, (2) rotating that chessboard by some amount around some axis, and (3) moving that oriented chessboard to a particular place.”

The book also mentions that if you want to get a rotation matrix out of the rvector you need to use the rodrigues transform: “Each of these rotation vectors can be converted to a 3-by-3 rotation matrix by calling cvRodrigues2()”

I strongly suggest you get “learning OpenCv” (http://www.amazon.com/Learning-OpenCV-Computer-Vision-Library/dp/0596516134/ref=sr_1_1?ie=UTF8&qid=1421834697&sr=8-1&keywords=learning+opencv). It has a whole chapter on how to do camera calibration. And it has all the mathematical explanation. Additionally, it points you to other papers that might be of interest.

Hope it helps

]]>With respect to which point on the chessboard is the rotation and translation calculated (since every point on the chessboard will have a different rotation and translation)? ]]>

I want to know how the rotational and translational matrices are calculated. A chessboard has many corners. The rotational matrix and translational matrix describe rotation and translation. Which point’s rotation and translation is considered here? Is it the center point of the image? ]]>