At what DPI should http://docs.opencv.org/3.1.0/pattern.png be printed?
Alternatively: Does the size of the printout not matter? I've read the relevant OpenCV pages with the tutorials but, amazingly, this is still not clear to me. Someone here must know the answer...
(Also, I notice that it has a note that it is "9x6". What about it is 9x6? Nothing, as far as I can tell...)
Thanks.
After much looking around I eventually found a partial answer in a few different sources, but primarily in samples/cpp/tutorial_code/calib3d/camera_calibration/in_VID5.xml . In the sample configuartion file it has:
<!-- Number of inner corners per a item row and column. (square, circle) -->
<BoardSize_Width> 9</BoardSize_Width>
<BoardSize_Height>6</BoardSize_Height>
So this answers one of the questions: 9x6 represent the inner corners.
Finally, it has:
<!-- The size of a square in some user defined metric system (pixel, millimeter)-->
<Square_Size>50</Square_Size>
Though this still doesn't fully answer the DPI question since it's not clear where we define the metric system (does anyone know)?
However, from what I have read, the real size of the printed squares does not matter for the calibration coefficients, but instead only matters for the extrinsic values used in calculating the pose. Therefore, for the purposes of simple calibration, it doesn't matter.
For 9x6 size, you would have had your answer in the tutorial and in the documentation.
In my opinion, you print your chessboard in whatever DPI / size (A4, A3, etc.) you want and you construct your 3D model according to the real size if you want to have correct extrinsics. You can use the sample code to do that.
Asked: 2016-06-09 11:00:26 -0600
Seen: 5,049 times
Last updated: Jun 14 '16
9x6 is the number of squares on cols x rows. it's unrelated to any dpi problem
Except that... It's not 9x6 squares. :)