I tried a camera calibration with python and opencv to find the camera matrix. I used the following code from this link

CodePudding user response：

Your misconception is about "focal length". It's an overloaded term.

• "focal length" (unit mm) in the optical part: it describes the distance between the lens plane and image/sensor plane
• "focal length" (unit pixels) in the camera matrix: it describes a scale factor for mapping the real world to a picture of a certain resolution

1750 may very well be correct, if you have a high resolution picture (Full HD or something).

The calculation goes:

f [pixels] = (focal length [mm]) / (pixel pitch [µm / pixel])

(take care of the units and prefixes, 1 mm = 1000 µm)

Example: a Pixel 4a phone, which has 1.40 µm pixel pitch and 4.38 mm focal length, has f = ~3128.57 (= fx = fy).

Another example: A Pixel 4a has a diagonal Field of View of approximately 77.7 degrees, and a resolution of 4032 x 3024 pixels, so that's 5040 pixels diagonally. You can calculate:

f = (5040 / 2) / tan(~77.7° / 2)

f = ~3128.6 [pixels]

And that calculation you can apply to arbitrary cameras for which you know the field of view and picture size. Use horizontal FoV and horizontal resolution if the diagonal resolution is ambiguous. That can happen if the sensor isn't 16:9 but the video you take from it is cropped to 16:9... assuming the crop only crops vertically, and leaves the horizontal alone.

Why don't you need the size of the chessboard squares in this code? Because it only calibrates the intrinsic parameters (camera matrix and distortion coefficients). Those don't depend on the distance to the board or any other object in the scene.

If you were to calibrate extrinsic parameters, i.e. the distance of cameras in a stereo setup, then you would need to give the size of the squares.