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I am trying to 3d motion track a scene from my Sony HDR-HC7e Camcorder. I know that he has a 1/2.9" ClearVid CMOS sensor in it but how do I get the horizontal sensor Size? and should I youse the 1440 * 1080 footage that comes from the camcorder to track and resize it after the rack and render or should I directly resize it? It was filmed with a 5.4 mm focal length. I know that for sure. The footage also appears to have a pixel aspect ratio of 1.33 what do I have to do with this?

I have transformed the Interlaced footage with a small AVSynth plus script. I had the idea to get all the parameters from the camera through a grid shot and some software analysis but the Meshroom Camera Calibration node doesn't detect the gird and Matlab's camera calibration software is too expensive. The shot that I am trying to track as well as some calibration shots, can be found here.

I also want to compute the K1 and K2 distortion parameters and Optical center with Nuke. Do I have to use the 14401080 footage or a 19201080 resized version for that?

The grid in the calib_raw shot has a cell size of 2020mm and the nuke_calib_raw shot has a cell size of 18.518.5mm

All info about the Sony HDR-HC7e comes from this manual.

I was also wondering what the sensor size for stills captured in 3 by 4 is because I was doing some photogrammetry In just used the 1/2.9 inch converted to mm. And for the K distortion coefficients and optical center I would just take grid image taken in the same mode an send it rough nuke. Or? Software use there is Meshroom.

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  • $\begingroup$ whoa, that's a lot of questions! Did you try to just track it with Blender without worrying too much about the details? might be suprising but, if it isn't a very long and complicated shot, you can get pretty good results with the default settings $\endgroup$
    – pevinkinel
    Commented Sep 19, 2020 at 19:29
  • $\begingroup$ 3.8 px error aint to good $\endgroup$
    – Phönix 64
    Commented Sep 19, 2020 at 19:34
  • $\begingroup$ Please ask one question per post. $\endgroup$
    – susu
    Commented Sep 19, 2020 at 20:18
  • $\begingroup$ As an alternative to mathlab you can use GNU Octave (free open source), there is a port for camera calibration. The camera matrix (lens and optical center) work fine, but the numbers for radial distortion (k1,k2) don't really coincide with those in blender's way to do camera reconstruction). $\endgroup$
    – susu
    Commented Sep 19, 2020 at 20:35

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There is no need to un-squeeze the anamorphic pixels as an extra step (specially if you have already de-interlaced it, so you are dealing with a degraded original to begin with), just enter the aspect ratio in the Track > Camera > Pixel Aspect section.

enter image description here

For the sensor width enter the value you know and set the lens to 5.4mm

You can try to find the radial distortion coefficients (k1, k2) using the annotation tool. The procedure is outlined in this post:

How to determine lens undistortion values for motion tracking?

But Blender will optimize those settings in the solve if you set the refine settings to Focal Length, Optical Center, K1, K2.

enter image description here

At the time of solving and refining blender usually gets those parameters quite accurately.

You should aim for an error average of less than 1/3 of a pixel (0.3)

For accurate tracking and solving, please read the very detailed post on motion tracking:

How can I get better results when doing camera motion tracking?

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  • $\begingroup$ Okay, one more Question. Blender wants the horizontal sensor width but I only know the vertikal how can I calculte the horizontal? $\endgroup$
    – Phönix 64
    Commented Sep 19, 2020 at 20:39
  • $\begingroup$ If you know the aspect ratio of the sensor then all you need is basic trigonometry to calculate it en.wikipedia.org/wiki/Trigonometry $\endgroup$
    – susu
    Commented Sep 19, 2020 at 21:28
  • $\begingroup$ and the aspect ratio is 1.33 so the formula would be: 1/2.9" are 8.75862068966 mm so 8.75862068966 ^2 = x^2 + (1.33*x)^2 so x = 5.26359 and x is the vertical so the horizontal is x*1.33 = 7.0005747 $\endgroup$
    – Phönix 64
    Commented Sep 19, 2020 at 21:31
  • $\begingroup$ No, the physical aspect ratio of the sensor. $\endgroup$
    – susu
    Commented Sep 19, 2020 at 21:33
  • $\begingroup$ hm, and where is that written down? $\endgroup$
    – Phönix 64
    Commented Sep 19, 2020 at 21:36

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