[FFmpeg-user] Size of 10-bit 4:2:2 video streams?

Bruchez Olivier olivier.bruchez at epfl.ch
Thu Sep 15 14:53:01 EEST 2022

Hi all,

I have a 10-bit 4:2:2 (yuv422p10le) AVI file containing only a video stream:

Input #0, avi, from 'test.avi':
  Duration: 01:28:00.80, start: 0.000000, bitrate: 221185 kb/s
  Stream #0:0: Video: v210 (v210 / 0x30313276), yuv422p10le, 720x576, 25 fps, 25 tbr, 25 tbn, 25 tbc

The file has a size of 146 GB = 146004770816 bytes.

Am I correct to infer that those 10-bit YUV values are grouped into 30-bit packets and encoded into 32-bit/4-byte packets?

With 20 bits per pixel (Y + either U or V), that would give a theoretical file size of:

88 * 60 * 25 * 720 * 576 * (10 + 10) / 30 * 32 / 8 = 145981440000 bytes

Which seems to more or less match the original file size.

Now here's my problem. When I extract the raw video stream:

ffmpeg -i test.avi -c:v rawvideo -pix_fmt yuv422p10le -f rawvideo test.raw

I get a RAW video file with a size of 96 GB = 96636395520 bytes. I’ve lost one third of the video stream.

Another strange thing: if I copy the video stream to another AVI file:

ffmpeg -i test.avi -c copy test.copy.avi

The resulting file has a size of 64 GB = 64425220302 bytes. This time I’ve lost 56% of the video stream.

What's happening here? It seems like I’m missing something obvious.

Thanks for any help,

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