[FFmpeg-user] What does "-vol" parameter mean in ffmpeg?

[littlebat]

On Wed, 18 May 2011 15:11:03 +0200
Nicolas George <nicolas.george at normalesup.org> wrote:

> L'octidi 28 floréal, an CCXIX, littlebat a écrit :
> > I can't understand how to use "-vol" parameter to increase volume in
> > ffmpeg. I know "-vol 256" mean keep volume unchanged. But I don't
> > know how to choose a proper number for a video file to increase
> > volume. For example, does "-vol 512" mean volume intensity is twice
> > as original file?
> 
> Just try it for yourself:
> 
> perl -e 'print pack("s*", 3000, 2000, 1000, 0, -1000, -2000)' |
> ffmpeg -f s16le -i - -vol 512 -f s16le - 2> /dev/null |
> perl -e '$/=\2; while(<>) { print unpack("s", $_), "\n" }'
> 
> Of course, you can do exactly the same with any GUI PCM editor.
> 
> By looking at the result, you can see that obviously each sample is
> multiplied by vol/256.

Maybe I have understood the "-vol volnumber" parameter in old version
ffmpeg and "volume=volumenumber" audio filter in new version ffmpeg.

According to the page: http://en.wikipedia.org/wiki/Decibel
"volnumber/256" in old ffmpeg and "volumenumber"(not dB number) in new
ffmpeg audio filter, is the "amplitude ratio" or "sound pressure
level" ratio.

So, "-vol 2560" in old ffmpeg or "volume=10" in new ffmpeg equal to
"volume=20dB" in new ffmpeg. The formula is dBnumber=20*lg(amplitude
ratio).