Hallo
[...]
> 3475 frames / 29.97 fps = 115.94 seconds of video
> and
> 1152kBit / 8 = 144KByte video bitrate
>
> so
>
> X (filesize in KBytes) / 115.94 seconds = 144Kbytes video rate
>
> X = 16695.36 Kbytes or (16695.36 * 1024) 17096048 bytes
>
> but what is have is less than that
>
> [EMA
On Tue, 25 Nov 2003, Bernhard Praschinger wrote:
> If you forget the -V option when multiplexing a VBR stream mpeg2enc will
> create a files as large as the maximal video bitrate + audio bitrate. If
> you calculate than the average bitrate it seems that you have a CBR
> stream.
Minor c
Hallo
> > > i was wondering how i could determine the average bitrate. ??
> > > ideally i would like program to tell me what it is.
> > You can only answer that question after you have encoded and MultiPLEXed
> > the stream with a audios stream.
> > That that two most interrestin lines look like t
