[Ffmpeg-devel] h263+: size of encoded frames after opening ffmpeg

[Martin]

Hi,

Martin wrote:
> Hi,
> 
> Michael Niedermayer wrote:
> 
>>Hi
>>
>>On Fri, Jan 19, 2007 at 11:23:59AM +0100, Martin wrote:
>>
>>
>>>Hi,
>>>
>>>Michael Niedermayer wrote:
>>>
>>>
>>>>Hi
>>>>
>>>>On Wed, Jan 17, 2007 at 11:14:40AM +0100, Martin wrote:
>>>>[...]
>>>>
>>>>
>>>>
>>>>>>>Even better, is there a
>>>>>>>possibility to change the bitrate without closing and reopening ffmpeg?
>>>>>>
>>>>>>
>>>>>>you can set the quality of frames which affects the bitrate,
>>>>>>closing and reopening within the same stream is generally not allowed
>>>>>>doing so means you are on your own
>>>>>
>>>>>Ok, how can I set the quality within the same stream without closing and
>>>>>reopening? I tried to set the AVCodecContext.crf parameter but it
>>>>>doesn't seem to have any effect on quality and size of outputted frames.
>>>>>Is there another parameter which I can use?
>>>>
>>>>
>>>>AVFrame.quality 
>>>>and set CODEC_FLAG_QSCALE
>>>
>>>Thank you very much, that worked. How is the relationship between
>>>quality and bitrate? Is there a formula which expresses it?
>>
>>
>>besides encode at quality X and look at the bitrate, no but theres some
>>approximation, and that is that quality and bitrate are approximatly
>>inversly proportional so double the quality variable and bitrate should
>>be approximately cut in half
> 
> 
> what I measured didn't give this behaviour (double quality -> half bitrate):
> I used a bitrate of 1280000 bits/sec in AVCodecContext. Measured Bitrate
> with different qualities:
> 
> ## Qual Bitrate
> 1) 600: 3012856
> 2) 1200: 1002368
> 3) 2400: 556024
> 4) 4800: 419112
> 
> A function is inversely proportional if x*y = a, where a is some
> constant. So the products of quality and bitrate are:
> 
> 1) 1807713600
> 2) 1202841600
> 3) 1334457600
> 4) 2011737600
> 
> This is far from approximately constant. There must be some computation
> that you are doing to get a bitrate from quality, right? I really need
> to know how to compute it and appreciate any help on this!
> 
> Thanks,
> 
> Martin
> 

is there nobody who can help me on this problem? I am really grateful
for any help!

Thanks,
Martin