View Full Version : Getting MPEG2 average bitrate from VOBs?

poochie2
6th February 2006, 19:38
Hello everybody.

I'm looking for a way to get the average bitrate of the video stream from concatenated VOBs. I found "Bitrate Viewer" but I must open each VOB by myself and calculate the weighted average, which is not something I like too much... any suggestion?
Guest
6th February 2006, 21:57
DGIndex. Do a preview over the project and the Info window will show the overall average video bitrate.
poochie2
6th February 2006, 22:13
DGIndex. Do a preview over the project and the Info window will show the overall average video bitrate.

I must be missing something: I made a project file, I reloaded it and previewed it. It still shows the instant bitrate, not the overall one! :confused:
Guest
6th February 2006, 23:27
The Info dialog shows both the instantaneous and average bitrates in different fields. Run the preview over the entire project and then look at the Bitrate (Avg) field of the Info dialog.
poochie2
7th February 2006, 00:22
The Iinfo dialog shows both the instantaneous and average bitrates in different fields. Run the preview over the entire project and then look at the Bitrate (Avg) field of the Info dialog.

I see what was missing... I had version 1.43 VS 1.46!
It works, but it takes about 15 minutes to get the bitrates for a typical 4.2 GB series of VOBs, which is a bit too much :(
foxyshadis
7th February 2006, 01:00
Maybe it's demuxing audio; writing would definitely kill performance, because it should go about as fast as your drive can read it. Audio->Output Method should be set to disable.
Mug Funky
7th February 2006, 05:51
this can also help you if it's only DVDs you're checking (this tool doesn't allow loading m2v files separately).

http://www.videohelp.com/tools?tool=DVD_Bit_Rate_Viewer
Guest
7th February 2006, 06:34
but it takes about 15 minutes to get the bitrates for a typical 4.2 GB series of VOBs Hey, you said you wanted the overall average. That means you have to parse the entire project.
poochie2
7th February 2006, 09:55
Maybe it's demuxing audio; writing would definitely kill performance, because it should go about as fast as your drive can read it. Audio->Output Method should be set to disable.

Nope, I disabled that. I think its the preview that is lowering the performance. Bitrate viewer only seeks for bitrate, so it's quite faster, it takes about 4 minutes more or less.
Mug Funky
7th February 2006, 09:55
hmm, actually, if all you want is the bitrate, you could divide total size by length... any media player plus windows explorer should be enough.
poochie2
7th February 2006, 09:57
this can also help you if it's only DVDs you're checking (this tool doesn't allow loading m2v files separately).

http://www.videohelp.com/tools?tool=DVD_Bit_Rate_Viewer

I'll try that!
poochie2
8th February 2006, 21:40
How funny:
Bitrate Viewer= 4255 kbps
DVD Bit Rate Viewer= 4820 kbps (maybe it's including the dual 384 audio tracks making 4052 kbps)
DGIndex= 4357 kbps

So I decided to check manually: DVD shrink claims there are 3493 MB of video stream, using VobEdit I demuxed the MPEG stream and got 3437 in 1h50'18". The calculation nakes 4257 kbps... and the winner is: Bitrate Viewer!

Using 3493 MB as size the result is 4324 kbps, almost close to DGIndex value...

Which is the correct one? I hope 4257 is... :angry:
Mug Funky
9th February 2006, 06:53
bitrateview gives wrongly scaled results... maybe DVD bitrate viewer does too?

4255 * 1.024 = 4357.12 = DGindex's result.

it's the old "kilo = 1024 or does it equal 1000?" problem again.
poochie2
9th February 2006, 14:26
bitrateview gives wrongly scaled results... maybe DVD bitrate viewer does too?

4255 * 1.024 = 4357.12 = DGindex's result.

it's the old "kilo = 1024 or does it equal 1000?" problem again.

But in my calculations I already used 1024 instead of 1000!

3437*1024*8/(3600*1+60*50+18)=4254 :confused:
Centurione
9th February 2006, 15:03
it seems to me:
3437 MB = 3437*1024*1024*8 = 28831645696 bits.

28831645696/((60+50)*60+18) = 4356549 bits/sec
so 4356.5 (Kbit ???).

So it seems that DGIndex results are correct, there is always the confusion on the base we are using (1024 or 1000).

regards, Centurione
poochie2
9th February 2006, 15:43
it seems to me:
28831645696/((60+50)*60+18) = 4356549 bits/sec
so 4356.5 (Kbit ???).

So it seems that DGIndex results are correct, there is always the confusion on the base we are using (1024 or 1000).

regards, Centurione

I now understood what you meant! Speaking of bitrates in general is it right to consider the 1024 the most correct value?