Is there some way of accurately calculating how long (in minutes) an mpeg can be to fit on 4.37GB DVD-R when resolution and average bitrate is known? This so that I can place just enough video in a Studio9 project without having it to be re-rendered.
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There are no stupid questions, only stupid people.
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Unless I have misunderstood your question, it is impossible to estimate permissible length in the absence of a value for average bitrate. It's like saying how far can I go on a tank of fuel in a Mercedes SLK 230 if I don't know the car's fuel consumption rate.Originally Posted by Ruude
I'd like to be able to give you the answer you appear to be reuesting, but I don't see how this is possible.
Arky ;o) -
Ooops! my sincere apologies, Ruude - I read your post too quickly and misread 'known' as 'unknown'. I was wrong, hands downOriginally Posted by jimmalenko
The following may help you:
http://www.dvddemystified.com/files/DVDcalc.xls
Again, please accept my apologies
If this calculator does not help you, then please repost and I'll do my best to help further.
Regards,
Arky ;o) -
90 minutes at 6500 kbps=4.37gb with 224 kbps audio
120 minutes at 4800 kbps=4.37gb with 224 kbps audio
150 minutes at 3800 kbps=4.37gb with 224 kbps audioI think,therefore i am a hamster. -
What about resolution in this estimate?Originally Posted by johns0There are no stupid questions, only stupid people.
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Sorry, resolution is not computed in this formula. It has nothing to do with how big space an MPEG will be versus how long it is.
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OK, then what about this:Originally Posted by Tommyknocker
a 4 hour video capped at 480x576 with 4.5 bitrate gives about 11gig
--------- -320x576- -------- 7gig
so resolution is definately a factor no?There are no stupid questions, only stupid people. -
Either the video is shorter or the bitrate is lower. Resolution is not a factor. Bitrate is exactly as the name suggests. The amount of bits of data over time, resolution does not effect this.Originally Posted by Ruude
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"Each problem that I solved became a rule which served afterwards to solve other problems." - Rene Descartes (1596-1650)
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This is nice, but I want to go the other way round.Originally Posted by BJ_MThere are no stupid questions, only stupid people.
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Originally Posted by Ruude
you can use it to figure out either way -- its not hard at all"Each problem that I solved became a rule which served afterwards to solve other problems." - Rene Descartes (1596-1650)
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