Size of rendered loop section?

tcbetka wrote on 9/28/2008, 4:20 PM
I just finished editing together 2.75 hours of volleyball footage, encompassing 12 games. These games are all on one track. I know how to render looped sections, but how do I know how big each loop section will be when rendered? Do I have to pre-render a section to see how large it will be...? That doesn't seem to be very efficient to me, so I must be missing something. I didn't see it in the interactive tutorial section, but I still need to go through Gary's DVDs (got them yesterday!!!).

I need to finish them tonight, so any help would be appreciated. I need to burn these to 4.7gb DVDs, as MPEG-2.

Thanks in advanced.

TB

Comments

Chienworks wrote on 9/28/2008, 5:38 PM
It depends on the bitrate. Add the video and audio bitrates together, divide by 8, multiply by the duration in seconds. For example, if you have a 10 minute section and you plan on using 6mbps for video and 240Kbps for audio then 6.234mbps / 8 x 600 seconds = 467.6 MB.
tcbetka wrote on 9/28/2008, 6:01 PM
AH...thanks! I wondered about that formula--I thought I saw something like it before.

But in the Custom > Video tabs of the render template, you must have to use "Constant bit rate" then? I know you can select the average bit rate you want if you use variable, but then you have to specify the mean bit rate--and this of course could be variable. Or could it? I mean on the one hand, if you never achieved a maximum bit rate of 9.5Mbps, then it wouldn't seem at all likely that you'd ever see a mean bit rate of 6Mbps.

But it's close enough, I suppose. At least it should get you close enough to the point where DVD Architect can adjust the compression to get your material onto the DVD.

But thanks again though--I will give it a shot now.

TB
Chienworks wrote on 9/28/2008, 6:16 PM
Average bitrate is just that ... the average. Over any reasonable amount of time 6Mbps average should be the same file size as using a constant bitrate of 6Mbps. Of course, it's not going to come out exact. High-motion high-detail video is going to use more bits. Lots of still images or low detail areas will use less bits. If your entire video is a slideshow of stills the resulting file will probably be much smaller than the average bitrate would indicate.
PeterWright wrote on 9/28/2008, 6:23 PM
If you can , avoid the situation where DVD Architect "adjusts the compression" - if it has to do this, it will recompress an already compressed file, reducing quality.

There are bitrate calculators - do a search on this forum and there are several links - which tell you what bitrate is necaessary to fit a certain duration of video onto a DVD. If you have several separate pieces, you'll need to total their duration up before deciding what bitrate is necessary.
tcbetka wrote on 9/28/2008, 6:47 PM
I understand all of that...

For this project I have about 2:30:00 of volleyball footage, edited together with cuts, fades, some text...etc. There are 12 games within this amount of time, and as luck would have it, I was able to render after about 1:10:00, which was right at the end of game 6--so it worked out well.

So using 6Mbps video and 224Kbps audio:

6.224 / 8 * (70 * 60) = ~3.27gb

I could have snuck another game into this first disk, but the next game was against a different team so it made sense not to--simply for continuity for the coach (basically each team was played twice). But the remainder of the footage should only take about 3.9gb or so, thus I am happy to have a nice buffer of space.

But the formula given in this thread is most useful--thanks again for posting it.

Item #2,356 learned this weekend on the Sony Vegas Video forums.!

(My brain really hurts...)

TB
Chienworks wrote on 9/28/2008, 8:11 PM
Another really handy one is to divide 600 by the length in minutes. 600 / 70 = 8.571. That means you'd be able to use an average video bitrate of 8Mbps and still have plenty of room left over for the audio. For the second disc, 600 / 80 = 7.500, so you could use 7.2Mbps for video. At those rates you could easily pick a constant bitrate and not muck about with variable.
tcbetka wrote on 9/28/2008, 8:35 PM
Neat...

Thanks!


TB

EDIT: I should ask--where are you guys coming up with these? Where are they referenced; the explanation of their basis, I mean. I'd like to read a bit more about them (maybe there isn't anything published...). Cool stuff though! I just got Gary Kleiner's incredible DVD set, but have only had time to get through about half of the first one. Maybe he talks about it some.
Chienworks wrote on 9/29/2008, 3:49 AM
It's all plain simple math, algebra and linear analysis. Convert megabits per second to megabytes per second to megabytes per minute and you're done. Any 5th grader could do it, once they know the measurements to work with! :-)
tcbetka wrote on 9/29/2008, 5:12 AM
Now that you mention it....you're right! It is simple algebra--and learning the definition of bitrate. We use the term so much, I guess maybe we sometimes forget what it really means.

Thanks again for pointing it out.

TB