View Full Version : How do you calculate the bitrate use for 2 pass WM9 encodings?
Sgt_Strider
27th June 2003, 23:18
I don't think there is a calculator that is available for calculating the right bitrate to use for a WM9 2 pass encode session. There is one for xvid, RV9, and divx5. Do anyone here know a formula to determining the right bitrate? I made a post in the doom9 forum and a microsoft employee name MingCl mention the following:
"The way to calculate the bitrate for 2-pass VBR is pretty much the same as CBR (except the factor of buffer size in CBR).
However, one key factor, which we didn't seem to document in the help of the VCM beta release (and I will make sure that we add this in RTM), is that there is no frame dropping allowed in the VBR mode on the encode side. That is, if the specified bitrate is too low and the ecoder can't meet this target even when using the lowest quality, it will exceed the bitrate. This is documented in WMEncoder 9, but not WMV 9 VCM beta.
But I am not sure whether this is the cause of Crabba's issue since it is strange that the bitrate is decreased when the key frame distance is reduced. It should be the other way. So I do need to get Crabba's settings and possibly source as well to repro the problem here."
I'm hoping that one of you at least have a clue in what he's saying and clearify this for me. Thx
SeeMoreDigital
28th June 2003, 00:04
Sgt_Strider
Manually calculating the bitrate you require is not so difficult when using WME9 VCM, as it's quite bitrate accurate.
For PAL & NTSC calculations, if say, you had a movie that is 131mins (7860secs) long and you wanted to get the entire thing onto 700MB CD (with the audio at 64kbps) you could use the following calculation as a guide: -
Audio: 64kbps x 7860secs = 503,040 / 8 = 62,880 / 1024 = 61.4MB
61.4MB - 700MB = 638.6MB remaining for the video....... call it 635MB to be safe!
635MB x 1024 = 650,240 x 8 = 5,201,920 / 7860secs = 661.82kbps. Call it 662kbps!
However, I have noticed that different encoding applications can create different finished file sizes, even though the same bitrate is used. So beware!
Works for me though. 20 movies later!
Shame WME9 VCM only encodes at 3 frames per second on the second pass - at full bar!
Sgt_Strider
28th June 2003, 00:54
Originally posted by SeeMoreDigital
Sgt_Strider
Manually calculating the bitrate you require is not so difficult when using WME9 VCM, as it's quite bitrate accurate.
For PAL & NTSC calculations, if say, you had a movie that is 131mins (7860secs) long and you wanted to get the entire thing onto 700MB CD (with the audio at 64kbps) you could use the following calculation as a guide: -
Audio: 64kbps x 7860secs = 503,040 / 8 = 62,880 / 1024 = 61.4MB
61.4MB - 700MB = 638.6MB remaining for the video....... call it 635MB to be safe!
635MB x 1024 = 650,240 x 8 = 5,201,920 / 7860secs = 661.82kbps. Call it 662kbps!
However, I have noticed that different encoding applications can create different finished file sizes, even though the same bitrate is used. So beware!
Works for me though. 20 movies later!
Shame WME9 VCM only encodes at 3 frames per second on the second pass - at full bar!
Is it accurate? Is your method base on the statement that came from MingCL? I'm just wondering...
SeeMoreDigital
28th June 2003, 12:09
Yes, it is accurate.
I did a couple of 'video only' tests (GoldMember 90min PAL and MIB 94min PAL) I wanted the 'video' to be 660MB. It was smack on.
In the last few days I managed to do a PAL -v- NTSC test using StarWars1. The PAL version is 131mins long and the NTSC version is 136mins long.
Both of the encodes came in on target (700MB) when using an Mp3 audio setting of 64kbps, which is low I know (for the audio), but I'm doing some low bit audio test comparisons too!
However, please be mindfull of my comment regarding different encoding applications. I use a little known application called MPEGmediator 1.5 and have to carry out all 2pass encodes manually.
I did some 60min DivX 2pass 'video only' tests recently with VirtualDubMod, Vidomi, DVD2AVI & Gordian Knot. And although the 'video' input and 'encoding' output setting were identical. The output MB sizes were all very different!
I have to admit that I have not replicated the same test using WMV9 VCM codec because it takes so long for WMV9 VCM files to encode. Especially if you set the 'performance setting' to max!
With regard to MingCL comments, I found them a bit odd. As I've never experienced any frame drop (well, you wouldn't expect too when encoding at just 3 frames per second!) I would like to think that any frame drop would be shuffled toward the end titles.
Crabba
29th June 2003, 00:17
Like I said in that thread, just use Gordian knot and set it to 'DivX3' and you will get perfect bitrate/size for WMV9 VCM!
Just add '000' to the bitrate in the WMV9 tab ofcourse... ie if bitrate is 1275 you enter 1275000 in the bitrate tab.
This has worked perfectly for me on several files, except for the WMV9 bug with really tiny files...
Sgt_Strider
29th June 2003, 02:27
Originally posted by Crabba
Like I said in that thread, just use Gordian knot and set it to 'DivX3' and you will get perfect bitrate/size for WMV9 VCM!
Just add '000' to the bitrate in the WMV9 tab ofcourse... ie if bitrate is 1275 you enter 1275000 in the bitrate tab.
This has worked perfectly for me on several files, except for the WMV9 bug with really tiny files...
I see, that's cool.
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