Fmazzanti
5th February 2003, 12:13
Hello everybody,
I'm trying to use DVD2SVCD to encode my DVD to DVD-R, but I'm afraid the question I post applies equally well to SVCD encoding with this WONDERFUL tool.
The story is that I want to find out what is the Average bitrate I must set in the Bitrate tab to get a video stream of a given size (only the video, not the audio). So let's imagine I want to encode a stuff containing 2 audio tracks and one video stream. Also I want the video stream to be of a given size, let's say, 3800Mb (remember I encode to DVD-r). Also let's say I know that the 2 audios is going to take 500Mb (I do not encode them, simply keep the .ac3 files). The question is: what's the appropiate Average bitrate I must use?
I think I can do it in the following way:
call 'x' the bitrate in Kbits per sec.
call 'l' the length of the movie in secs.
Then x*l = Kbits that will take the movie.
Now I multiply it by 1000 to get it in bits, divide it over 8 to convert that to bytes, and finally divide it by 1024*1024 to convert form bytes to Kbytes to Mb. This formula reads in the end:
Mb = x*l*1000/(8*1024*1024)
Now I want to invert this, since I know what I want the size of the video to be (the Mb in the lhs) and am looking for the bitrate (the x). That leads me to
x = 8*1024*1024*Mb/(l*1000)
which is the bitrate at which I must encode the video stream (only the video!) to recover the size I want (the 3800Mb I said before).
Now how do I do that in practice? I would fire up DVD2SVCD, and in the bitrate tab set the Average bitrate to the value 'x' I've just obtained. However, D2S will process also audio tracks. Well, then I only have to select the 2 audio streams I want to copy, and increase the 'CD size' value to 3800 (=size of video) + 500 (=size of two audios). In this way, D2S will first extract (once again) the audios, discount their size from the 'CD size' value, which would give me 3800Mv back for video encodeing at the bitrate I set to get exactly this size for the .mpv file.
Would you say that works in this way? I'd like to know your opinions...
Thanks.
I'm trying to use DVD2SVCD to encode my DVD to DVD-R, but I'm afraid the question I post applies equally well to SVCD encoding with this WONDERFUL tool.
The story is that I want to find out what is the Average bitrate I must set in the Bitrate tab to get a video stream of a given size (only the video, not the audio). So let's imagine I want to encode a stuff containing 2 audio tracks and one video stream. Also I want the video stream to be of a given size, let's say, 3800Mb (remember I encode to DVD-r). Also let's say I know that the 2 audios is going to take 500Mb (I do not encode them, simply keep the .ac3 files). The question is: what's the appropiate Average bitrate I must use?
I think I can do it in the following way:
call 'x' the bitrate in Kbits per sec.
call 'l' the length of the movie in secs.
Then x*l = Kbits that will take the movie.
Now I multiply it by 1000 to get it in bits, divide it over 8 to convert that to bytes, and finally divide it by 1024*1024 to convert form bytes to Kbytes to Mb. This formula reads in the end:
Mb = x*l*1000/(8*1024*1024)
Now I want to invert this, since I know what I want the size of the video to be (the Mb in the lhs) and am looking for the bitrate (the x). That leads me to
x = 8*1024*1024*Mb/(l*1000)
which is the bitrate at which I must encode the video stream (only the video!) to recover the size I want (the 3800Mb I said before).
Now how do I do that in practice? I would fire up DVD2SVCD, and in the bitrate tab set the Average bitrate to the value 'x' I've just obtained. However, D2S will process also audio tracks. Well, then I only have to select the 2 audio streams I want to copy, and increase the 'CD size' value to 3800 (=size of video) + 500 (=size of two audios). In this way, D2S will first extract (once again) the audios, discount their size from the 'CD size' value, which would give me 3800Mv back for video encodeing at the bitrate I set to get exactly this size for the .mpv file.
Would you say that works in this way? I'd like to know your opinions...
Thanks.