Quantcast DVD2SVCD: Predicting Audio File Size - digitalFAQ.com Forums [Archives]
  #1  
11-03-2002, 07:36 PM
black prince black prince is offline
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Hi All,

Since D2S (RC4) predicts video file size after encoding audio and using
what's left to predict video, would it make sense to find out the audio
file size in advance. I started with a crude approach using
Mpeg Mediator to extract the .wav from the vob's and Headac3he
to predict approximate audio file sizes for .mp2.

Next, I tried a formula:

For 16bit Stereo: ([total movie seconds * 44KHz] * 2) * 2 = file size (.wav).
For example: ([7272 * 48] * 2 ) * 2 = 1,396,224 (.wav) and
Headac3he estimated 1,339,000 (.wav). I would like to determine the
file size for .mp2 at various audio bitrates (i.e. 192, 128, etc.)
If someone has a formula to predict audio file size for .mp2 please
post it.

-black prince
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  #2  
11-03-2002, 07:54 PM
kwag kwag is offline
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Piece of cake black prince

If you're going to encode audio at say 192Kbps, each second of audio would hold 192,000 bits / 8 = 24,000 Bytes per second. So just multiply your movie playing time in seconds by 24,000 bytes, in this case, and that will be your total file size for the audio.

-kwag
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  #3  
11-03-2002, 09:05 PM
black prince black prince is offline
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Hi Kwag,

Kwag wrote:
Quote:
Piece of cake black prince

If you're going to encode audio at say 192Kbps, each second of audio would hold 192,000 bits / 8 = 24,000 Bytes per second. So just multiply your movie playing time in seconds by 24,000 bytes, in this case, and that will be your total file size for the audio.

-kwag
Thanks Kwag. I didn't know it was that simple.

-black prince
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11-04-2002, 02:58 AM
VILLA21 VILLA21 is offline
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"...If you're going to encode audio at say 192Kbps, each second of audio would hold 192,000 bits / 8 = 24,000 Bytes per second..."

For 192kb/s Stereo is (192,000 / * 2 or not?
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11-04-2002, 06:59 AM
kwag kwag is offline
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Quote:
Originally Posted by VILLA21
"...If you're going to encode audio at say 192Kbps, each second of audio would hold 192,000 bits / 8 = 24,000 Bytes per second..."

For 192kb/s Stereo is (192,000 / * 2 or not?
Hi VILLA21,

Doesn't matter if it's stereo, joint stereo, or dual channel. It's still the same. FileSize in Bytes = ( Kbps / 8 ) * total seconds.
You can double check this by opening a WAV file with WinAmp, and taking note of play time. Then open the WAV file with HeadAC3he and select .mp2 as output and your bit rate. See the output file size and compare to the formula. It will be the same.

-kwag
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  #6  
11-04-2002, 07:23 AM
black prince black prince is offline
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Hi Kwag and Villa21,

Kwag wrote:
Quote:
Hi VILLA21,

Doesn't matter if it's stereo, joint stereo, or dual channel. It's still the same. FileSize in Bytes = ( Kbps / 8 ) * total seconds.
You can double check this by opening a WAV file with WinAmp, and taking note of play time. Then open the WAV file with HeadAC3he and select .mp2 as output and your bit rate. See the output file size and compare to the formula. It will be the same.

-kwag
Kwag is correct. I compared the file sizes with Headac3he and the
formula. It's very close between the two. Headac3he is rounded and
the formula is more precise. I try to give 680 to 700 MB to video
and the rest to audio when encoding LBR on 74 minuted CD. I created
a spreadsheet in which I only enter the total movie time (seconds)
and all the audio bitrates are there with estimated file sizes. I choose
the one that comes close to how much space I want for video.

-black prince
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  #7  
11-04-2002, 10:59 AM
muaddib muaddib is offline
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Quote:
Originally Posted by VILLA21
"...If you're going to encode audio at say 192Kbps, each second of audio would hold 192,000 bits / 8 = 24,000 Bytes per second..."

For 192kb/s Stereo is (192,000 / * 2 or not?
Hi VILLA21,

I believe that when you encode stereo or dual channel, you split the available bitrate between the 2 channels. So, if you are doing a 192kbps stereo, you are actually doing ~96kbps for each channel.
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11-14-2002, 10:22 AM
serrabastien serrabastien is offline
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Quote:
Originally Posted by kwag
Piece of cake black prince

If you're going to encode audio at say 192Kbps, each second of audio would hold 192,000 bits / 8 = 24,000 Bytes per second. So just multiply your movie playing time in seconds by 24,000 bytes, in this case, and that will be your total file size for the audio.

-kwag
192Kbps = 192x1024 bps and not 192,000...
So you have to multiply your movie playing time by 24,576 bytes
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  #9  
11-14-2002, 10:44 AM
kwag kwag is offline
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Quote:
Originally Posted by 'Z
Quote:
Originally Posted by kwag
Piece of cake black prince

If you're going to encode audio at say 192Kbps, each second of audio would hold 192,000 bits / 8 = 24,000 Bytes per second. So just multiply your movie playing time in seconds by 24,000 bytes, in this case, and that will be your total file size for the audio.

-kwag
192Kbps = 192x1024 bps and not 192,000...
So you have to multiply your movie playing time by 24,576 bytes
No. A bit is a bit. 192Kbs is 192,000 bits. Not Bytes. If it was in Kilo Bytes, then it would be 1024. But it's already measured in bits. So it is 192,000 / 8 = 24,000.

Edit: Correction, but for Kilo Bytes it's 24,000 * 1.024 = 24, 576. You're right ( I hate Hard Disk manufacturers, with their 1024 marketing gimmick )

-kwag
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  #10  
11-15-2002, 12:26 PM
serrabastien serrabastien is offline
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It's not really HD manufacturers fault... It's the binary system (you know this 0 and 1 )...
1k = 2^10, 1M = 2^20 ...
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