manual bitrate calculations

[azurenights]

Hi I am just wondering if Anyone can tell me how to work out bitrate calculations manually??? I am encoding three mpeg files in svcd format to put on one CD, but I will have different video bitrates for each mpeg file. I want to work out how much space I have left for audio and whether I have to increase or decrease the bitrates for the video.

thanks

[Kedirekin]

It's actually fairly simple. Just multiply the number of seconds time the bitrate. The hard part is knowing if your encoder treats a kbps as 1024 bps or as 1000 bps.

As for calculating the amount of space on the CD, SVCD uses mode 2 form 2 for storing the video data. I believe mode2 form 2 gives 2324 bits of user data per sector, and there are 75 sectors in each second on the CD (74 minute CD has ~333,000 sectors (74*60*75), 80 minute has ~360,000 (80*60*75)). So again it's just math.

You should leave some slack for user data (added during muxing, scan offsets for example), for the data track (which holds PBC and extra stuff, like menus), and for intra-track waste (between the data track and all video tracks). 4 MB is usually fairly safe.

FYI: a bitrate calculator like fitCD will take all of this into account for you. Unfortunately, I do not know a bitrate calculator that'll do multiple tracks with different bitrates.

[azurenights]

okay well I am using cinema craft encoder to encode my video and toolame to encode my audio. Do these encoders treat a kbps as 1024bps or 1000 bps??

when you were referring to 333,000 sectors, don't you mean 333,000 bytes instead???

thanks

[Kedirekin]

You know, even though I use both CCE and toolame (via DVD2SVCD), I do not know what they equate 1 kbps to. If no one knows the answer (anyone?), you'll have to derive it emperically.

Actually, I did mean 333,000 sectors. 333,000 bytes would be ~325 KB, and we know CDRs hold more than that. The math is quite simple

74 minutes * 60 sec/minute * 75 sectors/sec = 333,000 sectors

and

333,000 sectors * 2324 bytes/sector = 773,892,000 bytes
= ~755,754 KB
= ~738 MB

[Bugs Bunny]

>Hi I am just wondering if Anyone can tell me how to work out bitrate calculations manually??? I am encoding three mpeg files in svcd format to put on one CD, but I will have different video bitrates for each mpeg file.


Ha OK yes I vill.. dont know no vcd dont know no svcd dont know no calcoolas dont know no algebrie BUT I KNOW BITRATES HAH! OK this be bitrate calc I use fer Divx..

(CD_size - (T * audio_spec)) * 8 / T / 60 / 1.024 = # Kbit/s simplifies to..

(CD_size - (T * audio_spec)) / T / 7.68 = # Kbit/s

CD_size = 734003 or 681574 (1000-byte chunks)
T = video length in minutes (round up)
audio_spec = size of audio per min, usually you can figure this out from bit rate, e.g. 64Kbit, 44.1KHz CBR WMA is 480 KByte/min. But if it's VBR or mono or multi-channel, you'll just have to do the audio first, then plug the whole size thing in for the (T * aud). Also remember to calculate in the interleave overhead.

The *8 is to convert from byte to bit. The /1.024 is to convert from decimal to binary. The /60 is to convert from minute to second.

This might seem a long-winded way to do things if yer used to use a software calc, but the good thing is that if ya want to do multiple sound tracks, you can easily mod the eq to suit yer needs.

Or, er, may be not. Since yer vcd bloke..