how much you get on a disc

[flaninacupboard]

ok, for an 80 min cdr, you can get 80 mins of VCD standard video and sound. agreed? good. now then:
1150 + 224 = 1374
1374 * 60(seconds) * 80(minutes) = 6595200
6595200 / 1024(megabits / 8(megabytes) = 805.0781
why is that? and why is the VCDbitrate calc woefully wrong?
you stick in 1 * 80 min CDR, 80 mins video, 224audio, it tells you you need a video bitrate of 1105 ?! anyone know of a better one?

[flaninacupboard]

no?
shall i just continue to use 1150 + 224 * 80 / video length - 224?
using the example of 60 mins i get
1150+224*80 = 109920
109920/60-224 = 1608
which seems to add up.
1608+224*60(seconds)*60(minutes) = 6595200
6595200/8/1024 = 805.0781 megabytes.

it's over 800 meg though, which seems wrong.
however
1150+224*60*80 = 6595200
6595200/8/1024 = 805.0781 megabytes, so i should be fine. so why are these bit-calcs wrong, and will my maths work out alright?

[Chopper Face]

You can fit more than the megage says. Something about not having error bits or something. You fit 740 mb on a 74 min CD and stuff.

[flaninacupboard]

yeah, i know. 740 on a 74min CDR and 800 on an 80 min CDR. my question is more about why it actually seems you can get 805megs on an 80 min CDR, and why the bitrate calcs always get it wrong.

[ktnwin]

We have learned that one minute of VCD takes 10MegaBytes of MPEG-1.
OK, 80 min takes 800 MBs, how does it fit on a 700MB CD-R

Because the MPEG file contains the ECC bytes in each of the data sectors, and the DAT file put on the VCD does not. By removing the ECC bytes (I remmebered vaguely that it's about 2240 - 2049 = 10%)

That's how we normally fit 800MB MPEG on a 80 min CD-R.

[flaninacupboard]

yeah, i thought it was the headers that went, but ok. so is it ECC removal the bitrate calcs dont do right?

[flaninacupboard]

nobody know? no?

then i'll find-y out-y.

[mrbass]

http://www.ping.be/satcp/realsize01.htm
answers all your questions