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Rio
Guys,

After searching the boards, I haven't found the answer to this small problem.

In this case, I have a USB drive with 128MB and I'm trying to encode a whole CD (Guns N' Roses Greatest Hits) which is around 80 minutes long (79:16 to be exact) to MP3 using LAME.

At what bitrate should I encode the whole album (that is, using --preset XXX) to fit almost exactly as 128MB?

Thanks!
smok3
>>> 128*8*1024
1048576
>>> float(_)/(80*60)
218.45333333333335
>>>
= 218 kbits/s

i think.

------------------

edit, to be exact;

>>> 128*8*1024
1048576
>>> float(_)/(79*60+16)
220.47434819175777
>>>
= 220 kbits/s

------------------

p.s. i would try '-V 2 --vbr-new'
http://www.hydrogenaudio.org/forums/index....showtopic=28124
SebastianG
128 MB * 1024^2 bytes/MB / 79.283 min / 60 sec/min / 125 bytes/kilobit = 225.7 kilobits/sec

Some care has to be taken because kilobits/sec usually means 1000 bits/sec where as 1 KiBiByte = 1024 bytes. So, it's 125 bytes/kilobit and not 128 bytes/kilobit.

I agree with smok3. Try LAME with -V2 or -V3 in case the V2 version requires more than 128 MB.

Cheers!
SG
Engywuck
QUOTE (SebastianG @ Oct 23 2007, 10:17) *
128 MB * 1024^2 bytes/MB / 79.283 min / 60 sec/min / 125 bytes/kilobit = 225.7 kilobits/sec

Some care has to be taken because kilobits/sec usually means 1000 bits/sec where as 1 KiBiByte = 1024 bytes. So, it's 125 bytes/kilobit and not 128 bytes/kilobit.


Sure... but at least hard disks use the same method, so it's probably more like 128.000.000 bytes (minus some for file system, usually) on the USB drive (at least my USB sticks work like hard disks in this regard smile.gif)

to be conservative let's use 125 "real" MB
125.000.000 bytes * 8 bit/byte / ( 79.283 min * 60 secs/min) = 210.217 bit/sec = 210 kbps (215kbps, if 128MB, 225kbps for 128 MiB)

The result is the same: use -V2 or perhaps -V1, if your music is really good compressible, -V3 if -V2 still doesn't fit. Or take 10 Euros (15 dollars?) and buy a 4 GB USB drive smile.gif

(BTW: this forum uses strange conversions, too... max post length is 1024000 chars...)
MedO
It's probably a 128MiB (=128*1024*1024 Byte)-Flash-Chip, but as has already been said you have to substract filesystem size. Also, there is probably memory reserved by the controller for its own data and for defect management.
Rio
Thanks for all your replies! What I failed to include in my computation was the 8 bit/byte.

(That's why I always get around 26-27 only)

Previously I have been trans-/encoding to V2 to fit an 80 min album in the flash drive for my "usb mp3 player+fm modulator" in the car, but I had to delete one file or two (usually the track I least like).

EDIT: Thanks smok3 and Sebastian, I also tried to compute using 125 bytes/kilobyte.
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