I MAY BE WRONG OR I MAY BE RIGHT... HARD DISK RECORDING COMPUTATIONS

[BALDO]

i was playing with my cellphone calculator and was thinking of "predicting the disk usage of wav files in recording and here is my observation.. the magic number is 5
in recording 44.1 k 16 bit.. use 5 x number of minutes ( 30 secs is 0.5 and 20 seconds is 0.4) x no. of tracks.. SO..
in a 3.30 min "stereo recording " it will 5x 3.5 x 2 = 35 MB.. now here is the fun part.. what if it is in 48 k or 96 or 192k? or what if it is 24 bit or 32 bit?.. ok here is my guess - mahina ako sa math ha hehehe 
in a 3.30 min recording of 48k and 24 bit stereo it will be 5x 3.5 x 1.088 (48k not 44.1) x 1.5 (24 bit)  x 2(stereo) =57.12 MB..there will be variances but i noticed it will be close.
SAMPLE RATE              BIT RATE
44.1 = x 1                  16 bit = x1    24 bit = x 1.5     32 bit = x2
48    =x 1.088             
96    =x 2.176
192  = x 4.3537
so we can now approximate a 3 min song recorded in 48k/24 bit and 24 tracks and it is
5 x 3 x 1.088 x 1.5 x 24 = 587.5 MB
and we can now approximate a 4 min song recorded in 192k / 32 bit and 24 tracks and it goes...
5 x 4 x 4.3537 x 2 x 24 = 4179.5 MB or 4.1 G... i hope this is right  me disclaimer naman eh 
happy computing and i hope this helps.. yes YOU .. yung nag iisip kung kasya pa ba ang 30 songs sa 20 gigs na HD .. 

[BALDO]

just to add.. wala pa palang processing yun like EQ's, reverbs, compression , automation etc etc ..raw tracks pa lang.. hmmm there maybe some additional numbers added to it but at least me idea na tayo sa possible disk use... back to the drawing board i guess hehehe..

[skunkyfunk]

Ganito lang man:

The number of samples in any given MONO or STEREO waveform is calculated this way:

(length of soundclip in seconds) x (sample rate) = no. of samples

Examples:

for a 5:00 song in 44,100KHz Sampling Rate

(5minutes x 60 seconds/min) x (44,100 samples/second) = 13,230,000  samples


for a 3:30 song in 88,200KHz Sampling Rate

(3.5 minutes x 60 seconds/min) x (88,200 samples/second) = 18,522,000  samples


Now, how do you calculate the NUMBER OF BYTES an audio file takes?

FOR MONO:

For 16-bit, it is simply computed by multiplying the number of samples by 2.  That is because 8 bits = 1 byte, THUS 16 bits =  2 bytes.    For 24-bit, you multiply the number of samples by 3.  For 32-bit, you multiply the number of samples by 4.

Example:

a 5:00 16-bit/44,100Hz wav file in MONO should take


13,230,000 samples x 2 bytes/sample = 26,460,000 bytes OR 25.23Mb (since 1Kb =  1,024 bytes)

a 5:00 24-bit/44,100Hz wav file in MONO should take

13,230,000 samples x 3 bytes/sample = 39,690,000 bytes OR 37.85Mb (since 1Kb =  1,024 bytes, and 1MB = 1024Kb)

a 5:00 32-bit/44,100Hz wav file in MONO should take

13,230,000 samples x 4 bytes/sample = 52,920,000 bytes OR 50.46Mb (since 1Kb =  1,024 bytes)


FOR STEREO:

Likewise, for 16-bit, it is simply computed by multiplying the number of samples by 2.  That is because 8 bits = 1 byte, THUS 16 bits =  2 bytes.    For 24-bit, you multiply the number of samples by 3.  For 32-bit, you multiply the number of samples by 4.  But for STEREO, simply double that of computing for MONO, since you have TWO signals with the same sampling rate and bit-depth.

Example:

a 5:00 16-bit/44,100Hz wav file in STEREO should take


13,230,000 samples x 2 bytes/sample X 2 = 52,920,000 bytes OR 50.46Mb (since 1Kb =  1,024 bytes)

Dapat siguro may quiz sa thread na ito...