Times 6 hours times 26 tracks is ? GB

is this a trick question?

Rates are for each channel of audio 16-bits per sample sampled at 44.100Hz (44.1Khz)
705,600 bits per second
705,600 divided into bytes(by 8) then kilobytes(by 1024) =86.132KB per second 24-bits per sample sampled at 4,8000Hz (48Khz)
1,152,000 bits per second
1,152,000 divided into bytes(by 8) then kilobytes(by 1024) =140.625KB per second 24-bits per sample sampled at 9,6000Hz (96Khz)
2,304,000 bits per second
2,304,000 divided into bytes(by 8) then kilobytes(by 1024) =281.25KB per second 24-bits per sample sampled at 192,000Hz (192Khz)
4,608,000 bits per second
4,608,000 divided into bytes(by 8) then kilobytes(by 1024) =562.50KB per second

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If the number of bytes exceeds 3 digits, then divide by 1024 to convert kiloBYTES to megaBYTES.
Example: 24-bit@96Khz audio gives 140.625KB per second per channel.
To calculate how much space is taken per minute of audio at that rate, multiply 140.625 times 60 for the amount of seconds.
140.625 (KiloBYTES) X (times) 60 (seconds for a minute) = 8437.50 kiloBYTES. To convert that to megaBYTES, divide that number by 1024 because there are 1024 bytes in a kilobyte, 1024 kilobytes in a megabyte, 1024 megabytes in a gigabyte, ect.
Divide 8437.50 kilobytes by 1024 to convert it to megabytes.
8437.50 (kilobytes / (divided by) 1024 = 8.239 megaBYTES.

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1 channel (mono) of 16-bit@44.1Khz=86.132KB per second (5.046MB per minute)
1 channel (mono) of 24-bit@48Khz=140.625KB per second (8.239MB per minute)
1 channel (mono) of 24-bit@96Khz=281.25KB per second (16.479MB per minute)
1 channel (mono) of 24-bit@192Khz=562.50KB per second (32.958MB per minute)

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2 channels (stereo) of 16-bit@44.1Khz=172.264KB per second (10.093MB per minute)
2 channels (stereo) of 24-bit@48Khz=281.25KB per second (16.479MB per minute)
2 channels (stereo) of 24-bit@96Khz=562.50KB per second (32.958MB per minute)
2 channels (stereo) of 24-bit@192Khz=1.098MB per second (65.88MB per minute) 4 channels (quadraphonic) of 16-bit@44.1Khz=344.528KB per second (20.187MB per minute)
4 channels (quadraphonic) of 24-bit@48Khz=562.50KB per second (32.958MB per minute)
4 channels (quadraphonic) of 24-bit@96Khz=1.098MB per second (67.50MB per minute)
4 channels (quadraphonic) of 24-bit@192Khz=2.197MB per second (131.835MB per minute)

Ouch. My brain. 

48kHz @ 16 bits = 5.49 MB/minute/track
At 24 tracks, 1 minute = 131.84MB

48kHz @ 24 bits = 8.24 MB/minute/track
At 24 tracks, 1 minute = 197.75MB

A formula for figuring out your MB vs song minutes would be:
A) 16 bit/44.1 kHz:
121.124MB x Total audio minutes = Total MB for a song
B) 24 bit/44.1 kHz:
181.686MB x Total audio minutes = Total MB for a song
C) 16 bit/48 kHz:
131.84MB x Total audio minutes = Total MB for a song
D) 24 bit/48 kHz:
197.75MB x Total audio minutes = Total MB for a song

The formula is:
Sample Rate (a) x Bit Resolution (b)= bits per second (c)

Now it's just a matter of getting bits per second to MB/min...

Bits per second (c) / 8 (bits per byte) = bytes per second (d)

Bytes per second (d) / 1024 (bytes per kilobyte) = kilobytes per second (e)

Kilobytes per second x 60 (seconds per minute) = kilobytes per minute (f)

Kilobytes per minute (f) / 1024 (kilobytes per megabyte) = megabytes per minute (g)

So the math formula using the letters I included would be:
a (Hz) x b (bit depth) = c bits/sec
c / 8 = d B/sec.
d / 1024 = e kB/sec
e x 60 = f kB/min
f / 1024 = g MB/min

alexoosthoek


Times 6 hours times 26 tracks is ? GB

many

@musicman... Thanks for that. I was semi kidding with the "ouch" thing. I knew there was an equation somewhere but never actually seen it. I'm just ultra braindead today. Too much running around in there at the moment. I also kind of suck at math but can usually muddle my way through stuff eventually.

Bookmarked for future nerdiness.

;-)

IT'S CALLED A HARD DRIVE.
GET TWO.

Not a trick question MM, I've been asked to do a recording at 24/48 for 6 hours. Thanks!

Er..... a lot?