WAV file too big?
- Thread starter TroyLoud
- Start date
garryknight
New member
One of my 8-minute tracks is 145 MB in size. That's because I limited it to 44100 Hz sample rate with a 16-bit depth. If I wanted higher quality (and I probably wouldn't be able to hear the difference) I could double the sample rate or go for 24 or 32-bit depth. Your 3:26 file is a reasonable size.
bandcoach
Zukatoku - Mod Scientist
doing the numbers backwards
3'26" @ 44100 @ 16 bits = 34.65MB
3'26" @ 48000 @ 16 bits = 37.19MB
3'26" @ 44100 @ 24 bits = 51.98MB
3'26" @ 48000 @ 24 bits = 56.58MB
so if your settings match either of the 24 bit bit depth settings then yes this is normal
the actual math goes like this
((((time in seconds x sample rate in Hz x bit depth x number of tracks in the resulting file)/8 bits)/1024)/1024)
number of tracks is based on content: 2 for stereo, 3 for 2.1, 6 for 5.1 and 8 for 7.1
Hz is simply s[sup]-1[/sup] ~ per second
seconds and s[sup]-1[/sup] cancel each other out leaving the units of the result as bits
the first division moves the result to bytes (/8 bits) ~ 8 bits to the byte
the second division moves the result to kilobytes (/1024) ~ 1024 bytes to the kilobytes
the third division moves the result to megabytes (/1024) ~ 1024 kilobytes to the megabyte
e.g.
3'26" = 206 seconds
sample rate of 44.1kHz/second = 44100/second
bit depth of 16 bits
assume stereo file
put these into the formula above and we get
file size = ((((206 seconds * 44100/seconds * 16 bits * 2) / 8 bits) / 1024) / 1024)
= ((290707200/8)/1024)/1024
= (36338400/1024)/1024
= 35486.71875kB/1024
= 34.65MB
Notes
The above formula is similar tot that required to calculate bitrate but is different in three ways
1) there is no time used in the calculations as bitrate is content length independent
2) we do not divide by 8 bits as we are interested in bits not bytes
3) we divide by 100 instead of 1024 (bitrate is a decimal concept not a binary concept)
so the bitrate formula is
sample rate x bit depth x number of channels / 1000
in our example above this becomes
44100*16*2/1000 = 1411.2 kbps - this is standard for all stereo wav files recorded at 44.1kHz sample rate and a bit depth of 16 bits
3'26" @ 44100 @ 16 bits = 34.65MB
3'26" @ 48000 @ 16 bits = 37.19MB
3'26" @ 44100 @ 24 bits = 51.98MB
3'26" @ 48000 @ 24 bits = 56.58MB
so if your settings match either of the 24 bit bit depth settings then yes this is normal
the actual math goes like this
((((time in seconds x sample rate in Hz x bit depth x number of tracks in the resulting file)/8 bits)/1024)/1024)
number of tracks is based on content: 2 for stereo, 3 for 2.1, 6 for 5.1 and 8 for 7.1
Hz is simply s[sup]-1[/sup] ~ per second
seconds and s[sup]-1[/sup] cancel each other out leaving the units of the result as bits
the first division moves the result to bytes (/8 bits) ~ 8 bits to the byte
the second division moves the result to kilobytes (/1024) ~ 1024 bytes to the kilobytes
the third division moves the result to megabytes (/1024) ~ 1024 kilobytes to the megabyte
e.g.
3'26" = 206 seconds
sample rate of 44.1kHz/second = 44100/second
bit depth of 16 bits
assume stereo file
put these into the formula above and we get
file size = ((((206 seconds * 44100/seconds * 16 bits * 2) / 8 bits) / 1024) / 1024)
= ((290707200/8)/1024)/1024
= (36338400/1024)/1024
= 35486.71875kB/1024
= 34.65MB
Notes
The above formula is similar tot that required to calculate bitrate but is different in three ways
1) there is no time used in the calculations as bitrate is content length independent
2) we do not divide by 8 bits as we are interested in bits not bytes
3) we divide by 100 instead of 1024 (bitrate is a decimal concept not a binary concept)
so the bitrate formula is
sample rate x bit depth x number of channels / 1000
in our example above this becomes
44100*16*2/1000 = 1411.2 kbps - this is standard for all stereo wav files recorded at 44.1kHz sample rate and a bit depth of 16 bits
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