volltreffer and Son of Mixerman:
What I ment was not that the least significant bits of the recorded sound gets unused (=truncation), but that the most significant bits of the Digital-to-Analog converter remains zero/unused. But this is however not true! It would be if negative/signed values was represented as I thought they were. Or for an A/D designed to only set positive values, to for example to set a voltage between 0 and 5 volts. But it isn't this way for audio data. Not for signed values and not even for waveforms represented with just unsigned values (having the 0 volt offset at half of the range). I thought specific bits was unused with signed values, but it isn't - because all the bits gets inverted when representing a negative value.
So IF a waveform was only made up of positive values, like "half" of an waveform, then you could use that "0 to 5 volts converter" and then bits would really get unused and always zero when all the values are lower than half of the resolution. So, you can say that by only playing at half volume, then at any time, the audio COULD have been represented with one bit less, even if not one specific bit always is unused.
That is if negative values are represented according to standards.
The most important thing is that you just get access to half of the dynamic resolution by playing at 50% of the total volume. So if you have your monitor knob at 33, then you are only using 33% of the total dynamic range of your converter!
/Anders