In article ,
Paul Dormer wrote:
My amplifier "allows you to vary the volume from 0 (silence) to 79
(maximum), where each step is equivalent to 0.5dB." How would one
relate this adjustment to actual listening levels?

It's only a rough guide as ears vary, but the accepted amount to double
the apparent 'loudness' is 10dB. But even this rough guide changes if you
start from a whisper or a roar. You could test this yourself. Tune into
some speech - say R4 - and set the level so it's about natural. Then see
if 10dB either side does indeed sound like a doubling or halving to you.

However, a 3dB increase doubles the power output from the amplifier,
which is why 'upgrading' from say a 30 watt to 50 watt amp makes little
difference to the perceived actual maximum level.

To further confuse things, if you wanted to double the output from your
say CD, that would be 6 dB since we're there dealing with voltage, not
current gain.

As you've hopefully gathered from this dBs ain't all the same...

I've copied the FAQ from rec.audio.pro which might also further confuse
things...

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Q3.3 - What is the difference between dBv, dBu, dBV, dBm, dB SPL, and
plain old dB? Why not just use regular voltage and power
measurements?
Our ears respond logarithmically to increases in sound pressure
level. In order to simplify the calculations of these levels, as
well as the electrical equivalents of them in audio systems, the
industry uses a logarithmic system to denote the values.
Specifically, the decibel is used to denote logarithmic level
above a given reference. For instance, when measuring sound
pressure level, the basic reference against which we take
measurements is the threshold of hearing for the average
individual, 10^-12 W/m^2. The formula for dB SPL then becomes:
10 Log X / 10^-12 where X is the intensity in watts per square
meter
The first people who were concerned about transmitting audio over
wires were, of course, the telephone company. Thanks to Ma Bell
we have a bunch of other decibel measurements. We can use the
decibel to measure electrical power as well. In this case, the
formula is referenced to 1 milliwatt in the denominator, and the
unit is dBm. 1 milliwatt was chosen as the canonical reference
by Ma Bell. Since P=V^2 / R, we can also express not only power
gain in dB but also voltage gain. In this case the equation
changes a bit, since we have the ^2 exponent. When we take the
logarithm, the exponent comes around into the coefficient, making
our voltage formula 20 log. In the voltage scenario, the
reference value becomes 0.775 V (the voltage drop across 600 ohms
that results in 1 mW of power). The voltage measurement unit is
dBv.
The Europeans, not having any need to abide by Ma Bell's choice
for a canonical value, chose 1V as their reference, and this is
reflected as dBV instead of dBv. To avoid confusion, the
Europeans write the American dBv as dBu. Confused yet? [Gabe]

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--
*Some days you're the dog, some days the hydrant.

Dave Plowman London SW
To e-mail, change noise into sound.