Johannes said:

small audible reduction of 3 dB require a 50% reduction in sound

energy. ....

A doubling of absolute signal power represents an increase in dB of 3dB.

Reference

http://ccrma-www.stanford.edu/~oded/M151/decibels.html ,

but this is also generally known from any textbook. ....

This sounds correct, but you said in your earlier post:

"o A reduction of 10dB(A) is a 50% reduction in sound energy"

So you mixed up 'sound energy' with 'Percieved noise level' .

Strictly, as 3dB reduction corresponds to 50% reduction in sound energy,

....

It seems to me that JHH is mixing up two different "dB" units and is

also mixing up sound energy with sound intensity or power. JHH seems

to be using a Sound Power Level "dB" while he quotes someone who is

obviously using a Sound Pressure Level "dB". And sound energy seems

inappropriate here since we don't care about that and have no

reference energy to form a ratio with in the "dB" calculation. I

suspect that he meant sound power or intensity (power per unit area)

which both have the 50% = 3 dB characteristic.

The unit "dBA" (AKA "dB(A)") always (AFAIK) corresponds to something

called a Sound Pressure Level (SPL), which is neither an energy or a

power or even a sound pressure; it is a non-linear function of an

average (RMS) sound pressure, based on one definition of "Bel" (the

common (base 10) logarithm of the ratio of a power and a reference

power). Since (see Note 1) the power transmitted by a pressure wave

(through a fixed area?) is proportional to the square of the RMS

pressure (i.e., it's proportional to the average of the square of the

pressure), we have

SPL = 10 * log( Power1/Power0 )

= 10 * log( (k*Pressure1^2)/(k*Pressure0^2) )

= 20 * log( Pressure1/Pressure0 )

where the pressures are measured at a standard distance. SPL is

favored because it is easily measured with a calibrated microphone.

(See Note 2.)

(The following all assumes a omni-directional (AKA isotropic) source

of sound in free air; a source transmitting constant power.)

From the above plus some even more basic physics, one can see that

doubling the distance quarters the power through a fixed area, which

can be expressed as a power (per unit area) reduction of 6 dB or (from

the above equations) a SPL reduction of 6 dBA. Note that for a

circular sound dispersal pattern (a disk), these figures are 3 dB and

3 dbA and for an (imaginary) linear "dispersal" pattern, they are 0 db

and 0 dBA.

I think that the doubling/halfing one's distance from an isotropic

sound source in free air should *define* the meaning of "half/twice as

loud", but it seem that sound experts have agreed that these terms

should be associated with a SPL change of 10 dBA, based (foolishly,

IMO) on what "half/twice as loud" meant to some experimental subjects,

approximately.

Note 1:

I've not seen a good discussion of the reason, but it apparently

stems from the fact that the power of a pressure wave is proportional

to the square of the maximum displacements of the air molecules which

is proportional to the maximum pressure. (I suspect that this is

because an increase in displacement causes an increase in the force

required to compress the air (F = k * D), so the work done, W = F * D

= k * D * D = k * D^2, where k is some constant.)

Note 2:

The "A" in "dBA" implies that the frequency response of the microphone

which measures pressure is roughly that of a human ear, so that

inaudible sounds are ignored, roughly speaking. It should also imply

that the measurement is made 3 feet from the source in a particular

test enclosure per some ANSI standard, but fan makers often use 1

meter from a source suspended by springs in an anechoic chamber

because it gives them better numbers. The 3ft/1m difference is about

0.8 dB and the suspension avoids enclosure vibrations.