If I have four computer fans of which each is said to run at 46 dB, and they run close to each other, how loud is the whole system?

I somehow recall from my physics course that 10 decibel more means twice the noise, is that right?

So would it be 66 dB for four such fans then?

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    $\begingroup$ 1st: 3 dBs are double, approximatly, and 2nd that would make about 52 dB, but only if those sources are totally uncorrelated. $\endgroup$ – Georg Dec 2 '11 at 13:09
  • $\begingroup$ if they are correlated, would they be louder or less loud? $\endgroup$ – Mat Dec 2 '11 at 14:16
  • $\begingroup$ That depends on kind of correlation, of course. Not easy to calculate unless some very simple cases. Explain kind of those noise sources, maybe someone here underatands the math. $\endgroup$ – Georg Dec 2 '11 at 14:43
  • $\begingroup$ See also physics.stackexchange.com/questions/9017/… $\endgroup$ – Cees Timmerman Dec 12 '13 at 10:25

First, dB means nothing by itself. You need to give a reference level, like dBW or dB SPL. We'll assume dB SPL.

Second, noise measurements from a point source like this require a distance measurement to be meaningful, since the level drops off with distance. We'll assume you're measuring at the same distance in both cases, and the fans are equidistant from the SPL meter.

Fans have some noise that is correlated between them (hum), and some that is uncorrelated (whoosh). The two types of noise combine differently. Without knowing what the noise is, it's impossible to know.

I think the sums would be:

  • correlated: $10 \log_{10}\left(\frac{{(4 p_{\mathrm{{rms}}}})^2}{{p_{\mathrm{ref}}}^2}\right)\mbox{ dB}$
  • uncorrelated: $10 \log_{10}\left(4\frac{{p_{\mathrm{{rms}}}}^2}{{p_{\mathrm{ref}}}^2}\right)\mbox{ dB}$
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    $\begingroup$ It is impossible for two fans to cancel out each other's noise realistically, since it requires coherence between the two fans which means that one fan is absorbing the noise energy of the other. The noise energy goes radially outward, and you can't absorb it all unless one fan is a sphere surrounding the other, and all the spherical fan noise is directed inward. $\endgroup$ – Ron Maimon Dec 3 '11 at 6:31
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    $\begingroup$ @RonMaimon: It's possible for them to cancel completely at the point that the SPL meter is measuring, though. $\endgroup$ – endolith Dec 3 '11 at 6:57
  • $\begingroup$ Ah yes, you're right. $\endgroup$ – Ron Maimon Dec 3 '11 at 7:27
  • $\begingroup$ @RonMaimon: Even if you had a point source and a spherical shell source, they would not cancel in the region between, would they? They would form a standing wave as the waves travel in opposite directions. $\endgroup$ – endolith Dec 3 '11 at 18:04
  • $\begingroup$ yes, I meant cancel in the exterior. $\endgroup$ – Ron Maimon Dec 3 '11 at 20:16

10 decibels means ten times the power density of sound. That is chosen so that ten decibels equals one bel, and a one bel increase simply adds a zero on the end of the sound power density. Twenty decibels multiplies by 10 twice, so twenty decibels is an overall 100-fold increase in the sound.

That means one decibel multiplies the sound by the smaller amount $10^{.1} = 1.259$.

For example, a two decibel increase multiplies the sound by $1.259^2 = 1.585$. If you keep going up to ten decibels, we recover $1.259^{10} = 10$.

With four fans, the power they put into sound is four times as much. If you work it out, you'll find that six decibels gives a noise energy increase of $1.259^6 = 3.98$. Four fans put out a sound 6 decibels higher than one fan, so you're left with 46+6 = 52 decibels.

If you want to calculate it for an arbitrary number of fans $n$, you would need to solve

$$1.259^x = n$$

or, leaving it in terms of a power of ten,

$$(10^{.1})^x = 10^{.1x} = n$$

This is solved by a logarithm.

$$.1x = \log_{10}n$$

$$x = 10 \log_{10}n$$

For example, if you had 1000 fans, you'd have $n=1000$. The logarithm base ten of 1000 is 3, so your sound level would be 30 decibels greater for 1000 fans than for one fan.

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    $\begingroup$ Why the downvote? Everything here looks correct to me...! $\endgroup$ – Daniel Chisholm Dec 2 '11 at 22:54

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