20dB fan X 2 = 40dB?

Noise is not a linear function. So overlaying two does not simply adds.

I've always heard a doubling of sound is +3db, in your case a total of 23 db

Yep, dB is a measure in 10*log10(sound energy), so you need 10 20 dB fans to get to 30 dB, or 2 will get you 23 dB. Also why going from 10 watts/channel to 100 watts/channel only gets you 10 dB more sound.

Actually I think it is 10*log(SPL2^2/SPL1^2), which by definition is 20*log(SPL2/SPL1).

So in this case it is 20*log(40/20) or 20*log(2) = 6dB increase or 26dB.

Or is my math wrong?

Guess my math is correct?
http://www.engineeringtoolbox.com/sound-pressure-d_711.html

I second that!

Sent from my HTC Sensation using Tapatalk 2

I think we're talking about 2 different things, power and pressure. Doubling the power from an amp is a 3 dB increase, but doubling the pressure is a 6 dB increase. THIS other page from engineeringtoolbox illustrates the difference. In any case, even two 20 dB fans are still going to be relatively quiet.

I also think that different concepts are being mixed here. In an audio system, doubling the perceived volume requires a 10db increase which requires a 10-fold increase in amplifier power. Which is a different thing than how much perceived volume increases through the addition of two independent sound sources, an answer to which I admit I do not know.

Yeah. We're not doubling the power as in a speaker but doubling the pressure which results in a 6dB increase in sound pressure level for every doubling of sound pressure.

Sound pressure is what you sense as it is a change in pressure over a reference level that is felt by your eardrums. Sound power isn't felt by your eardrums.

If the ambient noise in the room is 20dB and you have one fan producing 20dB then the SPL is 26dB. Adding another 20dB fan will result in an additional ~5dB increase in SPL so now the total SPL is 31dB or 11dB over the reference level of 20dB. Of course these are measured at 1meter so across the room on your couch it isn't really heard at all. The only time I can hear my HTPC is when the AVR is muted. Other than that, I never hear it becuase the ambient SPL is high enough that any additional sound pressure from the HTPC is negligable.
Edited by Sammy2 - 8/16/12 at 8:45am

You second that my math is wrong?

I did not attend engineering school for 6 years and spend the last 20+ years practicing engineering wiith bad math. Did I?

 

Does SPL translate directly into perceived volume? I'm not so sure.

How far away is "just over head"? A meter or more? I still think that it won't be heard over the reference level of your AVR and speakers. If you have audessey, run it with the fans and projector (has its own fan?) on for best results.

I don't think it does either. Also, most of what Sammy2 wrote is incorrect. Sound is a wave. You cannot just add sound pressure levels. You need to consider frequency and phase as well. And phase depends sensitively on position of the wave sources and the observation point.

But as a very rough approximation, if you want to add waves that don't have exactly the same frequency (there will be slight variations in fan frequency even when driven by a good controller), then you just add the power. So two fans produce about twice the sound power of one fan, and +3dB since we are looking at double the power (not double the pressure level).

I'm not adding pressure levels and of course this is at a common frequency which I assume two matched fans would operate at within a few percentage points. And of course the frequency is important. This is why audessey should be run with the fans on so it can correct for that frequency and pressure level that is now part of the ambient noise in the room. The distance to the source of the sound pressure is important too. I understand this. We are talking about 1meter which is where standard measurements are taken. The farther away the lower the "perceived" sound.

You hear sound pressure. We are not as interested in sound power level because it is not what we hear.

http://www.usmotors.com/TechDocs/ProFacts/Sound-Power-Pressure.aspx
Edited by Sammy2 - 8/16/12 at 10:28am

Unfortunately, you do not understand, because you are continuing to post incorrect information.

Decibels are defined as 10 log(Power ratio). Since sound power is proportional to the square of the pressure level, you can also write it as 20 log(level ratio).

However, YOU CANNOT JUST ADD WAVE AMPLITUDES AS SCALARS. Waves have amplitude, frequency, and phase. If you look at the superposition of two pressure waves of the same frequency and amplitude, then the pressure level of the superposition depends sensitively on position (look up wave interference for the details). The superposition is NOT just a wave with twice the amplitude. In some positions, the two waves will cancel out and result in zero power, and in other positions, the waves will constructively interfere and result in four times the power.

This is all fundamental physics of waves and is covered by any Physics 101 type course.

Now, as to why two 20dB fans will add to about 23dB (rather than 26dB as you incorrectly stated), it has to do with frequency variation and the definition of decibels.

As I already wrote, two fans will not have exactly the same frequency, even if they are both controlled by a good controller. No one tries to make fan controllers with extremly high frequency accuracy, and even if they did, the random variations in the fans motors and bearings, as well as the chaotic air flow, will usually result in pressure waves that have enough variation in frequency that you don't get much wave interference. If the fans were producing identical frequency pressure waves, you would be able to move your microphone around and find positions where there was no sound at all, and other positions where the sound power is 4 times that of a single fan alone. I don't know about you, but I've never noticed that effect with two fans. Two acoustic speakers, definitely. But two fans, no.

Anyway, the interference (or lack thereof) of the pressure waves produced by the fans is mostly irrelevant to the decibel question. As I already stated, decibels are defined as 10 log(Power ratio). Double the power gives a +3dB increase in decibels. And two fans create double the sound pressure power of one fan. So two 20dB fans will result in 23dB of noise.

I understand everything you state about waves, amplitude, frequency and phase. I'm talking in platitudes and making assumptions, as engineers always have to do, that these are equal for both fans which they may very well not be. Engineers make basic assumptions to avoid analysis paralysis otherwise we can get lost in endless computations and nothing practical gets accomplished. Maybe I over simplified. For example if these two fans were 180 degrees out of phase with each other they would cancel each other out assuming the frequencies were the same. Amplitude is, in a sense, related to the sound power but not the sound pressure so again that is assumed to be equal for each fan.

What I'm not understanding is why you are talking about sound power when we sense sound pressure and not power because it is the changes in pressure of the atmosphere due to the sound source is what moves our eardrums and is in turn interpreted by our brains as sound. A change in sound pressure is expressed as I indicated above. Changes in sound power do not square the terms. It is this square in terms that makes the difference in sound pressure two times the difference in sound power.

You clearly do not understand what I was explaining. I don't know how to explain it any differently. The noise I hear from two fans does NOT have significant constructive or destructive interference, since the sound does not vary significantly as I change my ear position (and therefore phase) relative to two fans. The reason is that the frequencies of the two fans are not exactly the same (or more technically, there is significant phase noise). But all of that was simply a sidebar (to explain why dB is the ratio of powers, not levels) that I only brought up to try to help you understand why you were incorrect about two 20dB fans adding to 26dB. The correct sum is 23dB. Obviously the explanation did not help.

So I'll just make it simple so I stop wasting both of our time.

Use the formula for adding multiple incoherent radiating sound sources:

http://en.wikipedia.org/wiki/Sound_pressure#Multiple_sources

Lsum = 10 x log10( 10^(20/10) + 10^(20/10) ) = 10 x log10( 100 + 100 ) = 23dB

You clearly are condensending.

Not bad math, just calculating the wrong thing. We are interested in doubling the sound energy [10log(10^(L1/10)+10^(L2/10))] not the pressure.

I have to say Jim2100, and lespurgeon are correct. You want to measure the sound energy not pressure. Sure fire way to check is to go to Radio Shack and pick up a cheap dB meter and measure 1, then 2 fans. You should see a +3dB increase in sound level.

As a low whisper in a library is 30 dB, can we all agree that either 23 dB or 26 dB are BOTH going to be rather quite in the application raised by the OP?

I do get the desire to be accurate in Engineering. What I'm suggesting to our little family here is that the "noise is drowning the signal"

Peace, y'all

What is the signal and what is the noise in your metaphor?

Yup. Thinking about this it is the energy in the sound wave that we hear. It is always about energy.
Edited by Sammy2 - 8/18/12 at 6:46am

Why California Is Broke LOL

Reference to Enron? It is the same $15,000,000,000.00 that they stole from the CA taxpayers in the "deregulation of energy" scam in CA that has been dragging the CA budget for 10 years now.

So all this math aside, can someone recommend a really quiet table/desk fan I can just put behind all my A/V equipment to improve air circulation?