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I read on Wikipedia:

Using both ear muffs (whether passive or active) and earplugs simultaneously results in maximum protection, but the efficacy of such combined protection relative to preventing permanent ear damage is inconclusive, with evidence indicating that a combined noise reduction ratio (NRR) of only 36 dB (C-weighted) is the maximum possible using ear muffs and earplugs simultaneously, equating to only a 36 - 7 = 29 dB(A) protection.[23]

The reference [23] (mirror) says:

For frequent shooting or even occasional high velocity rifle work, we suggest more protection. Your safest choice is a combination of plugs plus a muff worn over them. In combination, a good estimate of the maximum protection provided can be computed by adding 6 to the plug's rating: for example, an Ultimate 10 muff at 30 NRR, plus a MAX foam plug, will give you at least 36 NRR. Additionally, using this muffs-and-plugs system, you will have about as much protection as it is possible to get in a portable hearing protection device--at any price.

Why can't wearing both ear muffs (whether passive or active) and earplugs simultaneously result in more than 36 noise reduction ratio (NRR) at best?

I would have thought that wearing a custom-fitted earplugs with 30 dB NRR in addition to an earmuff with 33 dB NRR would result in a protection higher than 36 dB NRR.

11

Small disclaimer about decibels

For sound pressures, decibels are defined as follow:

XdB = 20 log (p1/p0)

with p1 being the amplitude of the pressure field of the sound, and p0 a reference (20microPascal of pressure).

This translates in terms of power/loudness as:

Pow_dB = 10 log(P1/P0)

Because of this definition, you cannot simply add sound powers:

10 log((P1+P2)/P0) different from 10 log (P1/P0) + 10 log (P2/P0) 

However thanks to the first formula here they will add up when using decibels for gain or attenuation: let's start with a sound pressure P1, and let's apply a two times gain. The resulting pressure is 2*P1. In dB:

X = 10 log(2*P1/P0) = 10 log(2) + 10 log(P1/P0) approx 10 log(P1/P0) + 3dB

It we multiply this last sound level by 2 another time, it will finally ends up that we added 3dB again, resulting in a final value of

 X = 10 log(4*P1/P0) = 10 log(4) + 10 log(P1/P0) approx 10 log(P1/P0) + 3dB + 3dB = 10 log(P1/P0) + 6dB 

Obviously this does work the other way around. Let's divide our sound by two:

X = 10 log(P1/(2*P0) = 10 log(1/2) + 10 log(P1/P0) approx 10 log(P1/P0) - 3dB

Dividing the power is equivalent to subtracting dB.

That is neat, because, it is much simpler for us human behind a mixer, to think in terms of addition or subtraction when behind a console.

About your problem

According to what I said, actually, you should be able to add up the dB reduction of protection…

But: these protection are protecting your ears, and, unfortunately, sound also travels into flesh. Let's have a look at the transmission coefficient of sound:

T = 2 Z2 / (Z1+Z2)

Where Z are the acoustic impedance of our two medias (let's say, air and water, mainly like flesh).

One can calculate these two impedances:

Zwater = 1.5 10^6 Pa s/m
Zair = 408 Pa s/m

Resulting in:

 T = 2Zair/(Zair + Zwater) = 0.000544

Let's transform that in terms of dB attenuation:

10 log (2*Z1/(Z1+Z2)) = -32.64 dB

You will note that this is not far for -36dB, especially considering that one's head is only composed of water! Things will be a bit different when considering the skull and so, but the general picture is here.

This is why an ear protection cannot reduce more than that: it does not prevent the sound to reach your inner ear from the flesh, or your skull…

Q.E.D.

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    Would this mean that you'd be able to get more than 36 dB of hearing protection by wearing a full-body anti-sound suit? – nick012000 Jul 6 at 9:02
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    @nick012000 Very likely. Not sure how practical this would be… – Tom Jul 6 at 9:47
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    @nick012000 Presumably anechoic chambers block sound by more than 36 dB, and they are full-body anti-sound suits, in cube shapes with lots of extra space. – user253751 Jul 6 at 13:31
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    Wikipedia says (quoting an ISO standard) that sound pressure level is defined as 20 times the log of a pressure ratio, and sound power level is defined as 10 times the log of a power ratio. Taking the log of a pressure ratio and then multiplying it by 10 is incorrect; it won't give you the sound pressure level. – Tanner Swett Jul 6 at 14:27
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    @nick012000: for certain values of "full-body anti-sound suit". You will get better than -36dB if you can surround yourself with a layer of hard enough vacuum, perhaps with mag-lev suspension or soft rubber soles! More realistically, some kind of helmet could perhaps do better than earmuffs and plugs. – Steve Jessop Jul 7 at 1:39
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In addition to what @Tom_C said, another reason is that both earplugs and earmuffs are very good at blocking high-frequency sounds, but very poor at blocking low frequency ones.

Presumably the "30dB reduction" assumes the sound is equally loud across some standard range of frequencies, but while wearing both pieces of gear, the "input" to the earplugs will consist largely of the low frequencies that the earmuffs weren't able to block.

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  • Fair point, but they attenuate low frequencies still quite well according to these graphs. – Franck Dernoncourt Jul 6 at 18:20
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    @FranckDernoncourt: Hmm, that graphic ends an order of magnitude above the lowest audible frequency. According to this report (pg 14), it varies wildly by brand, with some fairly expensive brands holding steady across the spectrum, while others behave as I described. – BlueRaja - Danny Pflughoeft Jul 6 at 18:51
  • Thanks, this is a very interesting document (mirror: archive.org/details/…)! The difference of attenuation between 63 dB and 250 dB isn't that huge on pages 14 and 15 though, but point taken and definitely good to be aware of. I wish more earplugs had measurements for 63 dB (and below). I have some customized ER25 (typo in the price in their table btw, should be ~150 EUR). I'd be curious to see a similar table for customized protectors for sleep or construction environments. – Franck Dernoncourt Jul 6 at 20:09
  • @frankdernoncourt I think you meant Hz and not dB ;) – Tom Jul 6 at 21:42
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Seems like a point of confusion on this question is actually the physics, not the math.

A simple experiment you can try at home that explains why earplugs + earmuffs doesn't double the dB of attenuation:

You need two powered speakers and a dB meter. Create a mono track in your favorite DAW of some sound source, pink noise would be perfect. Or if you have a noise generator, use that. Play the noise mixed equally through both powered speakers. Turn off one speaker and adjust the other one so that you have a clear reading in the dB meter - something like 70 dB would be good. Then turn off the speaker you just adjusted, turn the other one on, and adjust it to match the first (it helps to have a tripod to mount the dB meter so that it doesn't move between readings).

Now you should have two speakers each calibrated to play correlated white noise at 70 dB. When you turn on both speakers, the resulting SPL is not 140 dB, which is a damn good thing because that would damage your hearing. The SPL meter should show something close to 76 dB. In practice, it will probably be lower than 76 dB because even though the signal being fed to both speakers is correlated, the resulting sound arriving at the dB meter from each speaker is no longer correlated because of different path lengths and reflections from room surfaces.

If you want to explore further, calibrate both speakers to play the noise at 60 dB (instead of 70). Now when they are both on, you should get a level of approximately 66 dB. Doubling the amount of sound is an increase of 6 dB. When you turn off one of the speakers, the level drops by 6 dB. Halving the amount of sound is a drop of 6 dB.

It's not a perfect analogy, but the physics and math match up with NRRs. If you have 30 NRR earplugs and you add 30 NRR earmuffs, you get an extra 6 dB of reduction for a total of 36. If you instead had 20 NRR earplugs and added 20 NRR earmuffs, you would end up with an NRR of 26. If the two numbers don't match, then it gets a little messier with the math. The point is, you can't add NRRs, because that is not acoustically analogous to a 30 dB reduction followed by another 30 dB reduction. The same way adding a second speaker playing the same sound is not acoustically adding 70 dB of gain to another 70 dB of gain.


A simpler explanation:

Your intuition that starting with 30 dB of reduction and providing an additional 30 dB of reduction is doubling the reduction. That is correct. What's confusing is that when you double the reduction, you increase that reduction by 6 dB. The reason why 2x = 6 dB is what I wrote below:


This cuts right to heart of what decibels actually are. One way to think about them is as ratios. Another way, which helps us understand why they don't add or subtract linearly, is to think of them as exponents.

For example, 22 + 22 does not equal 24, it equals 23. So if we ignore the common base of all those exponents, it appears that 2 + 2 = 3! Also note that:

  • 23 + 23 = 24
  • 24 + 24 = 25
  • 25 + 25 = 26
  • etc.

A decibel is an exponent with the base removed, so combining two different decibel values does not result in a new value that is the sum of the two values. When we add an exponent of 2 to itself, the value we get is the exponent plus 1 of 2. So in "exponent of 2" math:

x (+) x := x + 1

Where (+) is the special "adding exponents of 2" operation and := is a special kind of "equals" for the operation.

Just like we can make that special "exponents of 2" math, there is special math for "adding" decibel values. In decibel math:

x (+) x := x + 6

(Experts in decibels will want to point that that this is not always true - there are actually two different kinds of decibels that have slightly different math for each of them. For this answer, let's just focus on the kind of decibel that obeys the above "formula".)

In other words, if we have a 30 dB sound and we add another 30 dB sound to it, the resulting sound level will be 36 dB, not 60 dB.

The same relationship holds in reverse for negative decibels. If we have 30 dB of sound reduction, and we add another 30 dB of sound reduction, the total reduction is 36 dB.

The reason why 36 dB is a soft limit on portable sound reduction is because you can't keep adding on more and bigger earmuffs.

You might be thinking, "Hey Todd, you said 30 dB plus 30 dB equals 36 dB, but what about 30 dB plus 33 db?" The answer is that those 3 extra dB aren't that much when combined with the other 30 dB, and we can easily round the total down a little to 36 dB.

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    Thanks for the great explanation, makes sense! (I wonder if it'd be preferable to replace "2 + 2 = 3!" by "2 + 2 = 3." so that people don't read the exclamation mark as a factorial.) – Franck Dernoncourt Jul 6 at 5:21
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    This is explanation is incorrect. Adding 30dB soundwave to another 30dB soundwave of the same frequency can indeed result in 36 dB soundwave (or anything between 0-36, depending on the relative phase of the waves). When talking about sound pressure the x dB means "10^x louder then reference loudness". Sound pressure is added together, which is not easily represented with the exponents. – ja2142 Jul 6 at 8:40
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    However using earplugs with attenuation of 20dB and 10dB ear muffs does yield a theoretical attenuation of 30db. Here x db of attenuation means "sound that comes out will be 10^x times less loud then the ones that comes in". When chaining attenuators one would have to multiply the ratios of attenuation to get the total. 10^x*10^y=10^(x+y). As ratios are multiplied, exponents are added. Check out the answer from @Tom_C - it's probably correct explanation. – ja2142 Jul 6 at 8:40
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    Why 6? Shouldn't it be 3? (Because 10^0.3 is approximately 2, as 10^3 is approximately 2^10.) (Added: Oh I remember that someone once explained to me that there is sometimes a square involved somewhere.) – Carsten S Jul 6 at 11:21
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    @ToddWilcox But earmuffs and plugs are chained attenuators.The sound between earmuffs and earplugs is only attenuated by the earmuffs. So as an 'input' to the earplugs you take 'output' of the earmuffs. It is very much like the case of two walls in succession - the attenuation multiplies (so adds in dB). As I stated, I do agree with two speakers not working like that, but here both pieces of hearing protection attenuate sound, they don't produce it. – ja2142 Jul 6 at 16:35

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