If the decibel scale is logarithmic, isn’t 33dB between 100 and 133 a much different amount in reality than 33dB is between 30dB & 63dB? So how can NRR 33 ear plugs reduce 60dB down to 30 while also reducing 140dB down to 110? Wouldn’t it only, say, reduce a 60dB noise by 33dB, but reduce a 140dB noise by a lesser amount? Does the ear plugs’ ability to reduce the sound increase with the volume?
If they really do subtract 33dB across the spectrum of dB ranges, then why are sub-33dB sounds still audible? Shouldn’t sounds around that range be effectively 0dB and inaudible?
If 60dB sounds are, say, 45dB with NRR33 earplugs, then are 100dB sounds actually 85 instead of 67? Or is there some mechanism enabling sounds that should be lowered to sub-40dB to be above 40, while also lowering 130dB to 100?
In: Physics
NRR is measured against a test sound profile, it is impossible to make a hearing protection device that uniformly reduces sound by the same amount across all frequencies and sound pressure levels. The rating applies only to the test profile, but is intended to be representative of protection performance for comparison to other hearing protection devices.
The ear plugs absorb ~99.97% of the sound and let 0.03% or 1/3000 through. If you have a louder sound they absorb more energy, but the fraction stays the same (approximately).
On a logarithmic scale, multiplying the power by a constant is converted to a constant difference in dB.
> then why are sub-33dB sounds still audible?
The fraction of the sound they absorb depends on the frequency. It’s technically 33 dB at some specific frequency and likely less elsewhere.
In a logarithmic scale, ‘subtracting’ from it is *really* just dividing the force.
This works because of how exponents work: 10^(a)/10^b = 10^(a-b). Because a log scale is just looking at the exponents, this means that dividing the *a* number by the *b* number would show up as *a-b* on the scale.
So if something ‘subtracts 30 dB’, then what it is doing is dividing the power of the noise by 10^3 or letting through .1% of the original noise. Blocking a percentage of force like this is pretty simple for most materials to do.
Of course, this only works from a math perspective. Earplugs will *generally* work like that, but because of how different frequencies work it will be more nuanced.
Think of the sound as an energy cannon pointed at your ear.
A 3 dB reduction is like putting up a shield that blocks 50% of the energy from hitting you
A 6 dB reduction is like putting up *two* 3dB shields, one after another, meaning that 50% of 50% (or 0.5 squared) makes it to you, so 25%.
A 33 dB reduction is like putting up 11 of those sheilds, meaning that 0.5 ^ 11, or 1 / 2048, or 0.05% of the energy makes it to you.
Logarithmic means the reduction is multiplicative
>If 60dB sounds are, say, 45dB with NRR33 earplugs, then are 100dB sounds actually 85 instead of 67?
No, dB is already on a log scale, so you subtract. Subtracting on a log scale is like multipling by a number less than one on a linear scale, if you want to see the earplugs take away less energy when the sound beam is weaker (they do), you have to convert dB to [something else like RMS](https://www.everythingrf.com/rf-calculators/dbm-to-vrms-calculator)
edit:
>why are quiet sounds still audible?
Because human hearing is very sensitive (meaning it can detect sounds at low dB) especially at frequencies indicative of human speech. Think of it like the earplugs, rather than mowing all the grass to the same height, just reduces the size of each blade of grass by 99.5%. You can still feel the taller ones if you run your hand over it though, or if you crouch down and look at the profile up close. The louder frequencies still poke out and are intelligible.
[https://en.wikipedia.org/wiki/Absolute_threshold_of_hearing](https://en.wikipedia.org/wiki/Absolute_threshold_of_hearing)
Because as you say it’s logarithmic, and substraction in logarithm equals to linear division.
Most people can hear a little bit below 0dB. But, 33dB is rather quiet, whisper level. Your hearing is limited by the noise, rather than the ear sensitivity.
Part of the reason that you can still hear quiet sounds while wearing earplugs is that earplugs only block sound waves from coming into your ear directly through the air. Just like you’ll still hear any vibrations that can travel through the earplug, you’ll also hear a lot of vibrations that travel through your body. This is the idea behind how bone-conduction headphones work. Also we’re mostly made of water, so it’s easy for a sound from the air to go into the body, much easier than the other way around.
0dB isnt no sound or no volume, 0dB is a factor of 1 compared to a reference value. Negative dB values then represent dampening compared to the reference.
It is exactly because dB are logarithmic that you can just subtract them from each other. That’s one of their main properties.
It’s easiest to think of dB as a RELATIVE measurement. It tells you how much more or less intense (louder) something is compared to something else. So when people give hearing protection a NRR of “30 dB” its saying it’ll make the noise 30 dB less intense. Less intense relative to *what* you may ask? Relative to the noise youre talking about. So if it’s a jet engine taking off (around 140 dB) it’ll reduce it to 110 dB.
But then you might ask “well, if dB is only a measure of increase or decrease in intensity, then what does it mean when someone says a jet engine is 140 dB at takeoff??? 140 dB relative to WHAT?!?!?”. When people talking about noise volumes in term of dB, that is STILL not an absolute measurement; it uses the dB scale so it must be a RELATIVE merasurement. But relative to what? When we’re discussing hearing/volume, the reference level (i.e. 0 dB) is the baseline hearing perception curve. It’s basically the quietest noise the average human ear can detect. Technically, if anyone is listing the volume of a noise and using dB, it should likely be correctly listed as dBA (somedtime dB-A). This denotes that you’re using the “A-Weighting” reference curve as your baseline (this A-weighted curve is that “quietest noise humans can hear”). More info on A-weighting can be found here: [https://en.wikipedia.org/wiki/A-weighting](https://en.wikipedia.org/wiki/A-weighting)
So put it all together. The dB scale is a way to represent RELATIVE levels of intensity (volume). When so hearing protection with an NRR of 30 dB means it lowers the nosie youre hearing by 30 dB. And when people talka bout the loudness of a thing in terms of dB, they usually mean dB-A, which shows how muhc louder something is than the quietest noise we can detect. So if someone says a jet engine is 140 dB-A loud, and you put on earmuffs with a 30 dB NRR, what that means is a jet engine is 140 dB louder than the quiestest noise you can hear, but the headphones reduce that to only 110 dB louder than the quietest noise you cna hear (140 dB – 130 dB = 110 dB).
zero dB does not represent no sound, it means the sound level is equal to the reference value. -10dB, for example, would be at 1/10 of the reference value. Likewise, 10dB would be 10 x the reference value.
decibels are a comparison of two values.
A part of the confusion could be that “dB” itself is not a unit. By this, scientists and engineers designate that a number (or more precisely a factor such as 0.5 in 0.5*x) is not just the raw number, but a number on a logarithmic scale. Others here have explained the math behind it, but basically what happens is that instead of multiplying (e.g. 0.5) with a number, you convert both numbers into the logarithmic scale and then calculate the sum. (This is a specific characteristic of this scale)
The loudness of the sound instead is given not as dB, but as dBa. You know that sound is basically just a pressure wave of air. dBa is the multiplier that you have if you measure the pressure in the wave and compare it to the atmospheric pressure.
So (1) there are no sounds of e.g. “3dB”, but of “3dBa” which means that the pressure is 2*[atmospheric pressure] and (2) as adding and subtracting dB is just multiplying by a number, you can only dampen the sound (i.e. have “a quarter of the loudness”).
The logical question to ask would be why it is not possible to multiply by zero. That would then turn off all sound. This is a bit more complicated and can only be answered with exceeding ELI5 even more. Just so much: There is a mathematics of how systems (e.g. speakers, but also many others) behave if there is an input (e.g. a sound from outside) and an output (the sound level on ear-side). This math together with physics does not allow this.
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