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
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.
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