3 ms·

Wow, I had no idea they were that bad. 22NRR for guns at close range is...not much. Most range shooters will be using at least 30+ (which is WAY MORE than ~40%

by TylerE 3y ago

Wow, I had no idea they were that bad. 22NRR for guns at close range is...not much. Most range shooters will be using at least 30+ (which is WAY MORE than ~40% effective - decibels are logarithmic). Some will even wear 30+ plugs under 35+ muffs. (Which don't add, btw, again, logarithms, but doubling up does get you a few extra decibels reduction over either component alone.

mkmk 3y ago
Tradeoff with retaining ability to hear commands/maintain situational awareness on the battlefield, perhaps?
kazinator 3y ago
Yes, it should add! If we cascade a 30 dB reduction with another 30 dB reduction, we should get 60 dB reduction. The only reason you likely won't get anywhere near 60 dB from a combination of plugs and muffs has nothing to do with logarithms, but with physics/topology. Namely, the reason why these devices have the rating they do and not something better is not only that sounds goes through them but that it goes around them, e.g. through your skull or Eustachian tube or whatever. When signal is leaking around the attenuators, then their decibel reductions cannot simply add. Even if the attenuators completely block signal, they will be bypassed, and so the best decibel reduction is limited by the floor of the bypassing signal. But decibels definitely add. E.g. in an electronic circuit where we have vanishingly low leakage around our attenuation stages, if we cascade a -6 dB drop with a -12 dB drop we will get -18 dB. Now suppose we have a parasitic leakage from input to output so that even if we disconnect our circuits signal path, some input still reaches the output: say -100 dB. It might be going through ground wires, or inductive or capacitive cross-talk between circuit traces or whatever. That then sets the upper bound on our attenuation; we will not be able to attenuate better than -100 dB, because then the leakage dominates. I mentioned the Eustachian tube. If your ears are stuffed to heck and you have muffs over that, and then yawn with an open mouth, you may suddenly hear a lot more noise. Oops!
- TylerE 3y ago
  No, because decibles are logarithmic. 30db + 30db = 33db https://www.noisemeters.com/apps/db-calculator/ https://www.noisemeters.com/apps/db-calculator/ With hearing protection the ratings are in isolation. They are not circuit components that work serially. Sound transmits through the skull physically.
  - kazinator 3y ago
    I'm afraid are seriously misunderstanding what that calculator is for. In this calculator, whenever you add any equal decibel values X, you get X + X = X + 3: three more decibels. There is a familiar meaning behind that: it means that if we have two independent noise sources of about equal intensity, and add them together, the result is 3dB more intense. E.g. if we have a loudspeaker playing white noise, and add an identical loudspeaker playing the same kind of of white noise at the same level, we get 3dB more white noise. It has nothing to do with cascaded boosts or attenuations. If we add together two sources that contain the same signal, in phase, then the rule is 6 dB. Like two identical loudspeakers in a cabinet instead of one, playing exactly the same signal. The reason we can add decibels for cascaded attenuations and boosts is exactly the same reason why we can add displacements on a slide rule to do multiplication. Attenuations (such as a voltage divider) multiply the signal. E.g. multiplying by 1/2 is -6db. In the logarithmic space, multiplication turns to addition. This is typically covered in high school. > They are not circuit components that work serially. That's true, but I addressed in it my comment (with several redundant remarks) in such a way that the above remark doesn't add more clarity. I have a sneaking suspicion you might not have read it in detail.
    - javajosh 3y ago
      Perhaps it would be helpful to consider the path of a phonon going through a barrier. The odds of that phonon making it through is 1 chance in N. If there are two barriers, the odds of a phonon making it through both in series is 1 chance in N^2. That is, the odds multiply, and you are absolutely correct that dB's add (since they are basically log(N) in this case and log(N^2) = 2log(N)).

by TylerE 3y ago

mkmk 3y ago

Tradeoff with retaining ability to hear commands/maintain situational awareness on the battlefield, perhaps?

kazinator 3y ago

Yes, it should add! If we cascade a 30 dB reduction with another 30 dB reduction, we should get 60 dB reduction. The only reason you likely won't get anywhere near 60 dB from a combination of plugs and muffs has nothing to do with logarithms, but with physics/topology. Namely, the reason why these devices have the rating they do and not something better is not only that sounds goes through them but that it goes around them, e.g. through your skull or Eustachian tube or whatever. When signal is leaking around the attenuators, then their decibel reductions cannot simply add. Even if the attenuators completely block signal, they will be bypassed, and so the best decibel reduction is limited by the floor of the bypassing signal. But decibels definitely add. E.g. in an electronic circuit where we have vanishingly low leakage around our attenuation stages, if we cascade a -6 dB drop with a -12 dB drop we will get -18 dB. Now suppose we have a parasitic leakage from input to output so that even if we disconnect our circuits signal path, some input still reaches the output: say -100 dB. It might be going through ground wires, or inductive or capacitive cross-talk between circuit traces or whatever. That then sets the upper bound on our attenuation; we will not be able to attenuate better than -100 dB, because then the leakage dominates. I mentioned the Eustachian tube. If your ears are stuffed to heck and you have muffs over that, and then yawn with an open mouth, you may suddenly hear a lot more noise. Oops!

TylerE 3y ago
No, because decibles are logarithmic. 30db + 30db = 33db https://www.noisemeters.com/apps/db-calculator/ https://www.noisemeters.com/apps/db-calculator/ With hearing protection the ratings are in isolation. They are not circuit components that work serially. Sound transmits through the skull physically.
- kazinator 3y ago
  I'm afraid are seriously misunderstanding what that calculator is for. In this calculator, whenever you add any equal decibel values X, you get X + X = X + 3: three more decibels. There is a familiar meaning behind that: it means that if we have two independent noise sources of about equal intensity, and add them together, the result is 3dB more intense. E.g. if we have a loudspeaker playing white noise, and add an identical loudspeaker playing the same kind of of white noise at the same level, we get 3dB more white noise. It has nothing to do with cascaded boosts or attenuations. If we add together two sources that contain the same signal, in phase, then the rule is 6 dB. Like two identical loudspeakers in a cabinet instead of one, playing exactly the same signal. The reason we can add decibels for cascaded attenuations and boosts is exactly the same reason why we can add displacements on a slide rule to do multiplication. Attenuations (such as a voltage divider) multiply the signal. E.g. multiplying by 1/2 is -6db. In the logarithmic space, multiplication turns to addition. This is typically covered in high school. > They are not circuit components that work serially. That's true, but I addressed in it my comment (with several redundant remarks) in such a way that the above remark doesn't add more clarity. I have a sneaking suspicion you might not have read it in detail.
  - javajosh 3y ago
    Perhaps it would be helpful to consider the path of a phonon going through a barrier. The odds of that phonon making it through is 1 chance in N. If there are two barriers, the odds of a phonon making it through both in series is 1 chance in N^2. That is, the odds multiply, and you are absolutely correct that dB's add (since they are basically log(N) in this case and log(N^2) = 2log(N)).