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where are the limitations of the distance law? Can I calculate a sound from the free field back to the near field?

For example: If I have an impulse sound which is 80 dB loud at 2m. Is it than 118 dB in 0.025m (calculated)? If this calculation is not possible, how much decibel would it be instead approximately?

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Sound sources are generally very complicated, and generate both evanescent fields (which decay exponentially with distance) and propagating fields. Far from the source the evanescent fields are completely negligible and the complicated diffraction patterns of the propagating field have simplified dramatically. It is only in this "far field" that any distance law can be valid.

If you are trying to determine the sound level at a point closer to the source but still in the far field, then yes you can use the further data to inform the nearer. However, if you are trying to reconstruct the near field (where the pressure field is complicated), no you cannot completely reconstruct it. The best you can do is to reconstruct the propagating portion, which is a non-trivial inverse problem (e.g., look up acoustic hologrophy).

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  • $\begingroup$ Ok. Thank you. The noise was calculated for measuring the noise from a normal slap on the face in combat sports (it was done on a dummy, not on a human). Are the 118 dB valid for this? If not, how much decibel are approximately not considered by this calculation (just an estimation). $\endgroup$
    – J. Scott
    Commented Mar 2, 2018 at 20:15
  • $\begingroup$ I'm confused about where you got the 118 dB estimate from, so I don't know if it is reasonable or not. The distance law I recall (still not valid here) is 3 dB down per doubling of distance, which would yield $80+3\log_2(2/0.025)\approx99dB$. What distance law are you referring to? $\endgroup$
    – Michael M
    Commented Mar 2, 2018 at 20:25
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    $\begingroup$ It depends on how accurate you want to be. If you are OK with a $\pm5$ dB uncertainty, then sure, go with 118 dB (you will probably be a little high in your estimate). If you want to be really accurate, you need to know a lot more about the problem. $\endgroup$
    – Michael M
    Commented Mar 2, 2018 at 20:34
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    $\begingroup$ Two points. (1) The far field is an average of the near field details, and so parts will be louder than the far field will suggest and parts will be quieter. Thus, as mentioned in the post you sent, the levels will vary widely. The graph you showed is just a schematic and should not be used for calculations. Also, I pulled the $\pm5$ dB out of my head, trying to give you an idea of what is happening - I don't know what the uncertainty really is. $\endgroup$
    – Michael M
    Commented Mar 5, 2018 at 12:44
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    $\begingroup$ (2) The inverse square law (6 dB per doubling of distance) assumes a point source. This is never truly the case. The source is always part of a distribution that is averaged out. If you were to take the inverse square law literally, then as you get closer to the source the level would go to infinity, which is not physical. Thus, at some distance the inverse square law will provide an over-estimate of the level, which is why I made my previous comment. $\endgroup$
    – Michael M
    Commented Mar 5, 2018 at 12:46

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