[Noisebridge-discuss] Loudspeaker SPL calculation

[Michael Shiloh]

I feel a little silly, because I'm sure the answer is simple and I'm 
just blind to it.



Master Handbook of Acoustics, 5th Edition, Everest and Pohlmann.
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Example 2:

Assume point source and free field, so we know inverse square law applies.

An input of 1 W produces SPL of 115 dB at 1 m. What is the SPL at 6.1 m?

Answer: SPL = 115 - 20 log 6.1/1
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I think I understand intuitively that the ratio of the distances, 
squared (hence the 20 in front of the log), represents the difference in 
pressure, but I don't understand how they got here. Or why they subtract 
from 115 to get the answer.

I do understand that the the exact relationship doesn't matter. What 
matters is how the pressure changes as we change distance.

We know that SPL = 20 log (pressure/(reference pressure, 20 uPa))

I could solve the above for the pressure at 1 m, since we know the SPL 
is 115, and then calculate the pressure drop when we move to 6.1 m 
(inverse square for intensity is inverse distance for pressure, so the 
pressure drops to 1/6.1 of the original) but I don't think that's what 
they are doing.

How did they avoid doing the tedious calculation I see?

Michael