[Noisebridge-discuss] Loudspeaker SPL calculation

[jim]

   i don't know either. i've interspersed my "understanding" 
and hope for clarification. nice write-up, by the way. 

On Thu, 2010-07-29 at 15:31 -0700, Michael Shiloh wrote:
> 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.
> --------------------
> Example 2:
> 
> Assume point source and free field, so we know inverse square law applies.
js: "point source" == one speaker in a closed back cab 
js: "free field" == something like the cab is hanging by 
    a thread 50 meters above flat ground that spreads out 
    for a km or so, or maybe it's hanging above the tip 
    of a pointy mountain top. 
> 
> An input of 1 W produces SPL of 115 dB at 1 m. What is the SPL at 6.1 m?
JS: i'm assuming 1W at 1m is a matter of unity-ing out the 
    baseline, i.e. all existing hard-won formulas have 
    been written wrt meters and Watts. 
> 
> Answer: SPL = 115 - 20 log 6.1/1
> ------------------------
> 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.
JS: mainly i'm developing a headache, but subtracting 
    from 115 is because 115 is the SPL where other parms 
    are at unity and subtracting is what you do if distance 
    is greater than unity (while adding is what you do if 
    distance is less than unity) and the thing you subtract 
    is the (20 log(distance_in_meters)/1(SOME_UNIT)) because 
    that's the formula. 
    SO what are the measurement units of 6.1 and 1 (the 
    denominator)? seems something to do with 
    Watts = ((SPL_thingies)(meters)) 
    (Watts/meters) = SPL_thingies 
    TROUBLE: watts is energy per time_unit; how to express 
    energy? joules? BTUs? pounds? and what happens when one 
    divides joules, say, by meters? 
JS: the headache's getting worse. i need a snack. 
> 
> 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
>