[Noisebridge-discuss] Loudspeaker SPL calculation

[Michael Shiloh]

By the way, I think the 1 Watt is a red herring and has nothing to do 
with the calculation. There is a part B that changes the wattage, but 
I'm not even sure it was needed there.



On 07/29/2010 05:43 PM, jim wrote:
>
>     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
>>
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