[Noisebridge-discuss] Loudspeaker SPL calculation

[Josh Myer]

On Thu, Jul 29, 2010 at 5:46 PM, Michael Shiloh
<michaelshiloh1010 at gmail.com> wrote:
> 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.
>

You're totally right.  And the answer to how to do this more
efficiently is: correct one gotcha, then the wonder of log identities.

Pressure is power over area.  The area of a 1m sphere is 4*pi*1^2, the
area at 6.1m is 4*pi*6.1^2.  The ratio of the power-per-unit area at
1m vs 6.1m is thus

4*pi*1^2/(4*pi*6.1^2) = (1/6.1)^2

If you solve for p_rms at 1m, then scale it down by the above factor
to get p_rms at 6.1m, and solve for SPL at 6.1m, you'll find the right
answer.

Then, ask yourself: what's the difference between the two pressures?
It's effectively division by 6.1^2, right?

BUT FIRST!  A quick review of some handy log identities:

log (a/b) = log(a) - log(b)
log(a^b) = b*log(a)

And let's apply those to a simple pattern, so we can see how they'll
be applicable here.

log(x*(1/6.1)^2) =  log(x) + log( (1/6.1)^2) = log(x) + 2*log(1/6.1) =
log(x) - 2*log(6.1/1)



AND WE'RE BACK: So, back to the question, can we be lazier in the
computation here?

Our original equation:

115 = 20 log(p_rms/p_ref)

And our pressure at 6.1m is going to be p_rms*(1/6.1)^2, so we jump to
that equation, with our new SPL an unknown:

L = 20 log(p_rms_at_6.1m/p_ref) = 20 log(p_rms/p_ref*(1/6.1)^2)

And now we bring it home with the log identities:

L = 20 log (p_rms/p_ref) - 2*log(6.1/1)

But!  115 = 20 log(p_rms/p_ref), so we can substitute that in and...

L = 115 - 2*log(6.1/1)

et voila.  Email is crap for formatting maths, so you may want to copy
the above down to something more legible.
--
/jbm