SPL vs. Frequency vs driver movement

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Hi folks,

I'm looking for a formula that can help me calculate the max SPL output for a given size cone with defined X-max (excursion?) at a set frequency.

My point is, I need to see how big a driver I need to obtain a given SPL at 200 Hz with +/- 3.5 mm movement.

Can anyone help?

Jennice
 
p=Sd*Xmax*f^2*pi*rho0/(r*sqrt(2))

where p=sound pressure in Pa, Sd=eqv. piston area in m2, Xmax=max peak excursion in m, f=frequency in Hz, rho0=1.2kg/m3, r=speaker-to-mic distance in m.

The sound pressure p can be converted to sound pressure level

Lp=20*log10(p/0,00002)

So with
Sd=500 cm2
Xmax=10 mm
f=30Hz
r=1

p=0.05*0.01*30^2*3.14*1.2/(1*1.41)=1.2 Pa

Lp = 20*log10(1.2/0.00002)=95.56 dB

Note: It is not certain that it is the cone amplitude that limits max SPL.
 
RobWells said:

Although the chart only goes up to 100 Hz, every time you go up one octave, the volume of air required to maintain the same SPL is divided by four.

OR:

The same volume of air produces 12 dB higher SPL than it did at the lower frequency.

Let's take a 5.75", (145 mm), Peerless woofer, for example, model # 850108. According to Peerless, this unit has a cone area of 91 sq cm. With an excursion of (+ or -) 0.35 cm, it has a volume displacement of 31.85 cu cm.

Looking at the chart, we see that 100 Hz, 31 cu cm produces about 100 dB SPL. So we raise that 12 dB when we move up an octave to 200 Hz. So, a 5.75", (145 mm), speaker, at an excursion of (+ or -) 3.5 mm, will produce 112 dB @ 200 Hz. 🙂
 
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