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Surface area, Excursion, Vd and Spl
Surface area, Excursion, Vd and Spl
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Old 14th July 2004, 02:33 PM   #1
mart34 is offline mart34  United Kingdom
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Default Surface area, Excursion, Vd and Spl

What is the relationship between surface area, excursion, spl and the volume of air displaced by a speaker?

If you double the surface area of a speaker it produces +3dB, but a doubling of excursion produces +6dB even though in both cases the volume of air displaced has doubled.

Example-

SYSTEM 1
1 speaker with an effective surface area of 1000cm2 having 1 watt of power fed through it produces 90dB
Add another speaker with the same surface area and same power going through it, and the total combined output will be 93dB

2000cm2 of air displaced, 2 Watts of power = 93dB


SYSTEM 2
1 speaker with an effective surface area of 1000cm2 having 1 watt of power fed through it produces 90dB
If you were to then double the power going through the speaker to 2 Watts the total output will be 93dB
A doubling of power through a speaker increases the excursion by a factor of 1.41.

1410cm2 of air displaced, 2 Watts of power = 93dB


Can anybody give an answer as to why the volume of air displaced is different between the two systems but the spl is the same?

Am I missing something very simple?, are the figures above incorrect?
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Old 14th July 2004, 03:59 PM   #2
andrikos is offline andrikos  Cyprus
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cm^2 = surface area
cm^3 = volume displaced.

The one variable of the equation that's missing is the woofer travel. This is HIGHLY dependent on the frequency.
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Old 14th July 2004, 04:24 PM   #3
markp is offline markp  United States
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You are forgetting that in your 2 driver situation the surface area is 2000cm2 but the displacement is not going to be 2000cm2. It will be .707 of the displacement of a single driver producing the same spl level.
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Old 14th July 2004, 04:37 PM   #4
BillFitzmaurice is offline BillFitzmaurice  United States
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You also aren't considering the 3dB rise in radiating efficiency when using two drivers. In the case of system 1 with dual drivers and 2 watts input the output will rise to 96dB, not 93dB.
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Old 14th July 2004, 05:50 PM   #5
454Casull is offline 454Casull  Canada
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http://woodartistry.com/linkwitzlab/faq.htm#Q21
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Old 14th July 2004, 09:33 PM   #6
Svante is offline Svante  Sweden
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Default Re: Surface area, Excursion, Vd and Spl

Quote:
Originally posted by mart34

If you double the surface area of a speaker it produces +3dB, but a doubling of excursion produces +6dB even though in both cases the volume of air displaced has doubled.
Nope! Doubled area and same displacement gives +6 dB.

The equation for sound pressure (in pascals) is

p=U*rho0*f / (2*r)

where U is the volume flow in m3/s
rho0=1.2 kg/m3
f is the frequency in Hz
r is the distance

U is also equal to V*2*pi*f, where V is the volume of air displaced in m3.

To get from sound pressure (p) to sound pressure level (SPL)

SPL = 20*log10(p/pref);

where pref=0.00002 Pa

Free space assumed (4 pi).
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Old 23rd October 2011, 07:31 PM   #7
454Casull is offline 454Casull  Canada
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Sorry to bump a dead thread.

www.linkwitzlab.com/spl_max1.xls

Why does SL say that 6 dB should be added for radiation into half-space vs full-space?
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Old 23rd October 2011, 08:31 PM   #8
speaker dave is offline speaker dave  United States
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The general assumption is that there will be a 6dB difference between LF level in half space vs. full space.

There are 2 factors involved. In half space the energy that typically goes into the back hemisphere is constrained to stay in the front. Secondly the air load on the woofer is doubled (radiation resistance). This factor gives a 3dB increase. The first factor gives a little less than 3dB, since most woofers will have some directivity, even at low frequencies.

In my experience the difference is usually about 4 1/2 dB from the 2 factors combined.

David S.
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Old 23rd October 2011, 10:42 PM   #9
john k... is offline john k...  United States
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Why 6dB? It's very simple. An acoustic wave expanding into 4Pi space expands into twice the volume that it expands into in 2Pi space. This means the acoustic pressure in 2Pi space must be twice the acoustic pressure in 4Pi space. Thus the ration of 2Pi pressure to 4Pi pressure is 2. The SPL increase goes like 20 Log of the pressure ratio, 20 Log(2) = 6dB.
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Old 12th May 2016, 02:38 PM   #10
shank1207 is offline shank1207  France
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Quote:
Originally Posted by Svante View Post
Nope! Doubled area and same displacement gives +6 dB.

The equation for sound pressure (in pascals) is

p=U*rho0*f / (2*r)

where U is the volume flow in m3/s
rho0=1.2 kg/m3
f is the frequency in Hz
r is the distance

U is also equal to V*2*pi*f, where V is the volume of air displaced in m3.

To get from sound pressure (p) to sound pressure level (SPL)

SPL = 20*log10(p/pref);

where pref=0.00002 Pa

Free space assumed (4 pi).
Hi thats a a very neat formula but if U is volume flow , then U=Surface*ExcursionMax*2*pi*f(Sd*Xmax*j*omega)?? so you are just derivating excursion with j*omega to get velocity is that correct?
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