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14th July 2004, 02:33 PM  #1 
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Join Date: Jul 2004
Location: England

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? 
14th July 2004, 03:59 PM  #2 
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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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14th July 2004, 04:24 PM  #3 
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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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14th July 2004, 04:37 PM  #4 
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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.

14th July 2004, 05:50 PM  #5 
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14th July 2004, 09:33 PM  #6  
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Re: Surface area, Excursion, Vd and Spl
Quote:
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). 

23rd October 2011, 07:31 PM  #7 
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Join Date: Jan 2004
Location: Toronto, ON, Canada

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 halfspace vs fullspace? 
23rd October 2011, 08:31 PM  #8 
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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. 
23rd October 2011, 10:42 PM  #9 
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Join Date: Aug 2004
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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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12th May 2016, 02:38 PM  #10  
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Quote:


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