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MultiWay Conventional loudspeakers with crossovers 

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26th May 2013, 01:02 AM  #1 
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Join Date: Sep 2012
Location: Maine, Bangorarea

puzzled about Vas
Without giving it much thought, I've always assumed that Vas is directly proportional to Sd or surface area of the driver cone. Not!
Vas = po*c^2*Cas where po= air density c = speed of sound Cas = acoustic compliance of driver suspension I don't see Sd in that equation. Another equation that I ran across is Vas = (1.4 X 10^5)*Cms*Sd^2 but that is the same thing as Cms = Cas/ Sd^2 So the equation is saying that Vas is directly proportional to the compliance of the driver's suspension (spider and surround) no matter what diameter size the driver is. Am I missing something or is this something that isn't generally understood? Sorry if you are more interested in the practical side, but I think that some of you like to discuss this kind of thing, at least I hope so. Pete Last edited by cT equals piD; 26th May 2013 at 01:04 AM. Reason: typo 
26th May 2013, 01:08 AM  #2 
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26th May 2013, 01:39 AM  #3  
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Quote:
Vas = Sd^2* rho * c^2 * Cms rho ~=1.2 c^2=344^2=118300 multiply these together and make a factor called k and you have: Vas = k*Sd^2*Cms Vas is proportional to the square of cone area times compliance.
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26th May 2013, 03:37 AM  #4  
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Quote:
Vas = k*Sd^2*Cms however, Cms which is mechanical compliance of the driver suspension, Cms = Cas/ Sd^2 Substituting for Cms in the above eq. for Vas, Vas = k*Cas That is, Sd drops out of the equation. Cas is the acoustic compliance of the driver suspension. 

26th May 2013, 03:41 AM  #5  
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Quote:
Pete 

26th May 2013, 04:05 AM  #6  
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Quote:
Quote:
The other equation, Vas = k*Sd^2*Cms, is more useful as Sd and Cms are what's given in the TS parameters. Cms being the mechanical compliance in mm/N i.e Cms = distance moved / force applied. Last edited by godfrey; 26th May 2013 at 04:09 AM. 

26th May 2013, 08:16 PM  #7  
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Location: Maine, Bangorarea

Quote:
However there is the often expressed opinion that Vas increases with increased Sd because the larger diaphragm compresses the enclosed volume of air more. This is I think is untrue. For a given enclosed volume of air, the pressure on the cone of the driver increases with increased area of the cone since the enclosed air exerts X gm of force per square cm. The X gm of force per square cm increases as the enclosed volume of air is decreased. But this is with the cone of the driver stationary. For example, from the spec. of a 12" woofer, Vd = Sd *Xmax =265 cm^3. Specified Vas = 204 X 10^3 cm^3. So then 0,1*Vas/ Vd = (204/265)*10^2 = 770 That is, Vas is 770 times Vd, even for a relatively small enclosed volume of air backloading the woofer equal to 1/10th of Vas. From Boyle's law, the woofer at max. excursion compresses the enclosed volume of air equal to 1/10th of Vas by only 1/10th of 1%. Yet the box resonance frequency equals 3,3 times free air resonance. 

27th May 2013, 01:08 AM  #8 
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There are some math errors in the calculation of how pressure is changed in the volume of air backloading the example woofer in the above post my apologies.
The ratio of 0,1 times Vas to Vd equals 77 not 770. The pressure increase/ decrease of the volume of air by the woofer is +/ 1%, not 0,1%. What I say above, that a smaller volume of air (at atmospheric pressure, obviously) will exert greater pressure isn't going to fly. So, um, you could say that I'm still puzzled, although I do believe that compliance of a backloaded driver is inversely proportional to Sd squared. But I do see an extremely low pressurization of the enclosed air in almost all cases, as in the above example of the 12" woofer. Pete 
27th May 2013, 01:52 AM  #9  
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Quote:
Vas = k * Cms * Sd^2 it is still dependent on Sd. You are confused because Cas has Sd factored in already as has already been pointed out to you. Go back to Small's paper and note that there are models with all of the parameters "pulled" to the electrical side, and also pulled to the acoustical side. You'll find that Cas on the acoustical side is related to Cms by the Sd factor. You need to read Small's paper more carefully, or perhaps go back to his earlier papers if you are looking at a later one.
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29th May 2013, 12:59 AM  #10  
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Quote:
Atmospheric pressure equals about 1 X 10^6 dynes/ cm^2. So with full excursion of the cone of the woofer backwards, the pressure inside the closed box is increased to atmospheric pressure plus 0,01 times 1 X 10^6 dynes/ cm^2, that is, net pressure = 1 X 10^6 dynes/ cm^2 times 0,01 = 1 X 10^4 dynes/ cm^2. 10,000 dynes/ cm^2 is a considerable force, especially in relation to the mechanical compliance of the woofer, I would think. Agree/ disagree anybody? Regards, Pete 

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