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 26th May 2013, 02:02 AM #1 diyAudio Member   Join Date: Sep 2012 Location: Maine, Bangor-area 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 02:04 AM. Reason: typo
 26th May 2013, 02:08 AM #2 diyAudio Member     Join Date: May 2005 Location: Victoria, B.C. This might help: Vas Question jeff
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Quote:
 Originally Posted by cT equals piD Without giving it much thought, I've always assumed that Vas is directly proportional to Sd or surface area of the driver cone.
Vas is fairly well explained in the wikipedia article on thiele small parameters.

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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Join Date: Sep 2012
Location: Maine, Bangor-area
Quote:
 Originally Posted by Ron E Vas is fairly well explained in the wikipedia article on thiele small parameters. 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.
Okay, but here's the thing. By the way, I'm mostly taking this from a paper by Small. As you say,

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.

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Join Date: Sep 2012
Location: Maine, Bangor-area
Quote:
 Originally Posted by vinylkid58 This might help: Vas Question jeff
It's fairly late at night now where I am, so I don't know if the answer to my question is in there. But nonetheless or whether or not, it looks interesting. Thanks!

-Pete

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Join Date: Nov 2009
Location: Cape Town
Quote:
 Originally Posted by cT equals piD 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.
Nope. The compliance of the driver's suspension (spider and surround) is Cms, not Cas.

Quote:
 Originally Posted by cT equals piD Cms = Cas/ Sd^2 Substituting for Cms in the above eq. for Vas, Vas = k*Cas That is, Sd drops out of the equation.
Well, yes, but only because you've wrapped it up into Cas.
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 05:09 AM.

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Join Date: Sep 2012
Location: Maine, Bangor-area
Quote:
 Originally Posted by godfrey Nope. The compliance of the driver's suspension (spider and surround) is Cms, not Cas. Well, yes, but only because you've wrapped it up into Cas. 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.
Yes, Cas must be the compliance when the driver is back-loaded by an enclosed volume of air. Cas would be the product of Cms and Sd^2. Cms is the purely mechanical compliance.

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, 02:08 AM #8 diyAudio Member   Join Date: Sep 2012 Location: Maine, Bangor-area There are some math errors in the calculation of how pressure is changed in the volume of air back-loading 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 back-loaded 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
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Quote:
 Originally Posted by cT equals piD Okay, but here's the thing. By the way, I'm mostly taking this from a paper by Small. As you say, 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.
You are just playing with the equations here and you have come to an incorrect conclusion. Cas is Cms "pulled" to the acoustical side of the model, it is Cms * Sd^2 and therefore this is how you want to look at it:
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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Join Date: Sep 2012
Location: Maine, Bangor-area
Quote:
 Originally Posted by cT equals piD 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 = 77 That is, Vas is 77 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%. Yet the box resonance frequency equals 3,3 times free air resonance.
For the inquisitive, here is my solution to the above apparent contradiction of claiming that the above 12" woofer pressurizes an enclosed volume of air equal to 0,1 times Vas when it only increases pressure of the enclosed volume by about 1%.

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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