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#### 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. 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

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#### Ron E

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.

#### cT equals piD

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.

#### godfrey

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.

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.

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#### cT equals piD

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.

#### cT equals piD

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

#### PB2

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.

#### 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

#### Ron E

Agree/ disagree anybody?

Rather than making an assertion and asking for comments, perhaps you would be better suited to ask a question. Are you attempting to calculate force on the cone due to the box compliance?

#### cT equals piD

Rather than making an assertion and asking for comments, perhaps you would be better suited to ask a question. Are you attempting to calculate force on the cone due to the box compliance?

Well, first of all I calculated that, for even a relatively high value of alpha (equal to Vas/ Vb) equal to 10, the maximum volume displacement, Vd, of the example 12" woofer is only 1/77th of the box volume Vb. Thus it would seem that such a small change of the volume of the enclosed air would have a negligible affect on compliance of the woofer. But we know from experience that alpha equal to 10 significantly raises the resonant frequency.

So my solution to the apparent contradiction is to realize that atmospheric pressure is a quite high figure. Even 1/100th of atmospheric pressure is quite a bit of force, so it is only necessary to slightly compress/ expand the air in the box to alter the compliance of the system of the driver mounted on the box.

What I'd like to know is if this line of reasoning is correct.

Regards,
Pete

#### Ron E

Speaker compliance and equivalent volume
Cms*Sd^2=Cas, Cas*rho*c^2=Vas

Box compliance derived from volume
Cmb*Sd^2=Cab, Cab*rho*c^2=Vab

Box+speaker combined compliance
Cmt*Sd^2=Cat, Cat*rho*c^2=Vat

where:
Cmt= Cms*Cmb/(Cms+Cmb)

Units of Cms, Cmb, Cmt are meters per newton.

Use Sd of speaker and you can calculate box compliance versus speaker compliance vs. total compliance for that system

#### cT equals piD

In the above thread, Posts by Speaker Dave, Post #2 except for the first paragraph and Post #4, and Post #11 by head unit I think give the best explanation of what Vas is.

The way that Small goes about defining it I would say is confusing, but perhaps not to an acoustics engineer.

A good definition of Vas I think would be that volume of air back-loading a given loudspeaker driver that results in the box resonance frequency Fcb equal to 1.414 times free air resonance Fs. This happens when the strictly acoustic compliance equals the mechanical compliance. If you consider mechanical compliance to be a constant, for increasing diameter of the driver, volume of the box must increase according to the square of the diameter increase simply because for a given excursion a greater quantity of air is compressed / decompressed by the cone.

Wikipedia on the Thiele Small parameters states that large values of Vas mean lower stiffness and generally require larger enclosures. I think that that is a misleading statement. The larger value of Vas can be due to the large diameter of the driver.

-Pete

#### Greebster

Wikipedia on the Thiele Small parameters states that large values of Vas mean lower stiffness and generally require larger enclosures. I think that that is a misleading statement. The larger value of Vas can be due to the large diameter of the driver.

-Pete

True, but in respect to compliance of a 15" driver that is relatively stiff that Vas would be much lower. Apples to apples, any driver of same size/type would have a relatively low Vas if it were stiff and relatively high if loose.

#### PB2

You are starting to get it now. I'm surprised that the better definition of Vas is not stated more often - its been a long time since I read the papers. I would put it this way, that Vas is the volume of air that provides an acoustical compliance, as seen by the mechanical circuit (or reflected into the mechanical circuit) that is equal to the driver suspension compliance Cms. And yes, that volume of air does provide an Fc equal to 1.41 * Fs because the resonant frequency is inverse square law with regard to compliance. A box volume equal to Vas provides a total compliance that is half of the Cms value, Sqrt(1/.5) = 1.41.
Another way to look at it is if you could remove the mechanical suspensions and provide all of the suspension stiffness from the box compliance, Vas is the volume of air that would provide the same Fs. I believe that this was the original intent, to represent Cms as box volume.

The box is a fairly linear air spring for small changes in volume and quite non-linear for very large displacements:

Example: 12" driver, at 10mm excursion in a 40l box produces 1.75% H2 simply due to the air non-linearity - 10mm is not large considering modern high output drivers:

Subwoofer design

I pointed out the non-linearity of air to the brainwashed AR enthusiasts about 15 or 20 years ago - thank you Mr. Siegfried Linkwitz for taking the time to explain it and provide an estimate.

Anyway, T&S analysis is mostly small signal where the driver is assumed to be linear - so back to your discussion.

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#### Ron E

What I'd like to know is if this line of reasoning is correct.

In short, no.
Alpha has nothing to do with the the ratio of box displacement to box volume.
Say you have a driver with a Vas of 25 liters in a 7 liter box, that driver has a Vd of half a liter - the percentage of Vd to box volume is 7%.

Alpha of 10 means that the box contributes 10 times more stiffness than the driver. That's all.

Oh and your 1.414 observation is just an elementary consequence of the fact that :
Fc/Fs = sqrt(Vas/Vb+1)
...so when Vas and Vb are equal, the square root of 2 is 1.414.

...the problem with defining Vas off that relation is that that relation is only approximately true because there are damping terms due to leakage and absorption even in a very meticulously crafted box, especially if it has damping material in it.

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#### cT equals piD

In short, no.
Alpha has nothing to do with the the ratio of box displacement to box volume.
Say you have a driver with a Vas of 25 liters in a 7 liter box, that driver has a Vd of half a liter - the percentage of Vd to box volume is 7%.

Alpha of 10 means that the box contributes 10 times more stiffness than the driver. That's all.

Oh and your 1.414 observation is just an elementary consequence of the fact that :
Fc/Fs = sqrt(Vas/Vb+1)
...so when Vas and Vb are equal, the square root of 2 is 1.414.

...the problem with defining Vas off that relation is that that relation is only approximately true because there are damping terms due to leakage and absorption even in a very meticulously crafted box, especially if it has damping material in it.

I would think that you would agree that alpha indirectly affects the ratio of box displacement to box volume. That is, if you consider excursion and diameter of the driver to be constant, then the ratio increases as alpha increases. Alpha increasing means that the internal volume of the box is decreasing.

My point in defining Vas the way that I did was to make it apparent that Vas is directly proportional to both mechanical compliance and Sd. Defining it that way I thought of as being just a way to make Vas understandable. Small gives an equation for finding Vas if you know Cms and Sd.

In part what I found puzzling was that an apparently very small change in the volume occupied by the enclosed air could exert enough force to significantly change free air resonance Fs. So I'm somewhat intrigued to understand all of the physics behind that. I might be posting more about that.

Regards,
Pete

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#### PB2

What you are trying to understand is covered by every basic text on acoustics. This is a good one:
Acoustics: Leo L. Beranek: 9780883184943: Amazon.com: Books

Alpha is simply the ratio of the suspension compliance expressed as Vas over the box volume Vb, it is that simple.
A higher Alpha indicates that most of the compliance or restoring force is provided by the box. Villchur stated that distortion could be reduced with high Alpha since the air spring was more linear than the suspensions of the time, and if the air spring dominated then the distortion would be lower.

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#### PB2

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