So, I'm having a problem with the definition of Vas.

I've read somewhere that Vas is the volume of air that, when compressed down to 1 m³, exerts the same force as the speaker suspension.

So I'd imagine the speaker suspension is also pushing it back too due to the displacement (the same one that compressed the air), and the two forces are equal, both pushing it back.

The problem with that is that, considering isothermal compression at least, the initial volume doesn't matter: there are only two volumes where the suspension and the force caused by pressure difference will be equal, one being the initial volume itself (as both will be zero) and the other one being a fixed value. So Vas could be literally anything, the final volume would be the same, and it would only be 1 m³ for some combinations of suspension stiffness and cone area.

I considered adiabatic compression too (which makes much more physical sense), and then the initial volume does matter and it's possible to find a "Vas", but the values I found were nonsensical.

But then I went for another interpretation: The volume of a sealed box where the air would behave as spring of constant K similar to that of the speaker suspension.

Considering adiabatic compression and expansion, and that the pressure in the outside of the cone is constant (a good approximation in most cases, as the variation of pressure inside is much greater than the variation outside), the force caused by the pressure difference is:

F = P

Where x is the displacement, V the volume, A the frontal area of the cone, P

Doing a Taylor expansion of the term [V/(V-A·x)]

F = -1.4·P

Where the spring constant would be 1.4·P

That approximation is very good for small x and large volumes, which is mostly the case for speakers, as the volume they displace is very small compared to the enclosure volume.

That interpretation gives more reasonable Vas results. For example, a 20 cm (effective cone diameter) speaker with a 400 N/m spring constant suspension would have a Vas of 0.35 m³ or 350 liters (considering P

Is that the correct meaning of Vas?

I've read somewhere that Vas is the volume of air that, when compressed down to 1 m³, exerts the same force as the speaker suspension.

So I'd imagine the speaker suspension is also pushing it back too due to the displacement (the same one that compressed the air), and the two forces are equal, both pushing it back.

The problem with that is that, considering isothermal compression at least, the initial volume doesn't matter: there are only two volumes where the suspension and the force caused by pressure difference will be equal, one being the initial volume itself (as both will be zero) and the other one being a fixed value. So Vas could be literally anything, the final volume would be the same, and it would only be 1 m³ for some combinations of suspension stiffness and cone area.

I considered adiabatic compression too (which makes much more physical sense), and then the initial volume does matter and it's possible to find a "Vas", but the values I found were nonsensical.

But then I went for another interpretation: The volume of a sealed box where the air would behave as spring of constant K similar to that of the speaker suspension.

Considering adiabatic compression and expansion, and that the pressure in the outside of the cone is constant (a good approximation in most cases, as the variation of pressure inside is much greater than the variation outside), the force caused by the pressure difference is:

F = P

_{atm}·A·{1-[V/(V*±*A·x)]^{1.4}}Where x is the displacement, V the volume, A the frontal area of the cone, P

_{atm}the atmospheric pressure and 1.4 the adiabatic index of air. Using SI units, the force will be in newtons. The symbol*±*was used because it will depend if it's an expansion or compression.Doing a Taylor expansion of the term [V/(V-A·x)]

^{1.4}near x=0, we can reduce everything to a spring-like force of:F = -1.4·P

_{atm}·A²·x/VWhere the spring constant would be 1.4·P

_{atm}·A²/VThat approximation is very good for small x and large volumes, which is mostly the case for speakers, as the volume they displace is very small compared to the enclosure volume.

That interpretation gives more reasonable Vas results. For example, a 20 cm (effective cone diameter) speaker with a 400 N/m spring constant suspension would have a Vas of 0.35 m³ or 350 liters (considering P

_{atm}= 101325 Pa). This is a speaker that would lower 25 mm if you put a 1 kg weight over it.Is that the correct meaning of Vas?

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