What is the actual interpretation of Vas?

guferr

Member
2018-02-24 6:55 pm
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 = Patm·A·{1-[V/(V±A·x)]1.4}

Where x is the displacement, V the volume, A the frontal area of the cone, Patm 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·Patm·A²·x/V

Where the spring constant would be 1.4·Patm·A²/V

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 Patm = 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?
 
Last edited:

guferr

Member
2018-02-24 6:55 pm
Vas = rho * (c ^ 2) * (Sd ^ 2) * Cms

Where:

rho = density of air
c = velocity of sound in air
Sd = driver diaphragm piston area
Cms = driver diaphragm suspension mechanical compliance

I wonder if the derivation was the same.

Because I used a taylor expansion in order to find an approximation, and even then my formula basically is the same of this one you presented.

First, rewriting it to put Vas in function of the spring constant K:

Vas = 1.4·P·A²/K

1/K is Cms, so:

Vas = 1.4·P·A²·Cms

The speed of sound in air can be approximated as:

c²=1.4·R·T/M

Where 1.4 is the adiabatic index of air, R is the ideal gas constant, T the absolute temperature and M the molar mass.

For an ideal gas you could rewrite R·T as P·V/n, so c²=1.4·(P·V/n)/M =1.4·P·V/(n·M)

V/(n·M) is simply the inverse of air density, so:

c²=1.4·P/rho

and

1.4·P = rho·

Replacing that, now what I wrote reads

Vas = rho··A²·Cms

The same formula again
 
Last edited:
I wonder if the derivation was the same.

Vas is simply a volume of air having the same acoustic compliance as the driver suspension. The acoustic compliance of the suspension (Cas) is given by (Sd ^ 2) * Cms (mechanical compliance Cms converted to acoustical compliance Cas)

Vas is given by rho * (c ^ 2) * Cas

Therefore Vas = rho * (c ^ 2) * (Sd ^2) * Cms

See technical papers by Thiele and Small if you would like to know more.