Calculation of Vas from Frequency Measurements

Status
Not open for further replies.
Hello,

I have just recently started learning about Thiele-Small parameters, and I have seen two different equations for calculating the Vas of a speaker when you use the "added volume" technique.

After getting fs, the free air frequency, and fc, the frequency in a sealed box of volume Vb, one way to calculate Vas is:

1) Vas = Vb * [(fc/fs)^2-1]

I have also seen a method that requires you to get the Q values:

2) Vas = Vb * [(fc/fs)*(Qec/Qes)-1]

The first equation seems more logical in that if you turn around and use that Vas to predict the fc of the speaker in that test enclosure, you get exactly the right answer.

However, I have seen the second equation uses in reputable sources, most notably W. Marshall Leach's book

Does anyone have a good explanation for the difference between these two equations?

Thanks!
 
Both of them are in Dickason's Loudspeaker Design Cookbook so definitely trustworthy.

The first equation is just a simpler/quicker, but less accurate way of doing it.

The best possible way to get Vas is if you know the moving mass of the drive unit, Mms. You can trust the manufacturer on this figure as it's not really subject to any real production tolerances.

Cms = sqrt [(6.283 * Fs)^2 * Mms]

Cms in metres/newton
Mms in kg

Vas = 1.42 * 10^5 * Sd^2 * Cms

Sd in m^2, again you can trust the manufacturer figure, or measure it at cone diameter plus a third of the surround.
 
The first is exactly what I would use. The second I have never seen. I assume that Qec is the value corresponding to Qes, but mounted in the box? If so Qec/Qes should be equal to fc/fs, if the box is lossless and the equations would be identical.

This and the fact that you got the equation from a reputable source, tells me that the second equation somehow tries to compensate for losses in the box. It is well known that stuffing inside the box makes it appear larger due to isothermal compression, so losses and apparent box volume has a relation. I cannot see, however, how this would more than approximate this relation. In fact I think it even goes the wrong way. Hmm...

Anyone else?
 
Paraphrased from Appendix 3 of "Theory and Design of Loudspeaker Enclosures" by J.E Benson

The second equation (A3-16 in reference) is independent of changes in the mechanical resistance of the driver between Fs and Fc.

Box losses are still assumed negligible.
 
I still don't understand this.

Vas is the box volume that would give the same spring constant as the suspension Cms. The first equation is derived from that
fs=1/(2*pi*sqrt(Mms*Cms))
fc=1/(2*pi*sqrt(Mms*(Cms//Cb)))
Cms=Vas/(rho0*c^2*Sd^2)
Cb=Vb/(rho0*c^2*Sd^2)

(a//b meaning ab/(a+b))

So, fc/fs=Sqrt(Vas/(Vas//VB)), which after some fiddling leads to Vas=Vb*((fc/fs)^2-1).

I realise that the above assumes that Mms and Sd is constant between the two cases, but the resistance? The resistance has nothing to do with the resonance frequencies. Hmm...
 
Status
Not open for further replies.