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-   -   Calculating Qec Qmc (http://www.diyaudio.com/forums/multi-way/145491-calculating-qec-qmc.html)

 MCPete 8th June 2009 03:38 AM

Calculating Qec Qmc

In books such as The Loudspeaker Design Cookbook by Dickason and that book by Weems, Designing, Building, and Testing...., I've seen the formula

Qtc = ((Vas/ Vb) + 1)^2 *Qts

However, I was wondering if putting a driver in a closed box only increased Qm or only increased Qe or both. Intuitively it would seem correct to suppose that Qe would remain the same and Qm would increase. That's my intuition, as putting the driver in a closed box isn't changing anything electrical, but only the mechanical side of its operation.

Then I manipulated some equations by Thiele in a paper of his to discover that

Qec = ((Vas/ Vb) + 1)^2 * Qes

Qmc = ((Vas/Vb) +1)^2 * Qms

This being in agreement with the standard equation relating Qts to Qtc that I started with.

Would all of you in the know with T/S calculations agree with my equations for Qec and Qmc?

Why should Qe change when the electrical condition of the driver is unchanged?

 MCPete 11th June 2009 02:18 AM

correction to equations

Sorry! -to all those who have looked at this. The equations are incorrect, I used the wrong symbolism for square root. The correct equations for Qec and Qmc are:

Qec = sqrt((Vas/Vb) +1) * Qes
Qmc = sqrt((Vas/Vb) +1) * Qms
Rather than reply to my own post, I did try to be able to edit my post, but was unsuccessful for what reason I don't know.

-Pete

 Ron E 11th June 2009 03:54 AM

Qxc/Qxs = Fc/Fs , so Q changes in proportion to the resonant frequency

Electrical damping is Bl^2/Re - unchanged

You can't edit because they lock posts for editing after perhaps 1/2 hour. They could leave it open, but choose not to.

 john k... 11th June 2009 11:20 AM

In terms of mechanical parameters

Qms = (1/Rms) x sqrt(Mms/Cms)

When placed in a box Cms is replaced by the combination of Cms in parallel with Cb, where Cb is the compliance of the box.

Similarly,

Qes = (Re/Bl)^2 x sqrt(Mms/Cms)

Again, in a box Cms must be replaced with Cms in parallel with Cb.

Since Cb in parallel with Cms is always less that Cms, both Qes and Qms increase by sqrt(Cms/Cab) where

Cab = Cb x Cms /(Cb + Cms)

If you work through the math,

Sqrt(Cms/Cab) = Sqrt( (Cb + Cms)/Cb) = Sqrt(1 + Cms/Cb) = sqrt(1 + alpha).

------

Alpha = Vas/Vb.

Since

Vas = Sd^2 x Cms x const.

and

Vb = Sd^2 x Cb x const.

Cms/Cb = Vas/ Vb = alpha.

The bottom line is that both Qe and Qm increase by a factor of

Sqrt(1 + alpha), as does Qt because of the change in compliance.

 Pano 11th June 2009 05:05 PM

Excelent! Thanks John.

I can see this in the calculations of Unibox, too.

 MCPete 12th June 2009 01:05 AM

Okay, so it DOES seem reasonable that decreasing compliance by back loading the driver with a closed box would increase Qm. But it DOES NOT seem reasonable at all that decreased compliance would increase Qe. Is this perhaps a case of Thiele/ Small as a simplication becoming stretched?

Now, I'm not going to lose sleep if this doesn't become reasonable. I'm just exercising my curiosity here. :)

Thanks for confirming my "discovery", anyway.

Thanks to Ron for the info about editing.

I did neglect to edit my first equation relating Qtc to Qts. I'll do that now just to make this more worthy of the archive.

Qtc = sqrt((Vas/ Vb) + 1) * Qts

Regards,
Pete

 Ron E 12th June 2009 02:41 AM

Quote:
 Originally posted by MCPete Is this perhaps a case of Thiele/ Small as a simplication becoming stretched?
No. Thiele small theory neglects radiation resistance, assumes the elements can be lumped and ignores changes in parameters with level and frequency, but still predicts your "discovery".

John proved both Qes and Qms are modified by compliance (the only variable which affects Fs). This is why I put my answer in terms of Fs - most never get past Fs, Qts, Vas modeling...

 john k... 12th June 2009 08:58 PM

Quote:
 Originally posted by MCPete Okay, so it DOES seem reasonable that decreasing compliance by back loading the driver with a closed box would increase Qm. But it DOES NOT seem reasonable at all that decreased compliance would increase Qe. Is this perhaps a case of Thiele/ Small as a simplication becoming stretched? Now, I'm not going to lose sleep if this doesn't become reasonable. I'm just exercising my curiosity here. :) Thanks for confirming my "discovery", anyway. Thanks to Ron for the info about editing. I did neglect to edit my first equation relating Qtc to Qts. I'll do that now just to make this more worthy of the archive. Qtc = sqrt((Vas/ Vb) + 1) * Qts Regards, Pete
Well you just have to look at the definitions of Qe and Qm. Both depend on moving mass and total compliance. The differences is that Qm includes mechanical damping and Qe electrical damping. As Ron noted eariler, the electrical damping doesn't change. Neither does the mechanical damping. Also note that Q is not just damping. It is actually a ratio of the ability to store energy to the ability to dissapate energy. You should be able to grasp that a stiffer spring can store more energy than a weak one. Thus when the compliance decreases Q increases because the system can store more energy.

Qm and Qe come directly out of the equation which governs the motion of the cone.

 MCPete 13th June 2009 01:39 AM

Yes, I see in John K's post that both Qes and Qms are directly proportional to sqrt(Mms/ Cms). But these are equations devised by Thiele and Small that define what Qes and Qms are. So merely stating their equations doesn't answer my question.

In other words Thiele and Small understood a connection between compliance and electrical Q. The equation for Qes states the connection, but the equation by itself doesn't provide any explanation as to why there is a connection.

In the case of Qms, Qms directly proportional to stiffness is fairly easy to comprehend. A spring with increased tension can store and release more energy as a result of being made more stiff.

But WhY should compliance affect electrical Q? Given the sophistication of the authors of these equations, I would think that there probably is an explainable link between electrical Q and compliance. -That is what I would maybe like to be able to grasp so long as it doesn't require being able to do Calculus.

-Pete

 Ron E 13th June 2009 03:04 AM

Quote:
 Originally posted by MCPete But WhY should compliance affect electrical Q? Given the sophistication of the authors of these equations, I would think that there probably is an explainable link between electrical Q and compliance.
What happens with a coil of wire when you move it through a magnetic field? What happens if you move it faster (Higher Fs)?

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