Formulas to calculate new TS values when mass added - diyAudio
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Old 14th February 2004, 01:41 AM   #1
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Default Formulas to calculate new TS values when mass added

Does anyone have formula's to calculate the new value of Qts/Qms/Qes/Fs as you add more mass to the cone?

I have some formulas here, but they are dependant on each other, so it doesn't work right for this. As you can see:

Click the image to open in full size.

Qms there relies on Rms as well, and to calculate Rms requires Qms. + I think Rms relies on Mms, as its related to Qms.

Any help highly appreciated!

Adrian

BTW: I know WinISD can model the effects of this, but I require the formulas.
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Old 14th February 2004, 04:36 AM   #2
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I cannot give you the formulas.

But if you download Bullock and White's freeware DOS program, you can change certain parameters and see how they affect each other.
http://www.hal-pc.org/~bwhitejr/

You cannot change Mms by itself in the program, but if you lower Fs you see that Mms is affected by most of all, and Rms is affected some. Bl will have to be increased to support the other parameters remaining the same.

Try it. It might open up the relationships for you.
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Old 14th February 2004, 05:33 AM   #3
GM is offline GM  United States
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If Mms is unknown:

mass (m) = 22.428*dia^4(cm)/(Fs^2*Vas (liters))

then:

m' = m + mass added

mass ratio (mr) = m'/m

Fs' = Fs/mr^0.5

Qes' = Qes*mr^0.5

Qms' = Qms*mr^0.5

then: Qts' = Qes'*Qms'/(Qes'+Qms')

Vas is unchanged

n0 = 9.614*10^-10*Fs'^3*Vas(liters)/Qes'

SPL = 112.02+10*Log(n0)

GM
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Old 14th February 2004, 09:25 AM   #4
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Thanks a lot for those formula's GM! It's exactly what I needed.

Quote:
Try it. It might open up the relationships for you.
I appreciate your input, kelticwizard. I've got the formula's now though, as you've probably noticed.

Adrian
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Old 14th February 2004, 04:53 PM   #5
Ron E is offline Ron E  United States
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Betcha got those from me on the Bass list, GM - or from Bob Stout's post paraphrasing me. I haven't seen that formula I derived for cone mass anywhere else.
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Old 14th February 2004, 06:33 PM   #6
qwad is offline qwad  Australia
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yes and we all know plagarism is rife in audio now don't we? cheers and good luck to all TOMCAT:
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Old 14th February 2004, 06:35 PM   #7
Svante is offline Svante  Sweden
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OK, I'll go through them and see if I end up with the same as the others:

ws=1/sqrt(Mms*Cms) (you lack the root)

Fs=1/(2*pi*sqrt(Mms*Cms))=ws/(2*pi) (you have it inverted)

Qes = ws*Mms / Res (mass reactance divided by resistance)

since Res=(Bl)^2 / Re

Qes= ws*Mms * Re/((Bl)^2) (As you have it)

Qms = ws*Mms / Rms (mass reactance divided by resistance)

Qts = ws*Mms / (Rms+Res) = Qes*Qms/ (Qes+Qms)

Rms does not change if you add mass to the speaker cone.

So, a mass load on the cone increases all Q values linearly (~Mms), and the resonace drops as ~1/root(Mms).

Did that answer your question?
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Old 14th February 2004, 07:00 PM   #8
Svante is offline Svante  Sweden
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Quote:
Originally posted by GM
If Mms is unknown:
mass (m) = 22.428*dia^4(cm)/(Fs^2*Vas (liters))
GM

Quote:
Originally posted by Ron E
Betcha got those from me on the Bass list, GM - or from Bob Stout's post paraphrasing me. I haven't seen that formula I derived for cone mass anywhere else.
Let's try right here:

Cms=Vas/(rho0*c^2*Ss^2)

Mms=1/(Cms*ws^2)=
=rho0*c^2*Ss^2/(Vas*ws^2)=
=rho0*c^2*(pi*dia^2/4)^2/(Vas*4*pi^2*Fs^2)=
=rho0*c^2*pi^2/(16*4*pi^2) * dia^4 / (Vas * Fs^2)=
=rho0*c^2 / 64 * dia^4 / (Vas * Fs^2).

rho0=1.2 kg/m3
c=345 m/s
gives

Mms=2231.7 * dia^4 / (Vas * Fs^2).

I assume that Mms is in kg, dia is in metres, Vas is in m3 and Fs in Hz.
Since 1 m3 is 1000 litres, 1 meter is 100 cm and 1kg is 1000g there would be a factor 0.01^4 / 10^-3 *1e3 = 10^-2 =0.01 between our equations.

They match!
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Old 14th February 2004, 07:36 PM   #9
GM is offline GM  United States
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At this point in time I don't recall thanks to a very poor memory. Because of this I have saved copious bits of info on a wide range of subjects gleaned from many sources over the decades, including the basslist.

As I've periodically stated on numerous forums, thanks to being severely math challenged, anytime I post math it's a given that I'm just sharing what others have already published somewhere or helped me with in private over the decades, you being one of them.

Anyway, if it was you, thanks for sharing! They are as accurate and much easier to understand/solve than the ones I used to have programmed in a spreadsheet a math whiz friend did for me decades ago.

GM
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