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Old 19th February 2008, 05:44 PM   #1
hooha is offline hooha  Canada
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Default Suspension Sag Equation

Hi guys,

Can someone please post up the equation used to calculate suspension sag of a driver?

Thanks a bunch.
Mark
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Old 19th February 2008, 06:03 PM   #2
AndrewT is offline AndrewT  Scotland
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http://www.diyaudio.com/forums/showt...377#post965377
This may help.
But note I am challenging that respected source.
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Old 19th February 2008, 06:10 PM   #3
hooha is offline hooha  Canada
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Thanks a lot.
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Old 19th February 2008, 07:40 PM   #4
GM is online now GM  United States
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Quote:
Originally posted by AndrewT

But note I am challenging that respected source.
Hmm, there's a suspended mass (Cms, Mms) and the gravitational force applying a constant force on it, so what's missing to make it more complex?

GM
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Old 19th February 2008, 08:01 PM   #5
kstrain is offline kstrain  Scotland
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Quote:
Originally posted by AndrewT
http://www.diyaudio.com/forums/showt...377#post965377
This may help.
But note I am challenging that respected source.
I'm not sure which aspect you are challenging, but it all looks OK to me (not going to the source, just what you wrote):

"Mms =1 / [(2*Pi*Fs)^2*Cms]"

The spring constant can be written k=1/Cms. Then writing the angular frequency ws = 2*Pi*Fs makes the equation match the standard form of the result for the resonant frequency of a "simple harmonic oscillator" - aka a mass on a spring - that looks like "w^2 = k/m".

"sag=Cms*Mms*g = g / [2*Pi*Fs]^2"

Again the spring constant is 1/Cms and the force stretching the spring is indeed the weight Mms*g so multiplying them gives the sag is as written.


"so sag depends of Fs squared, as I said it sounds too simple."

The result follows and - in case anyone is interested - is normal for systems of a mass on a spring under gravity (because the same "mass" is involved in calculating the gravitational and inertial effect - a deep, but mysterious, principle of physics). At which point I'd better stop or this is too much like the day job

Hope I did not misunderstand your challenge.

Ken

ps. posted here because here is where we are

pps. (edit) I'm always posting too slowly - already answered...
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Old 20th February 2008, 06:39 AM   #6
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http://www.subwoofertools.com/forum/sag.htm
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Old 20th February 2008, 09:36 AM   #7
AndrewT is offline AndrewT  Scotland
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Quote:
Originally posted by GM


Hmm, there's a suspended mass (Cms, Mms) and the gravitational force applying a constant force on it, so what's missing to make it more complex?

GM

Quote:
Originally posted by kstrain


I'm not sure which aspect you are challenging, but it all looks OK to me (not going to the source, just what you wrote):

"Mms =1 / [(2*Pi*Fs)^2*Cms]"

The spring constant can be written k=1/Cms. Then writing the angular frequency ws = 2*Pi*Fs makes the equation match the standard form of the result for the resonant frequency of a "simple harmonic oscillator" - aka a mass on a spring - that looks like "w^2 = k/m".

"sag=Cms*Mms*g = g / [2*Pi*Fs]^2"

Again the spring constant is 1/Cms and the force stretching the spring is indeed the weight Mms*g so multiplying them gives the sag is as written.


"so sag depends of Fs squared, as I said it sounds too simple."

The result follows and - in case anyone is interested - is normal for systems of a mass on a spring under gravity (because the same "mass" is involved in calculating the gravitational and inertial effect - a deep, but mysterious, principle of physics). At which point I'd better stop or this is too much like the day job

Hope I did not misunderstand your challenge.

Ken

ps. posted here because here is where we are

pps. (edit) I'm always posting too slowly - already answered...
Hi,
are you two guys saying that my interpretation is correct:- that sag depends ONLY on Fs^2 and has no place for stiffnesses in the simplified equation?
If the simplified equation is correct then that result comes directly from the source equations and then I agree completely with the originally quoted equations.
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Old 20th February 2008, 05:03 PM   #8
GM is online now GM  United States
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Well, these formulas are literally Greek to me, so all I'm saying is that a driver's Fs is a function of mass and compliance (stiffness), so down firing just adds the gravitational force component to my way of thinking:

Fs = [(1/Pi)/2]*{[1000/(Mms*Cms)]^0.5}

where:

Mms is in grams and Cms is in mm/N

GM
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Old 20th February 2008, 05:09 PM   #9
AKN is offline AKN  Sweden
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Hi,

Getting sag out of Mms or Fs should give slightly exaggerated numbers. Mms includes the mass of the air loading the driver. Air will not act as a weight contributing to sag.
I believe Mmd should give a more accurate sag value.
Any Cms nonlinearities will possibly slightly decrease sag.
That said, I think that we will only get an approximation of sag.
Long time creep du to plastic properties of the suspension materials may eventually come into play as time goes by,
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Old 20th February 2008, 05:27 PM   #10
AndrewT is offline AndrewT  Scotland
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Quote:
Originally posted by 4fun
Long time creep due to plastic properties of the suspension materials may eventually come into play as time goes by,
good point, I've not noticed before.
4fun,
if the simplified equation is true then Fs controls the amount of sag, That uses Mms. That would seem to indicate that Fs is the wrong value to use. Presumably on that basis, Fs' excluding the air load, would give a better estimate of sag.
Does the air load depend on cone velocity or cone displacement?
What proportion of the total dynamic mass comes solely from the actual mass of the moving components?
How high would Fs' rise above Fs?
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