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19th February 2008, 05:44 PM  #1 
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Join Date: Apr 2005
Location: Winnipeg, Manitoba

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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19th February 2008, 06:03 PM  #2 
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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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19th February 2008, 06:10 PM  #3 
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Thanks a lot.
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19th February 2008, 07:40 PM  #4  
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Quote:
GM
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19th February 2008, 08:01 PM  #5  
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Join Date: Apr 2006
Location: Glasgow

Quote:
"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... 

20th February 2008, 06:39 AM  #6 
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20th February 2008, 09:36 AM  #7  
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Join Date: Jul 2004
Location: Scottish Borders

Quote:
Quote:
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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20th February 2008, 05:03 PM  #8 
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Join Date: Jun 2003
Location: Chamblee, Ga.

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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20th February 2008, 05:09 PM  #9 
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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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/ Anders 
20th February 2008, 05:27 PM  #10  
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Join Date: Jul 2004
Location: Scottish Borders

Quote:
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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