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Damping coefficient formulae

  1. Aug 2, 2015 #1
    Hi, I have been looking for formulae for the damping coefficient, and I found two different formulae for it.

    http://www.xyobalancer.com/xyo-balancer-blog/viscous-damping-coefficient
    This webpage states that damping force, Fd is given by Fd=-cv, where c is the damping coefficient while v is the velocity.

    http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html
    However, this page states that damping coefficient is given by c/2m, where c=-Fd/v

    Is there a fixed definition for damping coefficient, or are both definitions acceptable?
     
  2. jcsd
  3. Aug 2, 2015 #2

    Hesch

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    No, they are different:
    In your first link, it's stated that damping ratio = c.
    The second link states that damping ratio = c/2m.

    I think there is some confusion as to terminology, but I prefer the second link:

    oscda2b.gif

    The last equation is the characteristic equation of the system, which could be written by Lapace:

    m * s2 + c * s + k = 0 ⇒

    s2 + (c/m)s + k/m = 0

    The last equation could also be written:

    s2 + 2ξωns + ωn2 = 0

    where ξ is the damping ratio in my definition, which regards the ratio that the amplitude of some oscillation has been reduced from one period to the following:

    Amplituden+1 = Amplituden * ( 1 - ξ )
     
    Last edited: Aug 2, 2015
  4. Aug 2, 2015 #3
    First of all, thank you for your reply.
    Which gives ξ=c/(2*√mk), so c is the damping coefficient & 2*√mk is the damping coefficient in the case of critical damping?
     
  5. Aug 2, 2015 #4

    Hesch

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    No,

    ξ = c / ( 2m * ωn ) , ωn = √( k/m )

    Try again, one step at a time.
     
    Last edited: Aug 2, 2015
  6. Aug 2, 2015 #5

    Hesch

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    Correction:
    This is not right, but the connection can be seen here:
    300px-2nd_Order_Damping_Ratios.svg.png
     
  7. Aug 2, 2015 #6
    The ga
    is the graph wrong because zeta should be 1 instead of 0.4 when critical damping occurs? I think I understood the concept.
     
  8. Aug 2, 2015 #7

    Hesch

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    No, I trust the graphs, but my expression in #2:

    Amplituden+1 = Amplituden * ( 1 - ξ )

    was simply wrong.

    Maybe you could calculate the correct expression? ( 1 - ξ ) must be substituted by something else ( f(ξ) ).

    Critical damping is when ξ = 1. No damping is when ξ = 0. In "typical" analog controllers as for temperature, motorspeed, a ξ = 0.65 . . 0.7 is often chosen, which compromises control speed and overshoot. In digital controllers a ξ = 1 is often chosen.
     
  9. Aug 4, 2015 #8
    Thank you about the graph, I understand that now :D

    I tried working the amplitude expression out,
    https://www.google.com/search?q=damping+ratio+formula&es_sm=91&source=lnms&tbm=isch&sa=X&ved=0CAcQ_AUoAWoVChMIn-qy07aPxwIV1Y-OCh2PVwZU&biw=1277&bih=637#imgrc=ArC96HvFuNB9TM%3A
    Since ξ is the ratio of damping coefficient of the system when critical damping occurs (Cc) to the damping coefficient of the oscillation (γ), ξ=γ/Cc. From this we have γ=ξ*Cc--(1)

    The expression for variation of amplitude with time is Ae^(-γt), so if we sub. (1) into the expression we get Amplituden+1 = Amplituden*e^(-ξ*Cc)t
     
  10. Aug 4, 2015 #9

    Hesch

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    Yes, that is right: If we damp an oscillation ( say sin(ωt) ) the sin(ωt)-function will be enveloped within a ±exponential-function. But if you look closely at the graphs in #6, you will see that the damped sin(ωt) has a somewhat lower frequency than the undamped sin(ωt): Thus the amplitude (peakvalue) will be somewhat smaller than
    Amplituden+1 = Amplituden*e^(-ξ*Cc)t.

    The ξ could be formulated as:

    Say you have a characteristic equation: s2 + 2s + 2 = 0 , you will find the roots z=( -1 + j1 ) and z=( -1 - j1 ). If you plot these roots in a root locus, and draw two lines through ( 0 , 0 ) and ( -1 ± j1 ) we could call the angles between these lines and the imaginary axis φ ( = 45° ). Then ξ = sin(φ) = 0.7071. So if you want a ξ = 0.6 , the roots of the characteristic equation must be on two lines with an angle = arcsin( 0.6 ) as to the imaginary axis ( = 36.87° ).

    Critical damping is exactly when the roots close up on the real axis ( φ = 90° , ξ = 1 ):
    module-basic_clip_image026_0002.gif
     
    Last edited: Aug 4, 2015
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