# Damping coefficient formulae

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1. Aug 2, 2015

### smokedvanilla

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. Aug 2, 2015

### Hesch

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:

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
3. Aug 2, 2015

### smokedvanilla

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?

4. Aug 2, 2015

### Hesch

No,

ξ = c / ( 2m * ωn ) , ωn = √( k/m )

Try again, one step at a time.

Last edited: Aug 2, 2015
5. Aug 2, 2015

### Hesch

Correction:
This is not right, but the connection can be seen here:

6. Aug 2, 2015

### smokedvanilla

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.

7. Aug 2, 2015

### Hesch

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.

8. Aug 4, 2015

### smokedvanilla

Thank you about the graph, I understand that now :D

I tried working the amplitude expression out,
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

9. Aug 4, 2015

### Hesch

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 ):

Last edited: Aug 4, 2015