PSN-L Email List Message

Subject: Re: Zero Length Spring
From: Brett Nordgren Brett3mr@.............
Date: Fri, 29 Dec 2006 17:50:03 -0500


The relationships between mechanical damping, Damping Factor and Q aren't 
too involved.

If you assume an undamped spring-mass system with a mass of M, kg, and a 
spring constant K, Newtons/m , it will vibrate with a natural frequency of 
omega-zero = 2 sqrt(K/M) radians per second. (You can divide radians per 
second by 2 Pi to get cps or Hz ).

If you then add velocity damping, which means you add a resistance force 
which is exactly proportional to the instantaneous velocity of the mass, of 
an amount = R, Newtons / meter/second, you can then calculate the Damping 
Factor, zeta and Quality factor, Q .

Quality Factor, Q, = M / R or Q = 1 / (2 * omega-zero * zeta)


Damping Factor, zeta, = R / ( 2 * sqrt (M * K)) or zeta = R / (2 * M * 
omega-zero) = 1 / (2 * omega-zero * Q)

Regarding dimensions, since one kilogram meter/sec^2 = 1 Newton, the units 
all cancel for both Q and zeta, so they are both dimensionless.

The zero in omega-zero signifies the *undamped* natural frequency.  As you 
add damping, looking at the frequency response of the spring-mass system, 
its amplitude peak gets lower and at the same time moves somewhat in 
frequency before the damping gets so great that the peak disappears altogether.

You can see the same equations in a much prettier form near the end of   (underscored 
spaces)   Though it's pretty big - a couple of MB, I think.


At 08:29 AM 12/28/2006 -0700, you wrote:
>I understand Damping and Q
>are simply inverses of each other.
>Q is quite complex in that it
>requires the knowledge of the
>3db (0.7071) cutoff points as
>relating to the frq of interest.
>Can anyone tell me the proper way
>to express damping ??
>I sort of understand Q but not
>how you arrive at the proper numbers
>for damping alone.


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