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I am not very good with physics terms, so please treat me as an ignorant.

I am trying to calculate a damping coefficient dynamically for a hydraulic-controlled door that opens and closes due to hydraulic pressure (opening/closing). The formula where I need my damping coefficient is:

viscous_fric_mom = $C \times \omega \times 2/\pi$;

The $\pi/2$ division is because the maximum angle that the door can be opened at is $\pi/2$ (i.e. 90°), $\omega$ is the angular velocity of the door which is in $\rm rad/s$. And $C$ is the damping coefficient, which I need to calculate dynamically.

The system specifies that my damping coefficient unit is in $\rm N m s$! I thought it would be $\rm N s /m$, because usually it is Force*time/distance. Apparently I am wrong. Could someone suggest what I should consider for calculating this damping coefficient? I am really bad at math and do not know any better of doing it.

UPDATE

I am trying to create a software model of a hydraulic-operated door. The door will open given that the effective hydraulic jack pressure have been applied. Same goes true when the door is closing i.e. an effective hydraulic jack retraction pressure must be applied. I use two part integration (integration of acceleration and velocity) to get the current angular position of the door, i.e. from the locked position. My current operating assumption is that the door will have either $\pi/2$ or $-\pi/2$ acceleration (i.e. opening or closing). If I integrate that, I get the velocity and double integration will give me the position. If I take the angular position and feed it back to my damping coefficient calculator, that should work, right?

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It looks to me as if you are calculating the torque from:

$$ T = C \frac{d\theta}{dt} $$

Damping

The angle $\theta$ is dimensionless, so the units of $d\theta/dt$ are $s^{-1}$. The units of torque are force times distance, $Nm$, so the units of $C$ must be $Nms$ so you get:

$$ Nm = Nms \times 1/s $$

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  • $\begingroup$ Thanks a lot for your answer. If I know my omega (angular velocity) and multiply it with -T which is torque I will be fine right? Sorry I am sounding like an idiot again. I am good at digital logic, but not very good in physical modelling :( $\endgroup$ – hagubear Oct 4 '13 at 11:47
  • $\begingroup$ @hagubear: It's not obvious that angular velocity multipled by torque is useful. $torque / \omega$ will give you $C$, or $c * \omega$ will give you torque. Incidentally your equation for viscous_fric_mom is torque multiplied by angular displacement, which is the work done i.e. the angular equivalent of force times distance. $\endgroup$ – John Rennie Oct 4 '13 at 12:11
  • $\begingroup$ Actually angular velocity multipled by torque is of course the rate of doing work or the power. Whether that's useful to you depends on the context I guess. $\endgroup$ – John Rennie Oct 4 '13 at 14:12
  • $\begingroup$ @JohnRennnie thanks. I have updated my question a bit to give you some more idea on what I am trying to do $\endgroup$ – hagubear Oct 4 '13 at 21:26
  • $\begingroup$ Also, I believe what you are trying the say that theoretically that is how I can achieve the coefficient. However, there might be some other moments involved that will affect the calculation? $\endgroup$ – hagubear Oct 4 '13 at 21:33

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