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Damping - square of velocity solution?

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DDP

Structural
Apr 23, 2002
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I am trying to model a simple dampened spring system with a mass under its own weight except I want the damping to be drag, thus proportional to velocity squared.

m y"(t) + c y'(t)^2 + k y(t) = W

When I graph the solution on MathCAD, it never decays, it is an endless oscillation. Any ideas?
 
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Yup. In the real world the damping force always opposes the velocity.

Your rather silly formulation fails to do so and hence pumps energy back into the system.



Cheers

Greg Locock

SIG:please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
 
Energy is not pumped into the system Locock. The amplitude does not grow it stays constant.

My c coefficient is positive (.5*Cd*Area*density), but so is the c when you assume a linear relationship between damping and velocity. They both should have the same effect I would think. When I take out the squared it behaves like it should.
 
I think I see what you're saying. Like Morison Equation where the drag term is Cd * w/2g * A * U * |U| ?
 
OK, the problem is that you need to set up a force that always opposes the velocity, but that is proportional to velocity sqd. Hence ishvaaag's idea is on the right track but wrong.

So, we want -sign(v)*constant*(v^2)

Otherwise, as I intimated, for half of each cycle the OP's formulation pumps energy back into the system, hence it is not destructive, hence the amplitude does not decay. The reason I was negative about it is that (D)^2 is a very unnatural ODE to write, which should have set alarm bells off. Graphing the contribution of each would have revealed that immediately.


Cheers

Greg Locock

SIG:please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
 
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