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Damping - square of velocity solution?

Damping - square of velocity solution?

Damping - square of velocity solution?

(OP)
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?   

RE: Damping - square of velocity solution?

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: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Damping - square of velocity solution?

Is the coefficient "c" positive or negative?
 

RE: Damping - square of velocity solution?

(OP)
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.

RE: Damping - square of velocity solution?

Then simply add to de second term a factor sign(y'(t)) to preserve the oposition.

RE: Damping - square of velocity solution?

(OP)
I think I see what you're saying.  Like Morison Equation where the drag term is Cd * w/2g * A * U * |U| ?

RE: Damping - square of velocity solution?

GregLocock.  Your rather abrupt response fails to answer the question in enough detail.
 

RE: Damping - square of velocity solution?

It does however point the original poster in the right direction, which nobody in the preceeding 6 hours had managed to do.

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Damping - square of velocity solution?

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: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

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