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?
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?
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?
RE: Damping - square of velocity solution?
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?
RE: Damping - square of velocity solution?
RE: Damping - square of velocity solution?
RE: Damping - square of velocity solution?
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?
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.