On 10/15/2018 05:27 AM, Jane Lee wrote: > > I see what you mean though I am confused how you got to the equation for > the displacement? Surely there needs to be some time dependence > somewhere at least for it to make sense dimensionally?
I may have used the wrong terminology and forgotten a factor of delta t. But if you start here: > > However, I realised that this moves each vertex by the velocity > at that point, > > but I need, for example, a vertex on the top boundary to have > moved as much as > > the ones below it has moved as well as the velocity at the points > itself. ...then you get to an equation like this: > d(x,y,z) = \int_z^0 v(x,y,zeta) dzeta ...where d(...) is the *cumulative displacement* by which you want to move the vertex and v(...) is the individual displacement you get. If you differentiate the formula by z, then you get d/dz d(x,y,z) = -v(x,y,z) which is the advection equation I mentioned in my previous post, and for which it is substantially easier to solve than to compute the integral. Best W. -- ------------------------------------------------------------------------ Wolfgang Bangerth email: [email protected] www: http://www.math.colostate.edu/~bangerth/ -- The deal.II project is located at http://www.dealii.org/ For mailing list/forum options, see https://groups.google.com/d/forum/dealii?hl=en --- You received this message because you are subscribed to the Google Groups "deal.II User Group" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. For more options, visit https://groups.google.com/d/optout.
