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Homework Help: Local Interface Curvature

  1. Feb 13, 2010 #1
    1. The problem statement, all variables and given/known data
    I am using the boundary element method to solve unknowns to the Laplace equation from classic potential flow theory for the time evolution of a fluid air interface. At each time step, I need to solve a material derivative equation numerically at every node along an interface to find the new velocity potential.

    In order to calculate the material derivative, I need to calculate the local interface curvature (et al.).


    2. Relevant equations

    From text, the local (mean) interface curvature can be calculated as 2H=div n. Where H is the mean interface curvature and n is the unit normal to the surface.

    3. The attempt at a solution
    The divergence of a vector field is a somewhat trivial calculation, e.g.:
    div F = (dF1/dx+dF2/dy+dF3/z) where each value of F is some function that can be differentiated (pde) (like x*y^2). So here is the problem/question, in the case of the unit normal, the values are scalar values (such as [1 0 1]'), therefore if I differentiate each of these values with respect to the independent variable the entire equation equals zero. No doubt I am missing something fundamental here, any advise would be greatly appreciated.
    I've attached a very simple sketch of a local discretized interface with nodes, and a unit normal just to help visualize what I am working with.
     

    Attached Files:

    Last edited: Feb 13, 2010
  2. jcsd
  3. Feb 13, 2010 #2
    I've figured this out and I'm going to answer my own question.

    Rather than find the unit normal for the point, it is easier to find the equation for the "curve" of interest. In the diagram I attached, I could take 3 points two either side of where I have annotated the normal. From the location of these points I can formulate an equation for the curve from the parabolic equation y=ax^2+bx+c. Since I have three points, I can find the values for a b and c for example;

    y_1=ax_1^2+bx_1+c
    y_2=ax_2^2+bx_2+c
    y_3=ax_3^2+bx_3+c

    which can be solved numerous ways for a, b and c which in turn are plugged back into the original equation.

    From this the curvature can be calculated as H(x)=abs(y")/[1+(y')^2]^(3/2) at any given point and for any curve equation.

    Side note: I spend quite a bit of time on here as a "spectator" generally I stumble across this site while looking for little tidbits of info. Even though I didn't receive any response for my first try here, I would say keep up the good work. This is a great resource.
     
    Last edited: Feb 13, 2010
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