Using a time gradient to compute velocity

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SUMMARY

This discussion focuses on computing the normal velocity of an evolving front in two dimensions using a time gradient approach. The user has collected x and y position data as a function of time and fitted a surface defined by t = f(x,y). The key question is whether evaluating the temporal gradient (grad(t)) of this surface and calculating the magnitude of the normal velocity as the reciprocal of grad(t) is physically meaningful. The consensus indicates that the term "reciprocal" should be replaced with "orthogonal" for clarity in defining the relationship between the velocity and the gradient.

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HughJass
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I need to compute the normal velocity of an evolving front in two dimensions (x,y). Let's say that I have collected numerous x and y position data as a function of time. If I plot these data on a set of x,y,t coordinate axes and fit a surface through them in a manner analogous to fitting a curve through 2D data, I've generated a smooth, parametric surface that expresses time as a function of the x and y coordinates, that is t = f(x,y). I can then visualize the evolution of the physical front by plotting time contours of the surface in the xy plane.

If I want to calculate the velocity of the front (the normal velocity associated with each time contour), is it at all physically meaningful to evaluate the temporal gradient (grad(t)) of my surface and calculate the magnitude of the normal velocity as the reciprocal of the magnitude of grad(t)?

Thanks in advance!
 
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If I understood your description correctly, then yes, if you change reciprocal by orthogonal. I'm not quite sure whether you can define a parametric curve ##y(t)=(x_0(t),y_0(t))## of the front point though.
 

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