I tried to explain figure below in mathematics form. As you can see I have got triangle (v1, v2, v3).

The signed shortest distance form red interface (level set value) is calculated for each vertex of the triangle. I applied linear interpolation to find a point (blue circle)on the edges with zero value of the level set.

It is clear on the below figure that for the edge v1-v3 the point has been detected correctly. In other words, the result of the linear interpolation is exactly on the interface.

although for the edge v1-v2 the result of the interpolation is not on the interface. the reason (as much as I know) is that level set values for each vertex of the edge v1-v2 is calculated from different parts of the interface (derivative of the interface is not continuous in our triangle).

My question is that, How I can show above explanation in mathematical format.

enter image description here

  • $\begingroup$ Welcome to MSE! After multiple readings, unfortunately, I don't understand your question. Is the red object given at the start? Are the vertices $v_{i}$ given, or constructed in terms of the red object? What are the two marked points, and why does only one lie on the red object? What notion of "shortest distance" are you using/looking for? Generally, it would help if you could explain what data you start with, what data or result you want to achieve, and what logical or geometric constraints the problem has. $\endgroup$ – Andrew D. Hwang Mar 6 '15 at 14:53
  • $\begingroup$ thanks for your comment I edited the question, I hope it is more clear now. it is same as level set method for creation of the iso-contour. $\endgroup$ – user2090491 Mar 6 '15 at 15:00
  • $\begingroup$ I don't understand the question. $\endgroup$ – copper.hat Mar 6 '15 at 17:14
  • $\begingroup$ I edited again :((((( $\endgroup$ – user2090491 Mar 6 '15 at 19:03
  • $\begingroup$ Here's my understanding of your question: Let $L$ be a curve in the plane (the red lines), and let $f$ denote the signed distance from $L$. If $v_{1}$ and $v_{2}$ are points not on $L$, and if $t = f(v_{1})/[f(v_{1} - f(v_{2})]$, when does $(1 - t)v_{1} + tv_{2}$ lie on $L$? (That is, when does the line through $\bigl(v_{1}, f(v_{1})\bigr)$ and $\bigl(v_{2}, f(v_{2})\bigr)$ cross $L$ at height $0$?) Does that seem like a reasonable interpretation? $\endgroup$ – Andrew D. Hwang Mar 6 '15 at 22:00

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