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One method could be to apply the (modified) Gram Schmidt process. To do this you would need to choose two vectors $v_1$ and $v_2$ not parallel to $nn$ such that $\operatorname{span}(v_1, v_2, nn) = \mathbb{R}^3$. Then set $nn$ as the first vector, $u_1$, in the Gram Schmidt process and the second and third vectors ($u_2$ and $u_3$) will be two orthogonal ...

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Easy: Decompose the polygons into (unoriented) line segments each of which is sorted by vertex index: $$A = \{[1,2], [2,3], [3,4], [4,5], [1,5]\}, \\ B = \{ [1,6], [6,7],[7,8],[3,8], [2,3], [1,2] \}.$$ Then you want to want to consider the union of all of these edges, but removing the duplicated ones. So you need to form  (A \cup B) \setminus (A \...

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Following there are some resources that might be useful. Wikipedia has a List of curves. They are listed according to some classification criteria and link to the article of each curve, where you can further read. Shikin, E. V. (1995). Handbook and atlas of curves. CRC Press. This books presents an atlas where the curves are listed in alphabetical order.

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As suggested in Kindlmann the curvature of a surface is defined by the relationship between positional changes in the neighborhood of a point placed on the surface and the change in the surface normal. Given a level-set $\Phi(\mathbf{x})$, we consider that the value of the level-set is positive inside the object, negative outside. Hence, we define the ...

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I ended up implementing the suggestion by @DanielShapero (or at least how I understood it): import numpy as np def distance(a, b): # a & b contains each the first 6 values of direction cosine matrix ax, ay = a # ax is the first vector, ay is the second vector bx, by = b # bx is the first vector, by is the second vector az = np.cross(ax, ay) # ...

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You can find an algorithm to find a vector that is perpendicular to your first vector $\vec{u}$ here . Starting with this couple of vectors (we will call the vector perpendicular to $\vec{u}$ as $\vec{v}$) you can simply do the cross product to find the vector that is perpendicular to both $\vec{w} = \vec{u} \times \vec{v}$ and hence obtain the couple of ...

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