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## Hot answers tagged computational-geometry

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The underlying problem is that the mesh really isn't the geometry. You want to simulate a bridge? It has a certain geometry, which you can approximate using a mesh, but the mesh is not the exact geometry. The same is true for most other curved objects. There are of course approaches to integrate the geometry into the finite element mesh. In particular, I'd ...

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I'm in the middle of doing this for myself. I think the most appropriate analogue to the Gaussian would be the heat kernel in hyperbolic space. Fortunately, this has been figured out before: https://www.math.uni-bielefeld.de/~grigor/nog.pdf (also available in a Bulletin of the London Mathematical Society). If you use the standard decay (e^{-dist^2/constant}... 3 Dont expand the divergence term, apply integration by parts without doing this $$\int_\Omega w \nabla\cdot(\alpha\nabla u) dx = - \int_\Omega \alpha \nabla u \cdot \nabla w dx + \int_{\partial\Omega} w \alpha \frac{\partial u}{\partial n}ds$$ In your notation, define a vector field $$\vec{A} = \alpha \nabla u$$ and then do the integration by parts on $$\... 3 Desmos seems to be picky about variable names. The moment I renamed \theta into t, the plotting started to work correctly. Desmos plot link 3 I don't know whether there is a way to do this with VTK functionality, but what you have here is a well-understood question. If you think of all of the vertices of the mesh as the nodes of a graph, where nodes are connected by the edges of the mesh, then your black line is a path through that graph. In essence, you are asking whether a node is on one side of ... 3 The duality between Voronoi cells and vertices of the triangulation is pretty clear: each vertex of the Delaunay triangulation is a site in the Voronoi diagram which gets associated with its Voronoi cell. To understand the duality between Delaunay triangles and Voronoi vertices, start by looking at this image from https://stackoverflow.com/questions/... 3 In the general case, this cannot be done in O(n+k) time, as there can be as many as 4\lfloor\frac{n}{2}\rfloor\lfloor\frac{k}{2}\rfloor, i.e., Θ(nk) intersection points. If the two polygons are convex, however, you can find the intersections in O(n+k) time. See Computational Geometry in C Second Edition, page 253, or Shamos (1978), page 116. 2 You can use FEniCS: from fenics import ( UnitSquareMesh, FunctionSpace, Expression, interpolate, assemble, sqrt, inner, grad, dx, TrialFunction, TestFunction, Function, solve, DirichletBC, DomainBoundary, MPI, XDMFFile, ) # Create mesh and define function space mesh = UnitSquareMesh(100, ... 2 What about this: Step 1: For each edge P_iP_j, look at the two triangles that share it: P_iP_jP_k and P_lP_jP_i. Step 2: Compute the counter-clockwise normal on each of the two triangle:$$\vec{n}_{ijk} = \frac{\vec{P_iP_j} \times \vec{P_jP_k}}{\|\vec{P_iP_j} \times \vec{P_jP_k}\|} \,\,\, \text{ and }\,\,\, \vec{n}_{lji} = \frac{\vec{P_lP_j} \times ... 2 It seems that the Jacobian ratio is defined as the ratio between the maximum and minimum Jacobian determinant in an element [1, 2]. And, that a value between 0.33333 and 1 is good-enough [3]. Nevertheless, for linear elements, the Jacobian is constant and thus the same over each element. As mentioned by @GustavoCosta, 3 descriptors commonly used for element ... 2 A possibility is to use Lloyd's method [1], it works as follows: Choose N points randomly in the input geometry Compute the Voronoi diagram of the points clipped by the input geometry Replace each point by the centroid of its Voronoi cell If not converged, goto 2 The algorithm is implemented in my geogram library [2] (works for an arbitrary geometry, ... 2 This is an interesting question, but also very problem dependent. For many scientific problems, you like to have properties like good angles (often preferring Delaunay meshes), and feature aligned triangulations, e.g., meshing edges in the domain as edges in the triangulation. In addition there are different requirements on the element size. Some problems ... 2 The infinite square well potential problem in non-relativistic quantum mechanics has energy eigenvaluesE_n=n^2\hbar^2\pi^2/2mL^2$,where$n^2=\sum_{k=1}^Nn_k^2$($N$=number of dimensions). A problem of interest is, given an energy eigenvalue, the degeneracy(i.e,number of eigenstates with same eigenvalue) are to be found.This amounts to counting the number ... 2 I am going to assume that we have arrays of the edges representing the top and bottom curves for the winding polygon with edges going from left to right. Also make$n$as the total number of edges in this polygon. Now consider the following visualization of the geometry where we construct some point using the two "sides" of the concave polygon: It ... 2 I'm sure there are better solutions than this, but since no one else has answered to this point, I'll throw out a this-is-what-I'd-do answer. Triangulate the polygon If your polygon doesn't have too many points, a simple$\mathcal{O}(N^2)$ear-clipping method could be viable. For large polygons, this might be an inefficient solution. It's important to the ... 2 You could use a medial axis transform if the transform is discretized, each point in the transform indicates the radius from that point to the nearest two edges. Doubling this gives the width. To deal with noise, you could take something like the 95th+ percentile of such points and then average. You could also look into rotating caliper methods: though I ... 2 "So I wonder why can't we use mesh to represent geometry like CG in scientific computation. The geometry is directly represented by a mesh, and then you can do the computation directly on it." ... This is exactly what we do using iso-parametric elements in FEM; we discretise the geometry and field variables using the same mesh. The concept of iso-... 1 Yes, you can use the same mesh to represent the geometry of your domain and to solve the PDE, that is something that you can do. For example, a square can be completely described by two triangles and you can use this mesh to solve your PDE. If the solution of your PDE is constant or linear this mesh would suffice but if you have higher gradients in your ... 1 If you want to create a mesh for scientific computing, is it feasible to start entirely from the perspective of computer graphics? -- slightly adapted from the OP Well, it depends. I mainly work in the field of computational fluid dynamics (CFD). My feeling from my experience in CFD and my limited experience in solid mechanics using the finite element ... 1 As a computer graphics person studying meshing, we care about two things. Element quality, and boundary fidelity. The boundary fidelity is as you mentioned for rendering purposes, but it is also needed for accurate collision detection in simulation. Element quality is required because you can bound error on numerical solutions to elliptic PDE based on how '... 1 @federicopolini is right in his answer: Introduce $$c= \sqrt{a}, d=\sqrt{b}$$ and your optimization problem will now read as follows: $$\min (x-c^2)^2+(y-d^2)^2$$ subject to the constraints $$c+d = 2, \\ c\ge 0,\\ d\ge 0.$$ The inequality constraints are important to ensure that you get a solution that makes sense. Now, you can eliminate$d=...

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From a comment: I suggest you to set $\sqrt{a}:=c$ and $\sqrt{b}:=d$ and then pass the problem in the variables c,d to whatever computational software you are using. I would avoid those non-smooth square roots in the constraints at all costs if it's possible. The general idea (from a very philosophical standpoint; this feels more like a comment than an ...

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The packing of spheres (or non-spherical particles) within a geometry can be achieved quite reliably using the Discrete Element Method. I refer you to the following introduction article (written by me and co-workers) on the topic: https://onlinelibrary.wiley.com/doi/full/10.1002/cjce.23501 Briefly, the discrete element method (DEM) is a molecular dynamics ...

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Based on our discussion in the comments, it seems vtkCellDataToPointData is what you want to convert volume shrinkage, which is stored as cell data, to nodal values or point data. I think it's possible cause volume shrinkage ($\frac{\Delta V}{V}$) is defined as trace of strain tensor and there is no reason that that parameter can't be interpolated to the ...

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The problem on which I originally made that comment is a linear algebra problem: consider the linear matrix equation $$\sum_{i=1}^k A_i X B_i = C,$$ where $A_i,B_i,C \in \mathbb{R}^{n\times n}$ are given, and $X\in \mathbb{R}^{n\times n}$ is the unknown. For $k=2$ this is a generalized Sylvester equation, and can be solved using a Bartels-Stewart-type ...

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