triangulation

问题 I am experimenting with scipy.spatial's implementation of Qhull's Delaunay triangulation. Is it possible to generate the triangulation in a manner that preserves the edges defined by the input vertices? (EDIT: i.e. a constrained Delaunay triangulation.) As can be done with the triangle package for Python. For example, in the picture below there are several triangles (blue) that disregard the location of the edges (red) that are defined by the vertices. Is there a way to enforce these edges

问题 I am trying to find a user through their Bluetooth strength (RSSI value). I have 3 Raspberry PIs, each gathering the signal strength of the user. Lets say the nodes returned: node1 = 65 node2 = 70 node3 = 75 How would I find the user through triangulation and pin point them on a map, and output the RSSI value? I have researched Trilateration and Ceva's Theorem but do not know how to implement them. I am unsure on how to locate the nodes in an environment, do I give the main node a location of

问题 I am trying to find a user through their Bluetooth strength (RSSI value). I have 3 Raspberry PIs, each gathering the signal strength of the user. Lets say the nodes returned: node1 = 65 node2 = 70 node3 = 75 How would I find the user through triangulation and pin point them on a map, and output the RSSI value? I have researched Trilateration and Ceva's Theorem but do not know how to implement them. I am unsure on how to locate the nodes in an environment, do I give the main node a location of

问题 I wanted to calculate the angle between two triangles in 3D space. The two triangles will always share exactly two points. e.g. Triangle 1: Point1 (x1, y1, z1), Point2 (x2, y2, z2), Point3 (x3, y3, z3). Triangle 2: Point1 (x1, y1, z1), Point2 (x2, y2, z2), Point4 (x4, y4, z4). Is there a way to calculate the angle between them efficiently in CUDA? 回答1: For each plane, you need to construct it's normal vector (perpendicular to all lines in that plane). The simple way to do that is to take the

问题 I am new to using CGAL, and I was wondering if CGAL supports 2D Delaunay triangulation of 3D points using an arbitrary plane. The example on CGAL's documentation lists only Projection_traits_xy_3<R> , Projection_traits_yz_3<R> , and Projection_traits_xz_3<R> , in other words, projection on the xy plane, the yz plane and the xz plane. Is there any way I can define an arbitrary projection plane instead of using the xy, yz and xz planes? thanks, 回答1: There is a template < class Kernel > class

问题 Given a 3D polyhedron defined by its boundary represented by a triangulated mesh, how can I implement an algorithm that determines whether a given 3D point belongs to the interior of the polyhedron ? 回答1: There are several ways of implementing this function. The simplest one is to create an infinite ray (or a very long segment) starting from the point and pointing towards an arbitrary direction, then counting the number of intersections between the ray and the triangles. If the number of

来源： https://stackoverflow.com/questions/64062242/numpy-and-matplotlib-tri-triangulation-to-geotiff-in-python

问题 I'm trying to write a function that when given two cameras, their rotation, translation matrices, focal point, and the coordinates of a point for each camera, will be able to triangulate the point into 3D space. Basically, given all the extrinsic/intrinsic values needed I'm familiar with the general idea: to somehow create two rays and find the closest point that satisfies the least squares problem, however, I don't know exactly how to translate the given information to a series of equations

问题 I'm using a triangulation library to compute the Constrained Delaunay Triangulation of a set of rectangles within some large boundary. The algorithm returns all the edges, but also adds edges inside of the rectangles that define the constraints. I want to be able to create a graph without the edges within any of the rectangles that are the constraints (with the exception of the large boundary of course) but removing these edges in the triangulation that is given to me takes longer than O(nlog

问题 Ok, so I have a polygon (simple but concave) that I'm trying to cut into triangles to make it collide with an other polygon. I knew my polygone was concave, so i decided to use LibGDX EarClippingTriangulator to manage to cut it into triangles. So, with this code, I get my triangles vertices : public void triangulate() { Vector<float[]> trianglesVertices = new Vector<float[]>(); ShortArray pointsCoords = new ShortArray(); EarClippingTriangulator triangulator = new EarClippingTriangulator(); //