I have a volume mesh which is actually a tetrahedral mesh. I would like to calculate the cross-section of this mesh given a plane function, saying z = 0. I can imagine that the cross section of a tetrahedron is either a triangle or a quadrilateral. For the first case, triangle, once I calculate the 3 cross points I can get it; but for the second case, how can I make the quadrilateral become 2 triangles? My problem is I cannot determine the diagonal of the quadrilateral.
Intersect all tetrahedron edges by the plane. You will get 3 or 4 intersection points.
If 3 points then a single triangle.
If 4 points, they form a convex quadrilateral. Take any 3 points, that form a first triangle. The other triangle if formed of the fourth point and the two endpoints of the edge that has this point to its right.
Alternatively (for a more general solution), tag the intersection points with the indexes of the faces incident on the edge, and reconstruct the ring of labels.
Ex: edges are common to faces AB, CD, DA and BC; then the section is ABCD.
This answer outlines a general volume-plane intersection algorithm. It will return the vertices of the intersection in order, so it's easy to determine the diagonal of your quadrilateral.
Related
Given a triangular mesh A in 3D space. Rotate and translate all its points to generate a new mesh B.
How to determine the equality of A and B, just by their vertices and faces?
Topology of the mesh is not important, I only care about the geometric equality, A and B should be equal even if their triangulation are changed. It is
something like the transform in-variance problem for triangular mesh, only translate and rotation is considered.
To complete #Spektre's answer, if the two meshes are not exactly the same, that is there is at least a pair of nodes or edges which does not perfectly overlap, You can use the Hausdorff distance to quantify the "difference" between the two meshes.
Assuming triangle faces only.
compare number of triangles
if not matching return false.
sort triangles by their size
if the sizes and order does not match between both meshes return false.
find distinct triangle in shapes
So either the biggest or smallest in area, edge length or whatever. If not present then you need other distinct feature like 2 most distant points etc ... If none present even that then you need the RANSAC for this.
Align both meshes so the matching triangles (or feature points) will have the same position in both meshes.
compare matching vertexes
so find the closest vertex form Mesh A to each vertex in mesh B and if the distance of any them cross some threshold return false
return true
In case meshes has no distinct features for 3 you need to either use brute force loop through all combinations of triangles form A and B until #4 returns true or all combinations tested or use RANSAC for this.
There are alternatives to #3 like find the centroid and closest and farthermost points to it and use them as basis vectors instead of triangle. that requires single vertex or close group of vertexes to be the min and max. if not present like in symmetrical meshes like cube icosahedron, sphere you're out of luck.
You can enhance this by using other features from the mesh if present like color, texture coordinate ...
[Edit1] just a crazy thinking on partial approach without the need of aligninig
compute average point C
compute biggest inscribed sphere centered at C
just distance from C to its closest point
compute smallest outscribed sphere centered at C
just distance from C to its farthest point
compare the radiuses between the shapes
if not equal shapes are not identical for sure. If equal then you have to check with approaches above.
The paper On 3D Shape Similarity (Heung-yeung Shum, Martial Hebert, Katsushi Ikeuchi) computes a similarity score between two triangular meshes by comparing semiregular sphere tessellations that have been deformed to approximate the original meshes.
In this case, the meshes are expected to be identical (up to some small error due to the transformation), so an algorithm inspired by the paper could be constructed as follows:
Group the vertices of each mesh A, B by the number of neighboring vertices they have.
Choose one vertex V_A from mesh A and vertex V_Bi from mesh B, both with same number of neighbors.
The vertex and its N neighbors V_n1...V_nN form a triangle fan of N triangles. Construct N transforms which take vertex V_Bi to V_A and each possible fan (starting from a different neighbor V_Bn1, V_Bn2, ..., V_BnN) to V_An1...V_AnN.
Find the minimum of the sums of distances from each vertex of B to the closest vertex to it in A, for each N transforms for each vertex V_Bi.
If a sum of near zero is found, the vertices of the transformed mesh B coincide with vertices of A, a mapping between them can be constructed, and you can do further topological, edge presence or direction checks, as needed.
I have a problem to prove if a number of 4 points can form a convex hull and if a 5th points is inside the hull or not. I managed to solve the problem of the convex hull using triangles, however I don't know how to prove if the 5th point is inside the hull created by the other 4 points.
Any ideas?
Thank you!
Assuming that you're asking about points on a plane - there is a standard approach for any polygon (convex or not) with any number of vertices.
Call your points a,b,c,d. They form a tetrahedron.
If the tetrahedron is not degenerated, the points are the vertex points (extreme points) of their convex hull.
To check this, check if a,b,c form a non-degenerated triangle. To do this compute the normal vector (call it n) of the triangle (b-a) cross (c-a). If it is zero then the triangle is degenerated, otherwise it is not..
The check if d is not in the plane of this triangle. It is if (d-a) dot product with n is zero. If so, the tetrahedron is degenerated.
Compute the normal vector for each of the 4 triangles of the tetrahedron.
Each normal vector together with (any) point of the triangle describes a half-space;
that is, all the points on one side of a plane.
Let n be the normal vector and p a point of the triangle.
A point q is said to be inside of the half-space, if n dot (q - p) is negative.
Check for all 4 triangles, if the forth point is inside. If not, adjust the sign of the normal vector.
A point is inside the tetrahedron if it is inside for all 4 half-spaces.
If you were able to find the convex hull of the four points, then you are able to split the quadrilateral in two triangles by a diagonal (unless one of the points was interior, then you have a single triangle left).
So it all amounts to a point-in-triangle test.
Again, being able to construct a convex hull, you must already have the area test handy (sign of the area of a triangle). You fifth point is inside a given triangle when the area test reports the same sign for the three triangles formed by the point and the three edges of the given triangle.
I want to determine whether a given triangle intersects a tetrahedron. I do not want to compute the solution polygon(if it exists) itself. Is there any library/package/published practical algorithm which can be used to solve this problem directly(unlike my attempt below)?
I think as a last resort I will have to use standard polygon-polygon intersection algorithm implementations to solve this problem indirectly.
My attempt on this problem:
I thought of breaking it into the problem of polygon-polygon intersection. So for each triangular face(say T1) of the tetrahedron and the given triangle T2, I thought of doing the following:
Compute intersection(a line) between planes corresponding to each triangle, say L1.
For each triangle:
For each edge of the triangle say L2, compute point of intersection P between L1 and L2.
Test(maybe using parametric form) of L2, if the point lies on the edge L2.
If for both triangles T1 and T2, there exists at least one edge on which the intersection point P lies, then it implies that the triangles(and hence the given tetrahedron and the triangle T2) do intersect.
Create an orthonormal frame based on the triangle (origin at some vertex, axis X using the direction of some side, axis Y and Z using the direction of another side and Gram-Schmidt formulas).
Transform the coordinates of the 3 + 4 vertices to the new frame.
If all Z of the 4 tetrahedron vertices are of the same sign, there is no intersection. Otherwise, find the 3 or 4 intersection point of the edges into XY, by linear interpolation on Z.
Now you need to check for intersections between a triangle and a triangle or (convex) quadrilateral, in 2D. You can solve that by means of a standard clipping algorithm, made simple by convexity of the polygons.
As an easy optimization, note that it is useless to compute the Z of the triangle vertices (=0), and the XY of the tetrahedron vertices before you know that there is a possible intersection.
You can also speedup the polygon intersection process by first using a bounding box test.
I just found a function in CGAL library CGAL::do_intersect(Type1< Kernel > obj1,Type2< Kernel > obj2 ) for computing intersection between two geometric objects. It permits Type1 and Type2 to be Triangle_3<Kernel> and Tetrahedron_3<Kernel> respectively.
the polygons image
All of the polygons are simply, there are no holes in them.
board polygon(P0 to P7)
Red polygon (R0 to R6)
Green polygon (G0 G1 P2 G3)
Yellow polygon(Y0 to Y3)
I want to got new four polygons marked as 1 to 4 , polygon 1's coordinates are(J7 J10 R5 R4).
When I use polygon clipping algorithm, I can got the results easy , board diff(red union green union yellow). But when I have more than 10,000 polygons, I need a long time to get my results. My polygons are simply and my result polygons are simply also, there are no holes in the result polygons also.
You know I can find out the four polygons form the image easy using eyes, but how to find them using algorithm.
Thanks.
If all vertices of your computed black polygons do not have more than 2 edges intersecting at the vertex, then there may be a faster way than a more general tool.
Since the number of polygons is on the order of 10000, first try computing the intersection points of all pairs of polygons, and hopefully the number of intersection points is small enough (like 10 million or less). Then, for each intersection point test to see if it is contained in the interior of another polygon (in case you have multiple polygons that jointly overlap). Testing to see if a point is contained in the interior of a polygon can be done quickly, you can read how online. Then, all intersection points that are not contained in another polygon, which note also contains all the original polygon vertices that are not contained in the interior of a polygon, these are the vertices for the "black" polygons you want to compute. These points should be stored with a secondary structure that for each polygon edge, it stores all the stored intersection points along that edge, in sorted order. Similarly, for each stored vertex and intersection point you should store the edges that intersect at that point, and the location of the intersection point in the previous structure. So then pick any stored intersection point that hasn't been used yet in a black polygon, and pick one edge that defines the intersection point. Then you move to the neighboring intersection point along the edge which has the property that the part of the edge between the two intersection points does not pass inside a polygon. Then continue by similarly moving along the other edge that defines the neighboring intersection point. Continue until you reach the vertex you started at; this defines one black polygon. Then you can pick a new unused stored intersection point and repeat. Since your polygons have no holes, this will find all black polygons.
Is there a computationally efficient way to determine the points of intersection of a straight line with all the edges of a given Voronoi tessellation in a rectangular plane area?
Thanks
Once you have your first intersection point, the rest is easy.
Prepare a database of edges: for each edge, list both polygons it belongs to, or say it is the outer edge (and so belongs to one polygon only). In your picture, the lower side of the rectangle would contain 4 edges of 4 different polygons.
Draw your line, find your first intersection point ([0, 0.25] in your picture, not circled). Say it's polygon A. Then the next intersection point (the lowest one circled in your picture) also belongs to A. You find the relevant edge with a binary search through the list of edges of A.
Now you have found the second edge of A, find out which other polygon it belongs to. Then use binary search to find which other edge of that other polygon the line intersects. And so on until you exit your rectangle.