In non-photorealistic rendering, line direction matters much of effects. In order to achieve a good effect, I wanna rotate texture to principal curvature directions.
I have 3D mesh model(consist of triangle set). I wanna estimate principle curvature directions for each vertices in the model. You can answer this question or tell me where can I find the methods(such as give out a paper).
The usual way to do this is determining the curvature tensor. See the resources at http://www-sop.inria.fr/members/Pierre.Alliez/demos/curvature/
Related
I understand how to use delaunay triangulation in 2d points?
But how to use delaunay triangulation in 3d points?
I mean I want to generate surface triangle mesh not tetrahedron mesh, so how can I use delaunay triangulation to generate 3d surface mesh?
Please give me some hint.
To triangulate a 3D point cloud you need the BallPivoting algorithm: https://vgc.poly.edu/~csilva/papers/tvcg99.pdf
There are two meanings of a 3D triangulation. One is when the whole space is filled, likely with tetrahedra (hexahedra and others may be also used). The other is called 2.5D, typically for terrains where the z is a property as the color or whatever, which doesn't influence the resulting triangulation.
If you use Shewchuk's triangle you can get the result.
If you are curious enough, you'll be able to select those tetrahedra that have one face not shared with other tetrahedra. These are the same tetrahedra "joined" with infinite/enclosing points. Extract those faces and you have your 3D surface triangulation.
If you want "direct" surface reconstruction then you undoubtly need to know in advance which vertices among the total given are in the surface. If you don't know them, perhaps the "maxima method" allows to find them out.
One your points cloud consists only of surface vertices, the triangulation method can be any one you like, from (adapted) incremental Chew's, Ruppert, etc to "ball-pivoting" method and "marching cubes" method.
The Delaunay tetrahedrization doesn't fit for two reasons
it fills a volume with tetrahedra, instead of defining a surface,
it fills the convex hull of the points, which is probably not what you expect.
To address the second problem, you need to accept concavities, and this implies that you need to specify a reference scale that tells what level of detail you want. This leads to the concept of Alpha Shapes, which are obtained as a subset of the faces.
Lookup "Alpha Shape" in an image search engine.
I am trying to find the the sharp corners of a NURBS curve. For this problem I define a limit curvature. I am trying to find the sections on the curve that has a curvature higher then this value. One option is to interpolate over the curve and calculate curvature for all values but it may take time and some sharp points are likely to be missed. Any ideas about how to find these sections in an effective way?
Computing the derivative of the curvature analytically, I guess that you will find a (terrible) expression with a polynomial at the numerator. A good polynomial solver will allow you to find the roots, hence the extrema, to split the curve in sections with a monotonic curvature, and from there find the precise solutions of k=c by regula falsi or similar.
A much simpler approach is by flattening the curve (convert to a smooth polyline) and estimating the local curvature on all triples of consecutive points (using their circumscribed circle). High curvature sections will probably also be detectable by anomalies in the point density while flattening.
The benefit of flattening over uniform sampling is that it auto-adjusts the point density.
Another idea is to resort to a method of approximation of curves by circular arcs (this can be compared to a second order flattening operation). You will find a few papers on the topic (do not confuse with circle approximation by curves), but usually these methods are complex.
Maybe it is also possible to devise an analytic formula for a lower bound on the NURBS curvature in a given interval and use that to implement a bisection approach.
Here is what I have so far.
I have a 3D model and I made a triangle mesh. Calculated and applied normals to the model too.
I want to apply different textures into the triangle. I also have the direction vector of all the texture I need.
For mapping, I do this:
I just calculate the Dot product of each triangle normal with the texture direction vector of each texture, and start comparing to see which texture could suitable BASED UPON the calculation of dot product.
But I realised that it is not as straight forward as I thought it was. Because two or more, different triangle could be in almost same orientation in 3D space, meaning one could be facing towards me and the other could be facing opposite direction (maybe parallel but different direction).
I think a better question is how do I use the calculated dot-product to distinguish the face of the triangle so I know I know which image/texture should be used ?
If the triangles are facing in opposite directions, the normals will also face in opposite directions, and the dot products will have opposite signs. Therefore the dot product gives you enough information to distinguish between the opposite faces. I can't think of a simple test which would give better results than the dot product.
If I have a mesh of triangles, how does one go about calculating the normals at each given vertex?
I understand how to find the normal of a single triangle. If I have triangles sharing vertices, I can partially find the answer by finding each triangle's respective normal, normalizing it, adding it to the total, and then normalizing the end result. However, this obviously does not take into account proper weighting of each normal (many tiny triangles can throw off the answer when linked with a large triangle, for example).
I think a good method should be using a weighted average but using angles instead of area as weights. This is in my opinion a better answer because the normal you are computing is a "local" feature so you don't really care about how big is the triangle that is contributing... you need a sort of "local" measure of the contribution and the angle between the two sides of the triangle on the specified vertex is such a local measure.
Using this approach a lot of small (thin) triangles doesn't give you an unbalanced answer.
Using angles is the same as using an area-weighted average if you localize the computation by using the intersection of the triangles with a small sphere centered in the vertex.
The weighted average appears to be the best approach.
But be aware that, depending on your application, sharp corners could still give you problems. In that case, you can compute multiple vertex normals by averaging surface normals whose cross product is less than some threshold (i.e., closer to being parallel).
Search for Offset triangular mesh using the multiple normal vectors of a vertex by SJ Kim, et. al., for more details about this method.
This blog post outlines three different methods and gives a visual example of why the standard and simple method (area weighted average of the normals of all the faces joining at the vertex) might sometimes give poor results.
You can give more weight to big triangles by multiplying the normal by the area of the triangle.
Check out this paper: Discrete Differential-Geometry Operators for Triangulated 2-Manifolds.
In particular, the "Discrete Mean Curvature Normal Operator" (Section 3.5, Equation 7) gives a robust normal that is independent of tessellation, unlike the methods in the blog post cited by another answer here.
Obviously you need to use a weighted average to get a correct normal, but using the triangles area won't give you what you need since the area of each triangle has no relationship with the % weight that triangles normal represents for a given vertex.
If you base it on the angle between the two sides coming into the vertex, you should get the correct weight for every triangle coming into it. It might be convenient if you could convert it to 2d somehow so you could go off of a 360 degree base for your weights, but most likely just using the angle itself as your weight multiplier for calculating it in 3d space and then adding up all the normals produced that way and normalizing the final result should produce the correct answer.
I have an array of points. I want to know if this array of point represents a circle, a square or a triangle.
Where should i begin? (i use C#)
Thanks
Jon
Depending on your problem, a good approach for this problem may be to use the Hough transform and all its derived algorithm
It consists in a transformation of the image space to an other space where the coordinate represents the objects parameters (angle and initial point for a line, coordinates of the center and radius for a circle)
The algorithm transforms each point of your array of points in points in the other space. Then you have to search in the new space if some points are prevailing. From these points, you will get the parameters of your object.
Of course, you need to do it once to recognize the lines (so you will know how many lines are in your bitmap and where they are) and to it to recognize the circles (it is not exactly the same algorithm)
You may have a look to this lecture (for Hough Circle Transform), but you could easily find the algorithm for line
EDIT: you can also have a look to these answers
Shape recognition algorithm(s)
Detecting an object on the image based on geometrical form
imagine it is each of these one-by-one and try to fit each of these shapes on the data.. for a square, you could find the four extreme points, and try charting out a square that goes through all of them..
Once you have got a shape in place.. you could measure the distance between each of the points and the part of the shape that is nearest to it.. then square these distances and add them up.. the shape which has the smallest sum-of-squares is probably your best bet
Use the Hough Transform.
I'm going to take a wild stab and say if you have 3 points the shape represents a triangle, 4 points is some kind of quadrilateral, any more than that is a circle.
Perhaps there's more information to your problem you could provide.