Calculate signed dihedral angle - graphics

OpenMesh has the calc_dihedral_angle() function to calculate the dihedral angle between two faces. Is there a signed/directed equivalent of this function? Halfedges of faces are directed, thus normals of faces are well defined. It is thus meaningful to talk about the convexity.
Consider the following simple case, of only two connected faces. Starting from a zero dihedral (in-plane neighboring faces) one can rotate one of the faces around the common edge in either direction. In one case, the surface will be convex, in the other case, it will be concave. calc_dihedral_angle() does not differentiate between the two. I am looking for a function which takes this directionality into account and gives either a positive or a negative dihedral, depending on the convexity.

Actually that is exactly what calc_dihedral_angle() does.

Related

Calculating average curvature of a polygonal chain with non-uniform segment length

I'm using a polygonal chain to approximate a curve. I want to approximate the average of a function of curvature of all points that lie on the curve. One function of curvature that I need is, for example, the square of curvature.
I can get near that by choosing some points on the chain, calculating the curvature in those points, applying the function on it (for example squaring it), and then averaging the calculated values.
I need both accuracy and speed. I appreciate both — fast, but approximate; as well as accurate, but slow solutions. I'm working in Java, but the answer doesn't need to be written in Java — it doesn't even need to contain any code at all.
Polygonal chain with uniform segment length
If the polygonal chain's segments all have equal length, I can just calculate the curvature in the vertices and then average that. I see two ways to get the curvature in a vertex.
One way is to get the circle that goes through the selected vertex, the vertex before it, and the one after it. The curvature is then 1/radius of the circle.
The other way is to calculate the external angle (in radians) of the two segments connected at the selected vertex and then divide its absolute value by the length of a segment. In the following image, φ marks the external angle:
I am not sure if this method is correct, as I haven't mathematically derived it, but I've noticed through experimentation that it gives similar results to the above method.
Polygonal chain with non-uniform segment length
Unfortunately, though, there's no guarantee that the segments have uniform length.
If I try using the first of the above methods, vertices connected to longer segments give lower curvatures, even if they are visibly sharper. I tried substituting previous and next vertices with a point x units before the selected vertex and a point x units after it. I don't know what to set the x constant to, to get accurate results. All the values I've tried seemed to give inaccurate results.
If I try using the second method, I don't know what length to divide the angle by. If I don't divide by anything at all, I actually get pretty good results for comparing two curves and determining which one is curvier, but I need to be able to determine the actual curvature in a point.
With both of these methods there's also the problem that parts with shorter segments (where points are denser) will affect the average more.
Another possible solution would be to ignore the vertices and instead use an array of points on the chain that are evenly spaced, treat them as a new polygonal chain (connect the points with straight lines), and then calculate curvatures on this new chain instead, using one of the methods I mentioned under the header titled "Polygonal chain with uniform segment length".
Finding such an array of points is not trivial, though, because I have to choose a segment length, and only after producing the points, I can see if the length of the resulting chain is divisible by the chosen segment length.
If you aren't short on space, the last solution you mentioned would be the best, because the "sphere" approximation, as you've perhaps realized, would give awful results in more extreme cases, especially if the curvature is large or changes sign quickly.
There are many ways to do interpolations, the simplest being quadratic and cubic splines. However if you have more pre-processing time, Lagrange polynomials produce very good results: https://en.wikipedia.org/wiki/Lagrange_polynomial.
Side note on your angle division method, consider this diagram:
(From simple geometry the inside angle there is also theta)
For a << l. So the curvature:
So your approximation is in fact correct for small curvatures.
An alternative is to use a local parabola approximation to estimate the curvature. Basically, to estimate the curvature at point P(i), you take P(i-1), P(i) and P(i+1) and construct a parabola from these 3 points. Then, you compute the curvature at P(i) from the parabola. Remember to use chord-length (or centripetal) parametrization when constructing the parabola.

Point Cloud - Principal Axes - Use of Inertia

I have got point clouds of different primitive objects (cone, plane, torus, cylinder, sphere, ellipsoid). The all vary in orientation, position and scaling. Furthermore all of them are initialized with a unique set of parameters (e.g. height, radius, etc.) so that their shape can be quiet different (some cones are tall, others are small and fat).
Now to my question:
I am trying to find the objects "principal components". Using PCA doesn't lead to good results, since rotated primitives can have their main variation in any direction (which doesn't have to be necessarily along the length of the objects).
The only chance that I see is to use somehow the symmetry of my primitives. Isn't there a method based on inertia? Maybe some way to find the main symmetry axis and two others perpendicular to it?
Can you give me some advice or point me to papers or implementations (maybe even python)?
Thanks a lot, Merlin.
PS: This is what I get if I only apply a PCA. Especially for cones this doesn't really work. Only cones that are almost identical in shape share the same orientation, but I need them all to point in one direction (e.g. up).
So you got cones and just need to rotate them all in the same direction?
If so you can fit a triangle to them and point the peak (e.g with the perpendicular bisectors of the sides) to your main axis.
You have an interesting problem. Normally used shape descriptors (VFH) that are invariant to shape but not pose (which is what you would want, really) would not be invariant to stretching in the shape.
I think to succeed at this you need to be clearer about the invariants that you are trying to maintain when a shape changes. Is it a topological invariant? If so, then here is a good starting point: https://www.google.com.tr/search?q=topologically+invariant+shape+descriptor
I decided to just stick to simple PCA since it's the only method that is totally generic and doesn't depend on prior (expert) knowledge about the data.

How do I check if a set of plane polygones create a watertight polyhedra

I am currently wondering if there is a common algorithm to check whether a set of plane polygones, not nescessarily triangles, contruct a watertight polyhedra. Each polygon has an oriantation (normal vector). A simple solution would just be to say yes or no. A more advanced version would be to point out the edges, where the polyhedron is "open". I am not really interesed on how to close to polyhedra.
I would like to point out, that my "holes" are not nescessarily small, e.g., one face of a cube might be missing. Thus, the "undersampling correction" algorithms dont seem to be the correct approach. Furthermore, I am talking of about 100 - 1000, not 1000000 polygons, so computation time should not really be a problem.
Any hints or tips?
kind regards,
curator
I believe you can use a simple topological test -- count the number of times each edge appears in the full list of polygons.
If the set of polygons define the surface of a closed volume, each edge should have count>=2, indicating that each edge is shared by (at least) two adjacent polygons. If the surface is manifold count==2 exactly.
Edges with count==1 indicate open regions of the surface.
The above answer does not cover many cases. A more correct (but not necessarily complete: I wouldn't know) algorithm is to ensure that every edge of every polygon (or of the mesh/polyhedron) has an even number of faces connected to it. Consider the following mesh:
The segment (line) between the closest vertex and the one below is attached to 3 faces (one one of the outer triangle and two of the inner triangle), which is greater than two faces. However this is clearly not closed.

"ray through vertex" special case when detecting point in polygon

To detect if a point is in a polygon, you project a line from the point, to infinity, and see how many of polygon's vertices it intersects with... simple enough. My problem is that if the ray intersects the polygon on one of the points, then it is counted as intersecting two segments, and considered outside the polygon. I altered my function to make it only count one of the segments when the ray intersects a point of the polygon, but there are cases where a line could intersect the point while still being outside as well. Take this image as an example:
If you assume the point in the top left is "infinity", and cast a ray to either of the other points, both intersect at a point of the polygon, and would count as intersecting the same number of vertices even though one is inside, and one is outside.
Is there a way to compensate for that, or do I just have to assume that those fringe cases won't pop up?
If the ray crosses a side exactly on a vertex, only count that side if the other vertex is above the ray. That will fix your corner case.
For example in the picture you posted, the lower ray crosses two sides of the square at the top-left vertex, but one side is above the ray and the other below, so that contributes 1 and the target point is found to be inside. The upper ray crosses two sides at the top-right vertex, both sides are below the ray, so they contribute 0 to the count and the target point is found to be outside.
Update:
I remembered reading an article which describes a technique for dealing with singular cases in general. Please read my other answer if interested.
While my first answer should do the trick for this simple problem, I can't help but mention that there exist general techniques for dealing with these kinds of special cases.
This article describes a technique for dealing with these kinds of issues in general. And one of the first examples they provide happens to be the algorithm you ask about!
The idea is to apply Automatic differentiation aka Dual numbers to compute symbolic perturbations.
By the way the same technique can also be used to avoid handling 0/0 as a special case in programs!
Here is the blog post I originally learned this from, it gives some great background to the technique, and the author blogs a lot about automatic differentiation (AD).
Despite appearances AD is a very practical technique especially in languages with good support for operator overloading (eg: C++, Haskell, Python ...) and I have used it in "real life" (industrial applications in C++).
Send ray in another direction.
If you try n+1 different directions (n is number of polygon points) one of them surely will not pass through any vertex.
This will simplify the code compared to consideration of corner cases.
Worst case becomes O(n)*CheckComplexity(n) which is likely O(n^2). If it's not acceptable, you can just sort all vertices by direction from the point to them and select middle of some interval. This will give O(n*log n).

Intersecting points with a polygon in OpenCV

My inputs
I have a vector<Point2f> that contains the contours of a polygon. I also have a list of points that need to be intersected with this polygon.
The problem
I want to calculate how much of these points intersect with the polygon. I want to repeat this calculation on a number of polygons to see which one contains the highest number of points.
Does OpenCV implement such intersection functionality of its own or will I need to implement an intersection function myself? I'm worried that if I try to implement it myself, the result will be unnecessarily slow. If OpenCV can't do it, are there other free graphics libraries that can perform this task?
pointPolygonTest does exactly what you're looking for, and it's pretty well optimized. The parameter is a Mat which you can make with the constructor that takes your vector of points.
The function determines whether the point is inside a contour, outside, or lies on an edge (or coincides with a vertex). It returns positive (inside), negative (outside) or zero (on an edge) value, correspondingly. When measureDist=false , the return value is +1, -1 and 0, respectively. Otherwise, the return value it is a signed distance between the point and the nearest contour edge.
Your problem seems easily parallelizable, though, i.e. each batch of candidate polygons could run on a different thread, so I'd definitely look into that if you're concerned about performance.

Resources