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.
Related
So I am using Kinect with Unity.
With the Kinect, we detect a hand gesture and when it is active we draw a line on the screen that follows where ever the hand is going. Every update the location is stored as the newest (and last) point in a line. However the lines can often look very choppy.
Here is a general picture that shows what I want to achieve:
With the red being the original line, and the purple being the new smoothed line. If the user suddenly stops and turns direction, we think we want it to not exactly do that but instead have a rapid turn or a loop.
My current solution is using Cubic Bezier, and only using points that are X distance away from each other (with Y points being placed between the two points using Cubic Bezier). However there are two problems with this, amongst others:
1) It often doesn't preserve the curves to the distance outwards the user drew them, for example if the user suddenly stop a line and reverse the direction there is a pretty good chance the line won't extend to point where the user reversed the direction.
2) There is also a chance that the selected "good" point is actually a "bad" random jump point.
So I've thought about other solutions. One including limiting the max angle between points (with 0 degrees being a straight line). However if the point has an angle beyond the limit the math behind lowering the angle while still following the drawn line as best possible seems complicated. But maybe it's not. Either way I'm not sure what to do and looking for help.
Keep in mind this needs to be done in real time as the user is drawing the line.
You can try the Ramer-Douglas-Peucker algorithm to simplify your curve:
https://en.wikipedia.org/wiki/Ramer%E2%80%93Douglas%E2%80%93Peucker_algorithm
It's a simple algorithm, and parameterization is reasonably intuitive. You may use it as a preprocessing step or maybe after one or more other algorithms. In any case it's a good algorithm to have in your toolbox.
Using angles to reject "jump" points may be tricky, as you've seen. One option is to compare the total length of N line segments to the straight-line distance between the extreme end points of that chain of N line segments. You can threshold the ratio of (totalLength/straightLineLength) to identify line segments to be rejected. This would be a quick calculation, and it's easy to understand.
If you want to take line segment lengths and segment-to-segment angles into consideration, you could treat the line segments as vectors and compute the cross product. If you imagine the two vectors as defining a parallelogram, and if knowing the area of the parallegram would be a method to accept/reject a point, then the cross product is another simple and quick calculation.
https://www.math.ucdavis.edu/~daddel/linear_algebra_appl/Applications/Determinant/Determinant/node4.html
If you only have a few dozen points, you could randomly eliminate one point at a time, generate your spline fits, and then calculate the point-to-spline distances for all the original points. Given all those point-to-spline distances you can generate a metric (e.g. mean distance) that you'd like to minimize: the best fit would result from eliminating points (Pn, Pn+k, ...) resulting in a spline fit quality S. This technique wouldn't scale well with more points, but it might be worth a try if you break each chain of line segments into groups of maybe half a dozen segments each.
Although it's overkill for this problem, I'll mention that Euler curves can be good fits to "natural" curves. What's nice about Euler curves is that you can generate an Euler curve fit by two points in space and the tangents at those two points in space. The code gets hairy, but Euler curves (a.k.a. aesthetic curves, if I remember correctly) can generate better and/or more useful fits to natural curves than Bezier nth degree splines.
https://en.wikipedia.org/wiki/Euler_spiral
Okay first I wasn't sure if this was better suited to the MathSO so apologies if it needs migrating.
I have a 3D grid of points (representing the centers of voxels) with pitch varying in each dimension, but regular. For example resolution may be 100 by 50 by 40 for a cube shaped object.
Giving me nVox = 200,000.
For each voxel - I would like to cast (nVox - 1) rays, ending at the center, and originating from each of the other voxels.
Now there is obviously a lot of overlap here but I am having trouble finding how to calculate the minimum set of rays required. This sounds like a problem that has an elegant solution, I am however struggling to find it.
As a start, it is obvious that you only need to compute
[nVox * (nVox - 1)] / 2
of the rays, as the other half will simply be in the opposite directions. It is also easy in the 2D case to combine all of those parallel to one of the grid axes (and the two diagonals).
So how do I find the minimum set of rays I need, to pass from all voxel centers, to all others?
If someone could point me in the right direction that'd be great. Any and all help will be much appreciated.
Your problem really isn't about three dimensions in any specific way. All the conceptual complexity is present in the two dimensional case.
Instead of connecting points individually, think about the set of lines that pass through at least two points on your grid. Thus instead of thinking about points initially, think about directions. For 2-D these directions are slopes of lines. These slopes have to be rational numbers, since they intersect points on an integer lattice. Since you have a finite lattice, the numerator and the denominator of the slope can be bounded by the size of the figure. So your underlying problem is enumerating possible slopes for rational numbers of bounded "height" (math jargon).
There's an algorithm for that. It's the one used to generate the Farey sequence of reduced fractions. If your figure is N pixels wide, there will (in general) be a slope with denominator N in the somewhere, but there can't be a slope in reduced form with denominator >N; it wouldn't fit.
It's easier to deal with slopes between 0 and 1 directly. You get the other directions by two operations: negating the slope and by interchanging axes. For three dimensions, you need two slopes to define a direction.
Given an arbitrary direction (no necessarily a rational one as above), there's a perpendicular linear space of dimension k-1; for 3-D that's a plane. Projecting a 3-D parallelpiped onto this plane yields a hexagon in general; two vertices project onto the interior, six project to the vertices of the hexagon.
For a given discrete direction, there's a minimal bounding box on the integer lattice such that two opposite vertices lie along that direction. As long as that bounding box fits within your original grid, each of the interior points of the projection each correspond to a line that intersects your grid in at least two points.
In summary, enumerate directions, then for each direction enumerate where that direction intersects your grid in at least two points.
I'm writing a data analysis program and part of it requires finding the volume of a shape. The shape information comes in the form of a lost of points, giving the radius and the angular coordinates of the point.
If the data points were uniformly distributed in coordinate space I would be able to perform the integral, but unfortunately the data points are basically randomly distributed.
My inefficient approach would be to find the nearest neighbours to each point and stitch the shape together like that, finding the volume of the stitched together parts.
Does anyone have a better approach to take?
Thanks.
IF those are surface points, one good way to do it would be to discretize the surface as triangles and convert the volume integral to a surface integral using Green's Theorem. Then you can use simple Gauss quadrature over the triangles.
Ok, here it is, along duffymo's lines I think.
First, triangulate the surface, and make sure you have consistent orientation of the triangles. Meaning that orientation of neighbouring triangle is such that the common edge is traversed in opposite directions.
Second, for each triangle ABC compute this expression: H*cross2D(B-A,C-A), where cross2D computes cross product using coordinates X and Y only, ignoring the Z coordinates, and H is the Z-coordinate of any convenient point in the triangle (although the barycentre would improve precision).
Third, sum up all the above expressions. The result would be the signed volume inside the surface (plus or minus depending on the choice of orientation).
Sounds like you want the convex hull of a point cloud. Fortunately, there are efficient ways of getting you there. Check out scipy.spatial.ConvexHull.
sorry for posting this in programing site, but there might be many programming people who are professional in geometry, 3d geometry... so allow this.
I have been given best fitted planes with the original point data. I want to model a pyramid for this data as the data represent a pyramid. My approach of this modeling is
Finding the intersection lines (e.g. AB, CD,..etc) for each pair of adjacent plane
Then, finding the pyramid top (T) by intersecting the previously found lines as these lines don’t pass through a single point
Intersecting the available side planes with a desired horizontal plane to get the basement
In figure – black triangles are original best fitted triangles; red
and blue triangles are model triangles
I want to show that the points are well fitted for the pyramid model
than that it fitted for the given best fitted planes. (Assume original
planes are updated as shown)
Actually step 2 is done using weighted least square process. Each intersection line is assigned with a weight. Weight is proportional to the angle between normal vectors of corresponding planes. in this step, I tried to find the point which is closest to all the intersection lines i.e. point T. according to the weights, line positions might change with respect to the influence of high weight line. That mean, original planes could change little bit. So I want to show that these new positions of planes are well fitted for the original point data than original planes.
Any idea to show this? I am thinking to use RMSE and show before and after RMSE. But again I think I should use weighted RMSE as all the planes refereeing to the point T are influenced so that I should cope this as a global case rather than looking individual planes….. But I can’t figure out a way to show this. Or maybe I should use some other measure…
So, I am confused and no idea to show this.. Please help me…
If you are given the best-fit planes, why not intersect the three of them to get a single unambiguous T, then determine the lines AT, BT, and CT?
This is not a rhetorical question, by the way. Your actual question seems to be for reassurance that your procedure yields "well-fitted" results, but you have not explained or described what kind of fit you're looking for!
Unfortunately, without this information, your question cannot be answered as asked. If you describe your goals, we may be able to help you achieve them -- or, if you have not yet articulated them for yourself, that exercise may be enough to let you answer your own question...
That said, I will mention that the only difference between the planes you started with and the planes your procedure ends up with should be due to floating point error. This is because, geometrically speaking, all three lines should intersect at the same point as the planes that generated them.
I'm looking for a way to determine the optimal X/Y/Z rotation of a set of vertices for rendering (using the X/Y coordinates, ignoring Z) on a 2D canvas.
I've had a couple of ideas, one being pure brute-force involving performing a 3-dimensional loop ranging from 0..359 (either in steps of 1 or more, depending on results/speed requirements) on the set of vertices, measuring the difference between the min/max on both X/Y axis, storing the highest results/rotation pairs and using the most effective pair.
The second idea would be to determine the two points with the greatest distance between them in Euclidean distance, calculate the angle required to rotate the 'path' between these two points to lay along the X axis (again, we're ignoring the Z axis, so the depth within the result would not matter) and then repeating several times. The problem I can see with this is first by repeating it we may be overriding our previous rotation with a new rotation, and that the original/subsequent rotation may not neccesarily result in the greatest 2D area used. The second issue being if we use a single iteration, then the same problem occurs - the two points furthest apart may not have other poitns aligned along the same 'path', and as such we will probably not get an optimal rotation for a 2D project.
Using the second idea, perhaps using the first say 3 iterations, storing the required rotation angle, and averaging across the 3 would return a more accurate result, as it is taking into account not just a single rotation but the top 3 'pairs'.
Please, rip these ideas apart, give insight of your own. I'm intreaged to see what solutions you all may have, or algorithms unknown to me you may quote.
I would compute the principal axes of inertia, and take the axis vector v with highest corresponding moment. I would then rotate the vertices to align v with the z-axis. Let me know if you want more details about how to go about this.
Intuitively, this finds the axis about which it's hardest to rotate the points, ie, around which the vertices are the most "spread out".
Without a concrete definition of what you consider optimal, it's impossible to say how well this method performs. However, it has a few desirable properties:
If the vertices are coplanar, this method is optimal in that it will always align that plane with the x-y plane.
If the vertices are arranged into a rectangular box, the box's shortest dimension gets aligned to the z-axis.
EDIT: Here's more detailed information about how to implement this approach.
First, assign a mass to each vertex. I'll discuss options for how to do this below.
Next, compute the center of mass of your set of vertices. Then translate all of your vertices by -1 times the center of mass, so that the new center of mass is now (0,0,0).
Compute the moment of inertia tensor. This is a 3x3 matrix whose entries are given by formulas you can find on Wikipedia. The formulas depend only on the vertex positions and the masses you assigned them.
Now you need to diagonalize the inertia tensor. Since it is symmetric positive-definite, it is possible to do this by finding its eigenvectors and eigenvalues. Unfortunately, numerical algorithms for finding these tend to be complicated; the most direct approach requires finding the roots of a cubic polynomial. However finding the eigenvalues and eigenvectors of a matrix is an extremely common problem and any linear algebra package worth its salt will come with code that can do this for you (for example, the open-source linear algebra package Eigen has SelfAdjointEigenSolver.) You might also be able to find lighter-weight code specialized to the 3x3 case on the Internet.
You now have three eigenvectors and their corresponding eigenvalues. These eigenvalues will be positive. Take the eigenvector corresponding to the largest eigenvalue; this vector points in the direction of your new z-axis.
Now, about the choice of mass. The simplest thing to do is to give all vertices a mass of 1. If all you have is a cloud of points, this is probably a good solution.
You could also set each star's mass to be its real-world mass, if you have access to that data. If you do this, the z-axis you compute will also be the axis about which the star system is (most likely) rotating.
This answer is intended to be valid only for convex polyhedra.
In http://203.208.166.84/masudhasan/cgta_silhouette.pdf you can find
"In this paper, we study how to select view points of convex polyhedra such that the silhouette satisfies certain properties. Specifically, we give algorithms to find all projections of a convex polyhedron such that a given set of edges, faces and/or vertices appear on the silhouette."
The paper is an in-depth analysis of the properties and algorithms of polyhedra projections. But it is not easy to follow, I should admit.
With that algorithm at hand, your problem is combinatorics: select all sets of possible vertexes, check whether or not exist a projection for each set, and if it does exists, calculate the area of the convex hull of the silhouette.
You did not provide the approx number of vertex. But as always, a combinatorial solution is not recommended for unbounded (aka big) quantities.