I am currently working on a hole detection problem in 3D point cloud data. I am referring to this paper "Detecting Holes in Point Set Surfaces" by Gerhard H Bendels, Ruwen Schnabel and Reinhard Klein. One of the criterions mentioned is an Angle Criterion in which we need to determine angles between consecutive points in a KD Tree(Radially Nearest Neighbors to a given point).
See Image:
Angle between points
I am using Open3D to extract a KD Tree but I believe it is giving me an unsorted list of points rather than a list of consecutive points.
See Image:
List of Nearest neighbors
The point below '______' is the point of interest and rest are it's neighbors. Now my question is,
How do I know which point is next to which point?
And if that's not possible to know, How can I find the angles as shown in the first image.
I just need the angles to find the boundary probability for each point, so an answer would really help me progress.
Thanks
What I've Tried so far..
I have tried generating vectors out of all the points and calculated the angles using dot product. But that seems wrong because I believe I may be calculating dot products between first and third point.
Related
I start this thread asking for your help in Excel.
The main goal is to determine the coordinates of the intersection point P=(x,y) between two curves (curve A, curve B) modeled by points.
The curves are non-linear and each defining point is determined using complex equations (equations are dependent by a lot of parameters chosen by user, as well as user will choose the number of points which will define the accuracy of the curves). That is to say that each curve (curve A and curve B) is always changing in the plane XY (Z coordinate is always zero, we are working on the XY plane) according to the input parameters and the number of the defining points is also depending by the user choice.
My first attempt was to determine the intersection point through the trend equations of each curve (I used the LINEST function to determine the coefficients of the polynomial equation) and by solving the solution putting them into a system. The problem is that Excel is not interpolating very well the curves because they are too wide, then the intersection point (the solution of the system) is very far from the real solution.
Then, what I want to do is to shorten the ranges of points to be able to find two defining trend equations for the curves, cutting away the portion of curves where cannot exist the intersection.
Today, in order to find the solution, I plot the curves on Siemens NX cad using multi-segment splines with order 3 and then I can easily find the coordinates of the intersection point. Please notice that I am using the multi-segment splines to be more precise with the approximation of the functions curve A and curve B.
Since I want to avoid the CAD tool and stay always on Excel, is there a way to select a shorter range of the defining points close to the intersection point in order to better approximate curve A and curve B with trend equations (Linest function with 4 points and 3rd order spline) and then find the solution?
I attach a picture to give you an example of Curve A and Curve B on the plane:
https://postimg.cc/MfnKYqtk
At the following link you can find the Excel file with the coordinate points and the curve plot:
https://www.mediafire.com/file/jqph8jrnin0i7g1/intersection.xlsx/file
I hope to solve this problem with your help, thank you in advance!
kalo86
Your question gave me some days of thinking and research.
With the help of https://pomax.github.io/bezierinfo/
§ 27 - Intersections (Line-line intersections)
and
§ 28 - Curve/curve intersection
your problem can be solved in Excel.
About the mystery of Excel smoothed lines you find details here:
https://blog.splitwise.com/2012/01/31/mystery-solved-the-secret-of-excel-curved-line-interpolation/
The author of this fit is Dr. Brian T. Murphy, PhD, PE from www.xlrotor.com. You find details here:
https://www.xlrotor.com/index.php/our-company/about-dr-murphy
https://www.xlrotor.com/index.php/knowledge-center/files
=>see Smooth_curve_bezier_example_file.xls
https://www.xlrotor.com/smooth_curve_bezier_example_file.zip
These knitted together you get the following results for the intersection of your given curves:
for the straight line intersection:
(x = -1,02914127711195 / y = 23,2340949174492)
for the smooth line intersection:
(x = -1,02947493047196 / y = 23,2370611219553)
For a full automation of your task you would need to add more details regarding the needed accuracy and what details you need for further processing (and this is actually not the scope of this website ;-).
Intersection of the straight lines:
Intersection of the smoothed lines:
comparison charts:
solution,
Thank you very much for the anwer, you perfectly centered my goal.
Your solution (for the smoothed lines) is very very close to what I determine in Siemens NX.
I'm going to read the documentation at the provided link https://pomax.github.io/bezierinfo/ in order to better understand the math behind this argument.
Then, to resume my request, you have been able to find the coordinates (x,y) of the intersection point between two curves without passing through an advanced CAD system with a very good precision.
I am starting to study now, best regards!
kalo86
I am currently working on a project that involves measuring distances all around a robot with a laser module, the robot then has to move based on the points that he gets.
I currently have access to 360 points that represent the distance from the center for each of the corresponding angles. (a distance for 0°, a distance for 1°, etc)
Here's an example of what the points look like when displayed on a 2D surface:
Circular representation of the points
What I'd like to be able to do is, rather than feeding the robot all 360 points, to feed it segments containing multiple points. For instance, the bottom part of the image would be a single segment even though the points are not completely aligned.
My question to you is, is there an existing algorithm that would help me achieve what I am trying to do?
(I'm working in python but that shouldn't really be a factor)
Thanks a lot.
Assuming your points are ordered:
For each point, look ahead by two points, if the middle point is less than some distance away from the segment between the two points, then push your endpoint 1 pt further, and check that now both of the middle points are still within some distance of your line segment. Proceed to do this until false, at which point roll back one pt and generate a segment, then set the end of that segment as the start of your next segment. Also, you could consider angles instead of just distances as there are some cases where that would be favorable. Also, if no segment can be made from a certain start point for several attempts, push the start point forward one (as not everything is going to simplify into segments)
Alternately, you could convert to Cartesian points and use the hough voting algorithm to detect lines from the resulting point-cloud.
I would like to find two vertices of two meshes (1 vertex per mesh) that define the closest distance between them. Or the two triangles would be fine I guess.
However I'm not sure how to search for this in CGAL's documentation, I'm sure that this is doable with some existing tool (probably based on a 3d distance field and/or AABBs). Could I please get a hint (keywords/link) on what to look for?
I've been pointed to the Optimal Distances CGAL package, but it's not exactly what I want, since it outputs the distance and the coordinates, so finding the vertex ID is an additional computational overhead.
I've already implemented a collision detection with CGAL to find the triangle-triangle intersection in a triangle-soup, using AABB-trees. I guess that I should be somehow close to this, although now a simple soup with all me object-triangles wouldn't do the job.
The solution found was this:
CGAL's Optimal Distances package can give an approximation of the closest distance between the convex hulls of two meshes, without explicitly computing the hulls. As a result one gets the shortest distance between these hulls, and the coordinates of the 2 points that lie on them and define this distance.
Then these coordinates can be used as a search-query in kd-trees that contains the original vertices of the meshes in order to find the closest vertices.
In case one mesh is non-convex, the hull that CGAL is using is very approximate, so convex decomposition might be necessary. In such a case one would have to check distances for each convex part and then take the shortest distance.
The above would result in something like this:
enter link description here
There are N points on a 2D grid (x,y). I need to find the shortest path, from point A to point B, but I can only travel from one point to another and I can't travel between two points if the distance between them is farther than a distance D. I thought it might be solved by using some kind of modified Dijkstra's algorithm, but I'm not sure how, because I've never implemented it before, just studied it on Wiki.
Well, Dijkstra finds shortest paths in graphs. So just consider the grid points to be nodes in a graph with edges between each node S and all other nodes T such that dist(S, T) <= D. You don't have to actually construct the graph because the edges are easily determined as needed by Dijkstra. Just check all nodes in a square around S with radius D. A S-T edge exists iff (Sx - Tx)^2 (Sy - Ty)^2 <= D^2.
Wiki explanation is sufficient for this.
Dijkstra's algorithm takes 3 inputs. The Graph, Starting node and Ending node.
To construct the graph just do this
For i 1..n in points
For j i+1..n in points
if(dist(points[i],points[j])<=D)
add j to childs of i
add i to childs of j
After constructing the graph, perform dijkstra.
The subtlety of a question like this lies in a critical definition - what is the measure of distance in your grid?
The are many different shortest path problems and solutions, and they are studied throughout mathematics. They are each characterised by the 'topology' of the area being searched. Consider a few distinct topologies with their own solutions:
A one sided piece of paper
Suppose your grid represents coordinates on a piece of paper - the shortest path is easy to find, as it is simply a straight line between those points.
The surface of the moon
If your grid represents locations on the moon in terms of latitude and longitude, the shortest path is an arc along the moon's surface - If you drove "in a straight line" between two points on the moon, you would be travelling in an arc, because of the moon's curvature.
Road Intersections
If you want to find the distance between two intersections in a grid of roads, where the traffic on each road has a different speed, and you can only travel along the roads, then you can find the shortest path using Dijkstra's algorithm.
One way road intersections
A slight variation of the above - we only need to consider roads in one direction. There might not be any paths in this case.
Summary
To give a good solution, we need to understand the topology of your grid. If the distance is pythagerous's theorem than that indicates euclidean geometry (like in the piece of paper example), so the solution is a straight line.
Is it possible you mean that you can travel between any two points if the are closer than D - like flying a plane between airports, for example?
EDIT: I didn't see your comment because you didn't use #. In your case your grid is like the airports a plane can fly between. The shortest path is found using Dijkstra's algorithm - the immediate neighbours of a point are all points closer than D. Find them, represent it all as a graph, and use Dijkstra's algorithm.
I would suggest using the formula to find the distance between 2 points i.e sqrt((x2-x1)^2+(y2-y1)^2). This distance is always the shortest between 2 points.
I am not good with math,and i just need someone to point me in the right direction.
Latitude Longitude
N 36° 13.488' W 095° 54.295'
N 36° 13.488' W 095° 53.805'
Assume that all three are located on a flat plane, at the same elevation.
Assume that the curvature of the earth is not a factor.
Assume that there are exactly 69.1691 miles per degree of latitude.
Assume that there are 55.9588 miles per degree of longitude (Tulsa area only)..
I am trying to figure out what the last points coordinate is.
Can anyone help. I just dont know where to begin
There are numerous ways to find the third point. The angles in an equilateral triangle are all 60 degrees. The third point lies on the bisector of the line connecting the two points you have. Expressing the points and lines using vectors, rather than coordinates, helps.
But the truth is, if you do not have enough mathematical knowledge for this kind of problem, you probably ought not to be tackling it as a programming problem. How can you know you've done it right?