How to find out distance between point along the surface from one point to all other point on the cylinder surface.
XYZ coordinates of points are known
e.g
suppose 4 points are there on the surface of the cylinder here i want to find out following data i.e distance between point(1-2),point(1-3),point(1-4),point(2-3),point(2-4),point(3,4) along the surface.H
What you are looking for is the geodesic of a cylinder. Since you mentioned Matlab in the tag, you can use this tool
Related
Building on the discussions here and here. I'm trying to compute the shortest distance between a 3D line and a 3D triangle.
I'm using barycentric coordinates to determine whether or not the point is inside the triangle. So given a triangle defined by vertices UVW and a line defined by point AB, I first compute the intersection of line AB with the plane defined by UVW. Let's call this intersection P and assume I've already done the checks to verify whether or not the point actually intersects the plane at all.
I then compute barycentric coordinates (S,T) such that S is defined along the edge UV and T is defined along the edge UW. Naturally, if 0≤S and 0≤T and S+T≤1 then P is on the triangle (or its edge) and my distance to the triangle is obviously zero.
If that's not true then P is outside the triangle and I need to compute the distance. The guidance of from the first link says to project point P onto all three edges to get three candidate points. Adding those points to the three triangle's vertices, you then have six points to test against.
Isn't it easier than that, though? If T<0, then don't you already know that UV is the closest edge and you only have to test against the projection of P onto that line? Similarly, if S<0 then UW would be the closest edge. If T>0 and S>0 then VW is the closest edge.
Thus based on the signs of S and T you already know the closest edge and only have to compute the distance from P to its projection onto that edge. If the projection isn't inside the triangle, then the closest point is either vertex. Thus your computations are about 1/3 of the proposed methods.
Am I missing something here, or is this a valid optimization? I'm fairly new to barycentric coordinates and their attributes.
It turns out that the problem of closest distance from a point and from a line are very similar and can both be reduced to a pure 2D problem.
Distance from a point
By Pythagoras, the squared distance from a point to a point of the triangle is the sum of the squared distance to the plane of support of the triangle and the squared distance of the projection of the point to that plane.
The latter distance is precisely the distance from the normal line to the triangle.
Distance from a line
Looking in the direction of the line, you see the projected triangle and the line is reduced to a single point. The requested 3D distance is equal to the 2D distance seen on the projection.
To obtain the desired coordinates, you use an auxiliary coordinate frame such that Z is in the direction of the line (and XY is a perpendicular plane); for convenience, choose the origin of the new frame to be on the line. Then by just ignoring Z, you get a planar problem in XY. The change of coordinates is an affine tranformation.
Point vs. triangle
Consider the three triangles formed by the origin (projection of the point/line) and a pair of triangle vertices (taken in cyclic order). The signed area of these triangles is a mere 2x2 determinant.
If the three areas have the same sign, the point is inside. Otherwise, the signs tell you where you are among the six surrounding regions, either past an edge or past a vertex.
On the upper figure, the point is inside (three positive areas). On the other figure, it is outside of the top-right edge (one negative area). Also note that dividing an area by the length of the corresponding side, you get the distance to the side. (Factor 2 omitted.)
The total work is
compute the affine frame;
convert the coordinates of the 3 or 4 points;
compute the three signed areas;
if inside, you are done;
otherwise, if in an edge region, compute a distance to a line and two distances to points;
otherwise you are in a vertex region, compute two distances to lines and one distance to vertex.
I'm using canny edge detection to detect the edges of a rope and eliminate the background, then using morphological filters I'm filling these edges then thinning them to be in pixel size. The plot indicates the xy coordinates of the rope. What I need to do is to get the intersection of the scattered data (blue *) with the red circle (get the coordinates of Points (1,2,3,4).
Then I need to get the whole points coordinates from the point of intersection (point1,2,3,4) to the center,grouped as A,B,C,D.
The center of the circle, the origin of the axes, and the radius are all known.
I've tried some Matlab functions to get the four intersections points like
InterX,intersections
and also I tried to manually find the intersections
idx=find(y1-y2<eps);
but no method gives me the 4 intersection points.
Thank you in Advance
You'll need a thick circle. I presume (from your previous question) that the rope points are at contiguous integer coordinates. Using a thick circle (donut) with a width of 1 ensures you'll find at least one point in each rope end. Connected component analysis will then tell you which of these points belong to the same rope end.
Basically, I have a many irregular circle on the ground in the form of x,y,z coordinates (of 200*3 matrix). but I want to fix a best circle in to the data of x,y,z coordinates (of 200*3 matrix).
Any help will be greatly appreciated.
I would try using the RANSAC algorithm which finds the parameters of your model (in your case a circle) given noisy data. The algorithm is quite easy to understand and robust against outliers.
The wikipedia article has a Matlab example for fitting a line but it shouldn't be too hard to adapt it to fit a circle.
These slides give a good introduction to the RANSAC algorithm (starting from page 42). They even show examples for fitting a circle.
Though this answer is late, I hope this helps others
To fit a circle to 3d points
Find the centroid of the 3d points (nx3 matrix)
Subtract the centroid from the 3D points.
Using RANSAC, fit a plane to the 3D points. You can refer here for the function to fit plane using RANSAC
Apply SVD to the 3d points (nx3 matrix) and get the v matrix
Generate the axes along the RANSAC plane using the axes from SVD. For example, if the plane norm is along the z-direction, then cross product between the 1st column of v matrix and the plane norm will generate the vector along the y-direction, then the cross product between the generated y-vector and plane norm will generate a vector along the x-direction. Using the generated vectors, form a Rotation matrix [x_vector y_vector z_vector]
Multiply the Rotation matrix with the centroid subtracted 3d points so that the points will be parallel to the XY plane.
Project the points to XY plane by simply removing the Z-axes from the 3d points
fit a circle using Least squares circle fit
Rotate the center of the circle using the inverse of the rotation matrix obtained from step 5
Translate back the center to the original location using the centroid
The circle in 3D will have the center, the radius will be the same as the 2D circle we obtained from step 8, the circle plane will be the RANSAC plane we obtained from step 3
I have a convex polygon in 3D. For simplicity, let it be a square with vertices, (0,0,0),(1,1,0),(1,1,1),(0,0,1).. I need to arrange these vertices in counter clockwise order. I found a solution here. It is suggested to determine the angle at the center of the polygon and sort them. I am not clear how is that going to work. Does anyone have a solution? I need a solution which is robust and even works when the vertices get very close.
A sample MATLAB code would be much appreciated!
This is actually quite a tedious problem so instead of actually doing it I am just going to explain how I would do it. First find the equation of the plane (you only need to use 3 points for this) and then find your rotation matrix. Then find your vectors in your new rotated space. After that is all said and done find which quadrant your point is in and if n > 1 in a particular quadrant then you must find the angle of each point (theta = arctan(y/x)). Then simply sort each quadrant by their angle (arguably you can just do separation by pi instead of quadrants (sort the points into when the y-component (post-rotation) is greater than zero).
Sorry I don't have time to actually test this but give it a go and feel free to post your code and I can help debug it if you like.
Luckily you have a convex polygon, so you can use the angle trick: find a point in the interior (e.g., find the midpoint of two non-adjacent points), and draw vectors to all the vertices. Choose one vector as a base, calculate the angles to the other vectors and order them. You can calculate the angles using the dot product: A · B = A B cos θ = |A||B| cos θ.
Below are the steps I followed.
The 3D planar polygon can be rotated to 2D plane using the known formulas. Use the one under the section Rotation matrix from axis and angle.
Then as indicated by #Glenn, an internal points needs to be calculated to find the angles. I take that internal point as the mean of the vertex locations.
Using the x-axis as the reference axis, the angle, on a 0 to 2pi scale, for each vertex can be calculated using atan2 function as explained here.
The non-negative angle measured counterclockwise from vector a to vector b, in the range [0,2pi], if a = [x1,y1] and b = [x2,y2], is given by:
angle = mod(atan2(y2-y1,x2-x1),2*pi);
Finally, sort the angles, [~,XI] = sort(angle);.
It's a long time since I used this, so I might be wrong, but I believe the command convhull does what you need - it returns the convex hull of a set of points (which, since you say your points are a convex set, should be the set of points themselves), arranged in counter-clockwise order.
Note that MathWorks recently delivered a new class DelaunayTri which is intended to superseded the functionality of convhull and other older computational geometry stuff. I believe it's more accurate, especially when the points get very close together. However I haven't tried it.
Hope that helps!
So here's another answer if you want to use convhull. Easily project your polygon into an axes plane by setting one coordinate zero. For example, in (0,0,0),(1,1,0),(1,1,1),(0,0,1) set y=0 to get (0,0),(1,0),(1,1),(0,1). Now your problem is 2D.
You might have to do some work to pick the right coordinate if your polygon's plane is orthogonal to some axis, if it is, pick that axis. The criterion is to make sure that your projected points don't end up on a line.
I want to plot 3D cube of size 2x2x5 (x,y,z) around an interest point to get the nearest points to it and inside the cube. The interest point may be at the center of cube.
How can I get the nearest points to the interest point?
There are several techniques for drawing cubes here. You will need to choose one that lets you specify size and origin. The sizes will be 2,2,5, and the origin will be the coordinates of the interest point.