Arrange the vertices of a 3D convex polygonal plane in counter clockwise direction in MATLAB

Question

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!

Answer 1

Below are the steps I followed.

1. 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.

2. 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.

3. 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);

4. Finally, sort the angles, [~,XI] = sort(angle);.

Answer 2

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.

Answer 3

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.

Answer 4

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!

Answer 5

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.