I have two independent 3D shapes; one is a square and another is a cone.
Let's assume the cone lies inside the square. How can I find out that the surface of the cone touches the square's surface when I move the cone in any direction?
It will be helpful if anyone can suggest an algorithm to check whether the surface touches another shape.
I am working with MATLAB, but the underlying logic will be appreciated in any language will be appriciated.
https://in.mathworks.com/matlabcentral/answers/367565-findout-surface-to-surface-intersection-between-two-3d-shapes
There is a relatively easy solution, thanks to the fact that the truncated cone is a convex shape, and finding its AABB is not so hard.
First rotate space so that the cube become axis aligned (and the cone in arbitrary position). Then to find the AABB of the base, is suffices to get the maxima an minima of the coordinates, using the parametric equation, C + R cos t + R' sin t, where C is the position vector of the center, and R, R' two orthogonal radii. You find the limit angles by canceling the derivative.
After finding the extrema on the three coordinates, the global bounding box is the one that surrounds these six points plus the apex.
By comparing the AABB to the cube, you can tell what distance remains before a collision, in any direction.
Related
Most of the times, I determine contour orientation generating 2D points and computing the closed polygon area. Depending on the area value sign I can understand if the contour is oriented clockwise or not (see How to determine if a list of polygon points are in clockwise order?).
Would it be possible to do the same computations without generating 2D points? I mean, relying only on geometric curve properties?
We are interested in determining the orientation of contours like these ones without sampling them with 2D points.
EDIT: Some interesting solutions can be found here:
https://math.stackexchange.com/questions/423718/general-way-to-find-out-whether-a-curve-is-positively-oriented
Scientific paper: Determining the orientation of closed planar curves, DJ Filip (1990)
How are those geometric curves defined?
Do you have an angle for them? The radius doesn't matter, only the difference between entry-angle and exit-angle of each curve.
In that case, a trivial idea crossing my mind is to just sum up all the angles. If the result is positive, you know you had more curves towards the right meaning it's a clockwise contour. If it was negative, then more curves were leftwards -> anti-clockwise contour. (assuming that positive angels determine a right-curve and vica versa)
After thinking about this for awhile, for polygons that contain arcs I think there are three ways to do this.
One, is to break the arcs into line segments and then use the area formula as described above. The success of this approach seems to be tied to how close the interpolation of the arcs is as this could cause the polygon to intersect itself.
A quicker way than the above would be to do the interpolation of the arcs and then find a vertex in the corner (minimal Y, if tie minimal X) and use the sign of the cross product for that vertex. Positive CCW, negative CW. Again, this is still tied to the accuracy of the interpolation.
I think a better approach would be to find the midpoint of the arc and create two line segments, one from the beginning of the arc to the midpoint and another from the midpoint to the end of the arc and replace the arc with these line segments. Now you have a polygon with only line segments. Then you can add up all the normalized cross products of all the vertices. The sign will tell you the direction. Positive is counter-clockwise, negative is clockwise. In this case it doesn't matter if the polygon self-intersects.
The picture below shows a triangular surface mesh. Its vertices are exactly on the surface of the original 3D object but the straight edges and faces have of course some geometric error where the original surface bends and I need some algorithm to estimate the smooth original surface.
Details: I have a height field of (a projectable part of) this surface (a 2.5D triangulation where each x,y pair has a unique height z) and I need to compute the height z of arbitrary x,y pairs. For example the z-value of the point in the image where the cursor points to.
If it was a 2D problem, I would use cubic splines but for surfaces I'm not sure what is the best solution.
As commented by #Darren what you need are patches.
It can be bi-linear patches or bi-quadratic or Coon's patches or other.
I have found no much reference doing a quick search but this links:
provide an overview: http://www.cs.cornell.edu/Courses/cs4620/2013fa/lectures/17surfaces.pdf
while this is more technical: https://www.doc.ic.ac.uk/~dfg/graphics/graphics2010/GraphicsHandout05.pdf
The concept is that you calculate splines along the edges (height function with respect to the straight edge segment itself) and then make a blending inside the surface delimited by the edges.
The patch os responsible for the blending meaning that inside any face you have an height which is a function of the point position coordinates inside the face and the values of the spline ssegments which are defined on the edges of the same face.
As per my knowledge it is quite easy to use this approach on a quadrilateral mesh (because it becomes easy to define on which edges sequence to do the splines) while I am not sure how to apply if you are forced to go for an actual triangulation.
Given a matrix nx3 that represents n points in 3D space. All points lie on a plane. The plane is given by its normal and a point lying on it. Is there a Matlab function or any Matlabby way to find the area directly from the matrix?
What i was trying to do is write a function that first computes the centroid,c, of the n-gon. Then form triangles : (1,2,c),(2,3,c),...,(n,1,c). Compute their area and sum up. But then i had to organise the polygon points in a cyclic order as they were unordered which i figured was hard. Is there a easy way to do so?
Is there a easier way in Matlab to just call some function on the matrix?
Here is perhaps an easier method.
First suppose that your plane is not parallel to the z-axis.
Then project the polygon down to the xy-plane simply by removing the 3rd coordinate.
Now compute the area A' in the xy-plane by the usual techniques.
If your plane makes an angle θ with the xy-plane, then your 3D
area A = A' / cos θ.
If your plane is parallel to the z-axis, do the same computation
w.r.t. the y-axis instead, projecting to the xz-plane.
To project from 3D to the plane normal to N, take some non-parallel vector A and compute the cross products U = N x A and V = N x U. After normalizing U and V, the dot products P.U and P.V give you 2D coordinates in the plane.
Joseph's solution is even easier (I'd recommend to drop the coordinate with the smallest absolute cosine).
You said the points all lie on a plane and you have the normal. You should then be able to reproject the 3-D points into 2-D coordinates in a new 2-D basis. I am not aware of a canned function in Matlab to do this , but coding it should not be difficult, this answer from Math.SE and this Matlab Central post should help you.
If you already solved the problem of finding the coordinates of the points in the 2-D plane they are in, you could use the Matlab boundary or convex hull function to compute the area of the boundary or convex hull enclosing the points.
[k,v]= boundary(x,y)
or
[k,v] =convhull(x,y)
where k is the vector of indices into points x,y, that define the boundary or convex hull, v is the area enclosed, and x, y are vectors of the x and y coordinates of your points.
What you were describing with trying to find triangles with the points sounds like a first attempt toward Delaunay triangulation. I think more recent versions of Matlab have functions to do Delaunay triangulation as well.
Using Matlab, how can I calculate a two points in Equilateral triangle if there are known one point and the Center of Gravity in a 3D ?
I know there is a infinite solutions but i need just a random one.
Thank you.
Take the vector pointing from the center of gravity to the point.
Create an orthogonal vector (this can be done in a few ways, I usually take the first vector, add 1.0 to each component until it is not parallel, then take the cross product with the original vector).
Rotate your vector 120 degrees about the orthogonal vector. (look up the rotation matrix about an arbitrary vector)
Create your second point by adding that vector to your center of gravity.
Create your third point by rotating it again or in the opposite direction.
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.