Shown Figure (1) is a typical Delaunay triangulation (blue) and it has a boundary line (black rectangle).
Each vertex in the Delaunay triangulation has a height value. So I can calculate the height inside convex hull. I am figuring out a method to calculate the height up to the boundary line (some sort of extrapolation).
There are two things associated with this task
Triangulate up to the boundary point
Figuring out the height at newly created triangle vertices
Anybody come across this issue?
Figure 1:
I'd project the convex hull points of the triangulation to the visible box segments and then insert the 4 box corners and the projected points into the triangulation.
There is no unique correct way to assign heights to the new points. One easy and stable method would be to assign to each new point the height of the closest visible convex hull vertex. Be careful with extrapolation: Triangles of the convex hull tend to have unstable slopes, see the large triangles in front of the below terrain image. Their projection to the xy plane has almost 0 area but due to the height difference they are large and almost 90 degrees to the xy plane.
I've had some luck with the following approach:
Find the segment on the convex hull that is closet to the extrapolation point
If I can drop a perpendicular onto the segment, interpolate between the two vertices of the segment.
If I can not construct the perpendicular, just use the closest vertex
This approach results in a continuous surface, but does not provide 1st derivative continuity.
You can find some code that might be helpful at TriangularFacetInterpolator.java. Look for the interpolateWithExteriorSupport method.
Related
I am working on image segmentation and I thought the convex hull can provide me with a simple solution to my problem. Currently I have polygons with for sides (see image below). Due to image processing issues, the shape does not have clean straight sides and hence when I use the standard convex hull (in Matlab) I may get more than the four main corners to define it.
My goal is to force the convex hull algorithm to find the best 4 vertices that will enclose my polygons (i.e. 4 best enclosing vertices per polygon). Is this possible? An example code will be appreciated.
Thanks
The problem of the minimum area bounding polygon is briefly mentioned in "Geometric applications of a matrix-searching algorithm" (see Applications section). It is not simple and is probably not the way for you.
For an easier (but approximate) answer to your question, you can consider the four cardinal directions and find the farthest points in these, which define a quadrilateral. (Also consider the four intermediate directions, which are more appropriate for an axis-aligned rectangle.)
If you insist having an enclosing quadrilateral, you can translate the four edges to the farthest points in the respective perpendicular directions, and find the pairwise intersections.
If you insist having a rectangle, compute the convex hull and find the minimum area or minimum perimeter bounding rectangle by the Rotating Calipers method. https://geidav.wordpress.com/tag/rotating-calipers/
I have a set of points(clusters) in higher dimension (30d to 100d). I need to identify concave hull of these points in an efficient manner.
Is there a way to do get the exact concave hull or atleast approximate concave hull of these set of points?
Further, if we have a set of points identified as a border point, is there a way to verify whether the points are actually border points?
In 100d, almost every point will be on the convex hull.
Just remember that a rectangle in 2d has 4 corners, but in 100d, it has 2^100 corners.
As an extremely rough approximation, take the minimum and maximum along each axis. If it is unique, the point is on the hull. For additional points, you can sample some random projections.
But again, the expected behaviour is that almost every point is on the hull, because it is the smallest or largest in some linear combination of features.
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.
sorry if this question might seem simple but I couldn't figure it out.
Imagine two arbitrary Rectangles that are randomly overlaped so that their outlines intersect at only two locations. Now you cut the area of overlap out of Rectangle 1.
This "bitten Rectangle 1" has now the following points (vertices): (1) All points of Rectangle 1 that lie outside of Rectangle 2, (2) All points of Rectangle 2 that lie inside Rectangle 1 and (3) the two intersection points.
The problem know is the following: How can I get the order of the new points so that the functions plot(...) or fill(...) would draw the right "Bitten Rectangle 1"?
What I did so far:
I determined the convex hull of all the points that lie outside of Rectangle 2 + the intersection points. Then one has to add the new points that lie inside rectangle 1 (due to overlap with Rectangle 2) between the indices of the first and second intersection point also in the right order.
The problem with this convex hull approach is that it only works if the intersection points lie on different lines of Rectangle 1, because then they are part of the convex hull.
If they lie on the same line they are no longer treated as part of the convex hull.
What I'd need is a method to get the order of all possible points that lie on the convex hull and not only the outer most ones.
Hope anybody can help me...
Thank you in advance,
Patrick
In Matlab, there is an amazing function called polybool that let you do any set operation on polygons: http://www.mathworks.com/help/toolbox/map/ref/polybool.html
All you have to do is to define four arrays describing the rectangles (rect1x, rect1y, rect2x and rect2y) and to call [resultx resulty] = polybool('subtraction', rect1x, rect1y, rect2x, rect2y). The resulting arrays will be describing the "Bitten Rectangle 1".
I have a polyhedron, with a list of vertices (v) and surfaces (s). How do I break this polyhedron into a series of tetrahedra?
I would particularly like to know if there are any built-in MATLAB commands for this.
For the convex case (no dents in the surface which cause surfaces to cover each other) and a triangle mesh, the simple solution is to calculate the center of the polyhedron and then connect the three corners of every face with the new center.
If you don't have a triangle mesh, then you must triangulate, first. Delaunay triangulation might help.
If there are holes or caves, this can be come arbitrarily complex.
I'm not sure the OP wanted a 'mesh' (Steiner points added) or a tetrahedralization (partition into tetrahedra, no Steiner points added). for a convex polyhedron, the addition of Steiner points (e.g. the 'center' point) is not necessary.
Stack overflow will not allow me to comment on gnovice's post (WTF, SO?), but the proof of the statement "the surfaces of a convex polyhedron are constraints in a Delaunay Tesselation" is rather simple: by definition, a simplex or subsimplex is a member in the Delaunay Tesselation if and only if there is a n-sphere circumscribing the simplex that strictly contains no point in the point set. for a surface triangle, construct the smallest circumscribing sphere, and 'puff' it outwards, away from the polyhedron, towards 'infinity'; eventually it will contain no other point. (in fact, the limit of the circumscribing sphere is a half-space; thus the convex hull is always a subset of the Delaunay Tesselation.)
for more on DT, see Okabe, et. al, 'Spatial Tesselations', or any of the papers by Shewchuk
(my thesis was on this stuff, but I remember less of it than I should...)
I would suggest trying the built-in function DELAUNAY3. The example given in the documentation link resembles Aaron's answer in that it uses the vertices plus the center point of the polyhedron to create a 3-D Delaunay tessellation, but shabbychef points out that you can still create a tessellation without including the extra point. You can then use TETRAMESH to visualize the resulting tetrahedral elements.
Your code might look something like this (assuming v is an N-by-3 matrix of vertex coordinate values):
v = [v; mean(v)]; %# Add an additional center point, if desired (this code
%# adds the mean of the vertices)
Tes = delaunay3(v(:,1),v(:,2),v(:,3)); %# Create the triangulation
tetramesh(Tes,v); %# Plot the tetrahedrons
Since you said in a comment that your polyhedron is convex, you shouldn't have to worry about specifying the surfaces as constraints in order to do the triangulation (shabbychef appears to give a more rigorous and terse proof of this than my comments below do).
NOTE: According to the documentation, DELAUNAY3 will be removed in a future release and DelaunayTri will effectively take its place (although currently it appears that defining constrained edges is still limited to only 2-D triangulations). For the sake of completeness, here is how you would use DelaunayTri and visualize the convex hull (i.e. polyhedral surface) as well:
DT = DelaunayTri(v); %# Using the same variable v as above
tetramesh(DT); %# Plot the tetrahedrons
figure; %# Make new figure window
ch = convexHull(DT); %# Get the convex hull
trisurf(ch,v(:,1),v(:,2),v(:,3),'FaceColor','cyan'); %# Plot the convex hull