Using plot::tube function in MuPAD, I can generate tubes with varying spine but the cross-section is always a circle, although the radius can vary along the spine. For instance:
plot::Tube([cos(t), sin(t), 0], 0.4 + 0.3*sin(t)*cos(t),t = -0.5*PI..0.5*PI,Mesh=[60,12]):
I want to plot tubes with (non-circular) elliptical cross-sections. Is there a way to achieve this in MuPAD?
P.S. I can generate tubes with arbitrary cross-sections using MATLAV, so please avoid including answers that use pure MATLAB commands.
As for your first question whether this can be done with plot::Tube: I don't think so. Consider the help of this function (emphasis mine):
plot::Tube creates generalized tubular plots, known as "canal
surfaces", with special cases known as "tube surface", "pipe surface"
or "tubular surfaces."
Intuitively, canal surfaces are space curves with thickness. More formally, a canal surface plot::Tube([x(t), y(t), z(t)], r(t), t = t_min..t_max) is the envelope of spheres with center [x(t), y(t), z(t)] and radius r(t), i.e., the thickness of the curve can vary with
the curve parameter t
The fact that a tube is defined as an envelope of a collection of spheres suggests that their cross-section is inherently circular.
I'm not familiar with MuPAD, so I don't know the elegant solution to your problem. For your simple example of a semi-circular base line I could put together this kludgy solution using the low-level plot::Surface, manually constructing the surface:
plot(
plot::Surface(
matrix([cos(t),sin(t),0])
+ numeric::rotationMatrix(t,[0,0,1]) * matrix([(0.2+0.1*sin(2*t))*cos(u),0,(0.2+0.1*cos(2*t))*sin(u)]),
t = -PI/2..PI/2,
u = 0..2*PI,
Mesh = [30,30]),
Scaling=Constrained)
Here t stands for the angular parameter along the semi-circular base line (toroidal direction), and u is the angle along the cross-section (poloidal direction). A given cross-section at angle t looks like this:
[(0.2+0.1*sin(2*t))*cos(u), 0, (0.2+0.1*cos(2*t))*sin(u)]
I.e. the semimajor axes are (0.2+0.1*sin(2*t)) and (0.2+0.1*cos(2*t)) along the cross-section of the tube.
You can see that I used a rotation matrix around the z axis to construct the surface. This made heavy use of the fact that the base line is a semi-circle. However, in general cases it should be possible to compute the derivative of your parameterized base line (with respect to t), and compute the necessary rotation angle (with parameter t) from that.
Related
I have a surface by the code below and another surface which is created by the exact same code. I want to see the height differences in another figure. How am I able to do that? Already operated with the Minus-operator but this won't work.
Furthermore the matrices have NOT the same size!
Appreciate your help!
x1 = Cx1;
y1 = Cy1;
z1 = Cz1;
tri1 = delaunay(x1,y1);
fig1 = figure%('units','normalized','outerposition',[0 0 1 1]);
trisurf(tri1,x2,y2,z2)
xlabel('x [mm] ','FontSize',30)
ylabel('y [mm] ','FontSize',30)
zlabel('z [mm] ','FontSize',30)
The simplest way to solve this problem is to interpolate from one mesh onto the other one. Such an approach works well when one is more highly resolved than the other, or when you're not as concerned with results at individual nodes, but rather the overall pattern across elements. If that's not the case, then you have a very complicated problem because you need to create a polygonal surface that fully captures all nodes and edges of both triangulations. Consider the following pair of triangular patterns:
A surface that captured all the variations would need to have all the vertices and edges that make up both of them, which is not a purely triangular surface. So, let us instead assume the easier case. To map results from one triangulation to the other, you simply need to formulate functions that define how the values vary along the triangles, which are more broadly called basis functions. It is often assumed that values betweeen the nodes (i.e. vertices) of the triangles vary linearly along the surfaces of the triangles. You can do it differently if you want, it just requires defining new basis functions. If we go for linear functions, then the equations in 2D are pretty simple. Let's say you make an array trimap that has which triangle each of the vertices of the other triangulation is inside of. This can be accomplished using the info here. Then, we set the coordinates of the vertices of the current triangle to (x1,y1), (x2,y2), and (x3,y3), and then do the math:
for cnt1=1,npoints
x1=x(tri1(trimap(cnt1),1));
x2=x(tri1(trimap(cnt1),2));
x3=x(tri1(trimap(cnt1),3));
y1=y(tri1(trimap(cnt1),1));
y2=y(tri1(trimap(cnt1),2));
y3=y(tri1(trimap(cnt1),3));
delta=x2*y3+x1*y2+x3*y1-x2*y1-x1*y3-x3*y2;
delta1=(x2*y3-x3*y2+xstat(cnt1)*(y2-y3)+ystat(cnt1)*(x3-x2));
delta2=(x3*y1-x1*y3+xstat(cnt1)*(y3-y1)+ystat(cnt1)*(x1-x3));
delta3=(x1*y2-x2*y1+xstat(cnt1)*(y1-y2)+ystat(cnt1)*(x2-x1));
weights(cnt1,1)=delta1/delta;
weights(cnt1,2)=delta2/delta;
weights(cnt1,3)=delta3/delta;
z1=z(tri1(trimap(cnt1),1));
z2=z(tri1(trimap(cnt1),2));
z3=z(tri1(trimap(cnt1),3));
valinterp(cnt1)=sum(weights(cnt1,:).*[z1,z2,z3]);
end
valinterp is the interpolated value for each point. Here and here are some nice slides explaining the mathematics behind all this. Note that I've not tested any of this code. Note also that you will need to do something to assign to values outside of the triangulation. Perhaps a null value, or an inverse-distance weighted value.
I'm looking for a matlab command that integrates a known function inside a triangle (which is known by his nodes).
I have 3 coordinates (x1,y1) (x2,y2) (x3,y3) that define a triangle and I know that inside the triangle there exist a known function f(x,y). I'm looking for a command that computes the integral of f over the triangle.
Use integral2 command in Matlab. The first example in the Documentation will cover your case. To get one side of your triangle parallel to the x-axis you can multiply your triangle with a rotation matrix and then "move" it on the x-axis.
Note that the integral is not influenced this transformations if your perform your substitutions correctly.
This transformations can make it easier to write your code but they're not necessary to use integral2.
So I have a 3 dimensional matrix of points that (presumably) define a surface. For my purposes, X and Y can be random values but when plotted along with their Z coordinates, they will define some undulating surface. I'd like to measure the local curvatures of said surface, and in order to do that, I need to be able to find the gradient of said surface, at which point calculating the curvature is trivial.
I have not yet found an implementation of how to measure this curvature that doesn't make use of Matlab's gradient function. The problem with Matlab's gradient function is that it assumes that the points are in some sort of order, similar to diff(X). This would suffice if my points were spaced along a grid, which is not necessarily the case.
One possible solution to measuring the gradient is to give in and assign each point to a discrete coordinate in a grid in the XY plane, thus overcoming this issue. However, this solution seems somewhat inelegant and was curious to see if anyone had suggestions. Thanks!
You can use griddata to interpolate from your scattered data points to grid spaced points and then calculate the gradient.
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 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