I have a cartesian connector element in ABAQUS to find a force-displacement diagram for cyclic analysis. The connector property is split into x and y directions and I have corresponding force and displacement data from both directions (blue and green dotted curves in the figure). Now I would like to combine both curves into a single force-displacement curve.
Force-displacement diagram
The resultant of forces and displacement I tried in the red curve is not correct.
Please suggest a solution or send a link for the solution.
excel sheet
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I have image of robot with yellow markers as shown
The yellow points shown are the markers. There are two cameras used to view placed at an offset of 90 degrees. The robot bends in between the cameras. The crude schematic of the setup can be referred.
https://i.stack.imgur.com/aVyDq.png
Using the two cameras I am able to get its 3d co-ordinates of the yellow markers. But, I need to find the 3d-co-oridnates of the central point of the robot as shown.
I need to find the 3d position of the red marker points which is inside the cylindrical robot. Firstly, is it even feasible? If yes, what is the method I can use to achieve this?
As a bonus, is there any literature where they find the 3d location of such internal points which I can refer to (I searched, but could not find anything similar to my ask).
I am welcome to a theoretical solution as well(as long as it assures to find the central point within a reasonable error), which I can later translate to code.
If you know the actual dimensions, or at least, shape (e.g. perfect circle) of the white bands, then yes, it is feasible and possible.
You need to do the following steps, which are quite non trivial to do, and I won't do them here:
Optional but extremely suggested: calibrate your camera, and
undistort it.
find the equation of the projection of a 3D circle into a 2D camera, for any given rotation. You can simplify this by assuming the white line will be completely horizontal. You want some function that takes the parameters that make a circle and a rotation.
Find all white bands in the image, segment them, and make them horizontal (rotate them)
Fit points in the corrected white circle to the equation in (1). That should give you the parameters of the circle in 3d (radious, angle), if you wrote the equation right.
Now that you have an analytic equation of the actual circle (equation from 1 with parameters from 3), you can map any point from this circle (e.g. its center) to the image location. Remember to uncorrect for the rotations in step 2.
This requires understanding of curve fitting, some geometric analytical maths, and decent code skills. Not trivial, but this will provide a solution that is highly accurate.
For an inaccurate solution:
Find end points of white circles
Make line connecting endpoints
Chose center as mid point of this line.
This will be inaccurate because: choosing end points will have more error than fitting an equation with all points, ignores cone shape of view of the camera, ignores geometry.
But it may be good enough for what you want.
I have been able to extract the midpoint by fitting an ellipse to the arc visible to the camera. The centroid of the ellipse is the required midpoint.
There will be wrong ellipses as well, which can be ignored. The steps to extract the ellipse were:
Extract the markers
Binarise and skeletonise
Fit ellipse to the arc (found a matlab function for this)
Get the centroid of the ellipse
hsv_img=rgb2hsv(im);
bin=new_hsv_img(:,:,3)>marker_th; %was chosen 0.35
%skeletonise
skel=bwskel(bin);
%use regionprops to get the pixelID list
stats=regionprops(skel,'all');
for i=1:numel(stats)
el = fit_ellipse(stats(i).PixelList(:,1),stats(i).PixelList(:,2));
ellipse_draw(el.a, el.b, -el.phi, el.X0_in, el.Y0_in, 'g');
The link for fit_ellipse function
Link for ellipse_draw function
Is it possible to graphically represent linear transformations in MATLAB like the one shown below?
In particular, I'm looking for a way to represent the way a coordinate grid might become squished, stretched, or rotated after a matrix multiplication has been performed. I've tried using meshgrid(x,y) to plot the grid but I've been having trouble getting it to display properly. Ideally, it would be nice to see the way that the transformation acts on a unit square (and, if possible, to see the transformation being performed in a continuous fashion, where you can see the graphs morphing into each other). However, I'm willing to sacrifice these last two requirements.
Here is the code I have used so far:
x = -10:2:10;
y = -10:2:10;
[X,Y] = meshgrid(x,y);
plot(X,Y)
For some reason, only vertical lines have been plotted. (The documentation says it will display a square grid.)
please read the plot doc.
the problem is that what you are doing is trying to plot mash matrixes in a plot not by using mesh or something like that.
you may find this very helpful.
I am currently doing some image segmentation on a bone qCT picture, see for instance images below.
I am trying to find the different borders in the picture for instance the outer border separating the bone to the noisy background. In this analysis I am getting a list of points (vec(1,:) containing x values and vex(2,:) containing the y values) in random order.
To get them into order I am using using a block of code which effectively takes the first point vec(1,1),vec(1,2) and then finds the closest point among the rest of the points in the vector. And then repeats.
Now my problem is that I want to smooth the data but how do I do that as the points lie in a circular formation? (I do have the Curve Fitting Toolbox)
Not exactly a smoothing procedure, but a way to simplify your data would be to compute the boundary of the convex hull of the data.
K = convhull(O(1,:), O(2,:));
plot(O(1,K), O(2,K));
You could also consider using alpha shapes if you want more control.
I have N 3D observations taken from an optical motion capture system in XYZ form.
The motion that was captured was just a simple circle arc, derived from a rigid body with fixed axis of rotation.
I used the princomp function in matlab to get all marker points on the same plane i.e. the plane on which the motion has been done.
(See a pic representing 3D data on the plane that was found, below)
What i want to do after the previous step is to look the fitted data on the plane that was found and get the curve of the captured motion in 2D.
In the princomp how to, it is said that
The first two coordinates of the principal component scores give the
projection of each point onto the plane, in the coordinate system of
the plane.
(from "Fitting an Orthogonal Regression Using Principal Components Analysis" article on mathworks help site)
So i thought that if i just plot those pc scores -plot(score(:,1),score(:,2))- i'll get the motion curve. Instead what i got is this.
(See a pic representing curve data in 2D derived from pc scores, below)
The 2d curve seems stretched and nonlinear (different y values for same x values) when it shouldn't be. The curve that i am looking for, should be interpolated by just using simple polynomial (polyfit) or circle fit in matlab.
Is this happening because the plane that was found looks like rhombus relative to the original coordinate system and the pc axes are rotated with respect to the basis of plane in such way that produce this stretch?
Then i thought that, this is happening because of the different coordinate systems of optical system and Matlab. Optical system's (ie cameras) co.sys. is XZY oriented and Matlab's default (i think) co.sys is XYZ oriented. I transformed my data to correspond to Matlab's co.sys through a rotation matrix, run again princomp but i got the same stretch in the 2D curve (the new curve just had different orientation now).
Somewhere else i read that
Principal Components Analysis chooses the first PCA axis as that line
that goes through the centroid, but also minimizes the square of the
distance of each point to that line. Thus, in some sense, the line is
as close to all of the data as possible. Equivalently, the line goes
through the maximum variation in the data. The second PCA axis also
must go through the centroid, and also goes through the maximum
variation in the data, but with a certain constraint: It must be
completely uncorrelated (i.e. at right angles, or "orthogonal") to PCA
axis 1.
I know that i am missing something but i have a problem understanding why i get a stretched curve. What i have to do so i can get the curve right?
Thanks in advance.
EDIT: Here is a sample data file (3 columns XYZ coords for 2 markers)
w w w.sendspace.com/file/2hiezc
I am trying to find the points with edge-adjacent Voronoi regions in a given dataset. I'm new to computational geometry, but from reading up online, it appeared that using the Delaunay tessellation would be an easy way to do this. This PDF in particular even has a lemma that states
Lemma 2.4 Two points of S are joined by a Delaunay edge iff their Voronoi regions
are edge-adjacent.
So, I found the delaunay tessellation of my dataset as
dt = delaunay(dataset); %using delaunayn() since dataset can be multidimensional
But now, when I plot this along with the voronoi diagram for this dataset, I find that the delaunay edges returned connect points whose regions are not actually edge-adjacent.
Here is the code I used to plot Voronoi and Delaunay together:
voronoi(dataset(:, 1),dataset(:, 2));
hold on;
dt = delaunayn(dataset);
triplot(dt, dataset(:, 1), dataset(:, 2), 'red');
And here is the output:
As an example of the problem, see the point X on the right end of the figure connected to point Y near the lower left corner.
Another example is in this SO question - the point 1 is connected to 2 and 3, even though they're not adjacent, and there doesn't seem to be any way 1 and 2 could share an edge even if extended to infinity. This question is actually what prompted me to test the delaunayn output with the above code.
Why is this happening, and how do I actually get the edge-adjacent neighbour regions that I need?
Note: For seeing the image in full size and clarity, please right click and choose 'View image' or similar.
As far as I can see (the quality of the diagram is not so good), the regions for X and Y should be adjacent below the plotted part. If you zoom out far enough, you should see them.
I.e. the edge where X and Y meet exists, but is just not shown on the plot.
The following diagram does not show the voronoi diagram outside of the plotting area, but how to find the intersection point described above (note, the bisector is not shown here):