I have a 3D volume and I need to set some parts of it to zero.
The problem is that I don't know how to refer to the parts I want. I know the shape of the area I want to crop, but I need that area revolved around the Z axis.
A picture's worth a thousand words:
The image shows the cross section of a 3D matrix in the XZ plane, in the middle is the Z axis. The area I want to select (or set to zero) in the volume is the red area, revolved around the green axis, the Z axis.
I'd rather have a solution that doesn't have a very big memory dependency (such as using meshgrid would) as these matrices can get quite big. Else, a meshgrid solution with cylindrical coordinates wouldn't be too difficult to make, as #Luis suggests.
How can I select this area?
Related
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
I am trying to pack same-radi circles on a sphere using matlab. I have figured out how to pack circles and spheres in a box or a cube but just can't figure out where to start when packing it on the surface of a sphere. I would love any hints or ideas in doing this.
Wish you all the best of luck
This is a really hard problem, not sure if there exists a generic solution.
To obtain perfectly regular distribution of circles, take any regular polyhedron, and put a circle at each vertex. The largest regular polyhedron, the dodecahedron has 20 vertices, so you would be able to fit 20 circles on a sphere with this method. If you are willing to have a tiling that is not perfectly regular, you can use shapes such as that of a buckyball (same as soccer ball).
Other ways to distribute points evenly around a sphere have been proposed.
A friend of mine proposed a way that allowed him to easily index each point. It was based on the icosahedron (dual of the dodecahedron, it has 20 faces), and tiled each triangular face. See his paper for more details.
That paper has references to some other methods known at that time. I think the most useful one would be the one where the sphere is cut vertically into slices of equal width (the surface of the sphere section has equal height, the cuts themselves are not spaced evenly), and each section is then a strip that can be tiled by circles evenly. This leads to one circle on the top and one on the bottom, and in between rows of circles. The distribution is not even, and the spacing depends on the ratio of the circle size and sphere size.
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.
I am doing work on symmetric images where I would like to define a symmetric (polar) coordinate space. Basically for the left image, I want 0 degrees to be defined along the right horizontal axis (as is the default). However, for the right image, I want 0 degrees to be defined along the left horizontal axis.
I know a phase shift of pi would do the trick. However, for comparison purposes, I am trying to keep the range of angles the same, [-pi : pi).
In the above color plot of the rotations in an object, note that they are both defined in the same direction. Ideally I'd like to see the colors of the right object flipped across its vertical axis.
I should note that these angles are calculated by taking the arctan(y/x) of the perimeter coordinates when measured from the centroid. Is there a different trig function that may result in the proper symmetry? I couldn't seem to come up with one while still claiming it was representative of direction.
Simple rounded corner rectangle code in Matlab can be written as follows.
rectangle('Position',[0,-1.37/2,3.75,1.37],...
'Curvature',[1],...
'LineWidth',1,'LineStyle','-')
daspect([1,1,1])
How to get the x and y coordinates arrays of this figure?
To get the axes units boundaries, do:
axisUnits = axis(axesHandle) % axesHandle could be gca
axisUnits will be an four elements array, with the following syntax: [xlowlim xhighlim ylowlim yhighlim], it will also contain the zlow and zhigh for 3-D plots.
But I think that is not what you need to know. Checking the matlab documentation for the rectangle properties, we find:
Position four-element vector [x,y,width,height]
Location and size of rectangle. Specifies the location and size of the
rectangle in the data units of the axes. The point defined by x, y
specifies one corner of the rectangle, and width and height define the
size in units along the x- and y-axes respectively.
It is also documented on the rectangle documentation:
rectangle('Position',[x,y,w,h]) draws the rectangle from the point x,y
and having a width of w and a height of h. Specify values in axes data
units.
See if this illustrate what you want. You have an x axis that goes from −100 to 100 and y axis that goes from 5 to 15. Suppose you want to put a rectangle from −30 to −20 in x and 8 to 10 in y.
rectangle('Position',[-30,8,10,2]);
As explained by the comments there appears to be no direct way to query the figure created by rectangle and extract x/y coordinates. On the other hand, I can think of two simple strategies to arrive at coordinates that will closely reproduce the curve generated with rectangle:
(1) Save the figure as an image (say .png) and process the image to extract points corresponding to the curve. Some degree of massaging is necessary but this is relatively straightforward if blunt and I expect the code to be somewhat slow at execution compared to getting data from an axes object.
(2) Write your own code to draw a rectangle with curved edges. While recreating precisely what matlab draws may not be so simple, you may be satisfied with your own version.
Whether you choose one of these approaches boils down to (a) what speed of execution you consider acceptable (b) how closely you need to replicate what rectangle draws on screen (c) whether you have image processing routines, say for reading an image file.
Edit
If you have the image processing toolbox you can arrive at a set of points representing the rectangle as follows:
h=rectangle('Position',[0,-1.37/2,3.75,1.37],...
'Curvature',[1],...
'LineWidth',1,'LineStyle','-')
daspect([1,1,1])
axis off
saveas(gca,'test.png');
im = imread('test.png');
im = rgb2gray(im);
figure, imshow(im)
Note that you will still need to apply a threshold to pick the relevant points from the image and then transform the coordinate system and rearrange the points in order to display properly as a connected set. You'll probably also want to tinker with resolution of the initial image file or apply image processing functions to get a smooth curve.