MATLAB: converting 3d plane into 3d matrix pixel locations - matlab

I am trying to get a slice of an elongated object in a 3d matrix (227x297x187 binary) that is orthogonal to its orientation..i.e. output image would be a 2d cross-section.
I was thinking the best way to do this would be to find a vector giving the orientation at some point on the object (so for a cross-section at the 40th slice I get the location of centroids at the 30th and 50th slices and find the vector between them), then get a list of pixels corresponding to a plane orthogonal to that vector at the desired slice (along centroid of object at 40th slice), and then retrieve the values in the matrix corresponding to those locations. Assuming the code I have attached finds the plane, I can plot it with surf() but don't know how to translate that into pixel locations in the matrix to generate the values for the 2d cross-section.
startslice=30;
endslice=50;
startnums=bwconncomp(binMATRIX(:,:,startslice));
startnums=regionprops(startnums,'Centroid');
endnums=bwconncomp(binMATRIX(:,:,endslice));
endnums=regionprops(endnums2,'Centroid');
midnums3=bwconncomp(binMATRIX(:,:,(startslice+(endslice-startslice)/2)));
midnums3=regionprops(midnums,'Centroid');
startstuff=[startslice, startnums.Centroid];
endstuff=[endslice, endnums.Centroid];
midstuff=[(startslice+(endslice-startslice)/2), midnums.Centroid];
nullstuff=null(endstuff-startstuff);
A=nullstuff(:,1);
B=nullstuff(:,2);
[x,y]=meshgrid(A,B);
z=1-x-y;
surf(x+midstuff(1),y+midstuff(2),z+midstuff(3))

Related

Crop 3D Mesh with plane Cut in Matlab

I have made 3D analysis code and I want to split or crop 3D mesh into 2 parts with 2D plane, what i expected: the final result I need is to find out what are the nodes on the left side and the right side, what you see on the image below is the nodes of the 3D object (bumpy), Do you know what method or library I need to use to solve this problem?
my problem
Here is my data structure from the 2D plane:
Column 1: Face
Column 2: X coordinate
Column 3: Y coordinate
column 4: Z coordinate
Column 5: Finite Element Value
data structure
.
The data structure from the 3D mesh is containing the same data as the table above, Thanks so much!
We can know the plane XYZ Coordinates, so I tried to find by using <= to find the axis value is larger or smaller than the plane coordinates:
find x,y,z 3D model coordinate is smaller than x,y,z cut plane coordinate
[r] = find((Name_OT(:,1)>=x) & (Name_OT(:,2)>=y) & Name_OT(:,3)>=z);
the blue line is the plane, and the coloured one is the result from my code, the ideal result is the coloured nodes full, but what happened here the colour node has a big hole or gap
not good result
You need to first decide whether you want to segment your data by a (linear) plane or not. If you prefer to keep a curved surface, please follow my earlier comment and edit your question.
To get a plane fit to the data for your cut, you can use fit.
Base on the plane, you can get a normal vector of the plane. That is reading coefficients of fit results and is in the documentation. Using that normal vector, you can rotate all your data so that the plane is normal to z axis. The rotation is matrix multiplication. From there, you can use logical indexing to segment your data set.
You can ofc also get the normal component of the data points relative to the plane cut and decide on a direction that way. You still need fit. From that point, it's basic matrix manipulation. An nx1 vector can multiply a 1xn vector in Matlab. So projectors can also be constructed from basic matrix manipulation. At a glance, this method is computationally inefficient.

Triangulated surface into 2D matrix - MATLAB

I have a 3D reconstruction that is of the format: vertices and faces.
It was read in by an STL, or an OBJ file, and I believe the terminology of this surface format is triangulated.
This is nice to visualise using trisurf, however I need the representation of my surface to be a 2D matrix.
Specifically: I have a 3 column matrix named vertices, and another 3 column matrix named faces. They do not have the same number of rows. I want a 2D matrix, where every cell in the matrix is the height of the surface.
Is this possible? How can this be done?

Embedding 2d plot into a 3d plot in matlab?

I have a set of data vectors z that has this 2d plot
How would I go about embed this set of data into a 3d plot like this in matlab? I'm asking for advice and suggestions. The theory I'm trying to employ is "for each data vector~zj, “copies” the data vector intothe first two entries of a 3D data vector~yjand then computes the squared length of~zj as the third entry of~yj. " or kernel trick.
Your 2d data will somehow be in the form, that you have x-coordinates and y-coordinates. Let's say you have a vector x and a vector y for simplification.
As you found out the plot3-function proivdes functionality to plot arbitrary points in 3d without the need of generating a mesh. What you need additionally is a third vector z with data for the 3rd dimension.
So what else can you do? The thing I am thinking about is rotating the plane you are plotting you "2d" data:
Rotational matrices can be seen here:
https://en.wikipedia.org/wiki/Rotation_matrix

I need to fit a best circle to the 3D data in matlab

Basically, I have a many irregular circle on the ground in the form of x,y,z coordinates (of 200*3 matrix). but I want to fix a best circle in to the data of x,y,z coordinates (of 200*3 matrix).
Any help will be greatly appreciated.
I would try using the RANSAC algorithm which finds the parameters of your model (in your case a circle) given noisy data. The algorithm is quite easy to understand and robust against outliers.
The wikipedia article has a Matlab example for fitting a line but it shouldn't be too hard to adapt it to fit a circle.
These slides give a good introduction to the RANSAC algorithm (starting from page 42). They even show examples for fitting a circle.
Though this answer is late, I hope this helps others
To fit a circle to 3d points
Find the centroid of the 3d points (nx3 matrix)
Subtract the centroid from the 3D points.
Using RANSAC, fit a plane to the 3D points. You can refer here for the function to fit plane using RANSAC
Apply SVD to the 3d points (nx3 matrix) and get the v matrix
Generate the axes along the RANSAC plane using the axes from SVD. For example, if the plane norm is along the z-direction, then cross product between the 1st column of v matrix and the plane norm will generate the vector along the y-direction, then the cross product between the generated y-vector and plane norm will generate a vector along the x-direction. Using the generated vectors, form a Rotation matrix [x_vector y_vector z_vector]
Multiply the Rotation matrix with the centroid subtracted 3d points so that the points will be parallel to the XY plane.
Project the points to XY plane by simply removing the Z-axes from the 3d points
fit a circle using Least squares circle fit
Rotate the center of the circle using the inverse of the rotation matrix obtained from step 5
Translate back the center to the original location using the centroid
The circle in 3D will have the center, the radius will be the same as the 2D circle we obtained from step 8, the circle plane will be the RANSAC plane we obtained from step 3

Extract arbitrarily rotated plane of data from 3D array as 2D array

I have a 3D matrix of data in matlab, but I want to extract an arbitrarily rotated slice of data from that matrix and store it as a 2D matrix, which I can access. Similar to how the slice() function displays data sliced at any angle, except I would also like to be able to view and modify the data as if it were an array.
I have the coordinates of the pivot-point of the plane as well as the angles of rotation (in x, y and z axis), I have also calculated the equation of the plane in the form:
Ax + By + Cz = D
and can extract a 3D matrix containing only the data that fall on that plane, but I don't know how to then convert that into a simple 2D array.
Another way of doing it would be to somehow rotate the source matrix in the opposite direction of the angle of the plane, so as to line up the plane of data with the XY axis, and simply extract that portion of the matrix, but I do not know if rotating a matrix like that is possible.
I hope this hasn't been answered elsewhere, I've been googling it all day, but none of the problems seem to exactly match mine.
Thanks
You can take a look at the code here. I think the function is similar to what you are trying to solve.
The function extracts an arbitrary plane from a volume given the size of the plane, the center point of the plane, and the plane normal, i.e. [A,B,C]. It also outputs the volumetric index and coordinate of each pixel on the plane.
Aha! May have just solved it myself.
To produce the plane equation I rotate a normal vector of (0,0,1) using rotation matrices and then find D. If I also rotate the following vectors:
(1,0,0) //step in the x direction of our 2D array
and
(0,1,0) //step in the y direction of our 2D array
I'll have the gradients that denote how much my coordinates in x,y,z have to change before I step to the next column in my array, or to the next row.
I'll mock this up ASAP and mark it as the answer if it works
EDIT: Ok slight alteration, when I'm rotating my vectors I should also rotate the point in 3D space that represents the xyz coordinates of x=0,y=0,z=0 (although I'm rotating around the centre of the structure, so it's actually -sizex/2,-sizey/2,-sizez/2, where size is the size of the data, and then I simply add size/2 to each coordinate after the rotations to translate it back to where it should be).
Now that I have the gradient change in 3D as I increase the x coordinate of my 2D array and the gradient change as I increase the y coordinate, I can simply loop through all possible x and y coordinates (the resulting array will be 50x50 for a 50x50x50 array, I'm not sure what it will be for irregular sizes, which I'll need to work out eventually) in my 2D array and calculate the resulting 3D coordinates on my plane in the data. My rotated corner value serves as the starting point. Hooray!
Just got to work out a good test for this encompassing all angles and then I'll approve this as an answer