I want to create a perspective projection of a 3D image onto the x,y plane with a focal length of 10 and a principal point (-1, -5).
I found:
view
viewmtx
But I do not get how to tell matlab that I want to use the x,y plane nor how to set the focal length? Can someone explain me how to do that?
I'm afraid you are not looking for the right functions.
The view function does only change the point of view on the current axis, while viewmtxreturn a transformation matrix.
You may want to do something similar to what it is discussed on this post How do I draw a texture-mapped triangle in MATLAB?, where maketform and imtransform are the key functions to get a plane image reprojected into another certain 3d plane.
Related
I have a set of control point pairs. One part of the pair is in world coordinates (3D). The other one is in pixel corrdinates of the image (2D).
My goal is to transform a surface you can see in this image onto a flat plane. The problem is that the surface is not perfectly flat, it kinda looks like a ribbon. Otherwise I could have used OpenCV's getPerspectiveTransform() or Matlab's fitgeotrans().
I know that I can use OpenCV's solvePnP() or Matlab's estimateWorldCameraPose() to get the pose of the camera. The camera matrix is known and the image is rectified. But what is the next step then? How can I transform my ribbon shaped surface onto a flat plane, i.e. get an orthographic top view? That is the step, I'm stuck on.
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.
I used the "imfreehand" to crop an irregular shape and save its positions into a variable. This position variable is a 85*2 double matrix (85 points, X and Y coordinates). Now, I want to use the same position to crop another image (different layer of the image, but the location of the objects is the same). The functions I can find all requires rectangle positions (X1,X2,Y1,Y2). In my situation, I have 82 different (X,Y) coordinates, how can I use the position information to crop a new image?
From what I understand, you want to take the coordinates created by imfreehand(...) to create a cropable object on another image. You can use the function impoly(hparent,position) for this purpose.
The MathWorks page provides an example to guide you on its usage.
I have a representation of 3D object as a a) cloud point and b) triangle mesh
My goal is to rotate this object and than obtain a surface which is visible from one specific view.
Then I would remove the points which are not visible from a given view. Does anyone know how to do this in MATLAB. What method is the fastest?
The file with the point cloud contains the coordinates of each point, and information about the color stored in three RGB channels.
First line:
`-35.4717 88.8637 -99.3782 97 78 46`
I will be grateful for any help.
One possible way would be to re-implement the pipeline of a graphic processor.
Transform your object and project all triangles into an image plane. In this image plane, the distances of each part of the triangle can be stored.
With that information you can check if a vertex is further away than the one you have painted into the image plane.
I have a shape which you can imagine as a lake in a field observed from the top (2D). I determined the border pixels of the shape after an image processing, so that I have the coordinates of each border point.
Now I want to calculate the perimeter of this shape. My problem is that I have the points not in following order that would give a closed loop, but unordered.
How can a problem like this be solved in Matlab? (including Curve-Fitting-Toolbox etc.)
Thank you for any suggestions!
You can use the function regionprops for this.
Turn your image into a binary image with 1 inside your 'lake' and 0 outside (which you should be easily able to do, as you mention you extracted the boundaries).
Then use:
props=regionprops(YourBinaryImage, 'Perimeter');
You can then access the perimeter as follows: props.Perimeter
If you have set of 3D points with (x,y,z) coordinates, you may set z to zero and use the 2D (x,y) points to find the convex hull using convhull regardless of their order .