How is this idea applied to a polygon model in order to rotate the entire object? Can this be connected to circles with a formula for performing rotation?
You probably work with matrices (if you don't, I'd suggest you do), so what you want to check out is rotation matrices.
Basically, by multiplying the vectors by the rotation matrix, you obtain the rotated vector, and by applying this to every point of the polygon, you get your rotated polygon.
Note that it also applies to 3D rotations, even if it's not exactly the same matrix.
Related
I am writing a program that captures real time images from a scene by two calibrated cameras (so the internal parameters of the cameras are known to us). Using two view geometry, I can find the essential matrix and use OpenCV or MATLAB to find the relative position and orientation of one camera with respect to another. Having the essential matrix, it is shown in Hartley and Zisserman's Multiple View Geometry that one can reconstruct the scene using triangulation up to scale. Now I want to use a reference length to determine the scale of reconstruction and resolve ambiguity.
I know the height of the front wall and I want to use it for determining the scale of reconstruction to measure other objects and their dimensions or their distance from the center of my first camera. How can it be done in practice?
Thanks in advance.
Edit: To add more information, I have already done linear trianglation (minimizing the algebraic error) but I am not sure if it is any useful because there is still a scale ambiguity that I don't know how to get rid of it. My ultimate goal is to recognize an object (like a Pepsi can) and separate it in a rectangular area (which is going to be written as a separate module by someone else) and then find the distance of each pixel in this rectangular area, i.e. the region of interest, to the camera. Then the distance from the camera to the object will be the minimum of the distances from the camera to the 3D coordinates of the pixels in the region of interest.
Might be a bit late, but at least for someone struggling with the same staff.
As far as I remember it is actually linear problem. You got essential matrix, which gives you rotation matrix and normalized translation vector specifying relative position of cameras. If you followed Hartley and Zissermanm you probably chose one of the cameras as origin of world coordinate system. Meaning all your triangulated points are in normalized distance from this origin. What is important is, that the direction of every triangulated point is correct.
If you have some reference in the scene (lets say height of the wall), then you just have to find this reference (2 points are enough - so opposite ends of the wall) and calculate "normalization coefficient" (sorry for terminology) as
coeff = realWorldDistanceOf2Points / distanceOfTriangulatedPoints
Once you have this coeff, just mulptiply all your triangulated points with it and you got real world points.
Example:
you know that opposite corners of the wall are 5m from each other. you find these corners in both images, triangulate them (lets call triangulated points c1 and c2), calculate their distance in the "normalized" world as ||c1 - c2|| and get the
coeff = 5 / ||c1 - c2||
and you get real 3d world points as triangulatedPoint*coeff.
Maybe easier option is to have both cameras in fixed relative position and calibrate them together by stereoCalibrate openCV/Matlab function (there is actually pretty nice GUI in Matlab for that) - it returns not just intrinsic params, but also extrinsic. But I don't know if this is your case.
Using Matlab, how can I calculate a two points in Equilateral triangle if there are known one point and the Center of Gravity in a 3D ?
I know there is a infinite solutions but i need just a random one.
Thank you.
Take the vector pointing from the center of gravity to the point.
Create an orthogonal vector (this can be done in a few ways, I usually take the first vector, add 1.0 to each component until it is not parallel, then take the cross product with the original vector).
Rotate your vector 120 degrees about the orthogonal vector. (look up the rotation matrix about an arbitrary vector)
Create your second point by adding that vector to your center of gravity.
Create your third point by rotating it again or in the opposite direction.
stereoParameters takes two extrinsic parameters: RotationOfCamera2 and TranslationOfCamera2.
The problem is that the documentation is a not very detailed about what RotationOfCamera2 really means, it only says: Rotation of camera 2 relative to camera 1, specified as a 3-by-3 matrix.
What is the coordinate system in this case ?
A rotation matrix can be specified in any coordinate system.
What does it exactly mean "the coordinate system of Camera 1" ? What are its x,y,z axes ?
In other words, if I calculate the Essential Matrix, how can I get the corresponding RotationOfCamera2 and TranslationOfCamera2 from the Essential Matrix ?
RotationOfCamera2 and TranslationOfCamera2 describe the transformation from camera1's coordinates into camera2's coordinates. A camera's coordinate system has its origin at the camera's optical center. Its X and Y-axes are in the image plane, and its Z-axis points out along the optical axis.
Equivalently, the extrinsics of camera 1 are identity rotation and zero translation, while the extrinsics of camera 2 are RotationOfCamera2 and TranslationOfCamera2.
If you have the Essential matrix, you can decompose it into the rotation and a translation. Two things to keep in mind. First, the translation is up to scale, so t will be a unit vector. Second, the rotation matrix will be a transpose of what you get from estimateCameraParameters, because of the difference in the vector-matrix multiplication conventions.
Out of curiosity, what is it that you are trying to accomplish? Are you working with a single moving camera? Otherwise, why not use the Stereo Camera Calibrator app to calibrate your cameras, and get rotation and translation for free?
Suppose for left camera's 1st checkerboard (or to any world reference) rotation is R1 and translation is T1, right camera's 1st checkerboard rotation is R2 and translation is T2, then you can calculate them as follows;
RotationOfCamera2 = R2*R1';
TranslationOfCamera2= T2-RotationOfCamera2*T1
But please note that this calculations are just for one identical checkerboard reference. Inside matlab these two parameters are calculated by all given pair of checkerboard images and calculate median values as initial guess. Later these parameters will be refine by nonlinear optimization. So after median calculations they might be sigtly differ. But if you have just one reference point tranfomation for both two camera, you should use above formula. Note Dima told, matlab's rotation matrix is transpose of normal usage. So I wrote it as how the literature tells not matlab's style.
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 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