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I am trying to get the rotation in quaternion relative to the world coordinates.
mat4 mat(1);
mat *= rotate(radians(45.0f), vec3(0, 1, 0));
quat q = conjugate(toQuat(mat));
log(degrees(eulerAngles(q)));
In this code, the matrix rotates 45 degrees around y-axis, but q represents a -45 degrees rotation around y-axis. How can I find the correct rotation?
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I have a line strip defined by 3 points, each having an x and y coordinate.
I'm trying to smooth out the middle (point 2) corner as shown in the picture below:
the gray line is the original line strip and the black one is the smoothed-out one.
The smoothed out area should be constant across multiple values (as in it is not dependent on length of the line between p1 and p2 or p2 and p3).
I've originally been using bezier curves and a simple spline, however that did not do the trick since the smooth curve was obviously not same across multiple values.
How can I do this?
Pick 2 points on each line that are the same distance to the corner. On those points draw two lines at right angles to the lines you already have (normal vectors pointing bottom left). They will cross at a point which will be the center of a circle, part this circle will then be the smoothed corner.
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I have been working on a opencv project that requires calculation of histogram for rotation invariant local binary pattern.I checked github and bytefish git repos but cant find what i am looking for.
Exactly what i want to say is that once i find the lbp image of any video frame or image, how to make it rotation invariant lbp and find its histogram as texture feature that can be further used for training or classification.
Can someone tell me how to exactly find the histogram for rotation invariant LBP and histogram for uniform rotation invariant LBP in opencv c++.
Thankyou!
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There's a triangle and a point lying outside it. We need to find the summary length of triangle sides that are visible from the point.
In this case the answer is AB+BC.
As input there are coordinates of the points.
So the question is how to solve it in C++?
P.S. In my opinion we need to find ρ(P;each of the points) and watch if this distance intersect with any of the triangle's sides. Then choose the two farthest points, which met the condition, and find the sum of the side(s).
Let's define an order for the triangle vertices. Let it be counter-clockwise. Your triangle is ACB (clockwise order would be ABC).
The point sees an edge if it's located in the right semi-plane that the edge defines.
This formula:
res = (y2 - y1)*(px - x1) - (x2 - x1)*(py - y1)
gives info about what semi-plane is (px,py) relative to (x1,y1)-(x2,y2) segment. Just get the sign of res.
You analize the three segments of the triangle an get those that the point is on the right side.
Choosing the other order changes the sign.
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Hey there I try to do a realtime visual odometry system for a monocular camera.
Now I'm looking for an equation to describe 3d points movement in 2d vectors. While researching I came across a very interesting looking equation. I'm refering to page 22. It basically makes a simplification under the assumption of a relatively small time step. But now I'm struggeling about the image coordinates x and y. It's said that x would be sth like x=(px-px0) and y=(py-py0). When I understand it right p0 is the center of rotation. But if this is the case the whole formular would make no sense for my case cause I would need a prior knowledge of the center of rotation. Which is based on the translation again.
So maybe can help understanding it or maybe point me to a better way to do it.
To use this equation, you must have calibrated your camera (with a pinhole model), so you have a set of distortion coefficients, a focal distance and the principal point, which is the intersection of the optical axis with the image plane, as illustrated here: http://docs.opencv.org/2.4/modules/calib3d/doc/camera_calibration_and_3d_reconstruction.html.
In the equation you mention, x and y coordinates are in pixels after distortion correction and relative to the center of projection, not the center of rotation. So, the px0 and py0 you are looking for, are the coordinates of the principal point, that is, cx0 and cy0 using the naming convention of the link above.
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I am trying to present a point cloud and its projection with OpenGL on the plane normal to the line connecting the two most distant points. I have succeed in presenting the point cloud on the scene with an orthonormal system. I have found the two farthest point in the cloud. And I found the plan on which I projected.
I tried to make this projection but envin.
I tried with the transformation matrices as GL_PROJECTION but envin.
Can someone give me hand?
You can calculate the coordinates of the projected points by mathematical formulas, then draw them with OpenGL.
Take a look at this link
You can drop a perpendicular from each point to your found plane, and compute its foot point on the plane. Then draw another point there.