I implement an astar path finding method based on quad tree(the grids are difference in size, it's the main case), now I can get the rough path, but I don't get a smooth method to optimize the path, anyone who has one or reference may helps, thanks.
haha,I find the method for the difference size path smooth. Just use the funnel algorithm too, but there is a difference: when add left and right point for two quads, find one intersection side, and sort the four points(a0, a1, b0, b1) like ([x, y0], [x, y1], [x, y2], [x, y3], and y0 <= y1 <= y2 <= y3) , and use the two points [x, y1], [x, y2].
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I've a radial distortion function which gives me relative distortion from 0 (image center) to the relative full image field (field height 1) in percent. For example this function would give me a distortion of up to 5% at the full relative field height of 1.
I tried to use this together with opencv undistort function to apply distortion but don't know how to fill the matrices.
As said, I've a source image only and don't know anything about the camera parameters like focal length, except that I know the distortion function.
How should I set the matrix in cv2.undistort(src_image, matrix, ...) ?
The OpenCv routine that's easier to use in your case is cv::remap, not undistort.
In the following I assume your distortion purely radial. Similar considerations apply if you have it already decomposed in (x, y).
So you have a distortion function d(r) of the distance r = sqrt((x - x_c)^2 + (y - y_c)^2) of a pixel (x, y) from the image center (x_c, y_c). The function expresses the relative change of the radius r_d of a pixel in the distorted image from the undistorted one r: (r_d - r) / r = d(r), or, equivalently, r_d = r * (1 - d(r)).
If you are given a distorted image, and want to remove the distortion, you need to invert the above equation (i.e. solve it analytically or numerically), finding the value of r for every r_d in the range of interest. Then you can trivially create two arrays, map_x and map_y, that represent the mapping from distorted to undistorted coordinates: for a given pair (x_d, y_d) of integer pixel coordinates in the distorted image, you compute the associated r_d = sqrt(((x_d - x_c)^2 + (y_d - y_c)^2), then the corresponding r as function of r_d from solving the equation, go back to (x, y), and assign map_x[y_d, x_d] = x; map_y[y_d, x_d] = y. Finally, you pass those to cv::remap.
In the picture of SVM from Wikipedia, at the lower left corner - pointed by the red arrow, there's b / ||w||. How is that calculated? In other words, why is the line in the picture b / ||w||? Thanks.
The line represents the affine subspace of points x, whose scalar product with the weight vector w yields b. As w is typically not a unit vector (i.e. need not have length 1), one has to divide b by the norm (the "length") of w to get the actual distance from the origin.
More precisely: imagine a vector x starting at the origin and reaching out to a point on the red line and let u be the unit vector in the direction of w, i.e. u = w / ||w||. Then the scalar product of x and u multiplied with u is the projection of x onto the unit vector u and its length corresponds to the distance of the red line from the origin. If you calculate the scalar product of <x,w> (written as x*w in the graphics) instead, you still get a projection on u, which has length b (that's actually how b is defined), so to get back the distance from the origin one has to calculate b/||w||.
I would like to construct a plane from a list of 3D points in OpenCV. I would like to obtain the result by finding the four parameters in the following form: Ax+By+Cz+D = 0. Would anyone suggest me a way to do it?
If the data does not contain outliers and does not contain more than one plane. Furthermore, all the points lay exactly on a plane (the data is not noisy), it is so simple:
Pick up three random points which are not lay on the same line.
solve this system of linear equations:
x1+by1+cz1+d = 0
x2+by2+cz2+d = 0
x3+by3+cz3+d = 0
then :
A= Choose any number you want in order to match your scale.
B= b*A
C= c*A
D= d*A
If the data is noisy or contains outliers or more than plane (or both) you need then some kind of Robust Estimation techniques. Search for RANSAC as a start.
if you are familar with RANSAC you can see this example here (it is about lines you can simply generlize it to deal with plane)
Answer by leveraging OpenCV
If you want to have equation solved by 3 points, just like follows:
ax + by + cz = 1
Example
you have three points: cv::Point3f p1, p2 and p3, and here is the code:
cv::Matx33f M(p1.x, p1.y, p1.z,
p2.x, p2.y, p2.z,
p3.x, p3.y, p3.z);
cv::Vec3f d(1, 1, 1);
cv::Vec3f coef = M.inv() * d;
Then, a, b, c are coef(0), coef(1), coef(2) sequentially.
I am applying a force and a torque on an node. This is my code:
myNode?.physicsBody?.applyForce(SCNVector3Make(0, -6, 4), atPosition: SCNVector3Make(0, 1, -1), impulse: true)
myNode?.physicsBody?.applyForce(SCNVector3Make(0, -2, 10), impulse: true)
myNode?.physicsBody?.applyTorque(SCNVector4Make(4, 2, 2.5, 1.6), impulse: true)
The object now falls down and moves from left to right afterwards. I want it fall down and move from right to the left(basically a reflection of the first movement across y-axis). I figured it out that there is very little I can do about the first 2 lines of code, because the force has no x-component. The last line, applyTorque, is the one I need to manipulate. How do you map across the y-axis if the vector has 4 components? I am a little rusty with math
The fuller version of the applyTorque function looks something like this:
.applyTorque(SCNVector4Make(x:, y:, z:, w:), impulse:)
So any numbers you put in the second position should be torque amounts around the y axis.
There's probably a relationship between the numbers and what they create in terms of rotational force on an object, but I've always just used trial-and-error to find what works. Sometimes it's HUGE numbers.
I am assuming that the x-axis is horizontal and the y-axis is vertical and the z-axis points straight at you (see the black arrows below):
I found evidence that this is indeed the case in SceneKit.
If
applyTorque(SCNVector4Make(x, y, z, w), impulse: boolean)
is the correct usage, then x is the amount of counter-clockwise rotation around the x-axis (see green circle arrow), and similarly for y and z. Again, this is my best guess, and it is possible that SceneKit uses clockwise rotation. Therefore, x, y, and z together determine the axis of rotation of the torsional force.
Here is a simpler way to think of it. x, y, and z create a vector in the 3D space described above. The object will rotate counter-clockwise around this vector.
w on the other hand, is the magnitude of the torque, and has nothing to do with the axis of rotation.
Your request to "map vector across the y-axis" is actually a reflection across the yz-plane. If you think about it, what you want is to rotate the opposite direction around the y-axis (negate y) and the same for z.
So the answer should be:
myNode?.physicsBody?.applyTorque(SCNVector4Make(4, -2, -2.5, 1.6), impulse: true)
According to the SceneKit documentation the SCNVector4 argument specifies the direction (x, y, z vector components) and magnitude (w vector component) of the force in newton-meters. To mirror the direction of the applied torque, all you have to do is invert the magnitude. (x, y, z, -magnitude)
I have a polyline figure, given as an array of relative x and y point coordinates (0.0 to 1.0).
I have to draw the figure with random position, scale and rotation angle.
How can I do it in the best way?
You could use a simple transformation with RT matrix.
Let X = (x y 1)^t be coordinates of one point of your figure. Let R be a 2x2 rotation matrix, and T be 2x1 translation vector of the transformation You plan to make. RT matrix A will have the form of A = [R T;0 0 1]. To get transformed coordinates of point X, You need to do this simple calculation AX = X', where X' are the new coordinates. Now, to get the whole figure transformed, instead of using a single column, You use a matrix where each column has x coordinate in first row, y in the second and 1 in the third row.
Of course You can try to use functions provided by OpenCV, shown in this tutorial, or ones intended for vectors of points instead of whole images, but the way above makes You actually understand what are You doing ;)