I wrote a program to display some pressure-measurements. I want to use NURBS for a detail visualisation. So I orientated me to
enter link description here
My field has a range of 40x48 squares. So 40 rows and 48 columns. The Z component(the height) should be variable.
But I don't understand how to define
glMap2f(GL_MAP2_VERTEX_3, 0, 1, 3, 4, 0, 1, 12, 4, &ctrlpoints[0][0][0]);
// Parameter:
// target:
// What the control points represent (e.g. MAP2_VERTEX_3).
//
// u1:
// Range of the variable 'u'.
//
// u2:
// Range of the variable 'u.
//
// ustride:
// Offset between beginning of one control point and the next.
//
// uorder:
// The degree plus one.
//
// v1:
// Range of the variable 'v'.
//
// v2:
// Range of the variable 'v'.
//
// vstride:
// Offset between beginning of one control point and the next.
//
// vorder:
// The degree plus one.
//
// points:
// The data for the points.
I don't know how to set the parameters in my case.
For example what is u1 and u2? OR what are my ControlPoints?
In this link you can find a more detailed description of the parameters:
https://msdn.microsoft.com/en-us/library/windows/desktop/ee872053(v=vs.85).aspx
In your surface X direction is given by u-coordinate and Y direction is given by v-coordinate. It is common to set u1-u2 and v1-v2 to the [0,1] interval.
The order of the surface is interpolated between points (you can do linear interpolation with order=1, quadratic interpolation with order=2 and so on. 2 or 3 should be ok for your needs.
With this said, I am afraid that the method "glMap2f" will not accurately represent your data as in general the "control points" are not on the surface itself (as shown in the following )
You have to look for an algorithm to interpolate a nurbs surface from a point cloud, and then use glMap2f with the control points calculated.
If you want to know more about Nurbs, look for "The Nurbs book" from L. Piegl
Related
In CGAL I need to compute the exact intersection points between a set of lines and a set o circles. Starting from the circles (which can have irrational radius but rational squared_radius) I should compute the vertical line passing through the x_extremal_points of each circle (not segment but lines) and calculate the intersection point of each circle with each line.
I’m using CircularKernel and Circle_2 for the circles and Line_2 for the lines.
Here’s an example of how I compute the circles and the lines and how I check if they intersect.
int main()
{
Point_2 a = Point_2(250.5, 98.5);
Point_2 b = Point_2(156, 139);
//Radius is half distance ab
Circular_k::FT aRad = CGAL::squared_distance(a, b);
Circle_2 circle_a = Circle_2(a, aRad/4);
Circular_arc_point_2 a_left_point = CGAL::x_extremal_point(circle_a, false);
Circular_arc_point_2 a_right_point = CGAL::x_extremal_point(circle_a, true);
//for example use only left extremal point of circle a
CGAL::Bbox_2 a_left_point_bb = a_left_point.bbox();
Line_2 a_left_line = Line_2(Point_2(a_left_point_bb.xmin(), a_left_point_bb.ymin()),
Point_2(a_left_point_bb.xmin(), a_left_point_bb.ymax()));
if ( do_intersect(a_left_line, circle_a) ) {
std::cout << "intersect";
}
else {
std::cout << " do not intersect ";
}
return 0;
}
This flow rises this exception:
CGAL error: precondition violation!
Expression : y != 0
File : c:\dev\cgal-4.7\include\cgal\gmp\gmpq_type.h
Line : 371
Explanation:
Refer to the bug-reporting instructions at http://www.cgal.org/bug_report.html
I can’t figure out how I can calculate the intersection points.
Also, Is there a better way to compute the lines? I know abot the x_extremal_point function but it returns the Circular_arc_point point and I’m not able to construct a vertical line passing through them directly without using Bounding box.
In your code, you seem to compute the intersection of a circle with the vertical line that passes through the extremal point of the circle (I forget the bounding box). Well, then the (double) intersection is the extremal point itself...
More globally, you say in your text of introduction that you want to compute exact intersections. Then you should certainly not use bounding boxes, which by definition introduce some approximation.
If I understand your text correctly,
* for testing the intersection of your vertical lines with the other circles, you don't need to construct the lines, you only need to compare the abscissae of the extremal points of two circles, which you can do with the CGAL circular kernel.
* for computing the intersection of a vertical line that has non-rational coefficients (since its equation is of the form x= +-sqrt(r)) with another circle, then the CGAL circular kernel will not give you a pre-cooked solution. That kernel will help, but you must still compute a few things by hand.
If you don't want to bother, then you can also just take a standard CGAL kernel with Core::Expr as underlying number type. It can do "anything", but it will be slower.
For efficiency, you should look at the underlying 1D problem: projecting the lines and the circle on the X axis, you have a set of points and a set of intervals [Xc-R, Xc+R].
If the L points are sorted increasingly, you can locate the left bound of an interval in time Lg(L) by dichotomy, and scan the list of points until the right bound. This results in a O(Lg(L).C + I) behavior (C circle intervals), where I is the number of intersections reported.
I guess that with a merge-like process using an active list, if the interval bounds are also sorted you can lower to O(L + C + I).
The extension to 2D is elementary.
I have a 3 channel Mat image, type is CV_8UC3.
I want to compare, in a loop, the intensity value of a pixel with its neighbours and then set 0 or 1 if the neighbour is greater or not.
I can get the intensity calling Img.at<Vec3b>(x,y).
But my question is: how can I compare two Vec3b?
Should I compare pixels value for every channel (BGR or Vec3b[0], Vec3b[1] and Vec3b[2]), and then merge the three channels results into a single Mat object?
Me again :)
If you want to compare (greater or less) two RGB values you need to project the 3-dimensional RGB space onto a plane or axis.
Of course, there are many possibilities to do this, but an easy way would be to use the HSV color space. The hue (H), however, is not appropriate as a linear order function because it is circular (i.e. the value 1.0 is identical with 0.0, so you cannot decide if 0.5 > 0.0 or 0.5 < 0.0). However, the saturation (S) or the value (V) are appropriate projection functions for your purpose:
If you want to have colored pixels "larger" than monochrome pixels, you will prefer S.
If you want to have lighter pixels larger than darker pixels, you will probably prefer V.
Also any combination of S and V would be a valid projection function, e.g. S+V.
As far as I understand, you want a measure to calculate distance/similarity between two Vec3b pixels. This can be reflected to the general problem of finding distance between two vectors in an n-mathematical space.
One of the famous measures (and I think this is what you're asking for), is the Euclidean distance.
If you are using Opencv then you can simply use:
cv::Vec3b a(1, 1, 1);
cv::Vec3b b(5, 5, 5);
double dist = cv::norm(a, b, CV_L2);
You can refer to this for reading about cv::norm and its options.
Edit: If you are doing this to measure color similarity, it's recommended to use the LAB color space as it's proved that Euclidean distance in LAB space is a good approximation for human perception of colors.
Edit 2: I see what you mean, for this you can get the magnitude of each vector and then compare them, something like this:
double a_magnitude = cv::norm(a, CV_L2);
double b_magnitude = cv::norm(b, CV_L2);
if(a_magnitude > b_magnitude)
// do something
else
// do something else.
I'm trying to program a simulation. Originally I'd randomly create points like so...
for (int c = 0; c < number; c++){
for(int d = 0; d < 3; d++){
coordinate[c][d] = randomrange(low, high);
}
}
Where randomrange() is an arbitrary range randomizer, number is the amount of created points, and d represents the x,y,z coordinate. It works, however I want to take things further. How would I define a known shape? Say I want 80 points on a circle's circumference, or 500 that form the edges of a cube. I can explain well on paper, but have a problem describing the process as coding. This doesn't pertain to the question, but I end up taking the points to txt file and then use matlab, scatter3 to plot the points. Creating the "shape" points is my issue.
Both a circle and a cube edges set are 1-dimensional sets, so you can represent them as real intervals. For a circle it's straightforward: use an interval (0, 2pi) and transform a random value phi from the interval into a point:
xcentre + R cos(phi), ycentre + R sin(phi)
For a cube you have 12 segments, so use interval (0, 12) and split a random number from the interval into an integer part and a fraction. Then use the integer as an edge number and the fraction as a position inside the edge.
Easy variant:
First think of the min/max x/y values (separately; to reduce the faulty values for the step below), generate some coordinates matching this range, and then check if it fulfills eg. a^2+b^2=r^2 (circle)
If not, try again.
Better, but only possible for certain shapes:
Generate a radius between (0-max) and an angle (0-360)
(or just an angle if it should be on the circle border)
and use some math (sin/cos...) to transform it into x and y.
http://en.wikipedia.org/wiki/Polar_coordinate_system
I'm trying to calculate the cameras position for an image. I have 2 images of a rubiks cube. The first image is considered to be the base image and the next image is the image after the camera has moved. So for the first image I assume that the camera is at (0,0,0). On this image I then identify the 4 corners of the front face of the rubiks cube as shown here (4 corners identified by the 4 blue circles).
Then for the next image (after camera movement), I identify the same face of the rubiks cube as show here
So by assuming the first image as the base image, does anyone know if/how i can calculate how much the camera has moved for image 2 as shown here:
I would suggest you use OpenCV for this. I also think, this question would be more suited to StackOverflow.
The textbook on this subject would be "Multiple-View Geometry" by Hartley and Zisserman. http://www.robots.ox.ac.uk/~vgg/hzbook/ (There is a sample chapter on the Fundamental Matrix on that website.)
Basically, first find the Fundamental Matrix, then by knowing the intrinsic parameters of the camera, find a solution to the position.
Fundamental Matrix: http://en.wikipedia.org/wiki/Fundamental_matrix_%28computer_vision%29
Intrinsic Parameters: Stuff like the focal length and where the principal point is on the image plane. If you have F, then E = K^t * F * K, if K is the intrinsic matrix and the same for both images.
How to find a solution to the camera position: http://en.wikipedia.org/wiki/Essential_matrix#Determining_R_and_t_from_E
Algorithm
This is how I would do it in OpenCV. I have done this before, so it ought to work.
1. Run Feature Detection and Detector Extractor on both images.
2. Match Features.
3. Use F = cv::findFundamentalMatrix with Ransac.
4. E = K.t() * F * K. // K needs to be found beforehand.
5. Do SingularValueDecomposition of E such that E = U * S * V.t()
6. R = U * W.inv() * V.t() // W = [[0, -1, 0], [1, 0, 0], [0, 0, 1]]
7. Tx = V * Z * V.t() // Z = [[0, -1, 0], [1, 0, 0], [0, 0, 0]]
8. get t from Tx (matrix version of cross product)
9. Find the correct solution. R.t() and -t are possiblities.
10. Get overall scale by knowing the length of the size of the Rubrik's cube.
Alternative Solutions
I am certain that a more straightforward approach can also work. The benefit of this approach is that no human input is needed (unsupervised). This is not true for the optional step 10 (determining scale).
A different solution would exploit the knowledge of the geometry of the Rubrik's cube. For example, six (5.5) points are needed to estimate the position of the camera, if the point's 3D position is known.
Unfortunatly, I am not aware of any software that does this for you automatically.
So here is the alternative algorithm:
Write down the coordinates of the corners of the cube as (X_i, Y_i, Z_i), and possibly also points with other knowable positions.
Mark the corresponding points u_i = (x_i, y_i).
For every correspondence create two lines in a matrix A.
(X_i, Y_i, Z_i, 1, 0, 0, 0, 0, -x_iX_i, -x_iY_i, -x_iZ_i -x_i)
(0, 0, 0, 0, X_i, Y_i, Z_i, 1, -y_iX_i, -y_iY_i, -y_iZ_i -y_i)
Then find p such that Ap = 0. I.e. p is the right kernel of A, or the least-squared solution to Ap=0.
De-flatten p, to create a 3x4 matrix. P.
I am trying to calculate the centre of mass (x,y,z) coordinates of an object defined in an STL file (stereo lithography, not to be confused with the standard template library). The STL file contains a closed object (or objects) defined by a boundary made of triangles. The triangles themselves are not necessarily in any order, the file is simply the coordinates 3 vertices of each triangle floating in 3D space plus a normal vector to the triangle (the normal should be disregarded as it is not always done properly). There is nothing that links each triangle to one another, it is assumed that the object is closed.
One simple approach would be to divide a volume (in this case, a box) into millions of elements and determine if each element is inside the object defined in the STL file or not, then sum up the moments and calculate the centre of mass. This would work but its far from elegant and extremely slow.
Another method would be to convert the boundary representation into a number of packed tetrahedron solids. Form that I could calculate the centre of mass of each tetrahedron, its volume, and resulting moment and thus calculate the overall centre of mass from the sum of all tetrahedrons. The problem with this is that I don't know how to convert a surface representation of triangles into a volume representation of tetrahedrons (I'm assuming its a fairly non trivial task).
Dose anyone know of any methods or can think up of any methods that I could try? Or maybe even any reference material that talks about this?
For more information about STL files (only the first 2 sections are important, everything else is useless): http://en.wikipedia.org/wiki/STL_%28file_format%29
After a lot of thinking and experimentation I have the answer!
First we add a 4th point to each triangle in to make them into tetrahedrons with a volume centroid. We calculate the volumes and centres of masses and multiply them by each other to get our moments. We sum the moments and divide by total volume to get our overall centroid.
We calculate volumes using the determinate method shown here (equation 32): http://mathworld.wolfram.com/Tetrahedron.html
The centroids of each of the tetrahedrons is simply the average of the 4 points.
The trick here is that due to the way the STL file is created, the triangles have a normal that point outwards from the part surface, following the right hand rule of the 3 verticies used to create the triangle. we can use this to our advantage by allowing us to have a consistent convention in which to determine if a volume of the tetrahedron should be added or subtracted from our net part (this is because the reference point we chose may not necessarily be inside the part and the overall part is not necessarily convex, it is, however a closed object).
Using the determine method to calculate the volume, the first three coordinate points will represent the three points of our triangle. The fourth point would be our common origin. If the normal created by the triangle (following the right hand rule going from point 1, 2, 3) points towards our common reference point, that volume will be calculated as not part of our overall solid, or negative volume (by pointing towards, i mean the vector created by the triangle's normal is pointing loosely towards the same side as a normal plane created by the vector from our reference point to the centroid of the tetrahedron). If the vector is pointing away from the reference point, it is then positive volume or inside the part. If it is normal, then the volume goes to zero as the triangle is in the same plane as the reference point.
We don't need to worry about actually keeping track of any of this as if we are consistent with our inputs (as in the triangles follow the right hand rule with normal facing outwards from the part) the determine will give us the correct sign.
Anyways, heres the code (its even more simple than the explanation).
class data // 3 vertices of each triangle
{
public:
float x1,y1,z1;
float x2,y2,z2;
float x3,y3,z3;
};
int main ()
{
int numTriangles; // pull in the STL file and determine number of triangles
data * triangles = new triangles [numTriangles];
// fill the triangles array with the data in the STL file
double totalVolume = 0, currentVolume;
double xCenter = 0, yCenter = 0, zCenter = 0;
for (int i = 0; i < numTriangles; i++)
{
totalVolume += currentVolume = (triangles[i].x1*triangles[i].y2*triangles[i].z3 - triangles[i].x1*triangles[i].y3*triangles[i].z2 - triangles[i].x2*triangles[i].y1*triangles[i].z3 + triangles[i].x2*triangles[i].y3*triangles[i].z1 + triangles[i].x3*triangles[i].y1*triangles[i].z2 - triangles[i].x3*triangles[i].y2*triangles[i].z1) / 6;
xCenter += ((triangles[i].x1 + triangles[i].x2 + triangles[i].x3) / 4) * currentVolume;
yCenter += ((triangles[i].y1 + triangles[i].y2 + triangles[i].y3) / 4) * currentVolume;
zCenter += ((triangles[i].z1 + triangles[i].z2 + triangles[i].z3) / 4) * currentVolume;
}
cout << endl << "Total Volume = " << totalVolume << endl;
cout << endl << "X center = " << xCenter/totalVolume << endl;
cout << endl << "Y center = " << yCenter/totalVolume << endl;
cout << endl << "Z center = " << zCenter/totalVolume << endl;
}
Extremely fast for calculating centres of mass for STL files.
EDIT: Look up "winding number algorithm" or "crossing number algorithm" - what I try to describe below is a 3-d crossing number algorithm.
I've got a feeling that something like this will work, but I don't have the ability to test it out, right now:
Build the filled-in 3-d structure from the triangles in the STL file iteratively. Start by picking a single point to use as a basis for the 3-d structure. Then, begin your structure by creating a triangular pyramid, with base defined by the first triangle in the STL file, and vertex your chosen point. Each such component of your iteratively built volume would also contain an "intersection parity" - initialize it to 0.
For each subsequent triangle in the STL file, create a similar pyramid, and see if it intersects with the 3-d structure that you've built so far. If it does, calculate the intersection, and segment the existing structure and the new pyramid so that no two components overlap. Keep the "intersection parity" of the outermost part of the new polyhedron 0, but toggle it on all inner portions of the intersection -- if it was 0, make it 1, if it was 1, make it 0.
At the end, you'll have the closed polyhedron defined by all the portions of your structure that have intersection parity 0. Calculate the moments of all of these polyhedrons, and average them together to get your center of mass. I think the complexity would be something like O(n^2).