My friend and I are making a 3d rendering engine from scratch in our VB class at school, but I am not sure how the math to form the cube would work. Given six variables:
rotX
rotY
rotZ
lenX
lenY
lenZ
Which represent the rotation on x,y,z and the length on x,y,z respectively, what would be the formulas to make the cube? I know that all I have to do is calculate three segments and from those segments just create three parallelograms, so I just need the math to find what the three segments are.
Thanks!
there are 2 basic 3D object representations for both are your data is insufficient.
surface representation
objects are set of surface polygons/vertexes/...
for cube its a set of 8 points + the triangles/quads for 6 faces
analytical representation
objects are set of equations describing the object
for cube its a intersection of 6 planes
I think you are using option 1 so what you need is:
- position
- orientation
- size
usually an axis aligned cube looks like this:
const double a=1.0; //cube size;
double pnt[8][3]= //cube points
{
+a,-a,+a,
+a,+a,+a,
-a,+a,+a,
-a,-a,+a,
+a,-a,-a,
+a,+a,-a,
-a,+a,-a,
-a,-a,-a
};
int tab[24]=
{
0,1,2,3, // 1st.quad
7,6,5,4, // 2nd.quad
4,5,1,0, // 3th.quad ...
5,6,2,1,
6,7,3,2,
7,4,0,3
};
well for size and orientation you can apply transformation matrix
or directly recompute points by direction vectors
so you need to remember position (point) and orientation (3 vectors) and size (scalar)
all above can be stored in single transformation matrix 4x4
but if you want the vectors then points will be like this:
P(+a,-a,+a) -> +a*I -a*J +a*K
where I,J,K are the orientation vectors
a is cube size
P(+a,-a,+a) is original axis aligned point in table above
Option 2 is more tricky to implement and unless you really need it (ray-tracing renders) then forget about it.
Related
I want to fill a 2D image along its polar radius, the data are stored in a image where each row or column corresponds to the radius in target image. How can I fill the target image efficiently? Such as with iradius or some functions? I do not prefer a pix-pix operation.
Are you looking for something like this?
number maxR = 100
image rValues := realimage("I(r)",4,maxR)
rValues = 10 + trunc(100*random())
image plot :=realimage("Ring",4,2*maxR,2*maxR)
rValues.ShowImage()
plot.ShowImage()
plot = rValues.warp(iradius,0)
You might also want to check out the relevant example code from the F1 help documentation of GMS itself:
Explaining warp a bit:
plot = rValues.warp(iradius,0)
Assigns values to plot based on a value-lookup in rValues.
For each pixel in plot a coordinate position in rValues is computed, and the value is simply looked up. If the computed coordinate is non-integer, bilinear interpolation between the 4 closest points is used.
In the example, the two 'formulas' for the coordinate calculation are simple x' = iradius and y' = 0 where iradius is an expression computed from the coordinate in plot, for convenience.
You can feed any expression into the parameters for warp( ) and the command is closely related to just using the square bracket notation of addressing values. In fact, the only difference is that warp performs the bilinear interpolation of values instead of truncating the coordinates to integer values.
Recently I'm struggling with a pose estimation problem with a single camera. I have some 3D points and the corresponding 2D points on the image. Then I use solvePnP to get the rotation and translation vectors. The problem is, how can I determine whether the vectors are right results?
Now I use an indirect way to do this:
I use the rotation matrix, the translation vector and the world 3D coordinates of a certain point to obtain the coordinates of that point in Camera system. Then all I have to do is to determine whether the coordinates are reasonable. I think I know the directions of x, y and z axes of Camera system.
Is Camera center the origin of the Camera system?
Now consider the x component of that point. Is x equavalent to the distance of the camera and the point in the world space in Camera's x-axis direction (the sign can then be determined by the point is placed on which side of the camera)?
The figure below is in world space, while the axes depicted are in Camera system.
========How Camera and the point be placed in the world space=============
|
|
Camera--------------------------> Z axis
| |} Xw?
| P(Xw, Yw, Zw)
|
v x-axis
My rvec and tvec results seems right and wrong. For a specified point, the z value seems reasonable, I mean, if this point is about one meter away from the camera in the z direction, then the z value is about 1. But for x and y, according to the location of the point I think x and y should be positive but they are negative. What's more, the pattern detected in the original image is like this:
But using the points coordinates calculated in Camera system and the camera intrinsic parameters, I get an image like this:
The target keeps its pattern. But it moved from bottom right to top left. I cannot understand why.
Yes, the camera center is the origin of the camera coordinate system, which seems to be right following to this post.
In case of camera pose estimation, value seems reasonable can be named as backprojection error. That's a measure of how well your resulting rotation and translation map the 3D points to the 2D pixels. Unfortunately, solvePnP does not return a residual error measure. Therefore one has to compute it:
cv::solvePnP(worldPoints, pixelPoints, camIntrinsics, camDistortion, rVec, tVec);
// Use computed solution to project 3D pattern to image
cv::Mat projectedPattern;
cv::projectPoints(worldPoints, rVec, tVec, camIntrinsics, camDistortion, projectedPattern);
// Compute error of each 2D-3D correspondence.
std::vector<float> errors;
for( int i=0; i < corners.size(); ++i)
{
float dx = pixelPoints.at(i).x - projectedPattern.at<float>(i, 0);
float dy = pixelPoints.at(i).y - projectedPattern.at<float>(i, 1);
// Euclidean distance between projected and real measured pixel
float err = sqrt(dx*dx + dy*dy);
errors.push_back(err);
}
// Here, compute max or average of your "errors"
An average backprojection error of a calibrated camera might be in the range of 0 - 2 pixel. According to your two pictures, this would be way more. To me, it looks like a scaling problem. If I am right, you compute the projection yourself. Maybe you can try once cv::projectPoints() and compare.
When it comes to transformations, I learned not to follow my imagination :) The first thing I Do with the returned rVec and tVec is usually creating a 4x4 rigid transformation matrix out of it (I posted once code here). This makes things even less intuitive, but instead it is compact and handy.
Now I know the answers.
Yes, the camera center is the origin of the camera coordinate system.
Consider that the coordinates in the camera system are calculated as (xc,yc,zc). Then xc should be the distance between the camera and
the point in real world in the x direction.
Next, how to determine whether the output matrices are right?
1. as #eidelen points out, backprojection error is one indicative measure.
2. Calculate the coordinates of the points according to their coordinates in the world coordinate system and the matrices.
So why did I get a wrong result(the pattern remained but moved to a different region of the image)?
Parameter cameraMatrix in solvePnP() is a matrix supplying the parameters of the camera's external parameters. In camera matrix, you should use width/2 and height/2 for cx and cy. While I use width and height of the image size. I think that caused the error. After I corrected that and re-calibrated the camera, everything seems fine.
I'm trying to figure out the most efficient/fast way to add a large number of convex quads (four given x,y points) into an array/list and then to check against those quads if a point is within or on the border of those quads.
I originally tried using ray casting but thought that it was a little overkill since I know that all my polygons will be quads and that they are also all convex.
currently, I am splitting each quad into two triangles that share an edge and then checking if the point is on or in each of those two triangles using their areas.
for example
Triangle ABC and test point P.
if (areaPAB + areaPAC + areaPBC == areaABC) { return true; }
This seems like it may run a little slow since I need to calculate the area of 4 different triangles to run the check and if the first triangle of the quad returns false, I have to get 4 more areas. (I include a bit of an epsilon in the check to make up for floating point errors)
I'm hoping that there is an even faster way that might involve a single check of a point against a quad rather than splitting it into two triangles.
I've attempted to reduce the number of checks by putting the polygon's into an array[,]. When adding a polygon, it checks the minimum and maximum x and y values and then using those, places the same poly into the proper array positions. When checking a point against the available polygons, it retrieves the proper list from the array of lists.
I've been searching through similar questions and I think what I'm using now may be the fastest way to figure out if a point is in a triangle, but I'm hoping that there's a better method to test against a quad that is always convex. Every polygon test I've looked up seems to be testing against a polygon that has many sides or is an irregular shape.
Thanks for taking the time to read my long winded question to what's prolly a simple problem.
I believe that fastest methods are:
1: Find mutual orientation of all vector pairs (DirectedEdge-CheckedPoint) through cross product signs. If all four signs are the same, then point is inside
Addition: for every edge
EV[i] = V[i+1] - V[i], where V[] - vertices in order
PV[i] = P - V[i]
Cross[i] = CrossProduct(EV[i], PV[i]) = EV[i].X * PV[i].Y - EV[i].Y * PV[i].X
Cross[i] value is positive, if point P lies in left semi-plane relatively to i-th edge (V[i] - V[i+1]), and negative otherwise. If all the Cross[] values are positive, then point p is inside the quad, vertices are in counter-clockwise order. f all the Cross[] values are negative, then point p is inside the quad, vertices are in clockwise order. If values have different signs, then point is outside the quad.
If quad set is the same for many point queries, then dmuir suggests to precalculate uniform line equation for every edge. Uniform line equation is a * x + b * y + c = 0. (a, b) is normal vector to edge. This equation has important property: sign of expression
(a * P.x + b * Y + c) determines semi-plane, where point P lies (as for crossproducts)
2: Split quad to 2 triangles and use vector method for each: express CheckedPoint vector in terms of basis vectors.
P = a*V1+b*V2
point is inside when a,b>=0 and their sum <=1
Both methods require about 10-15 additions, 6-10 multiplications and 2-7 comparisons (I don't consider floating point error compensation)
If you could afford to store, with each quad, the equation of each of its edges then you could save a little time over MBo's answer.
For example if you have an inward pointing normal vector N for each edge of the quad, and a constant d (which is N.p for one of the vertcies p on the edge) then a point x is in the quad if and only if N.x >= d for each edge. So thats 2 multiplications, one addition and one comparison per edge, and you'll need to perform up to 4 tests per point.This technique works for any convex polygon.
I am new to GLUT and opengl. I need to draw a scatterplot matrix for n dimensional array.
I have saved the data from csv to a vector of vectors and each vector corresponds to a row. I have plotted just one scatterplot. And used GL_LINES to draw the grid. My questions
1. How do I draw points in a particular grid? Using GL_POINTS I can only draw points in the entire window.
Please let me know need any further info to answer this question
Thanks
What you need to do is be able to transform your data's (x,y) coordinates into screen coordinates. The most straightforward way to do it actually does not rely on OpenGL or GLUT. All you have to do is use a little math. Determine the screen (x,y) coordinates of the place where you want a datapoint for (0,0) to be on the screen, and then determine how far apart you want one increment to be on the screen. Simply take your original data points, apply the offset, and then scale them, to get your screen coordinates, which you then pass into glVertex2f() (or whatever function you are using to specify points in your API).
For instance, you might decide you want point (0,0) in your data to be at location (200,0) on your screen, and the distance between 0 and 1 in your data to be 30 pixels on the screen. This operation will look like this:
int x = 0, y = 0; //Original data points
int scaleX = 30, scaleY = 30; //Scaling values for each component
int offsetX = 100, offsetY = 100; //Where you want the origin of your graph to be
// Apply the scaling values and offsets:
int screenX = x * scaleX + offsetX;
int screenY = y * scaleY + offsetY;
// Calls to your drawing functions using screenX and screenY as your coordinates
You will have to determine values that make sense for the scalaing and offsets. You can also have your program use different values for different sets of data, so you can display multiple graphs on the same screen. But this is a simple way to do it.
There are also other ways you can go about this. OpenGL has very powerful coordinate transformation functions and matrix math capabilities. Those may become more useful when you develop increasingly elaborate programs. They're most useful if you're going to be moving things around the screen in real-time, or operating on incredibly large data sets, as they allow you to perform these mathematical calculations very quickly using your graphics hardware (which is able to do them much faster than the CPU). However, the time it takes for the CPU to do simple calculations like those where you only are going to do them once or very infrequently on limited sets of data is not a problem for computers today.
I'm trying to implement a geometry templating engine. One of the parts is taking a prototypical polygonal mesh and aligning an instantiation with some points in the larger object.
So, the problem is this: given 3d point positions for some (perhaps all) of the verts in a polygonal mesh, find a scaled rotation that minimizes the difference between the transformed verts and the given point positions. I also have a centerpoint that can remain fixed, if that helps. The correspondence between the verts and the 3d locations is fixed.
I'm thinking this could be done by solving for the coefficients of a transformation matrix, but I'm a little unsure how to build the system to solve.
An example of this is a cube. The prototype would be the unit cube, centered at the origin, with vert indices:
4----5
|\ \
| 6----7
| | |
0 | 1 |
\| |
2----3
An example of the vert locations to fit:
v0: 1.243,2.163,-3.426
v1: 4.190,-0.408,-0.485
v2: -1.974,-1.525,-3.426
v3: 0.974,-4.096,-0.485
v5: 1.974,1.525,3.426
v7: -1.243,-2.163,3.426
So, given that prototype and those points, how do I find the single scale factor, and the rotation about x, y, and z that will minimize the distance between the verts and those positions? It would be best for the method to be generalizable to an arbitrary mesh, not just a cube.
Assuming you have all points and their correspondences, you can fine-tune your match by solving the least squares problem:
minimize Norm(T*V-M)
where T is the transformation matrix you are looking for, V are the vertices to fit, and M are the vertices of the prototype. Norm refers to the Frobenius norm. M and V are 3xN matrices where each column is a 3-vector of a vertex of the prototype and corresponding vertex in the fitting vertex set. T is a 3x3 transformation matrix. Then the transformation matrix that minimizes the mean squared error is inverse(V*transpose(V))*V*transpose(M). The resulting matrix will in general not be orthogonal (you wanted one which has no shear), so you can solve a matrix Procrustes problem to find the nearest orthogonal matrix with the SVD.
Now, if you don't know which given points will correspond to which prototype points, the problem you want to solve is called surface registration. This is an active field of research. See for example this paper, which also covers rigid registration, which is what you're after.
If you want to create a mesh on an arbitrary 3D geometry, this is not the way it's typically done.
You should look at octree mesh generation techniques. You'll have better success if you work with a true 3D primitive, which means tetrahedra instead of cubes.
If your geometry is a 3D body, all you'll have is a surface description to start with. Determining "optimal" interior points isn't meaningful, because you don't have any. You'll want them to be arranged in such a way that the tetrahedra inside aren't too distorted, but that's the best you'll be able to do.