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 ;)
Related
Following is the conversion for spherical to cartesian coordinate
X = r cosθ sinΦ
Y = r sinθ sinΦ
Z = rcosΦ
we are using the reverse computation to compute spherical coordinate from cartesian coordinate which is defined as
r = √(x^2+y^2+z^2 )
θ = atan(Y./X)
Φ = atan(√(X^2+Y^2 )./Z)
The problem arises when Y and X are zero so θ can take any arbitrary value so during Matlab computations this results in NAN(not a number ) which makes θ discontinuous. Is there any interpolation technique to remove this discontinuity and how to interpret θ in this case.
θ is a matrix at various point and it gives following result it has jumps and black patched that represent discontinuity whereas I need to generate the following image with smooth variation. Please see the obtained theta and correct theta variation by clicking on the link and suggest some changes.
Discontinuous_Theta_variation
Correct Theta variation
While doing conversion from Cartesian to Spherical coordinate system, however the formulas which are written here are correct but you first need to understand their physical significance.
'r' is the distance of the point from origin. θ is the angle from the positive x axis to the line which is made by projecting the given point to XY plane. And Φ being the angle from positive z-axis to the line which joins origin and given point.
http://www.learningaboutelectronics.com/Articles/Cartesian-rectangular-to-spherical-coordinate-converter-calculator.php#answer
So say, for a point which has X and Y coordinates as 0, that means it lies on z axis and hence, its projection on XY plane lies on the origin. So we cannot exactly determine the angle of origin from X axis. But please note that, since the point lies on Z axis, so Φ=0 or pi (depending whether Z is positive or negative).
So while coding this problem, you may adapt this approach that you first check for Φ, if it is 0 or pi then theta = 0 (by default).
I hope this serves the purpose.
I am searching lots of resources on internet for many days but i couldnt solve the problem.
I have a project in which i am supposed to detect the position of a circular object on a plane. Since on a plane, all i need is x and y position (not z) For this purpose i have chosen to go with image processing. The camera(single view, not stereo) position and orientation is fixed with respect to a reference coordinate system on the plane and are known
I have detected the image pixel coordinates of the centers of circles by using opencv. All i need is now to convert the coord. to real world.
http://www.packtpub.com/article/opencv-estimating-projective-relations-images
in this site and other sites as well, an homographic transformation is named as:
p = C[R|T]P; where P is real world coordinates and p is the pixel coord(in homographic coord). C is the camera matrix representing the intrinsic parameters, R is rotation matrix and T is the translational matrix. I have followed a tutorial on calibrating the camera on opencv(applied the cameraCalibration source file), i have 9 fine chessbordimages, and as an output i have the intrinsic camera matrix, and translational and rotational params of each of the image.
I have the 3x3 intrinsic camera matrix(focal lengths , and center pixels), and an 3x4 extrinsic matrix [R|T], in which R is the left 3x3 and T is the rigth 3x1. According to p = C[R|T]P formula, i assume that by multiplying these parameter matrices to the P(world) we get p(pixel). But what i need is to project the p(pixel) coord to P(world coordinates) on the ground plane.
I am studying electrical and electronics engineering. I did not take image processing or advanced linear algebra classes. As I remember from linear algebra course we can manipulate a transformation as P=[R|T]-1*C-1*p. However this is in euclidian coord system. I dont know such a thing is possible in hompographic. moreover 3x4 [R|T] Vector is not invertible. Moreover i dont know it is the correct way to go.
Intrinsic and extrinsic parameters are know, All i need is the real world project coordinate on the ground plane. Since point is on a plane, coordinates will be 2 dimensions(depth is not important, as an argument opposed single view geometry).Camera is fixed(position,orientation).How should i find real world coordinate of the point on an image captured by a camera(single view)?
EDIT
I have been reading "learning opencv" from Gary Bradski & Adrian Kaehler. On page 386 under Calibration->Homography section it is written: q = sMWQ where M is camera intrinsic matrix, W is 3x4 [R|T], S is an "up to" scale factor i assume related with homography concept, i dont know clearly.q is pixel cooord and Q is real coord. It is said in order to get real world coordinate(on the chessboard plane) of the coord of an object detected on image plane; Z=0 then also third column in W=0(axis rotation i assume), trimming these unnecessary parts; W is an 3x3 matrix. H=MW is an 3x3 homography matrix.Now we can invert homography matrix and left multiply with q to get Q=[X Y 1], where Z coord was trimmed.
I applied the mentioned algorithm. and I got some results that can not be in between the image corners(the image plane was parallel to the camera plane just in front of ~30 cm the camera, and i got results like 3000)(chessboard square sizes were entered in milimeters, so i assume outputted real world coordinates are again in milimeters). Anyway i am still trying stuff. By the way the results are previosuly very very large, but i divide all values in Q by third component of the Q to get (X,Y,1)
FINAL EDIT
I could not accomplish camera calibration methods. Anyway, I should have started with perspective projection and transform. This way i made very well estimations with a perspective transform between image plane and physical plane(having generated the transform by 4 pairs of corresponding coplanar points on the both planes). Then simply applied the transform on the image pixel points.
You said "i have the intrinsic camera matrix, and translational and rotational params of each of the image.” but these are translation and rotation from your camera to your chessboard. These have nothing to do with your circle. However if you really have translation and rotation matrices then getting 3D point is really easy.
Apply the inverse intrinsic matrix to your screen points in homogeneous notation: C-1*[u, v, 1], where u=col-w/2 and v=h/2-row, where col, row are image column and row and w, h are image width and height. As a result you will obtain 3d point with so-called camera normalized coordinates p = [x, y, z]T. All you need to do now is to subtract the translation and apply a transposed rotation: P=RT(p-T). The order of operations is inverse to the original that was rotate and then translate; note that transposed rotation does the inverse operation to original rotation but is much faster to calculate than R-1.
I'm trying to calculate a new camera position based on the motion of corresponding images.
the images conform to the pinhole camera model.
As a matter of fact, I don't get useful results, so I try to describe my procedure and hope that somebody can help me.
I match the features of the corresponding images with SIFT, match them with OpenCV's FlannBasedMatcher and calculate the fundamental matrix with OpenCV's findFundamentalMat (method RANSAC).
Then I calculate the essential matrix by the camera intrinsic matrix (K):
Mat E = K.t() * F * K;
I decompose the essential matrix to rotation and translation with singular value decomposition:
SVD decomp = SVD(E);
Matx33d W(0,-1,0,
1,0,0,
0,0,1);
Matx33d Wt(0,1,0,
-1,0,0,
0,0,1);
R1 = decomp.u * Mat(W) * decomp.vt;
R2 = decomp.u * Mat(Wt) * decomp.vt;
t1 = decomp.u.col(2); //u3
t2 = -decomp.u.col(2); //u3
Then I try to find the correct solution by triangulation. (this part is from http://www.morethantechnical.com/2012/01/04/simple-triangulation-with-opencv-from-harley-zisserman-w-code/ so I think that should work correct).
The new position is then calculated with:
new_pos = old_pos + -R.t()*t;
where new_pos & old_pos are vectors (3x1), R the rotation matrix (3x3) and t the translation vector (3x1).
Unfortunately I got no useful results, so maybe anyone has an idea what could be wrong.
Here are some results (just in case someone can confirm that any of them is definitely wrong):
F = [8.093827077399547e-07, 1.102681999632987e-06, -0.0007939604310854831;
1.29246107737264e-06, 1.492629957878578e-06, -0.001211264339006535;
-0.001052930954975217, -0.001278667878010564, 1]
K = [150, 0, 300;
0, 150, 400;
0, 0, 1]
E = [0.01821111092414898, 0.02481034499174221, -0.01651092283654529;
0.02908037424088439, 0.03358417405226801, -0.03397110489649674;
-0.04396975675562629, -0.05262169424538553, 0.04904210357279387]
t = [0.2970648246214448; 0.7352053067682792; 0.6092828956013705]
R = [0.2048034356172475, 0.4709818957303019, -0.858039396912323;
-0.8690270040802598, -0.3158728880490416, -0.3808101689488421;
-0.4503860776474556, 0.8236506374002566, 0.3446041331317597]
First of all you should check if
x' * F * x = 0
for your point correspondences x' and x. This should be of course only the case for the inliers of the fundamental matrix estimation with RANSAC.
Thereafter, you have to transform your point correspondences to normalized image coordinates (NCC) like this
xn = inv(K) * x
xn' = inv(K') * x'
where K' is the intrinsic camera matrix of the second image and x' are the points of the second image. I think in your case it is K = K'.
With these NCCs you can decompose your essential matrix like you described. You triangulate the normalized camera coordinates and check the depth of your triangulated points. But be careful, in literature they say that one point is sufficient to get the correct rotation and translation. From my experience you should check a few points since one point can be an outlier even after RANSAC.
Before you decompose the essential matrix make sure that E=U*diag(1,1,0)*Vt. This condition is required to get correct results for the four possible choices of the projection matrix.
When you've got the correct rotation and translation you can triangulate all your point correspondences (the inliers of the fundamental matrix estimation with RANSAC). Then, you should compute the reprojection error. Firstly, you compute the reprojected position like this
xp = K * P * X
xp' = K' * P' * X
where X is the computed (homogeneous) 3D position. P and P' are the 3x4 projection matrices. The projection matrix P is normally given by the identity. P' = [R, t] is given by the rotation matrix in the first 3 columns and rows and the translation in the fourth column, so that P is a 3x4 matrix. This only works if you transform your 3D position to homogeneous coordinates, i.e. 4x1 vectors instead of 3x1. Then, xp and xp' are also homogeneous coordinates representing your (reprojected) 2D positions of your corresponding points.
I think the
new_pos = old_pos + -R.t()*t;
is incorrect since firstly, you only translate the old_pos and you do not rotate it and secondly, you translate it with a wrong vector. The correct way is given above.
So, after you computed the reprojected points you can calculate the reprojection error. Since you are working with homogeneous coordinates you have to normalize them (xp = xp / xp(2), divide by last coordinate). This is given by
error = (x(0)-xp(0))^2 + (x(1)-xp(1))^2
If the error is large such as 10^2 your intrinsic camera calibration or your rotation/translation are incorrect (perhaps both). Depending on your coordinate system you can try to inverse your projection matrices. On that account you need to transform them to homogeneous coordinates before since you cannot invert a 3x4 matrix (without the pseudo inverse). Thus, add the fourth row [0 0 0 1], compute the inverse and remove the fourth row.
There is one more thing with reprojection error. In general, the reprojection error is the squared distance between your original point correspondence (in each image) and the reprojected position. You can take the square root to get the Euclidean distance between both points.
To update your camera position, you have to update the translation first, then update the rotation matrix.
t_ref += lambda * (R_ref * t);
R_ref = R * R_ref;
where t_ref and R_ref are your camera state, R and t are new calculated camera rotation and translation, and lambda is the scale factor.
I need to perform a 3D rotation of a 2D image on x and y axis.
I read that i have to use the Homographic matrix on OpenCV , but i don't know how to set the matrix to perform a common rotation angle. For example 30 degree on x axis or 45° on y axis.
I read this post : Translating and Rotating an Image in 3D using OpenCV. I have tried different values of the f but it doesn't work.
I want to know which parameters of the matrix i have to change and how (formula).
Thank you!
Follow that same post, but replace your rotation matrix. Familiarize yourself with the Rorigues() function. You can send it a 1 x 3 array of the x, y, and z rotations. It will give you a a 3 x 3 rotation matrix. Plug this matrix in as the first 3 columns and 3 rows of R (leave the rest the same). If you don't want any translation, make sure you set the variable dist to 0 in the code on that page.
I am playing with the affine transform in OpenCV and I am having trouble getting an intuitive understanding of it workings, and more specifically, just how do I specify the parameters of the map matrix so I can get a specific desired result.
To setup the question, the procedure I am using is 1st to define a warp matrix, then do the transform.
In OpenCV the 2 routines are (I am using an example in the excellent book OpenCV by Bradski & Kaehler):
cvGetAffineTransorm(srcTri, dstTri, warp_matrix);
cvWarpAffine(src, dst, warp_mat);
To define the warp matrix, srcTri and dstTri are defined as:
CvPoint2D32f srcTri[3], dstTri[3];
srcTri[3] is populated as follows:
srcTri[0].x = 0;
srcTri[0].y = 0;
srcTri[1].x = src->width - 1;
srcTri[1].y = 0;
srcTri[2].x = 0;
srcTri[2].y = src->height -1;
This is essentially the top left point, top right point, and bottom left point of the image for starting point of the matrix. This part makes sense to me.
But the values for dstTri[3] just are confusing, at least, when I vary a single point, I do not get the result I expect.
For example, if I then use the following for the dstTri[3]:
dstTri[0].x = 0;
dstTri[0].y = 0;
dstTri[1].x = src->width - 1;
dstTri[1].y = 0;
dstTri[2].x = 0;
dstTri[2].y = 100;
It seems that the only difference between the src and the dst point is that the bottom left point is moved to the right by 100 pixels. Intuitively, I feel that the bottom part of the image should be shifted to the right by 100 pixels, but this is not so.
Also, if I use the exact same values for dstTri[3] that I use for srcTri[3], I would think that the transform would produce the exact same image--but it does not.
Clearly, I do not understand what is going on here. So, what does the mapping from the srcTri[] to the dstTri[] represent?
Here is a mathematical explanation of an affine transform:
this is a matrix of size 3x3 that applies the following transformations on a 2D vector: Scale in X axis, scale Y, rotation, skew, and translation on the X and Y axes.
These are 6 transformations and thus you have six elements in your 3x3 matrix. The bottom row is always [0 0 1].
Why? because the bottom row represents the perspective transformation in axis x and y, and affine transformation does not include perspective transform.
(If you want to apply perspective warping use homography: also 3x3 matrix )
What is the relation between 6 values you insert into affine matrix and the 6 transformations it does? Let us look at this 3x3 matrix like
e*Zx*cos(a), -q1*sin(a) , dx,
e*q2*sin(a), Z y*cos(a), dy,
0 , 0 , 1
The dx and
dy elements are translation in x and y axis (just move the picture left-right, up down).
Zx is the relative scale(zoom) you apply to the image in X axis.
Zy is the same as above for y axis
a is the angle of rotation of the image. This is tricky since when you want to rotate by 'a' you have to insert sin(), cos() in 4 different places in the matrix.
'q' is the skew parameter. It is rarely used. It will cause your image to skew on the side (q1 causes y axis affects x axis and q2 causes x axis affect y axis)
Bonus: 'e' parameter is actually not a transformation. It can have values 1,-1. If it is 1 then nothing happens, but if it is -1 than the image is flipped horizontally. You can use it also to flip the image vertically but, this type of transformation is rarely used.
Very important Note!!!!!
The above explanation is mathematical. It assumes you multiply the matrix by the column vector from the right. As far as I remember, Matlab uses reverse multiplication (row vector from the left) so you will need to transpose this matrix. I am pretty sure that OpenCV uses regular multiplication but you need to check it.
Just enter only translation matrix (x shifted by 10 pixels, y by 1).
1,0,10
0,1,1
0,0,1
If you see a normal shift than everything is OK, but If shit appears than transpose the matrix to:
1,0,0
0,1,0
10,1,1