I'm trying to find the angle between two points in an image. The angle is with reference to the centre line of the camera.
In this image the center point is along the center of the image (assumption, I still have to figure out how to actually calculate it) and I want to find the angle between the line connecting point 1 and the camera center and the line connecting the desired point and the camera center
Now I want to know two things about finding the angle
- Is it possible to know the angle if the distance is not known exactly (but can be estimated by a human at run time) Assuming both points lie in the same plane in the image
- If the points are not in the same plane, how should I handle the angle calculation?
It can be achieved by inner product.
If you are talking in 3D space so your x(vector) and y(vector) should be in the form [a,b,f](a and b are points) where f is the distance of image plane from the camera center and a and a are the corresponding coordinates in the camera frame.
If it is in 2D space, so you have to specify the origin of your frame and find a and b according to that frame and your x and y vectors are in the form [a,b].
It can be found by using this formula:
Angle between two rays
K is the camera matrix. x1 and x2 are the image points given in homogeneous form like [u,v,1] and d1 and d2 are the corresponding 3D points.
See Richard Hartley, Australian National University, Canberra, Andrew Zisserman, University of Oxford: “Multiple View Geometry in Computer Vision”, 2nd edition, p. 209 for more details.
Inverting the camera matrix is quite simple. See
https://www.imatest.com/support/docs/pre-5-2/geometric-calibration-deprecated/projective-camera/ for more details.
Related
For my undergraduate paper I am working on a iPhone Application using openCV to detect domino tiles. The detection works well in close areas, but when the camera is angled the tiles far away are difficult to detect.
My approach to solve this I would want to do some spacial calculations. For this I would need to convert a 2D Pixel value into world coordinates, calculate a new 3D position with a vector and convert these coordinates back to 2D and then check the colour/shape at that position.
Additionally I would need to know the 3D positions for Augmented Reality additions.
The Camera Matrix i got trough this link create opencv camera matrix for iPhone 5 solvepnp
The Rotationmatrix of the Camera I get from the Core Motion.
Using Aruco markers would be my last resort, as I woulnd't get the decided effect that I would need for the paper.
Now my question is, can i not make calculations when I know the locations and distances of the circles on a lets say Tile with a 5 on it?
I wouldn't need to have a measurement in mm/inches, I can live with vectors without measurements.
The camera needs to be able to be rotated freely.
I tried to invert the calculation sm'=A[R|t]M' to be able to calculate the 2D coordinates in 3D. But I am stuck with inverting the [R|t] even on paper, and I don't know either how I'd do that in swift or c++.
I have read so many different posts on forums, in books etc. and I am completely stuck and appreciate any help/input you can give me. Otherwise I'm screwed.
Thank you so much for your help.
Update:
By using the solvePnP that was suggested by Micka I was able to get the Rotation and Translation Vectors for the angle of the camera.
Meaning that if you are able to identify multiple 2D Points in your image and know their respective 3D World coordinates (in mm, cm, inch, ...), then you can get the mechanisms to project points from known 3D World coordinates onto the respective 2D coordinates in your image. (use the opencv projectPoints function).
What is up next for me to solve is the translation from 2D into 3D coordinates, where I need to follow ozlsn's approach with the inverse of the received matrices out of solvePnP.
Update 2:
With a top down view I am getting along quite well to being able to detect the tiles and their position in the 3D world:
tile from top Down
However if I am now angling the view, my calculations are not working anymore. For example I check the bottom Edge of a 9-dot group and the center of the black division bar for 90° angles. If Corner1 -> Middle Edge -> Bar Center and Corner2 -> Middle Edge -> Bar Center are both 90° angles, than the bar in the middle is found and the position of the tile can be found.
When the view is Angled, then these angles will be shifted due to the perspective to lets say 130° and 50°. (I'll provide an image later).
The Idea I had now is to make a solvePNP of 4 Points (Bottom Edge plus Middle), claculate solvePNP and then rotate the needed dots and the center bar from 2d position to 3d position (height should be irrelevant?). Then i could check with the translated points if the angles are 90° and do also other needed distance calculations.
Here is an image of what I am trying to accomplish:
Markings for Problem
I first find the 9 dots and arrange them. For each Edge I try to find the black bar. As said above, seen from Top, the angle blue corner, green middle edge to yellow bar center is 90°.
However, as the camera is angled, the angle is not 90° anymore. I also cannot check if both angles are 180° together, that would give me false positives.
So I wanted to do the following steps:
Detect Center
Detect Edges (3 dots)
SolvePnP with those 4 points
rotate the edge and the center points (coordinates) to 3D positions
Measure the angles (check if both 90°)
Now I wonder how I can transform the 2D Coordinates of those points to 3D. I don't care about the distance, as I am just calculating those with reference to others (like 1.4 times distance Middle-Edge) etc., if I could measure the distance in mm, that would even be better though. Would give me better results.
With solvePnP I get the rvec which I could change into the rotation Matrix (with Rodrigues() I believe). To measure the angles, my understanding is that I don't need to apply the translation (tvec) from solvePnP.
This leads to my last question, when using the iPhone, can't I use the angles from the motion detection to build the rotation matrix beforehand and only use this to rotate the tile to show it from the top? I feel that this would save me a lot of CPU Time, when I don't have to solvePnP for each tile (there can be up to about 100 tile).
Find Homography
vector<Point2f> tileDots;
tileDots.push_back(corner1);
tileDots.push_back(edgeMiddle);
tileDots.push_back(corner2);
tileDots.push_back(middle.Dot->ellipse.center);
vector<Point2f> realLivePos;
realLivePos.push_back(Point2f(5.5,19.44));
realLivePos.push_back(Point2f(12.53,19.44));
realLivePos.push_back(Point2f(19.56,19.44));
realLivePos.push_back(Point2f(12.53,12.19));
Mat M = findHomography(tileDots, realLivePos, CV_RANSAC);
cout << "M = "<< endl << " " << M << endl << endl;
vector<Point2f> barPerspective;
barPerspective.push_back(corner1);
barPerspective.push_back(edgeMiddle);
barPerspective.push_back(corner2);
barPerspective.push_back(middle.Dot->ellipse.center);
barPerspective.push_back(possibleBar.center);
vector<Point2f> barTransformed;
if (countNonZero(M) < 1)
{
cout << "No Homography found" << endl;
} else {
perspectiveTransform(barPerspective, barTransformed, M);
}
This however gives me wrong values, and I don't know anymore where to look (Sehe den Wald vor lauter Bäumen nicht mehr).
Image Coordinates https://i.stack.imgur.com/c67EH.png
World Coordinates https://i.stack.imgur.com/Im6M8.png
Points to Transform https://i.stack.imgur.com/hHjBM.png
Transformed Points https://i.stack.imgur.com/P6lLS.png
You see I am even too stupid to post 4 images here??!!?
The 4th index item should be at x 2007 y 717.
I don't know what I am doing wrongly here.
Update 3:
I found the following post Computing x,y coordinate (3D) from image point which is doing exactly what I need. I don't know maybe there is a faster way to do it, but I am not able to find it otherwise. At the moment I can do the checks, but still need to do tests if the algorithm is now robust enough.
Result with SolvePnP to find bar Center
The matrix [R|t] is not square, so by-definition, you cannot invert it. However, this matrix lives in the projective space, which is nothing but an extension of R^n (Euclidean space) with a '1' added as the (n+1)st element. For compatibility issues, the matrices that multiplies with vectors of the projective space are appended by a '1' at their lower-right corner. That is : R becomes
[R|0]
[0|1]
In your case [R|t] becomes
[R|t]
[0|1]
and you can take its inverse which reads as
[R'|-Rt]
[0 | 1 ]
where ' is a transpose. The portion that you need is the top row.
Since the phone translates in the 3D space, you need the distance of the pixel in consideration. This means that the answer to your question about whether you need distances in mm/inches is a yes. The answer changes only if you can assume that the ratio of camera translation to the depth is very small and this is called weak perspective camera. The question that you're trying to tackle is not an easy one. There is still people researching on this at PhD degree.
I've got a question related to multiple view geometry.
I'm currently dealing with a problem where I have a number of images collected by a drone flying around an object of interest. This object is planar, and I am hoping to eventually stitch the images together.
Letting aside the classical way of identifying corresponding feature pairs, computing a homography and warping/blending, I want to see what information related to this task I can infer from prior known data.
Specifically, for each acquired image I know the following two things: I know the correspondence between the central point of my image and a point on the object of interest (on whose plane I would eventually want to warp my image). I also have a normal vector to the plane of each image.
So, knowing the centre point (in object-centric world coordinates) and the normal, I can derive the plane equation of each image.
My question is, knowing the plane equation of 2 images is it possible to compute a homography (or part of the transformation matrix, such as the rotation) between the 2?
I get the feeling that this may seem like a very straightforward/obvious answer to someone with deep knowledge of visual geometry but since it's not my strongest point I'd like to double check...
Thanks in advance!
Your "normal" is the direction of the focal axis of the camera.
So, IIUC, you have a 3D point that projects on the image center in both images, which is another way of saying that (absent other information) the motion of the camera consists of the focal axis orbiting about a point on the ground plane, plus an arbitrary rotation about the focal axis, plus an arbitrary translation along the focal axis.
The motion has a non-zero baseline, therefore the transformation between images is generally not a homography. However, the portion of the image occupied by the ground plane does, of course, transform as a homography.
Such a motion is defined by 5 parameters, e.g. the 3 components of the rotation vector for the orbit, plus the the angle of rotation about the focal axis, plus the displacement along the focal axis. However the one point correspondence you have gives you only two equations.
It follows that you don't have enough information to constrain the homography between the images of the ground plane.
This is a continuation of the question from Here-How to find angle formed by the blades of a wind turbine with respect to a horizontal imaginary axis?
I've decided to use the following methodology for this-
Getting a frame from a camera and putting it in a loop.
Performing Canny edge detection.
Perform HoughLinesP to detect lines in the image.
Finding Blade Angle:
Perform Probabilistic Hough Lines Transform on the image. Restrict the blade lines to the length of the blades, as known already.
The returned value will have the start and end points of the lines detected. Since there are no background noises, this gives the starting and end point of the blade lines and the image will have the blade lines.
Now, find the dot product with a vector (1,0) by finding the vectors of the blade lines detected or we can use atan2 to find the relative angle of all the points detected with respect to a horizontal.
Problem:
When the yaw angle of the turbine is changed and it is not directly facing the camera, how do I calculate the blade angle formed?
The idea is to basically map the angles when rotated back into the form when viewed head on. From what I've been able to understand, I thought I'd find the homography matrix, decompose the matrix to get rotation, convert to Euler angles to calculate shift from the original axis, then shift all the axes with that angle. However, it's just a vague idea with no concrete planning to go upon.
Or I begin with trying to find the projection matrix, then get camera matrix and rotation matrix? I am lost on this account completely and feel overwhelmed with the many functions...
Other things I came across was the perspective transform,solvepnp..
It would be great if anyone could suggest another way to deal with this? Any links of code snippets would be helpful. I'm not that familiar with OpenCV and would be grateful for any help.
Thanks!
Edit:
[Edit by Spektre]
Assume the tip of the blades plus the center (or the three "roots" of the blades") lie on a common plane.
Fit a homography between those points and the corresponding ones in a reference pose for the turbine (cv::findHomography in OpenCv)
Decompose the homography into rotation and translation using an estimated or assumed camera calibration (cv::decomposeHomographyMat).
Convert the rotation into Euler angles.
I have a set of points on the unit sphere and a corresponding set of values being equal, for simplicity, to 0 and 1. Thus I'm constructing the characteristic function of a set on the sphere. Typically, I have several such sets, which form a partition of the sphere. An example is given in the figure.
I was wondering if paraview can find boundaries between the cells and compute the length and the curvature of the boundaries.
I read in a paper that using gradient reconstruction the guys managed to find the curvature of such contours. I imagine that if the curvature can be found, the length should be somewhat simpler. If the answer to the above question is yes, where should I look for the corresponding documentation?
For points on the sphere if they are build based on great-circle distance principle, it means all lines connecting points are of a shortest distance and plane goes through sphere center. In such case angle could be computed as arccos of scalar product.
R = 1;
angle = arccos(x1*x2 + y1*y2 + z1*z2);
length = R*angle;
And parametric line from p1 to p2 could be build using slerp interpolation.
slerp(t) = sin((1.0-t)*angle)/sin(angle)*p1 + sin(t*angle)/sin(angle)*p2;
where t is in [0...1] range
In such case curvature is 1/R for all great circle lines. That would be first thing I would try - try to match actual boundaries with those made from great-circle approach. If they match, that's the answer
Links
https://en.wikipedia.org/wiki/Great_circle
https://en.wikipedia.org/wiki/Great-circle_distance
https://en.wikipedia.org/wiki/Slerp
UPDATE
In case of non-great arcs I would propose following modification. Build great arc plane which goes through sphere center and on intersection with surface makes great arc between the points. Fix axis as a line going through those two points. Start rotating great arc plane along above mentioned axis till you get the exactly your arc of circle connecting two points. At this moment you could get rotation angle, compute your circle plane position and radius r, curvature as 1/r etc
I have a webcam pointed at a table at a slant and with it I track markers.
I have a transformationMatrix in OpenSceneGraph and its translation part contains the relative coordinates from the tracked Object to the Camera.
Because the Camera is pointed at a slant, when I move the marker across the table the Y and Z axis is updated, although all I want to be updated is the Z axis, because the height of the marker doesnt change only its distance to the camera.
This has the effect when when project a model on the marker in OpenSceneGraph, the model is slightly off and when I move the marker arround the Y and Z values are updated incorrectly.
So my guess is I need a Transformation Matrix with which I multiply each point so that I have a new coordinate System which lies orthogonal on the table surface.
Something like this: A * v1 = v2 v1 being the camera Coordinates and v2 being my "table Coordinates"
So what I did now was chose 4 points to "calibrate" my system. So I placed the marker at the top left corner of the Screen and defined v1 as the current camera coordinates and v2 as (0,0,0) and I did that for 4 different points.
And then taking the linear equations I get from having an unknown Matrix and two known vectors I solved the matrix.
I thought the values I would get for the matrix would be the values I needed to multiply the camera Coordinates with so the model would updated correctly on the marker.
But when I multiply the known Camera Coordinates I gathered before with the matrix I didnt get anything close to what my "table coordinates" were suposed to be.
Is my aproach completely wrong, did I just mess something up in the equations? (solved with the help of wolframalpha.com) Is there an easier or better way of doing this?
Any help would be greatly appreciated, as I am kind of lost and under some time pressure :-/
Thanks,
David
when I move the marker across the table the Y and Z axis is updated, although all I want to be updated is the Z axis, because the height of the marker doesnt change only its distance to the camera.
Only true when your camera's view direction is aligned with your Y axis (or Z axis). If the camera is not aligned with Y, it means the transform will apply a rotation around the X axis, hence modifying both the Y and Z coordinates of the marker.
So my guess is I need a Transformation Matrix with which I multiply each point so that I have a new coordinate System which lies orthogonal on the table surface.
Yes it is. After that, you will have 2 transforms:
T_table to express marker's coordinates in the table referential,
T_camera to express table coordinates in the camera referential.
Finding T_camera from a single 2d image is hard because there's no depth information.
This is known as the Pose problem -- it has been studied by -among others- Daniel DeMenthon. He developed a fast and robust algorithm to find the pose of an object:
articles available on its research homepage, section 4 "Model Based Object Pose" (and particularly "Model-Based Object Pose in 25 Lines of Code", 1995);
code at the same place, section "POSIT (C and Matlab)".
Note that the OpenCv library offers an implementation of the DeMenthon's algorithm. This library also offers a convenient and easy-to-use interface to grab images from a webcam. It's worth a try: OpenCv homepage
If you know the location in the physical world of your four markers and you've recorded the positions as they appear on the camera, you ought to be able to derive some sort of transform.
When you do the calibration, surely you'd want to put the marker at the four corners of the table not the screen? If you're just doing the corners of the screen, I imagine you're probably not taking into acconut the slant of the table.
Is the table literally just slanted relative to the camera or is it also rotated at all?