OpenCV has capapabilities to compensate for distortion in patterns, such as a this board, for example:
Every example I ever saw for this process does it with grids or squares. I would like to know if something similar exists for a single circle. My practical case is that I detect an ellipse, and I need to calculate the angle between the plane of this ellipse and the projection plane where the ellipse is projected as a circle. I managed to achieve that in my own code, but I would like to know if there is something built into the library to that purpose.
Use the ellipse axes to your advantage
I don't know of any "circular projection" as you name it, but I'm thinking that you can rephrase your problem into having the solution already.
Images make any answer SO cool.
Forget the ellipse, take the axes
A circle can be thought of as 2 vectors with unit norm defining a plane.
The projected circle's axes you estimate are the projection of the unit referential into the 3D plane
Then for projecting back and forth is just an affair of applying the transformation described by the estimated axes vectors
Related
For a project, I need to store circles detected on some photos. The problem is that some of these photos are taken from an angle, meaning the circles are ellipses. Is it possible to somehow turn the ellipses into circles?
I thought of rectifying the ellipse, then transforming the rectangle to a square. Indeterminate problem comes to my mind, meaning there are too many possible variations for my approach, and the results are different for each approach.
To find perspective transform, you need to have 4 pairs of corresponding coordinates: points at distorted picture and their ideal positions after correction of perspective.
In this case you can calculate matrix of perspective transform with getPerspectiveTransform function and apply it to correct all the picture. Example
Given a photo containing a circle, for example this photo of a fountain:
is it possible to define the 3D position and rotation of the fountain in relation to the camera?
I realise we have to define the scale, so lets say the fountain is 2m wide (the diameter of the circle consisting of the inner rim of the fountain is 2m).
So assuming the circle is a perfect circle, and defining the diameter to 2m, is it possible to determine how the circle and the camera relate spatially? I dont know any camera matrix or anything, the only information i have is the picture.
I specifically want to determine the 3D coordinates of a given pixel on the rim of the fountain.
What would be the math and/or OpenCV code to do this?
Circle with perspective is an ellipse. So you basicly you need an ellipse detector.
This algorithm should work:
Detect all ellipses in the given image.
Filter ellipses that you think they are not a circles in origin. (This is not possible using just 1 Camera so you have to depend on previous knowledge. Something like that you knows that you are taking a photo for a circle).
mmm I stopped typing here and bring a paper&pen and started figuring how to estimate the Homography and it is not that easy! you should deal with the circle a special case of an ellipse and then try to construct a linear system of equations. However, I made quick googling :
https://www.researchgate.net/publication/265212988_Homography_estimation_using_one_ellipse_correspondence_and_minimal_additional_information
http://www.macs.hw.ac.uk/bmvc2006/papers/306.pdf
Seems very interesting topic, I am going to spare sometimes on it later!
I need to find the size or coordinates of a rectangle that is displayed as a quadrilateral in a 3D image. The quadrilateral is on a plane that lines up with 3d world vanishing points. To clarify, the quadrilateral IS a rectangle in the 3D world, and that's the rectangle I want the size of.
I do not need to get all the textures and make a new image. I also do not know the coordinates of the target rectangle as required by the homography (perspective transformation) solutions I've seen, because I don't know the aspect ratio it's supposed to have.
I've read through this thread: proportions of a perspective-deformed rectangle and the guy seemed to find an algorithm that works. However I've read other research papers that claim to calculate a homography yet they don't say how they did it. Also it seems such a basic function there would be something in the existing openCV library.
Thanks.
I have to detect the pattern of 6 circles using opencv. I have detected the circles and their centroids by using thresholding and contour function in opencv.
Now I have to define the relation between these circles in a way that should be invariant to scale and rotation. With this I would be able to detect this pattern in various views. I have to use this pattern for determining the object pose.
How can I achieve scale/rotation invariance? Do you have any reference I could read about it?
To make your pattern invariant toward rotation & scale, you have to normalize the direction and the scale when detecting your pattern. Here is a simple algorithm to achieve this
detect centers and circle size (you say you have already achieved this - good!)
compute the average center using a simple mean. Express all the centers from this mean
find the farthest center using a simple norm (euclidian is good enough)
scale the center position and the circle sizes so that this maximum distance is 1.0
rotate the centers so that coordinates of the farthest one is (1.0, 0)
you're done. You are now the proud owner of a scale/rotation invariant pattern detector!! Congratulations!
Now you can find patterns, transform them as suggested, and compare center position & circle sizes.
It is not entirely clear to me if you need to find the rotation, or merely get rid of it, or detect if the circles actually form the pattern you linked. Either way, the answer is much the same.
I would start by finding the two circles that have only one neighbour. For each circle centroid calculate the distance to the closest two neighbours. If the distances differ in more than say 10%, the centroid belongs to an "end" circle (one of the top ones in your link).
Now that you have found the two end circles, rotate them so that they are horizontal to each other. If the other centroids are now above them, rotate another 180 degrees so that the pattern ends up in the orientation you want.
Now you can calculate the scaling from the average inter-centroid distance.
Hope that helps.
Your question sounds exactly like what the SURF algorithm does. It finds groups of interest and groups them together in a way invarant to rotation and scale, and can find the same object in other pictures.
Just search for OpenCV and SURF.
Suppose cameras are calibrated therefore Metric projection matrices M_i(3x4) are there for view i from multiple views. As well, K_i(3x3) the camera matrix of each view is available. Can we calculate the location of plane at infinity in projective space?
Sure, the plane at infinity is always the plane where w = 0. If you are applying affine transformations, it remains fixed. It only shifts if you use a homography.
Yes, it is theoretically possible. The plane at infinity always remains fixed in terms of the actual projective 3D world. However, it is imaged differently by a moving camera upon each view and in these cases we say that the plane at infinity is not at its canonical position. Instead of thinking that the camera moved, it is more convenient to think that the entire 3D projective space has moved! Thus, we invent a 3D homography to "blame" for this change. Mathematically, the 3D homography tags along the left side of the projection matrix:
x = (P*H)*X
So, to answer the question: Although tricky, yes you may recover it, given that you have sufficiently enough reconstructed views of the scene. This is a process called auto calibration and it essentially involves one equation (that comes in many flavors) but unfortunately, gives non-linear equations. I suggest you look at the following:
http://nguyendangbinh.org/Proceedings/CVPR/1999/DATA/03_34.PDF
I believe it contains the most up-to-date method to iteratively compute the plane at infinity.