I currently have two images of a plane in real life from straight above. One to use as a reference image, and another when the plane has undergone a rotation fixed at the centre of the plane thus changing its orientation. The camera stays at a constant position.
I was wondering if I found the homography matrix of this rotation in opencv and then decomposed the homography matrix in order to find the rotation matrix whether this would yield accurate results and I would be able to find the three angles needed to describe the planes rotation in euclidean coordinates to a reasonable degree of accuracy.
Thanks
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Apologies in advance for my feeble maths.
I'm trying to be able to find the corners of a plane in space based on the equation of that plane. Here's what I know. I know three points on the plane and I know where they fall in the 2d coordinate space of the plane (x,y) and where they are in 3d space. I know the width and height of the plane and I can now calculate the equation of the plane. The plane sits on the inside of a large sphere that surrounds the origin so, in theory, it should more or less face where the camera is (though in my diagram it doesn't face the origin as it's just for illustrative purposes)
But it's not clear to me how I can use that to figure out another point. One thought I had was to find the transform that moves the plane parallel to the xy axis and rotate it round one of the points (so it stays in the same place), find the position of the new point, and then rotate it by the inverse of that transform. But it's not clear to me how I would find that transform matrix or how to use it. Could I do this using the normal and vector maths? I understand what normals are, but I'm fuzzy about how to use them.
I have a projected view of a 3D scene. The 2D points are computed by multiplying the 3D points in homogenous coordinates by a view matrix (which includes a translation and rotation) and a perspective matrix. I want to allow the user to move control points which describe the three axes, and update the rotation matrix based on this.
How do I compute the new rotation matrix given a change in projected 2D coordinates, assuming rotation around the origin? Solving for the position of the end of the single axis has a large degeneracy in the set of possible, but maybe solving for rotation in the axes perpendicular to the moved axis might work.
I am learning camera matrix stuff. I already known that I can get the homography of the camera (3*3 matrix) by using four points in a plane in object space. I want to know if we can get the homagraphy with four points not in a plane? If yes, how can I get the matrix? What formulas should I look at?
I also confused homography with another concept: I only need to know three points if I want to convert from points from one coordinate to another coordinate system. So why we need four points in computing homography?
Homography maps points
1. On plane to points at another plane
2. Projections of points in 3D (no obligatory lying on the same plane) during a pure camera rotation or zoom.
The latter can be easily verified if you look at the rays that connect points while sensor plane rotates: green are two sensor positions and black is a 3d object
Since Homography is between projections and not between objects in 3D you don’t care what these projections represent. But this can be confusing, I agree. For example you can point your camera at 3D scene (that is not flat!), then rotate your camera and the two resulting pictures of the scene will be related by homography. This is, by the way, a foundation for image panoramas.
Three point correspondences you mentioned may be reladte to a transformation called Affine (happens during large zooms when a perspective effects disappears) or to the finding a rigid rotation and translation in 3D space. Both require 3 point correspondences but the former needs only 2D points while the latter needs 3D points. The latter case has 6DOF ( 3 for rotation and 3 for translation) while each correspondence provides 2DOF, hence 6/2=3 correspondences. Homography has 8 DOF so there should be 8/2=4 correspondences;
Below is a little diagram that explains the difference between affine and homographs transformation when the original square tilts forward. In affine case the perspective effect is negligible that is far side has the same length as a near one. In the case of Homography the far side is shorter.
If you only have 4 points - and they're not on the same plane - then computing a homography will not work.
If you have a loads of points, and 4 of them do lie on a plane but some don't, there are filters you can use to try to remove the ones not lying on a plane. The filters implemented by OpenCV are called RANSAC and LMeDs.
Also as Hammer says in a comment under your question - The 4th point is there to figure out perspective.
Homography is a 3X3 matrix, which consists of 8 independent unknowns which means it requires 4 equations to solve these unknowns. So, in order to calculate homography we need at least 4 points.
In homography we assume that Z=0 in world scene, so the image projected is assumed as 2D. In a very famous journal named ORB-SLAM, the author formulated a scene-selective approach depending on motion parallax in scene.
Homography is the relation between two planes and the degree of freedom in case of homography transform is 7; hence you need minimum 4 corresponding points.
4 points will give you 4 pair of (x,y) hence you can calculate 7 variables. Homography is homogines transfrom hence the (3,3) value in homography matrix is always 1.
So your first question that can you calculate homography with 3 points in the plane and 4th not on the plane : it's not possible. You need projection of that point on the plane and then you can calculate the homography.
Your 2nd question about how to calculate homography matrix, you can see implemetation of findHomography() in opencv.
I have an image of a 3D rectangle (which due to the projection distortion is not a rectangle in the image). I know the all world and image coordinates of all corners of this rectangle.
What I need is to determine the world coordinate of a point in the image inside this rectangle. To do that I need to compute a transformation to unproject that rectangle to a 2D rectangle.
How can I compute that transform?
Thanks in advance
This is a special case of finding mappings between quadrilaterals that preserve straight lines. These are generally called homographic transforms. Here, one of the quads is a rectangle, so this is a popular special case. You can google these terms ("quad to quad", etc) to find explanations and code, but here are some sites for you.
Perspective Transform Estimation
a gaming forum discussion
extracting a quadrilateral image to a rectangle
Projective Warping & Mapping
ProjectiveMappings for ImageWarping by Paul Heckbert.
The math isn't particularly pleasant, but it isn't that hard either. You can also find some code from one of the above links.
If I understand you correctly, you have a 2D point in the projection of the rectangle, and you know the 3D (world) and 2D (image) coordinates of all four corners of the rectangle. The goal is to find the 3D coordinates of the unique point on the interior of the (3D, world) rectangle which projects to the given point.
(Do steps 1-3 below for both the 3D (world) coordinates, and the 2D (image) coordinates of the rectangle.)
Identify (any) one corner of the rectangle as its "origin", and call it "A", which we will treat as a vector.
Label the other vertices B, C, D, in order, so that C is diagonally opposite A.
Calculate the vectors v=AB and w=AD. These form nice local coordinates for points in the rectangle. Points in the rectangle will be of the form A+rv+sw, where r, s, are real numbers in the range [0,1]. This fact is true in world coordinates and in image coordinates. In world coordinates, v and w are orthogonal, but in image coordinates, they are not. That's ok.
Working in image coordinates, from the point (x,y) in the image of your rectangle, calculate the values of r and s. This can be done by linear algebra on the vector equations (x,y) = A+rv+sw, where only r and s are unknown. It will boil down to a 2x2 matrix equation, which you can solve generally in code using Cramer's rule. (This step will break if the determinant of the required matrix is zero. This corresponds to the case where the rectangle is seen edge-on. The solution isn't unique in that case. If that's possible, make special exception.)
Using the values of r and s from 4, compute A+rv+sw using the vectors A, v, w, for world coordinates. That's the world point on the rectangle.
So we have such situation:
In this illustration, the first quadrilateral is shown on the Image Plane and the second quadrilateral is shown on the World Plane. [1]
In my particular case the Image Plane has 3 quadrilaterals - projections of real world squares, which, as we know, have same size, lying on the same plane, with same rotation relative to the plane they are lying on, and are not situated on same line on plane.
I wonder if we can get rotation angles of Image Plane to World Plane knowing stuff described?
In my case as input I have such data structures: original image (RGB pixels), objects (squares) with angles points in pixels (x,y) on Image Plane.
Take a look at Sections 2 and 3 of Algorithms for plane-based pose estimation.
The methods described there assume that you know the (x,y) coordinates of the features in question - in this case the red squares.
The problem you are describing is generally known as pose estimation - determining the 3D orientation and position of an object relative to a camera from a 2D view. For you, the object is a plane. Googling 'pose estimation plane' should give you more sources.