how can you get the correct R and t from H when you have 4 solutions.
From my understanding you can only eliminate two.
Is there really no way to get the correct solutions if you only have points on a single plane?
Thanks for any help.
There is one approach to best select solution from two possible solutions. When you decompose Homography,after eliminating 2 possible solutions from 4,you get two set of
rotations, translations and normals.
Normals are nothing but possible camera_normals wrt scene of first image(when you estimate Homography between two images). So having known camera_normal wrt planar scene when first image was captured, you can choose solution corresponding to closest normal from n1,n2 by computing dot product dot(camera_normal,n1) and dot(camera_normal,n2), whichever is larger.
Thanks
Edit: camera_normal explanation
Planar Homography assumes that features used for computing Homography between tow images are in a plane. So camera_normal is direction of z-axis of camera in plane's frame.
Related
I would like to find the best fit plane from a list of 3D points. It means the plane has the least square distance from all the points. I read the article
Best fit plane by minimizing orthogonal distances
and
3D Least Squares Plane
I fully understand the solutio but it turns other to be impractical in my situation. I need to read a very very large list of 3d points, direcltly impementation would result in ill posed problem. Even I subtract the data with their average,(refere to the document here-> part3 : http://www.geometrictools.com/Documentation/LeastSquaresFitting.pdf) the number is still very large. So what can I do?
Is there an iterative way to implement it ?
I have changed the way to ask the question, I hope may be there are someone can give me more advices on it ?
Given a list of 3D Points
{(x0,y0,z0),
(x1,y1,z1)...
(xn-1,yn-1,zn-1)}
I would like to construct a plane by fitting all the 3D points. In this sense, I mean to find the plane with format (Ax+By+Cy+D = 0), thus its uses four parameters(A,B,C,D) to characterize a plane. The sum of distance between each point and the plane should be minimium.
I do try the menthod provided in the below link
http://www.geometrictools.com/Documentation/LeastSquaresFitting.pdf
But there are two problems:
-During calculation, the above algorithm needs to do summation of all points value, which lead to overflow problem if my number of points increases
-given newly added points, it has to do all the calculation again, is there a way to use the before calculated plane parameter and the newly given points to somehow fine tune the planes parameters?
PS:I am a bit greedy, if we need to involve all the points, it is possible that the plane finally obtained isn't good enough.I am thinking of using random sample consensus(RANSAC), is it the right direction?
If you are expecting a plane then most of the points are not that useful since even a handful should give you a good approximation of the final solution (module a bit more noise).
So here's the solution. Sample down your data set to something that works and run the smaller set through the fitting algorithm.
If you are not expecting that the points are on a plane then sub-sampling should still work, but you must consider error ranges for any solution (since they will likely be fairly big).
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'm writing an application which detect squares. I wrote algorithm of quad detection and now I must select squares of this quads list. How can I do this in perspective projection? Can I transform this 4 coordinates of quad to parallel projection? If I'd able to do this, I can simply check angles and length of edges.
Maybe it solves somehow different? Can you advice me tricks to solve this task.
Thanks!
Is it possible to get a rectangle distortion from few fixed points?
This example will explain better what I mean:
Suppose I've got this image with a rectangle and two points, the two points are recognized in the other image where the image is distorted
How can I reproduce the distortion knowing the position of the two(or maybe three) previous points??
My purpose is to get the distorted rectangle border. It's not an easy image as the one in the example so I can't just filter colors, I need to find a way to get the distorted image border.
I believe what you're looking for can be described as an affine transform. If you want general transform of a planar surface, you may want perspective transform instead.
You can find the OpenCV implementation here. The relevant functions are cv::getAffineTransform which requires 3 pairs of points or cv::getPerspectiveTransform which requires 4 pairs of points.
Note: if you're using an automatic feature detector/matcher, it would be best to use far more point pairs than the minimum and use a robust outlier rejection algorithm like RANSAC.
shift and rotation need - 2 points
Affine tranform need - 3 points
Perspective tranform need - 4 points
We have a grid with red squares on it. Meaning we have an array of 3 squares (with angles == 90 deg) which as we know have same size, lying on the same plane and with same rotation relative to the plane they are lying on, and are not situated on same line on plane.
We have a projection of the space which contains the plane with squares.
We want to turn our plane projection with squares so that we would see it like it's facing us, in general we need a formula for turning each point of that original plane projection so that it would be facing us like on the image below.
What formulas can be used for solving such problem, how to solve it, has any one faced something like this before?
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 square, 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 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.