Suppose you have an homography H relating a planar surface depicted in two different images, Ir and I. Ir is the reference image, where the planar surface is parallel to the image plane (and practically occupy the entire image). I is a run-time image (a photo of the planar surface taken at an arbitrary viewpoint). Let H be such that:
p = Hp', where p is a point in Ir and p' is the corresponding point in I.
Suppose you have two points p1=(x1,y) and p2=(x2,y), with x1 < x2, relative to the image Ir. Note that they belong to the same row (common y). Let H'=H^(-1). Using H', you can compute the corresponding points in I of the following points: (x1,y),(x1+1,y),...,(x2,y).
The question is: is there a way to avoid the matrix-vector multiplication to compute all those points? The easiest way that comes to me is to use the homography to compute the corresponding point of p1 and p2 (call them p1' and p2'). To obtain the others (that is: (x1+1,y), (x1+2,y),...,(x2-1, y)), linear interpolate p1' and p2' in the image I.
But since there is a projective transformation between Ir and I, i think that this method is quite imprecise.
Any other idea? This question is relative to the fact that i need a computational efficient way to extract a lot of (small) patches (of around 10x10 pixels) around a point p in Ir in a real-time software.
Thank you.
Ps.
Maybe the fact that i am using smal patches would make using linear interpolation a suitable approach?
You have a projective transform and, unfortunately, ratio of lengths are not invariant under this type of transformation.
My suggestion: explore the cross ratio because it is invariant under projective transformations. I think that for each 3 points you can get a "cheaper" 4th, avoiding the matrix-vector computation and using the cross ratio instead. But I have not put all the stuff in paper to verify if the cross ratio alternative is that much "computationally cheaper" than the matrix-vector multiplication.
Related
I want to know about why do we normalize the homography or fundamental matrix? Here is the code in particular.
H = H * (1.0 / H[2, 2]) # Normalization step. H is [3, 3] matrix.
I can understand that we have to normalize the data before computing SVD because of instability caused by linear least squares but why do we normalize it in end?
A homography in 3D space has 8 degrees of freedom by definition, mapping from one plane to another using perspective. Such a homography can be defined by giving four points, which makes eight coordinates (scalars).
A 3x3 matrix has 9 elements, so it has 9 degrees of freedom. That is one degree more than needed for a homography.
The homography doesn't change when the matrix is scaled (multiplied by a scalar). All the math works the same. You don't need to normalize your homography matrix.
It is a good idea to normalize.
For one, it makes the arithmetic somewhat tamer. Have some wikipedia links to fields of study because weaving all these into a coherent sentence... doesn't add anything:
Numerical analysis, Condition number, Floating-point arithmetic, Numerical error, Numerical stability, ...
Also, normalization makes the matrix easier for humans to interpret. The most common normalization is to scale the matrix such that the last element becomes 1. That is convenient because this whole math happens in a projective space, where the projection causes points to be mapped to the w=1 plane, making vectors have a 1 for the last element.
How is the homography matrix provided to you?
For example, in the scene that some library function calculates and provides the homography matrix to you,
if the function specification doesn't mention about the scale...
In an extreme case, the function can be implemented as:
Matrix3x3 CalculateHomographyMatrix( some arguments )
{
Matrix3x3 H = ...; //Homogoraphy Calculation
return Non_Zero_Random_Value * H; //Wow!
}
Element values may become very large or very small and using such values to your process may cause problems (floating point computation errors).
let's say I am placing a small object on a flat floor inside a room.
First step: Take a picture of the room floor from a known, static position in the world coordinate system.
Second step: Detect the bottom edge of the object in the image and map the pixel coordinate to the object position in the world coordinate system.
Third step: By using a measuring tape measure the real distance to the object.
I could move the small object, repeat this three steps for every pixel coordinate and create a lookup table (key: pixel coordinate; value: distance). This procedure is accurate enough for my use case. I know that it is problematic if there are multiple objects (an object could cover an other object).
My question: Is there an easier way to create this lookup table? Accidentally changing the camera angle by a few degrees destroys the hard work. ;)
Maybe it is possible to execute the three steps for a few specific pixel coordinates or positions in the world coordinate system and perform some "calibration" to calculate the distances with the computed parameters?
If the floor is flat, its equation is that of a plane, let
a.x + b.y + c.z = 1
in the camera coordinates (the origin is the optical center of the camera, XY forms the focal plane and Z the viewing direction).
Then a ray from the camera center to a point on the image at pixel coordinates (u, v) is given by
(u, v, f).t
where f is the focal length.
The ray hits the plane when
(a.u + b.v + c.f) t = 1,
i.e. at the point
(u, v, f) / (a.u + b.v + c.f)
Finally, the distance from the camera to the point is
p = √(u² + v² + f²) / (a.u + b.v + c.f)
This is the function that you need to tabulate. Assuming that f is known, you can determine the unknown coefficients a, b, c by taking three non-aligned points, measuring the image coordinates (u, v) and the distances, and solving a 3x3 system of linear equations.
From the last equation, you can then estimate the distance for any point of the image.
The focal distance can be measured (in pixels) by looking at a target of known size, at a known distance. By proportionality, the ratio of the distance over the size is f over the length in the image.
Most vision libraries (including opencv) have built in functions that will take a couple points from a camera reference frame and the related points from a Cartesian plane and generate your warp matrix (affine transformation) for you. (some are fancy enough to include non-linearity mappings with enough input points, but that brings you back to your time to calibrate issue)
A final note: most vision libraries use some type of grid to calibrate off of ie a checkerboard patter. If you wrote your calibration to work off of such a sheet, then you would only need to measure distances to 1 target object as the transformations would be calculated by the sheet and the target would just provide the world offsets.
I believe what you are after is called a Projective Transformation. The link below should guide you through exactly what you need.
Demonstration of calculating a projective transformation with proper math typesetting on the Math SE.
Although you can solve this by hand and write that into your code... I strongly recommend using a matrix math library or even writing your own matrix math functions prior to resorting to hand calculating the equations as you will have to solve them symbolically to turn it into code and that will be very expansive and prone to miscalculation.
Here are just a few tips that may help you with clarification (applying it to your problem):
-Your A matrix (source) is built from the 4 xy points in your camera image (pixel locations).
-Your B matrix (destination) is built from your measurements in in the real world.
-For fast recalibration, I suggest marking points on the ground to be able to quickly place the cube at the 4 locations (and subsequently get the altered pixel locations in the camera) without having to remeasure.
-You will only have to do steps 1-5 (once) during calibration, after that whenever you want to know the position of something just get the coordinates in your image and run them through step 6 and step 7.
-You will want your calibration points to be as far away from eachother as possible (within reason, as at extreme distances in a vanishing point situation, you start rapidly losing pixel density and therefore source image accuracy). Make sure that no 3 points are colinear (simply put, make your 4 points approximately square at almost the full span of your camera fov in the real world)
ps I apologize for not writing this out here, but they have fancy math editing and it looks way cleaner!
Final steps to applying this method to this situation:
In order to perform this calibration, you will have to set a global home position (likely easiest to do this arbitrarily on the floor and measure your camera position relative to that point). From this position, you will need to measure your object's distance from this position in both x and y coordinates on the floor. Although a more tightly packed calibration set will give you more error, the easiest solution for this may simply be to have a dimension-ed sheet(I am thinking piece of printer paper or a large board or something). The reason that this will be easier is that it will have built in axes (ie the two sides will be orthogonal and you will just use the four corners of the object and used canned distances in your calibration). EX: for a piece of paper your points would be (0,0), (0,8.5), (11,8.5), (11,0)
So using those points and the pixels you get will create your transform matrix, but that still just gives you a global x,y position on axes that may be hard to measure on (they may be skew depending on how you measured/ calibrated). So you will need to calculate your camera offset:
object in real world coords (from steps above): x1, y1
camera coords (Xc, Yc)
dist = sqrt( pow(x1-Xc,2) + pow(y1-Yc,2) )
If it is too cumbersome to try to measure the position of the camera from global origin by hand, you can instead measure the distance to 2 different points and feed those values into the above equation to calculate your camera offset, which you will then store and use anytime you want to get final distance.
As already mentioned in the previous answers you'll need a projective transformation or simply a homography. However, I'll consider it from a more practical view and will try to summarize it short and simple.
So, given the proper homography you can warp your picture of a plane such that it looks like you took it from above (like here). Even simpler you can transform a pixel coordinate of your image to world coordinates of the plane (the same is done during the warping for each pixel).
A homography is basically a 3x3 matrix and you transform a coordinate by multiplying it with the matrix. You may now think, wait 3x3 matrix and 2D coordinates: You'll need to use homogeneous coordinates.
However, most frameworks and libraries will do this handling for you. What you need to do is finding (at least) four points (x/y-coordinates) on your world plane/floor (preferably the corners of a rectangle, aligned with your desired world coordinate system), take a picture of them, measure the pixel coordinates and pass both to the "find-homography-function" of your desired computer vision or math library.
In OpenCV that would be findHomography, here an example (the method perspectiveTransform then performs the actual transformation).
In Matlab you can use something from here. Make sure you are using a projective transformation as transform type. The result is a projective tform, which can be used in combination with this method, in order to transform your points from one coordinate system to another.
In order to transform into the other direction you just have to invert your homography and use the result instead.
There are many posts about 3D reconstruction from stereo views of known internal calibration, some of which are excellent. I have read a lot of them, and based on what I have read I am trying to compute my own 3D scene reconstruction with the below pipeline / algorithm. I'll set out the method then ask specific questions at the bottom.
0. Calibrate your cameras:
This means retrieve the camera calibration matrices K1 and K2 for Camera 1 and Camera 2. These are 3x3 matrices encapsulating each camera's internal parameters: focal length, principal point offset / image centre. These don't change, you should only need to do this once, well, for each camera as long as you don't zoom or change the resolution you record in.
Do this offline. Do not argue.
I'm using OpenCV's CalibrateCamera() and checkerboard routines, but this functionality is also included in the Matlab Camera Calibration toolbox. The OpenCV routines seem to work nicely.
1. Fundamental Matrix F:
With your cameras now set up as a stereo rig. Determine the fundamental matrix (3x3) of that configuration using point correspondences between the two images/views.
How you obtain the correspondences is up to you and will depend a lot on the scene itself.
I am using OpenCV's findFundamentalMat() to get F, which provides a number of options method wise (8-point algorithm, RANSAC, LMEDS).
You can test the resulting matrix by plugging it into the defining equation of the Fundamental matrix: x'Fx = 0 where x' and x are the raw image point correspondences (x, y) in homogeneous coordinates (x, y, 1) and one of the three-vectors is transposed so that the multiplication makes sense. The nearer to zero for each correspondence, the better F is obeying it's relation. This is equivalent to checking how well the derived F actually maps from one image plane to another. I get an average deflection of ~2px using the 8-point algorithm.
2. Essential Matrix E:
Compute the Essential matrix directly from F and the calibration matrices.
E = K2TFK1
3. Internal Constraint upon E:
E should obey certain constraints. In particular, if decomposed by SVD into USV.t then it's singular values should be = a, a, 0. The first two diagonal elements of S should be equal, and the third zero.
I was surprised to read here that if this is not true when you test for it, you might choose to fabricate a new Essential matrix from the prior decomposition like so: E_new = U * diag(1,1,0) * V.t which is of course guaranteed to obey the constraint. You have essentially set S = (100,010,000) artificially.
4. Full Camera Projection Matrices:
There are two camera projection matrices P1 and P2. These are 3x4 and obey the x = PX relation. Also, P = K[R|t] and therefore K_inv.P = [R|t] (where the camera calibration has been removed).
The first matrix P1 (excluding the calibration matrix K) can be set to [I|0] then P2 (excluding K) is R|t
Compute the Rotation and translation between the two cameras R, t from the decomposition of E. There are two possible ways to calculate R (U*W*V.t and U*W.t*V.t) and two ways to calculate t (±third column of U), which means that there are four combinations of Rt, only one of which is valid.
Compute all four combinations, and choose the one that geometrically corresponds to the situation where a reconstructed point is in front of both cameras. I actually do this by carrying through and calculating the resulting P2 = [R|t] and triangulating the 3d position of a few correspondences in normalised coordinates to ensure that they have a positive depth (z-coord)
5. Triangulate in 3D
Finally, combine the recovered 3x4 projection matrices with their respective calibration matrices: P'1 = K1P1 and P'2 = K2P2
And triangulate the 3-space coordinates of each 2d point correspondence accordingly, for which I am using the LinearLS method from here.
QUESTIONS:
Are there any howling omissions and/or errors in this method?
My F matrix is apparently accurate (0.22% deflection in the mapping compared to typical coordinate values), but when testing E against x'Ex = 0 using normalised image correspondences the typical error in that mapping is >100% of the normalised coordinates themselves. Is testing E against xEx = 0 valid, and if so where is that jump in error coming from?
The error in my fundamental matrix estimation is significantly worse when using RANSAC than the 8pt algorithm, ±50px in the mapping between x and x'. This deeply concerns me.
'Enforcing the internal constraint' still sits very weirdly with me - how can it be valid to just manufacture a new Essential matrix from part of the decomposition of the original?
Is there a more efficient way of determining which combo of R and t to use than calculating P and triangulating some of the normalised coordinates?
My final re-projection error is hundreds of pixels in 720p images. Am I likely looking at problems in the calibration, determination of P-matrices or the triangulation?
The error in my fundamental matr1ix estimation is significantly worse
when using RANSAC than the 8pt algorithm, ±50px in the mapping between
x and x'. This deeply concerns me.
Using the 8pt algorithm does not exclude using the RANSAC principle.
When using the 8pt algorithm directly which points do you use? You have to choose 8 (good) points by yourself.
In theory you can compute a fundamental matrix from any point correspondences and you often get a degenerated fundamental matrix because the linear equations are not independend. Another point is that the 8pt algorithm uses a overdetermined system of linear equations so that one single outlier will destroy the fundamental matrix.
Have you tried to use the RANSAC result? I bet it represents one of the correct solutions for F.
My F matrix is apparently accurate (0.22% deflection in the mapping
compared to typical coordinate values), but when testing E against
x'Ex = 0 using normalised image correspondences the typical error in
that mapping is >100% of the normalised coordinates themselves. Is
testing E against xEx = 0 valid, and if so where is that jump in error
coming from?
Again, if F is degenerated, x'Fx = 0 can be for every point correspondence.
Another reason for you incorrect E may be the switch of the cameras (K1T * E * K2 instead of K2T * E * K1). Remember to check: x'Ex = 0
'Enforcing the internal constraint' still sits very weirdly with me -
how can it be valid to just manufacture a new Essential matrix from
part of the decomposition of the original?
It is explained in 'Multiple View Geometry in Computer Vision' from Hartley and Zisserman. As far as I know it has to do with the minimization of the Frobenius norm of F.
You can Google it and there are pdf resources.
Is there a more efficient way of determining which combo of R and t to
use than calculating P and triangulating some of the normalised
coordinates?
No as far as I know.
My final re-projection error is hundreds of pixels in 720p images. Am
I likely looking at problems in the calibration, determination of
P-matrices or the triangulation?
Your rigid body transformation P2 is incorrect because E is incorrect.
I do have two sets of points and I want to find the best transformation between them.
In OpenCV, you have the following function:
Mat H = Calib3d.findHomography(src_points, dest_points);
that returns you a 3x3 Homography matrix, using RANSAC. My problem is now, that I only need translation and rotation (& maybe scale), I don't need affine and perspective.
The thing is, my points are only in 2D.
(1) Is there a function to compute something like a homography but with less degrees of freedom?
(2) If there is none, is it possible to extract a 3x3 matrix that does only translation and rotation from the 3x3 homography matrix?
Thanks in advance for any help!
Isa
OpenCV estimateRigidTransform function is exactly what you need: it returns Translation, Rotation and Scale (use false value for fullAffine flag). And it DOES use RANSAC (see source code to be sure of it).
Homography is for 2D points, the third dimension is just for casting points in 3 dim homogeneous coordinates and performing perspective effects. You can always cast points back:
homogeneous [x, y, w]
cartesian [x/w, y/w]
However since you calculate 6DOF instead of 4DOF (similarity) you result is pretty different from what you expect with 4DOF. More flexible transformation will fit more points in RANSAC at the expense of distortions in transformations you care about. Bottom line - don’t try to decompose H, instead fit similarity or isometry (also called rigid or euclidean). The reason why they are absent in the library - they are expressed in closed form even with correct least squared metric in point coordinates and thus don't require non-linear optimization. In other words, they are very simple.
If you only have rotation and translation, I wrote a quick functions to find them (no RANSAC though). It is probably similar to a rigidTransform but more understandable (hopefully)
https://stackoverflow.com/a/18091472/457687
With scale there is still a closed form solution, but slightly different formulas for translation and scaling. See Learning similarity parameters, p. 25
I have two images that are taken from different positions. The 2nd camera is located to the right, up and backward with respect to 1st camera.
So I think there is a perspective transformation between the two views and not just an affine transform since cameras are at relatively different depths. Am I right?
I have a few corresponding points between the two images. I think of using these corresponding points to determine the transformation of each pixel from the 1st to the 2nd image.
I am confused by the functions findFundamentalMat and findHomography. Both return a 3x3 matrix. What is the difference between the two?
Is there any condition required/prerequisite to use them (when to use them)?
Which one to use to transform points from 1st image to 2nd image? In the 3x3 matrices, which the functions return, do they include the rotation and translation between the two image frames?
From Wikipedia, I read that the fundamental matrix is a relation between corresponding image points. In an SO answer here, it is said the essential matrix E is required to get corresponding points. But I do not have the internal camera matrix to calculate E. I just have the two images.
How should I proceed to determine the corresponding point?
Without any extra assumption on the world scene geometry, you cannot affirm that there is a projective transformation between the two views. This is only true if the scene is planar. A good reference on that topic is the book Multiple View Geometry in Computer Vision by Hartley and Zisserman.
If the world scene is not planar, you should definitely not use the findHomography function. You can use the findFundamentalMat function, which will provide you an estimation of the fundamental matrix F. This matrix describes the epipolar geometry between the two views. You may use F to rectify your images in order to apply stereo algorithms to determine a dense correspondence map.
I assume you are using the expression "perspective transformation" to mean "projective transformation". To the best of my knowledge, a perspective transformation is a world to image mapping, not an image to image mapping.
The Fundamental matrix has the relation
x'Fu = 0
with x in one image and u in the other iff x and u are projections of the same 3d point.
Also
l = Fu
defines a line (lx' = 0) where the correponding point of u must be on, so it can be used to confine the searchspace for the correspondences.
A Homography maps a point on one projection of a plane to another projection of the plane.
x = Hu
There are only two cases where the transformation between two views is a projective transformation (ie a homography): either the scene is planar or the two views were generated by a camera rotating around its center.