If I have 2 sets of points I can rotate one around with Procrustes analysis to align one with the other.
But suppose these 2 sets of points each are attached to images and I would like to rotate the images as well. Is there any way I can also rotate the image, instead of rotating just the points? The tutorial there uses a dot product for rotation (solve u, s, v = svd(p1', p2) and then do p2 . v . u', p' is transposed p)
However that doesn't tell me what the angle between the images is.
The page on wikipedia calculates an angle between each pair of points I think.
Maybe what I'm asking is impossible? If I rotate the first set of points to align it with the first, can't I also rotate the respective images by an angle as well? Point being, which angle is that?
I noticed that v . u' gives me a 2 x 2 matrix which seems to be the rotation matrix (there's a wikipedia page but I can't link there due to posting priviledges). I got the sin and cos of the third and first elements and then used arctan2, but the results I'm getting are kind of weird. I know they have to be transformed from radians but I'm not convinced what I'm doing is right. Trying the rotation it gives me on gimp makes it seem like it's not what I want, but I'll test some more.
It seems like your approach is mostly correct. Two things which come to mind:
1) The paper you linked to (Procrustes analysis) involves a translation and a scaling in addition to rotation. Depending on whether or not those operations are also performed on your images, you may end up with strange results that don't appear to match.
2) I think you may be overcomplicating your angle calculation. v * u' appears to be the correct rotation matrix, but I believe the correct angle only requires one of the matrix entries in the 2x2 matrix. For instance, just use acos() of the first matrix entry. As you've noticed, this will (depending on the program) give you an answer in radians which you'll have to convert to degrees if you want to try out the rotation in gimp, etc.
Hope this helps!
Related
From I've understood, the transform order (rotate first or translate first) yield different [R|t].
So I want to know what's the order of the 4 possible poses you got from essential matrix SVD.
I've read a code implementation of Pose From Essential Matrix(Hartley and Zisserman's multiple view geometry (page 259)). And the author seems to interpret it as rotate first then translate, where he retrieve camera position by using p = -R^T * t.
Also, opencv seems to use trasnlate first then rotate rule. Because the t vector I got from calibrating camera is the position of camera.
Or maybe I have been wrong and the order doesn't matter?
You shouldn't use SVD to decompose a transformation into rotation and translation components. Viewed as x' = M*x = T*R*x, the translation is just the fourth column of M, and the rotation is in the upper-left 3x3 submatrix.
If you feed the whole 4x4 matrix into SVD, I'm not sure what you'll get out, but it won't be useful. (If nothing else, U and V will not be affine.)
I am using matlab's built in function called Procrustes to see the rotation translation and scale between two images. But, I am just using coordinates of the brightest points in the image and rotating these coordinates about the center of the image. Procrustes compares two matrices and gives you the rotation, translation, and scale. However, procrustes only works correctly if the matrices are in the same order for comparison.
I am given an image and a separate comparison coordinate matrix. The end goal is to find how much the image has been rotated, translated, and scaled compared to the coordinate matrix. I can just use Procrustes for this, but I need to correctly order the coordinates found from the image to match the order in the comparison coordinate matrix. My thought was to compare the distance between every possible combination of points in the coordinate matrix and compare it to the coordinates that I find in the picture. I just do not know how to write this code due to the fact if there is n coordinates, there will be n! possible combinations.
Just searching for the shortest distance is not so hard.
A = rand(1E4,2);
B = rand(1E4,2);
tic
idx = nan(1,1E4);
for ct = 1:size(A,1)
d = sum((A(ct,:)-B).^2,2);
idx(ct) = find(d==min(d));
end
toc
plot(A(1:10,1),A(1:10,2),'.r',B(idx(1:10),1),B(idx(1:10),2),'.b')
takes half a second on my PC.
The problems can start when two points in set A are matched to the same location in set B.
length(unique(idx))==length(idx)
This can be solved in several ways. The best (imho) is to determine a probability that point B matches with point A based on the distance (usually something that decreases exponentially), and solve for the most probable situation.
A simpler method (but more error prone) is to remove the matched point from set B.
I am writing 3D app for OpenGL ES 2.0 where the user sets a path and flies over some terrain. It's basically a flight simulator on rails.
The path is defined by a series of points created from a spline. Every timeslice I advance the current position using interpolation i.e. I interpolate between p0 to p1, then when I reach p1 I interpolate between p1 and p2, then finally back from pN to p0.
I create a view matrix with something analogous to gluLookAt. The eye coord is the current position, the look at is the next position along the path and an up (0, 0, 1). So the camera looks towards where it is flying to next and Z points towards the sky.
But now I want to "bank" as I turn. i.e. the up vector is not necessarily directly straight up but a changes based on the rate of turn. I know my current direction and my last direction so I could increment or decrement the bank by some amount. The dot product would tell me the angle of turn, and the a cross product would tell me if its to the left or right. I could maintain a bank angle and keep it within the range -/+70 degrees, incrementing or decrementing appropriately.
I assume this is the correct approach but I could spend a long time implementing it to find out it isn't.
Am I on the right track and are there samples which demonstrate what I'm attempting to do?
Since you seem to have a nice smooth plane flying in normal conditions you don't need much... You are almost right in your approach and it will look totally natural. All you need is a cross product between 3 sequential points A, B, C: cross = cross(A-B, C-B). Now cross is the vector you need to turn the plane around the "forward" vector: Naturally the plane's up vector is (-gravitation) usually (0,0,1) and forward vector in point B is C-B (if no interpolation is needed) now "side" vector is side = normalized(cross(forward, up)) here is where you use the banking: side = side + cross*planeCorrectionParameter and then up = cross(normalized(side), normalized(forward)). "planeCorrectionParameter" is a parameter you should play with, in reality it would represent some combination of parameters such as dimensions of wings and hull, air density, gravity, speed, mass...
Note that some cross operations above might need swap in parameter order (cross(a,b) should be cross(b,a)) so play around a bit with that.
Your approach sounds correct but it will look unnatural. Take for example a path that looks like a sin function: The plane might be "going" to the left when it's actually going to the right.
I can mention two solutions to your problem. First, you can take the derivative of the spline. I'm assuming your spline is a f(t) function that returns a point (x, y, z). The derivative of a parametric curve is a vector that points to the rotation center: it'll point to the center of a circular path.
A couple of things to note with the above method: the derivative of a straight line is 0, and the vector will also be 0, so you have to fix the up vector manually. Also, you might want to fix this vector so it won't turn upside down.
That works and will look better than your method. But it will still look unnatural for some curves. The best method I can mention is quaternion interpolation, such as Slerp.
At each point of the curve, you also have a "right" vector: the vector that points to the right of the plane. From the curve and this vector, you can calculate the up vector at this point. Then, you use quaternion interpolation to interpolate the up vectors along the curve.
If position and rotation depends only on spline curvature the easiest way will be Numerical differentiation of 3D spline (you will have 2 derivatives one for vertical and one for horizontal components). Your UP and side will be normals to the tangent.
So first of all I have such image (and ofcourse I have all points coordinates in 2d so I can regenerate lines and check where they cross each other)
(source: narod.ru)
But hey, I have another Image of same lines (I know thay are same) and new coords of my points like on this image
(source: narod.ru)
So... now Having points (coords) on first image, How can I determin plane rotation and Z depth on second image (asuming first one's center was in point (0,0,0) with no rotation)?
What you're trying to find is called a projection matrix. Determining precise inverse projection usually requires that you have firmly established coordinates in both source and destination vectors, which the images above aren't going to give you. You can approximate using pixel positions, however.
This thread will give you a basic walkthrough of the techniques you need to use.
Let me say this up front: this problem is hard. There is a reason Dan Story's linked question has not been answered. Let provide an explanation for people who want to take a stab at it. I hope I'm wrong about how hard it is, though.
I will assume that the 2D screen coordinates and projection/perspective matrix is known to you. You need to know at least this much (if you don't know the projection matrix, essentially you are using a different camera to look at the world). Let's call each pair of 2D screen coordinates (a_i, b_i), and I will assume the projection matrix is of the form
P = [ px 0 0 0 ]
[ 0 py 0 0 ]
[ 0 0 pz pw]
[ 0 0 s 0 ], s = +/-1
Almost any reasonable projection has this form. Working through the rendering pipeline, you find that
a_i = px x_i / (s z_i)
b_i = py y_i / (s z_i)
where (x_i, y_i, z_i) are the original 3D coordinates of the point.
Now, let's assume you know your shape in a set of canonical coordinates (whatever you want), so that the vertices is (x0_i, y0_i, z0_i). We can arrange these as columns of a matrix C. The actual coordinates of the shape are a rigid transformation of these coordinates. Let's similarly organize the actual coordinates as columns of a matrix V. Then these are related by
V = R C + v 1^T (*)
where 1^T is a row vector of ones with the right length, R is an orthogonal rotation matrix of the rigid transformation, and v is the offset vector of the transformation.
Now, you have an expression for each column of V from above: the first column is { s a_1 z_1 / px, s b_1 z_1 / py, z_1 } and so on.
You must solve the set of equations (*) for the set of scalars z_i, and the rigid transformation defined R and v.
Difficulties
The equation is nonlinear in the unknowns, involving quotients of R and z_i
We have assumed up to now that you know which 2D coordinates correspond to which vertices of the original shape (if your shape is a square, this is slightly less of a problem).
We assume there is even a solution at all; if there are errors in the 2D data, then it's hard to say how well equation (*) will be satisfied; the transformation will be nonrigid or nonlinear.
It's called (digital) photogrammetry. Start Googling.
If you are really interested in this kind of problems (which are common in computer vision, tracking objects with cameras etc.), the following book contains a detailed treatment:
Ma, Soatto, Kosecka, Sastry, An Invitation to 3-D Vision, Springer 2004.
Beware: this is an advanced engineering text, and uses many techniques which are mathematical in nature. Skim through the sample chapters featured on the book's web page to get an idea.
As a followup to my previous question about determining camera parameters I have formulated a new problem.
I have two pictures of the same rectangle:
The first is an image without any transformations and shows the rectangle as it is.
The second image shows the rectangle after some 3d transformation (XYZ-rotation, scaling, XY-translation) is applied. This has caused the rectangle to look a trapezoid.
I hope the following picture describes my problem:
alt text http://wilco.menge.nl/application.data/cms/upload/transformation%20matrix.png
How do determine what transformations (more specifically: what transformation matrix) have caused this tranformation?
I know the pixel locations of the corners in both images, hence i also know the distances between the corners.
I'm confused. Is this a 2d or a 3d problem?
The way I understand it, you have a flat rectangle embedded in 3d space, and you're looking at two 2d "pictures" of it - one of the original version and one based on the transformed version. Is this correct?
If this is correct, then there is not enough information to solve the problem. For example, suppose the two pictures look exactly the same. This could be because the translation is the identity, or it could be because the translation moves the rectangle twice as far away from the camera and doubles its size (thus making it look exactly the same).
This is a math problem, not programming ..
you need to define a set of equations (your transformation matrix, my guess is 3 equations) and then solve it for the 4 transformations of the corner-points.
I've only ever described this using German words ... so the above will sound strange ..
Based on the information you have, this is not that easy. I will give you some ideas to play with, however. If you had the 3D coordinates of the corners, you'd have an easier time. Here's the basic idea.
Move a corner to the origin. Thereafter, rotations will take place about the origin.
Determine vectors of the axes. Do this by subtracting the adjacent corners from the origin point. These will be a local x and y axis for your world.
Determine angles using the vectors. You can use the dot and cross products to determine the angle between the local x axis and the global x axis (1, 0, 0).
Rotate by the angle in step 3. This will give you a new x axis which should match the global x axis and a new local y axis. You can then determine another rotation about the x axis which will bring the y axis into alignment with the global y axis.
Without the z coordinates, you can see that this will be difficult, but this is the general process. I hope this helps.
The solution will not be unique, as Alex319 points out.
If the second image is really a trapezoid as you say, then this won't be too hard. It is a trapezoid (not a parallelogram) because of perspective, so it must be an isosceles trapezoid.
Draw the two diagonals. They intersect at the center of the rectangle, so that takes care of the translation.
Rotate the trapezoid until its parallel sides are parallel to two sides of the original rectangle. (Which two? It doesn't matter.)
Draw a third parallel through the center. Scale this to the sides of the rectangle you chose.
Now for the rotation out of the plane. Measure the distance from the center to one of the parallel sides and use the law of sines.
If it's not a trapezoid, just a quadralateral, then it'll be harder, you'll have to use the angles between the diagonals to find the axis of rotation.