When I have the rotation matrix or quaternion representation of a camera's pose, is there a way to obtain the orientation vector of the camera?
Here the orientation vector means a 3D vector in the world coordinate (WC) that represents an orientation.
I read through the commonly used representations like euler angles and axis-angle, but I didn't find any representations that can represent the orientation of the camera in WC.
Could anyone help? Thank you!
You probably want the 3x1 Rodrigues vector. Just plug in the SO(3) rotation matrix of the camera orientation in world coordinates, and you will get a vector representation. Just to be clear, pose and orientation are different. Pose is orientation + position. If you want the position as well, that can be represented as a 3x1 vector of t = [x y z]' (using Matlab notation).
A typical representation of the pose is a 4x4 matrix in SE(3) (Special Euclidean Group), which is just:
T = [R t; 0 0 0 1]
Where R is the rotation matrix in SO(3).
Related
I am writing a program. I have, say, a grid of dots on a piece of paper. I fix one end and bend the paper toward the screen, giving me a trapezoidal shape from the camera's point of view. I have the (x,y) camera coordinate of each dot. Is there a simple way I can change these (x,y) to real life (x,y) which should give me a rectangle? I have the camera/real (x,y) of the original flat sheet of paper pre-bend if that helps.
I have looked at 3D Camera coordinates to world coordinates (change of basis?) and Transforming screen coordinates from security camera to real world coordinates.
Look up "homography". The transformation from a plane in 3D space to its image as captured by an ideal pinhole camera is a homography. It can be represented as a 3x3 matrix H that transforms the 3D coordinates X of points in the world to their corresponding homogeneous image coordinates x:
x = H * X
where X is a 3x1 vector of the world point coordinates, and x = [u, v, w]^T is the image point in homogeneous coordinates.
Given a minimum of 4 matches between world and image points (e.g. the corners of a rectangle) you can estimate the parameters of the matrix H. For details, look up "DLT algorithm". In OpenCV the routine to use is findHomography.
Recently I'm struggling with a pose estimation problem with a single camera. I have some 3D points and the corresponding 2D points on the image. Then I use solvePnP to get the rotation and translation vectors. The problem is, how can I determine whether the vectors are right results?
Now I use an indirect way to do this:
I use the rotation matrix, the translation vector and the world 3D coordinates of a certain point to obtain the coordinates of that point in Camera system. Then all I have to do is to determine whether the coordinates are reasonable. I think I know the directions of x, y and z axes of Camera system.
Is Camera center the origin of the Camera system?
Now consider the x component of that point. Is x equavalent to the distance of the camera and the point in the world space in Camera's x-axis direction (the sign can then be determined by the point is placed on which side of the camera)?
The figure below is in world space, while the axes depicted are in Camera system.
========How Camera and the point be placed in the world space=============
|
|
Camera--------------------------> Z axis
| |} Xw?
| P(Xw, Yw, Zw)
|
v x-axis
My rvec and tvec results seems right and wrong. For a specified point, the z value seems reasonable, I mean, if this point is about one meter away from the camera in the z direction, then the z value is about 1. But for x and y, according to the location of the point I think x and y should be positive but they are negative. What's more, the pattern detected in the original image is like this:
But using the points coordinates calculated in Camera system and the camera intrinsic parameters, I get an image like this:
The target keeps its pattern. But it moved from bottom right to top left. I cannot understand why.
Yes, the camera center is the origin of the camera coordinate system, which seems to be right following to this post.
In case of camera pose estimation, value seems reasonable can be named as backprojection error. That's a measure of how well your resulting rotation and translation map the 3D points to the 2D pixels. Unfortunately, solvePnP does not return a residual error measure. Therefore one has to compute it:
cv::solvePnP(worldPoints, pixelPoints, camIntrinsics, camDistortion, rVec, tVec);
// Use computed solution to project 3D pattern to image
cv::Mat projectedPattern;
cv::projectPoints(worldPoints, rVec, tVec, camIntrinsics, camDistortion, projectedPattern);
// Compute error of each 2D-3D correspondence.
std::vector<float> errors;
for( int i=0; i < corners.size(); ++i)
{
float dx = pixelPoints.at(i).x - projectedPattern.at<float>(i, 0);
float dy = pixelPoints.at(i).y - projectedPattern.at<float>(i, 1);
// Euclidean distance between projected and real measured pixel
float err = sqrt(dx*dx + dy*dy);
errors.push_back(err);
}
// Here, compute max or average of your "errors"
An average backprojection error of a calibrated camera might be in the range of 0 - 2 pixel. According to your two pictures, this would be way more. To me, it looks like a scaling problem. If I am right, you compute the projection yourself. Maybe you can try once cv::projectPoints() and compare.
When it comes to transformations, I learned not to follow my imagination :) The first thing I Do with the returned rVec and tVec is usually creating a 4x4 rigid transformation matrix out of it (I posted once code here). This makes things even less intuitive, but instead it is compact and handy.
Now I know the answers.
Yes, the camera center is the origin of the camera coordinate system.
Consider that the coordinates in the camera system are calculated as (xc,yc,zc). Then xc should be the distance between the camera and
the point in real world in the x direction.
Next, how to determine whether the output matrices are right?
1. as #eidelen points out, backprojection error is one indicative measure.
2. Calculate the coordinates of the points according to their coordinates in the world coordinate system and the matrices.
So why did I get a wrong result(the pattern remained but moved to a different region of the image)?
Parameter cameraMatrix in solvePnP() is a matrix supplying the parameters of the camera's external parameters. In camera matrix, you should use width/2 and height/2 for cx and cy. While I use width and height of the image size. I think that caused the error. After I corrected that and re-calibrated the camera, everything seems fine.
For modelviewMatrix I understand how to form translate and scale Matrix. But I am unable to understand how to form viewMatrix using GLKMatrix4MakeLookAt. Can anyone explain how to it works and how to give value to parameters(eye center up X Y Z).
GLK_INLINE GLKMatrix4 GLKMatrix4MakeLookAt(float eyeX, float eyeY, float eyeZ,
float centerX, float centerY, float centerZ,
float upX, float upY, float upZ)
GLKMatrix4MakeLookAt creates a viewing matrix (in the same way as gluLookAt does, in case you look at other OpenGL code). As the parameters suggest, it considers the position of the viewer's eye, the point in space they're looking at (e.g., a point on an object), and the up vector, which specifies which direction is "up" (e.g., pointing towards the sky). The viewing matrix generated is the combination of a rotation matrix (composed of a set of orthonormal bases [basis vectors]) and an translation.
Logically, the matrix is basically constructed in a few steps:
compute the line-of-sight vector, which is the normalized vector going from the eye's position to the point you're looking at, the center point.
compute the cross product of the line-of-sight vector with the up vector, and normalize the resulting vector.
compute the cross product of the vector computed in step 2. with the line-of-sight to complete the orthonormal basis.
create a 3x3 rotation matrix by setting the first row to the vector created in step 2., the middle row with the vector from step 3., and the bottom row to the negated, normalized line-of-sight vector.
those three steps produce a rotation matrix that will rotate the world coordinate system into eye coordinates (a coordinate system where the eye is located at the origin, and the line-of-sight is down the -z axis. The final viewing matrix is computed by multiplying a translation to the negated eye position, which moves the "world coordinate positioned eye" to the origin for eye coordinates.
Here's a related question showing the code of GLKMatrix4MakeLookAt, and here's a question with more detail about eye coordinates and related coordinate systems: (What exactly are eye space coordinates?) .
I am building an augmented reality application and I have the yaw, pitch, and roll for the camera. I want to start placing objects in the 3D environment. I want to make it so that when the user clicks, a 3D point pops up right where the camera is pointed (center of the 2D screen) and when the user moves, the point moves accordingly in 3D space. The camera does not change position, only orientation. Is there a proper way to recover the 3D location of this point? We can assume that all points are equidistant from the camera location.
I am able to accomplish this independently for two axes (OpenGL default orientation). This works for changes in the vertical axis:
x = -sin(pitch)
y = cos(pitch)
z = 0
This also works for changes in the horizontal axis:
x = 0
y = -sin(yaw)
z = cos(yaw)
I was thinking that I should just make combine into:
x = -sin(pitch)
y = sin(yaw) * cos(pitch)
z = cos(yaw)
and that seems to be close, but not exactly correct. Any suggestions would be greatly appreciated!
It sounds like you just want to convert from a rotation vector (pitch,yaw,roll) to a rotation matrix. The conversion can bee seen on the Wikipedia article on rotation matrices. The idea is that once you have constructed your matrix, to transform any point simply.
final_pos = rot_mat*initial_pose
where final and initial pose are 3x1 vectors and rot_mat is a 3x3 matrix.
I am currently trying to figure out a way to calcute the size of a given object with kinect
since I have the following data
angular field of view of the lens
distance
and width in pixels from a 800*600 resolution
I believe this can be possible to calculate. Does anyone has math skills to give me a little help?
With some trigonometry, it should be possible to approximate.
If you draw a right trangle ABC, with the camera at one of the legs (A), and the object at the far end (edge BC), where the right angle is (C), then the height of the object is going to be the height of leg BC. the distance to the pixel might be the distance of leg AC or AB. The Kinect sensor specifications are going to regulate that. If you get distance to the center of a pixel, then it will be AC. if you have distances to pixel corners then the distance will be AB.
With A representing the angle at the camera that the pixel takes up, d is the distance of the hypotenuse of a right angle and y is the distance of the far leg (edge BC):
sin(A) = y / d
y = d sin(A)
y is the length of the pixel projected into the object plane. You calculate it by multiplying the sin of the angel by the distance to the object.
Here I confess I do not know the API of the kinect, and what level of detail it provides. You say you have the angle of the field of vision. You might assume each pixel of your 800x600 pixel grid takes up an equal angle of your camera's field of vision. If you do, then you can break up that field of vision into equal pieces to measure the linear size of your object in each pixel.
You also mentioned that you have the distance to the object. I was assuming that you have a distance map for each pixel of the 800x600 grid. If this is incorrect, some calculations can be done to approximate a distance grid for the pixels involving the object of interest if you make some assumptions about the object being measured.