I want to convert the detected face rectangle into 3D coordinates. I have the intrinsic parameters of my webcam and my head dimension, how can I determine the depth Z using the projection equation?
x = fx X / Z + u
y = fy Y / Z + v
I understand that fx fy and u v are intrinsic parameters, and that X Y are given by my head dimension, x y are given by the detected face rectangle. It seems that only one equation is enough to determine Z. How to use both of them? Or I am wrong?
You are correct that you do not strictly need both of them to compute the depth. However, you may want to use both to improve accuracy.
Another thing to keep in mind is that if your camera is not looking perpendicularly at the planar object (e.g. face) you measured, one or both of the measurements may not be useful to compute the depth. For example if your camera is looking up at a rectangle, only the width will be a good measure for the depth, because the height is compressed by the viewing angle. I don't think this matters for your face detector though, because the proportions of the face are assumed fixed anyway?
Related
I'm trying to obtain the orientation of a square in the real world from an image. I know the projection of each vertex in the image and with this and a depth camera I can obtain the position of the centroid in the real world.
I need the orientation of the square (actually, the normal vector to the plane) and the depth camera has not enough resolution. The camera parameters are also known.
I've search and I've only found estimation algorithms too overkill for problems with much less information. But in this case, I have a lot of data of the shape, distance, camera, image, etc. but I am not being able to get it.
Thanks in advance.
I assume the image is captured with an ordinary camera, and that your "square" is well approximated by an actual geometrical rectangle, with parallel opposite sides and orthogonal adjacent ones
If you only need the square's normal, and the camera is calibrated (in particular, the nonlinear lens distortion is removed from the image), then it can trivially be obtained from the vanishing points and the center. The algorithm is as follows:
Express the images of the four vertices p_i, i=1..4, in homogeneous coordinates: p_i = (u_i, v_i, 1). The ordering of i is unimportant, but in the following I assume it's clockwise starting from any one vertex. Also, for convenience, where in the following I write, say, i + n, it's assumed that the addition is modulo 4, so that, e.g., i + 1 = 1 when i = 4.
Compute the equations of the lines covering the square sides: l_i = p_(i+1) X p_i, where X represents the cross product.
Compute the equations of the diagonals: d_13 = p_1 X p_3, d_24 = p_2 X p_4.
Compute the center: c = d_13 X d_24.
Compute the vanishing points of the pairs of parallel sides: v_13 = l_1 X l_3, v_24 = l_2 X l_4. They represent the directions of the images of two lines which, in 3D, are orthogonal to each other.
Compute the images of the axes the 3D orthogonal coordinate frame rooted at the square center, and with two of the axes parallel to the square sides: x = c X v_13, y = c X v_24.
Lastly, the plane normal, in 3D camera coordinate frame, is their cross product: z = x X y .
Note that removing the distortion is important, because even a small amount of distortion can greatly affects the location of the vanishing points when the square sides are nearly parallel.
If you want to know why this works, the following excerpt from Hartley and Zisserman's "Multiple View Geometry in Computer Vision" should sufice:
I'd like to get the pose (translation: x, y, z and rotation: Rx, Ry, Rz in World coordinate system) of the overhead camera. I got many object points and image points by moving the ChArUco calibration board with a robotic arm (like this https://www.youtube.com/watch?v=8q99dUPYCPs). Because of that, I already have exact positions of all the object points.
In order to feed many points to solvePnP, I set the first detected pattern (ChArUco board) as the first object and used it as the object coordinate system's origin. Then, I added the detected object points (from the second pattern to the last) to the first detected object points' coordinate system (the origin of the object frame is the origin of the first object).
After I got the transformation between the camera and the object's coordinate frame, I calculated the camera's pose based on that transformation.
The result looked pretty good at first, but when I measured the camera's absolute pose by using a ruler or a tape measure, I noticed that the extrinsic calibration result was around 15-20 millimeter off for z direction (the height of the camera), though almost correct for the others (x, y, Rx, Ry, Rz). The result was same even I changed the range of the object points by moving a robotic arm differently, it always ended up to have a few centimeters off for the height.
Has anyone experienced the similar problem before? I'd like to know anything I can try. What is the common mistake when the depth direction (z) is inaccurate?
I don't know how you measure the z but I believe that what you're measuring with the ruler is not z but the euclidean distance which is computed like so:
d=std::sqrt(x*x+y*y+z*z);
Let's take an example, if x=2; y=2; z=2;
then d will be d~3,5 so 3.5-2=1.5 is the difference you get between z and the ruler when you said around 15-20 millimeter off for z direction.
Given an object's 3D mesh file and an image that contains the object, what are some techniques to get the orientation/pose parameters of the 3d object in the image?
I tried searching for some techniques, but most seem to require texture information of the object or at least some additional information. Is there a way to get the pose parameters using just an image and a 3d mesh file (wavefront .obj)?
Here's an example of a 2D image that can be expected.
FOV of camera
Field of view of camera is absolute minimum to know to even start with this (how can you determine how to place object when you have no idea how it would affect scene). Basically you need transform matrix that maps from world GCS (global coordinate system) to Camera/Screen space and back. If you do not have a clue what about I am writing then perhaps you should not try any of this before you learn the math.
For unknown camera you can do some calibration based on markers or etalones (known size and shape) in the view. But much better is use real camera values (like FOV angles in x,y direction, focal length etc ...)
The goal for this is to create function that maps world GCS(x,y,z) into Screen LCS(x,y).
For more info read:
transform matrix anatomy
3D graphic pipeline
Perspective projection
Silhouette matching
In order to compare rendered and real image similarity you need some kind of measure. As you need to match geometry I think silhouette matching is the way (ignoring textures, shadows and stuff).
So first you need to obtain silhouettes. Use image segmentation for that and create ROI mask of your object. For rendered image is this easy as you van render the object with single color without any lighting directly into ROI mask.
So you need to construct function that compute the difference between silhouettes. You can use any kind of measure but I think you should start with non overlapping areas pixel count (it is easy to compute).
Basically you count pixels that are present only in one ROI (region of interest) mask.
estimate position
as you got the mesh then you know its size so place it in the GCS so rendered image has very close bounding box to real image. If you do not have FOV parameters then you need to rescale and translate each rendered image so it matches images bounding box (and as result you obtain only orientation not position of object of coarse). Cameras have perspective so the more far from camera you place your object the smaller it will be.
fit orientation
render few fixed orientations covering all orientations with some step 8^3 orientations. For each compute the difference of silhouette and chose orientation with smallest difference.
Then fit the orientation angles around it to minimize difference. If you do not know how optimization or fitting works see this:
How approximation search works
Beware too small amount of initial orientations can cause false positioves or missed solutions. Too high amount will be slow.
Now that was some basics in a nutshell. As your mesh is not very simple you may need to tweak this like use contours instead of silhouettes and using distance between contours instead of non overlapping pixels count which is really hard to compute ... You should start with simpler meshes like dice , coin etc ... and when grasping all of this move to more complex shapes ...
[Edit1] algebraic approach
If you know some points in the image that coresponds to known 3D points (in your mesh) then you can along with the FOV of the camera used compute the transform matrix placing your object ...
if the transform matrix is M (OpenGL style):
M = xx,yx,zx,ox
xy,yy,zy,oy
xz,yz,zz,oz
0, 0, 0, 1
Then any point from your mesh (x,y,z) is transformed to global world (x',y',z') like this:
(x',y',z') = M * (x,y,z)
The pixel position (x'',y'') is done by camera FOV perspective projection like this:
y''=FOVy*(z'+focus)*y' + ys2;
x''=FOVx*(z'+focus)*x' + xs2;
where camera is at (0,0,-focus), projection plane is at z=0 and viewing direction is +z so for any focal length focus and screen resolution (xs,ys):
xs2=xs*0.5;
ys2=ys*0.5;
FOVx=xs2/focus;
FOVy=ys2/focus;
When put all this together you obtain this:
xi'' = ( xx*xi + yx*yi + zx*zi + ox ) * ( xz*xi + yz*yi + zz*zi + ox + focus ) * FOVx
yi'' = ( xy*xi + yy*yi + zy*zi + oy ) * ( xz*xi + yz*yi + zz*zi + oy + focus ) * FOVy
where (xi,yi,zi) is i-th known point 3D position in mesh local coordinates and (xi'',yi'') is corresponding known 2D pixel positions. So unknowns are the M values:
{ xx,xy,xz,yx,yy,yx,zx,zy,zz,ox,oy,oz }
So we got 2 equations per each known point and 12 unknowns total. So you need to know 6 points. Solve the system of equations and construct your matrix M.
Also you can exploit that M is a uniform orthogonal/orthonormal matrix so vectors
X = (xx,xy,xz)
Y = (yx,yy,yz)
Z = (zx,zy,zz)
Are perpendicular to each other so:
(X.Y) = (Y.Z) = (Z.X) = 0.0
Which can lower the number of needed points by introducing these to your system. Also you can exploit cross product so if you know 2 vectors the thirth can be computed
Z = (X x Y)*scale
So instead of 3 variables you need just single scale (which is 1 for orthonormal matrix). If I assume orthonormal matrix then:
|X| = |Y| = |Z| = 1
so we got 6 additional equations (3 x dot, and 3 for cross) without any additional unknowns so 3 point are indeed enough.
I am searching lots of resources on internet for many days but i couldnt solve the problem.
I have a project in which i am supposed to detect the position of a circular object on a plane. Since on a plane, all i need is x and y position (not z) For this purpose i have chosen to go with image processing. The camera(single view, not stereo) position and orientation is fixed with respect to a reference coordinate system on the plane and are known
I have detected the image pixel coordinates of the centers of circles by using opencv. All i need is now to convert the coord. to real world.
http://www.packtpub.com/article/opencv-estimating-projective-relations-images
in this site and other sites as well, an homographic transformation is named as:
p = C[R|T]P; where P is real world coordinates and p is the pixel coord(in homographic coord). C is the camera matrix representing the intrinsic parameters, R is rotation matrix and T is the translational matrix. I have followed a tutorial on calibrating the camera on opencv(applied the cameraCalibration source file), i have 9 fine chessbordimages, and as an output i have the intrinsic camera matrix, and translational and rotational params of each of the image.
I have the 3x3 intrinsic camera matrix(focal lengths , and center pixels), and an 3x4 extrinsic matrix [R|T], in which R is the left 3x3 and T is the rigth 3x1. According to p = C[R|T]P formula, i assume that by multiplying these parameter matrices to the P(world) we get p(pixel). But what i need is to project the p(pixel) coord to P(world coordinates) on the ground plane.
I am studying electrical and electronics engineering. I did not take image processing or advanced linear algebra classes. As I remember from linear algebra course we can manipulate a transformation as P=[R|T]-1*C-1*p. However this is in euclidian coord system. I dont know such a thing is possible in hompographic. moreover 3x4 [R|T] Vector is not invertible. Moreover i dont know it is the correct way to go.
Intrinsic and extrinsic parameters are know, All i need is the real world project coordinate on the ground plane. Since point is on a plane, coordinates will be 2 dimensions(depth is not important, as an argument opposed single view geometry).Camera is fixed(position,orientation).How should i find real world coordinate of the point on an image captured by a camera(single view)?
EDIT
I have been reading "learning opencv" from Gary Bradski & Adrian Kaehler. On page 386 under Calibration->Homography section it is written: q = sMWQ where M is camera intrinsic matrix, W is 3x4 [R|T], S is an "up to" scale factor i assume related with homography concept, i dont know clearly.q is pixel cooord and Q is real coord. It is said in order to get real world coordinate(on the chessboard plane) of the coord of an object detected on image plane; Z=0 then also third column in W=0(axis rotation i assume), trimming these unnecessary parts; W is an 3x3 matrix. H=MW is an 3x3 homography matrix.Now we can invert homography matrix and left multiply with q to get Q=[X Y 1], where Z coord was trimmed.
I applied the mentioned algorithm. and I got some results that can not be in between the image corners(the image plane was parallel to the camera plane just in front of ~30 cm the camera, and i got results like 3000)(chessboard square sizes were entered in milimeters, so i assume outputted real world coordinates are again in milimeters). Anyway i am still trying stuff. By the way the results are previosuly very very large, but i divide all values in Q by third component of the Q to get (X,Y,1)
FINAL EDIT
I could not accomplish camera calibration methods. Anyway, I should have started with perspective projection and transform. This way i made very well estimations with a perspective transform between image plane and physical plane(having generated the transform by 4 pairs of corresponding coplanar points on the both planes). Then simply applied the transform on the image pixel points.
You said "i have the intrinsic camera matrix, and translational and rotational params of each of the image.” but these are translation and rotation from your camera to your chessboard. These have nothing to do with your circle. However if you really have translation and rotation matrices then getting 3D point is really easy.
Apply the inverse intrinsic matrix to your screen points in homogeneous notation: C-1*[u, v, 1], where u=col-w/2 and v=h/2-row, where col, row are image column and row and w, h are image width and height. As a result you will obtain 3d point with so-called camera normalized coordinates p = [x, y, z]T. All you need to do now is to subtract the translation and apply a transposed rotation: P=RT(p-T). The order of operations is inverse to the original that was rotate and then translate; note that transposed rotation does the inverse operation to original rotation but is much faster to calculate than R-1.
All i know is that the height and width of an object in video. can someone guide me to calculate distance of an detected object from camera in video using c or c++? is there any algorithm or formula to do that?
thanks in advance
Martin Ch was correct in saying that you need to calibrate your camera, but as vasile pointed out, it is not a linear change. Calibrating your camera means finding this matrix
camera_matrix = [fx,0 ,cx,
0,fy,cy,
0,0, 1];
This matrix operates on a 3 dimensional coordinate (x,y,z) and converts it into a 2 dimensional homogeneous coordinate. To convert to your regular euclidean (x,y) coordinate just divide the first and second component by the third. So now what are those variables doing?
cx/cy: They exist to let you change coordinate systems if you like. For instance you might want the origin in camera space to be in the top left of the image and the origin in world space to be in the center. In that case
cx = -width/2;
cy = -height/2;
If you are not changing coordinate systems just leave these as 0.
fx/fy: These specify your focal length in units of x pixels and y pixels, these are very often close to the same value so you may be able to just give them the same value f. These parameters essentially define how strong perspective effects are. The mapping from a world coordinate to a screen coordinate (as you can work out for yourself from the above matrix) assuming no cx and cy is
xsc = fx*xworld/zworld;
ysc = fy*yworld/zworld;
As you can see the important quantity that makes things bigger closer up and smaller farther away is the ratio f/z. It is not linear, but by using homogenous coordinates we can still use linear transforms.
In short. With a calibrated camera, and a known object size in world coordinates you can calculate its distance from the camera. If you are missing either one of those it is impossible. Without knowing the object size in world coordinates the best you can do is map its screen position to a ray in world coordinates by determining the ration xworld/zworld (knowing fx).
i don´t think it is easy if have to use camera only,
consider about to use 3rd device/sensor like kinect/stereo camera,
then you will get the depth(z) from the data.
https://en.wikipedia.org/wiki/OpenNI