I'm trying to get top view image from the image captured by a camera in perspective. Already got the intrinsic and extrinsic parameters of the camera to the plane , the ground. The camera is positioned on a robot pointed to the ground at a certain height and tilt. For this camera position got:
Intrinsic matrix (3X3):
KK= [624.2745,0,327.0749;0,622.0777,232.3448;0,0,1]
The translation vector and rotation matrix:
T = [-323.708466;-66.960728;1336.693284]
R =[0.0029,1.0000,-0.0034;0.3850,-0.0042,-0.9229;-0.9229,0.0013,-0.3850]
How can I get a Homography matrix to get top view image of the ground (chess set in the ground), with the information I have?
I'm using matlab. I've done the code to apply the matrix H to the captured image, but I still haven´t i way the get this H matrix properly.
Thanks in advance.
Related
I am trying to solve a question related to transformation of coordinates in 3-D space but not sure how to approach it.
Lets a vertex point named P is drawn at the origin with a 4x4 transformation matrix. It's then views through a camera that's positioned with a model view matrix and then through a simple projective transform matrix.
How do I calculate the new screen coordinates of P' (x,y,z)?
Before explain of pipeline, you need to know is how pipeline do process to draw on screen.
Everything between process is just matrix multiplication with vector
Model - World - Camera - Projection(or Nomalized Coordinate) - Screen
First step, we call it 'Model Space' because of (0,0,0) is based in model.
And we need to move model space to world space because of we are gonna place model to world so
we need to do transform will be (translate, rotation, scale)TRS * Model(Vector4) because definition of world transform will be different
After do it, model place in world.
Thrid, need to render on camrea space because what we see is through the camera. in world, camera also has position, viewport size and
rotation.. It needs to project from the camera. see
General Formula for Perspective Projection Matrix
After this job done, you will get nomalized coordinate which is Techinically 0-1 coordinates.
Finaly, Screen space. suppose that we are gonna make vido game for mobile. mobile has a lot of screen resolution. so how to get it done?
Simple, scale and translate to get result in screen space coordinate. Because of origin and screen size is different.
So what you are tring to do is 'step 4'.
If you want to get screen position P1 from world, formula will be "Screen Matrix * projection matrix * Camera matrix * P1"
If you want to get position from camera space it would be "Screen Matrix * projection matrix * P1".
There are useful links to understand matrix and calculation so see above links.
https://answers.unity.com/questions/1359718/what-do-the-values-in-the-matrix4x4-for-cameraproj.html
https://www.google.com/search?q=unity+camera+to+screen+matrix&newwindow=1&rlz=1C5CHFA_enKR857KR857&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjk5qfQ18DlAhUXfnAKHabECRUQ_AUIEigB&biw=1905&bih=744#imgrc=Y8AkoYg3wS4PeM:
I'm creating a 3D object in Three.js with 6 faces. Each face has a mesh which uses a THREE.PlaneGeometry(width and height both are 256). On the mesh I'm using a JPEG picture which is 256 by 256 for the texture. I'm trying to find a way to find the world coordinate of a pixel coordinate(for example 200,250 is the pixel coordinate) on the Object3D's PlaneGeometry corresponding to where that picture was used as texture.
Object hierarchy:-
Object3D-->face(object3d) (total 6 faces)-->Each face has a mesh(planegeometry) and uses a jpeg file as texture.
Picture1 pixel coordinate-->Used to create texture for Plane1-->World Coordinate corresponding to that pixel coordinate.
Can someone please help me.
Additional information:-
Thanks for the answer. I'm trying to compare 2 results.
Method 1:- One yaw/pitch is obtained by clicking on a specific point in the 3d object(e.g, center of a particular car headlight which is the front face) using a mouse and getting the point of intersection with the front face using raycasting.
Method 2:-The other yaw/pitch is obtained by taking the pixel coordinate of the same point(center of a particular car headlight) and calculating the world space coordinate for that pixel point. Pls note that pixel coordinate is taken from the JPEG file that was used as texture to create the PlaneGeometry for the mesh(which is a child of the front face).
Do you think the above comparison approach is supposed to produce the same results, assuming all other parameters are identical between the 2 approaches?
Well assuming your planes are PlaneGeometry(1,1) then the local coordinate X/Y/ZZ for a given pixel is pixelX / 256, pixelY / 256 and the Z is 0.5
so something like:
var localPoint = new THREE.Vector3(px/256,py/256,0.5)
var worldPoint = thePlaneObject.localToWorld(localPoint)
I use a OpenCV fisheye model function to perform fisheye calibration work. My image is a Circular fisheye (example), but I'm getting this result from the OpenCV fisheye model function.
I have the following problems:
I don't know why the result is an oval and not a perfect circle. Is this as expected?
Can OpenCV fisheye model be calibrated for a Circular fisheye image?
I don't understand why the image is not centered when using the cv::fisheye::calibrate function to get the Cx Cy parameter in K?
What tips (picture number, angle and position...) can be used on the checkboard to get the corrent camera matrix and Distortion factor?
Expected Result
My Result
First at all cv::fisheye uses a very simple idea. To remove radial distortion it will move points of fisheye circle in direction from circle center to circle edge.
Points near center will be moved a little. Points near edges will be moved on much larger distance.
In other words distance of point movement is not a constant. It's a function f(x)= 1+K1*x3+K2*x5+K3*x7=K4*x9. K1-K4 are coefficients of radial distortion of opencv fisheye undistortion model. In normal case undistorted image is always larger then initial image.
As you can see your undistorted image is smaller then initial fisheye image. I think the source of problem is bad calibration.
I don't know why the result is an oval and not a perfect circle. Is this as expected?
-> Tangential parameter of the calibration model can make it look like oval. It could be either your actual lens is tilted or calibration is incorrect. Just try to turn off tangential parameter option.
Can OpenCV fisheye model be calibrated for a Circular fisheye image?
-> No problem as far as I know. Try ocam as well.
I don't understand why the image is not centered when using the cv::fisheye::calibrate function to get the Cx Cy parameter in K?
-> It is normal that optical center does not align with the image center. It is a matter of degree, however. Cx, Cy represents the actual optical center. Low-quality fisheye camera manufactures does not control the quality of this parameter.
What tips (picture number, angle and position...) can be used on the checkboard to get the corrent camera matrix and Distortion factor?
-> clear images only, different distances, different angles, different positions. As many as possible.
I have a 2D camera defined by Direct2D 3x2 matrix like this :
ViewMatrix = ScaleMatrix * TranslationMatrix;
But when trying to do hit testing, I get at a point where I need to know my X,Y camera coordinate. I tried to keep track of hit in a vector but without success, scaling with offset center complicate a lot the work..
So I guess it should be possible to find my camera coordinate from these two matrix right ? But how ?
Thank a lot for help.
I am trying to build a simple camera matching (or match moving) application. The functionality is the same as that in most 3d applications like 3ds Max or Maya. Given an image of a cube and a 3d model of the cube, the user selects points on the image corresponding to each vertex of the model. The application must then generate a camera view that displays the 3d cube model from the same angle as shown in the image.
Can anyone point me in the direction of an algorithm for that?
PS: The camera is calibrated and the camera calibration matrix is available to the program
You can try with the algorithm illustrated step-by-step on http://www.offbytwo.net/camera-matching/. The Octave source code is provided, too.
As a plus, you don't need to start with a cube, but just with any two edges parallel to the x axis and two in the y direction.