view for points that lie on a same plane

Calibration:

I have calibrated the camera using this vision toolbox in Matlab. I used checkerboard images to do so. After calibration I get the cameraParams which contains:

Camera Extrinsics
RotationMatrices: [3x3x18 double]
TranslationVectors: [18x3 double]

and

 Camera Intrinsics
 IntrinsicMatrix: [3x3 double]
 FocalLength: [1.0446e+03 1.0428e+03]
 PrincipalPoint: [604.1474 359.7477]
 Skew: 3.5436

Aim: I have recorded trajectories of some objects in motion using this camera. Each object corresponds to a single point in a frame. Now, I want to project the points such that I get a top-view.

-101.7000  -77.4040
-102.4200  -77.4040

Code (Ref:https://stackoverflow.com/a/27260492/3646408 and answer by @Dima below):

function generate_homographic_matrix()
%% Calibrate camera
% Define images to process
path=['.' filesep 'Images' filesep];
list_imgs=dir([path '*.jpg']);
list_imgs_path=strcat(path,{list_imgs.name});

% Detect checkerboards in images
[imagePoints, boardSize, imagesUsed] = detectCheckerboardPoints(list_imgs_path);
imageFileNames = list_imgs_path(imagesUsed);

% Generate world coordinates of the corners of the squares
squareSize = 27;  % in units of 'mm'
worldPoints = generateCheckerboardPoints(boardSize, squareSize);

% Calibrate the camera
[cameraParams, imagesUsed, estimationErrors] = estimateCameraParameters(imagePoints, worldPoints, ...
    'EstimateSkew', true, 'EstimateTangentialDistortion', true, ...
    'NumRadialDistortionCoefficients', 3, 'WorldUnits', 'mm');
%% Compute homography for peripendicular plane to checkerboard
% Detect the checkerboard 
im=imread(['.' filesep 'Images' filesep 'exp_19.jpg']); %exp_19.jpg is the checkerboard orthogonal to the floor
[imagePoints, boardSize] = detectCheckerboardPoints(im);

% Compute rotation and translation of the camera.
[Rc, Tc] = extrinsics(imagePoints, worldPoints, cameraParams);

% Rc(rotation of the calibration view w.r.t the camera) = [x y z])
%then the floor has rotation Rf = [z x -y].(Normal vector of the floor goes up.)
Rf=[Rc(:,3),Rc(:,1),Rc(:,2)*-1];

% Translate it to the floor
H=452;%distance btw origin and floor
Fc = Rc * [0; H; 0];
Tc = Tc + Fc';

% Combine rotation and translation into one matrix:
Rf(3, :) = Tc;

% Compute the homography between the checkerboard and the image plane:
H = Rf * cameraParams.IntrinsicMatrix;

save('homographic_matrix.mat','H')
end

%% Transform points
function [x_transf,y_transf] =transform_points(xcor_i,ycor_i)
% creates a projective2D object and then transforms the points forward to
% get a top-view
% xcor_i and ycor_i are 1d vectors comprising of the x-coordinates and
% y-coordinates of trajectories. 
data=load('homographic_matrix.mat');
homo_matrix=data.H;
tform=projective2d(inv(homo_matrix));
[x_transf,y_transf] = transformPointsForward(tform,xcor_i,ycor_i);
end

Quoting text from OReilly Learning OpenCV Pg 412: "Once we have the homography matrix and the height parameter set as we wish, we could then remove the chessboard and drive the cart around, making a bird's-eye view video of the path..." This what I essentially wish to achieve.

Abhishek,

I don't entirely understand what you are trying to do. Are your points on a plane, and are you trying to create a bird's eye view of that plane?

If so, then you need to know the extrinsics, R and t, describing the relationship between that plane and the camera. One way to get R and t is to place a checkerboard on the plane, and then use the extrinsics function.

After that, you can follow the directions in the question you cited to get the homography. Once you have the homography, you can create a projective2D object, and use its transformPointsForward method to transform your points.

Since you have the size of squares on the grid, then given 2 points that you know are connected by an edge of size E (in real world units), you can calculate their 3D position.

Taking the camera intrinsic matrix K and the 3D position C and the camera orientation matrix R , you can calculate a ray to each of the points p by doing:

D = R^T * K^-1 * p

Each 3D point is defined as:

P = C + t*D

and you have the constraint that ||P1-P2|| = E ||P1-P2|| = E then it's a matter of solving for t1,t2 and finding the 3D position of the two points.

In order to create a top view, you can take the 3D points and project them using a camera model for that top view to generate a new image.

If all your points are on a single plane, it's enough to calculate the position of 3 points, and you can extrapolate the rest.

If your points are located on a plane that you know one coordinate of, you can do it simply for each point. For example, if you know that your camera is located at height h=Cz , and you want to find the 3D location of points in the frame, given that they are on the floor (z=0), then all you have to do is calculate the direction D as above, and then:

t=abs( (h-0)/D.z )

The 0 represent the height of the plane. Substitute for any other value for other planes.

Now that you have the value of t , you can calculate the 3D position of each point: P=C+t*D .

Then, to create a top view, create a new camera position and rotation to match your required projection, and you can project each point onto this camera's image plane. If you want a full image, you can interpolate positions and fill in the blanks where no feature point was present.

For more details, you can always read: http://www.robots.ox.ac.uk/~vgg/hzbook/index.html

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