Calibration of images to obtain a top-view for points that lie on a same plane

asked 2016-02-10 13:03:37 -0500

abhigenie92 gravatar image

Same question: http://stackoverflow.com/questions/34... Though, the below code is from Matlab. I am open to any solutions using opencv. Calibration:

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

    >> cameraParams

cameraParams = 

  cameraParameters with properties:

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

   Lens Distortion
        RadialDistortion: [0.0397 0.0798 -0.2034]
    TangentialDistortion: [-0.0063 -0.0165]

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

   Accuracy of Estimation
    MeanReprojectionError: 0.1269
       ReprojectionErrors: [48x2x18 double]
        ReprojectedPoints: [48x2x18 double]

   Calibration Settings
                        NumPatterns: 18
                        WorldPoints: [48x2 double]
                         WorldUnits: 'mm'
                       EstimateSkew: 1
    NumRadialDistortionCoefficients: 3
       EstimateTangentialDistortion: 1

Extrinstic: enter image description here

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.

Data sample:

K>> [xcor_i,ycor_i ]


ans =
    -101.7000  -77.4040
     -102.4200  -77.4040
     -103.6600  -77.4040
     -103.9300  -76.6720
     -103.9900  -76.5130
     -104.0000  -76.4780
     -105.0800  -76.4710
     -106.0400  -77.5660
     -106.2500  -77.8050
     -106.2900  -77.8570
     -106.3000  -77.8680
     -106.3000  -77.8710
     -107.7500  -78.9680
     -108.0600  -79.2070
     -108.1200  -79.2590
     -109.9500  -80.3680
     -111.4200  -80.6090
     -112.8200  -81.7590
     -113.8500  -82.3750
     -115.1500  -83.2410
     -116.1500  -83.4290
     -116.3700  -83.8360
     -117.5000  -84.2910
     -117.7400  -84.3890
     -118.8800  -84.7770
     -119.8400  -85.2270
     -121.1400  -85.3250
     -123.2200  -84.9800
     -125.4700  -85.2710
     -127.0400  -85.7000
     -128.8200  -85.7930
     -130.6500  -85.8130
     -132.4900  -85.8180
     -134.3300  -86.5500
     -136.1700  -87.0760
     -137.6500  -86.0920
     -138.6900  -86.9760
     -140.3600  -87.9000
     -142.1600  -88.4660
     -144.7200  -89.3210

Code(Ref:http://stackoverflow.com/a/27260492/3646408):

load('C:\Users\sony\Dropbox\calibration_images\matlab_calibration_data.mat');
R = cameraParams.RotationMatrices(:,:,1);
t = cameraParams.TranslationVectors(1, :);
% combine rotation and translation into one matrix:
R(3, :) = t;
%Now compute the homography between the checkerboard and the image plane:

H = R * cameraParams.IntrinsicMatrix;
%Transform the image using the inverse of the homography:
I=imread('C:\Users\sony\Dropbox\calibration_images\Images\exp_0.jpg');
J = imwarp(I, projective2d(inv(H)));
imshow(J);

How can I do the same for points?

Edit 1:

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.

New info:

  1. Note all these points(in data sample) are the on the same plane.
  2. Also, this plane is perpendicular to one of images of checkerboard used for calibration. For that image(below), I know the height of origin of the ...
(more)
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