I am working on a project which attempts to remove the perspective distortion from an image based on the known orientation of the camera. My thinking is that I can create a rotational matrix based on the known X, Y, and Z orientations of the camera. I can then apply those matrices to the image via the WarpPerspective method.
In my script (written in Python) I have created three rotational matrices, each based on an orientation angle. I have gotten to a point where I am stuck on two issues. First, when I load each individual matrix into the WarpPerspective method, it doesn't seem to be working correctly. Whenever I warp an image on one axis it appears to significantly overwarp the image. The contents of the image are only recognizable if I limit the orientation angle to around 1 degree or less.
Secondly, how do I combine the three rotational matrices into a single matrix to be loaded into the WarpPerspective method. Can I import a 3x3 rotational matrix into that method, or do I have to create a 4x4 projective matrix. Below is the code that I am working on.
Thank you for your help.
CR
from numpy import *
import cv
#Sets angle of camera and converts to radians
x = -14 * (pi/180)
y = 20 * (pi/180)
z = 15 * (pi/180)
#Creates the Rotational Matrices
rX = array([[1, 0, 0], [0, cos(x), -sin(x)], [0, sin(x), cos(x)]])
rY = array([[cos(y), 0, -sin(y)], [0, 1, 0], [sin(y), 0, cos(y)]])
rZ = array([[cos(z), sin(z), 0], [-sin(z), cos(z), 0], [0, 0, 1]])
#Converts to CVMat format
X = cv.fromarray(rX)
Y = cv.fromarray(rY)
Z = cv.fromarray(rZ)
#Imports image file and creates destination filespace
im = cv.LoadImage("reference_image.jpg")
dst = cv.CreateImage(cv.GetSize(im), cv.IPL_DEPTH_8U, 3)
#Warps Image
cv.WarpPerspective(im, dst, X)
#Display
cv.NamedWindow("distorted")
cv.ShowImage("distorted", im)
cv.NamedWindow("corrected")
cv.ShowImage("corrected", dst)
cv.WaitKey(0)
cv.DestroyWindow("distorted")
cv.DestroyWindow("corrected")
You are doing several things wrong. First, you can't rotate on the x or y axis without a camera model. Imagine a camera with an incredibly wide field of view. You could hold it really close to an object and see the entire thing but if that object rotated its edges would seem to fly towards you very quickly with a strong perspective distortion. On the other hand a small field of view (think telescope) has very little perspective distortion. A nice place to start is setting your image plane at least as far from the camera as it is wide and putting your object right on the image plane. That is what I did in this example (c++ openCV)
The steps are
//1
float x = -14 * (M_PI/180);
float y = 20 * (M_PI/180);
float z = 15 * (M_PI/180);
cv::Matx31f rot_vec(x,y,z);
cv::Matx33f rot_mat;
cv::Rodrigues(rot_vec, rot_mat); //converts to a rotation matrix
cv::Matx33f translation1(1,0,-image.cols/2,
0,1,-image.rows/2,
0,0,1);
rot_mat(0,2) = 0;
rot_mat(1,2) = 0;
rot_mat(2,2) = 1;
//2 and 3
cv::Matx33f trans = rot_mat*translation1;
//4
trans(2,2) += image.rows;
cv::Matx33f camera_mat(image.rows,0,image.rows/2,
0,image.rows,image.rows/2,
0,0,1);
//5
cv::Matx33f transform = camera_mat*trans;
//6
cv::Mat final;
cv::warpPerspective(image, final, cv::Mat(transform),image.size());
This code gave me this output
I did not see Franco's answer until I posted this. He is completely correct, using FindHomography would save you all these steps. Still I hope this is useful.
Just knowing the rotation is not enough unless your images are taken either using a telecentric lens, or with a telephoto lens with very long focal (in which cases the images are nearly orthographic, and there is no perspective distortion).
Besides, it's not necessary. True, you can undo the perspective foreshortening of one plane in the image by calibrating the camera (i.e. estimating the intrinsic and extrinsic parameters to form the full camera projection matrix).
But you achieve the same result much more simply if you can identify in the image a quadrangle which is the image of a real-world square (or rectangle with known width/height ratio). If you can do that, you can trivially compute the homography matrix that maps the square (rectangle) to the quadrangle, then warp using its inverse.
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