3D矩阵透视变换

Question

I am using shape from shading to generate a Digital Terrain Model (DTM) of an image taken using a camera mounted on a mobile platform. The algorithm written in Python seems to work reasonably well however the output is at an incline and a bit spherical so I'm suspecting that I need to remove perspective distortion and barrelling from the DTM.

The data is available here in case anyone is interested in having a go at this.

The camera is mounted at an inclination of 41 degrees and has the following camera and distortion matrices:

    cam_matrix = numpy.matrix([[246.00559,0.00000,169.87374],[0.00000,247.37317,132.21396],[0.00000,0.00000,1.00000]])
    distortion_matrix = numpy.matrix([0.04674, -0.11775, -0.00464, -0.00346, 0.00000])

How can I apply perspective transform and remove the barreling distortion from this matrix to obtain a flattened DTM?

I have attempted this using OpenCV but it doesn't work as OpenCv is expecting an image and the transforms simply move pixels around rather than manipulate their value. I have also researched Numpy and Scipy but haven't arrived to a conclusion or a solution yet. I am somewhat familiar with the theory behind these transforms but have mostly worked on 2D versions.

Any ideas?

Answer 1

You can use a 4 x 4 transformation matrix which is inversible and allows a bi-directional tranformation between the two coordinate systems that you want.

If you know the three rotations a , b and g , about x , y , z respectively, using the right-hand rule. The x0 , y0 , z0 are the translations between the origins of the two coordinate systems.

The transformation matrix is defined as:

T = np.array([[ cos(b)*cos(g), (sin(a)*sin(b)*cos(g) + cos(a)*sin(g)), (sin(a)*sin(g) - cos(a)*sin(b)*cos(g)), x0],
              [-cos(b)*sin(g), (cos(a)*cos(g) - sin(a)*sin(b)*sin(g)), (sin(a)*cos(g) + cos(a)*sin(b)*sin(g)), y0],
              [        sin(b), -sin(a)*cos(b), cos(a)*cos(b), z0]
              [ 0, 0, 0, 1])

To use it efficiently you should put your points in a 2-D array like:

orig = np.array([[x0, x1, ..., xn],
                 [y0, y1, ..., yn],
                 [z0, z1, ..., zn],
                 [ 1,  1, ...,  1]])

Then:

new = T.dot(orig)

will give you the transformed points.

Answer 2

Saullo GP Castro has a great explanation. This is the code I used with Python 3.8. If you play around with the T = make_matrix(90,0,0,1,0,0) line you can see the changes to the XYZ positions in the orig matrix.

#!/usr/bin/env python3
import numpy as np
import math
'''If you know the three rotations a, b and g, about x, y, z respectively, using the 
right-hand rule.
The x0, y0, z0 are the translations between the origins of the two coordinate 
systems.'''
#https://stackoverflow.com/questions/22175385/3d-matrix-perspective-transform/22311973#22311973

def make_matrix(roll,pitch,heading,x0,y0,z0):
    a = math.radians(roll)
    b = math.radians(pitch)
    g = math.radians(heading)

    T = np.array([[ math.cos(b)*math.cos(g), (math.sin(a)*math.sin(b)*math.cos(g) + 
    math.cos(a)*math.sin(g)), (math.sin(a)*math.sin(g) - 
    math.cos(a)*math.sin(b)*math.cos(g)), x0],
    [-math.cos(b)*math.sin(g), (math.cos(a)*math.cos(g) - 
    math.sin(a)*math.sin(b)*math.sin(g)), (math.sin(a)*math.cos(g) + 
    math.cos(a)*math.sin(b)*math.sin(g)), y0],
    [        math.sin(b), -math.sin(a)*math.cos(b), math.cos(a)*math.cos(b), z0],
    [ 0, 0, 0, 1]])
    return T

def load_orig():
    orig = np.array([[0, 1, 0],
                [0, 0, 1],
                [1, 0, 0],
                [ 1,  1, 1]])
    return orig

if __name__ == '__main__':
    T = make_matrix(90,0,0,1,0,0)
    orig = load_orig()
    new = T.dot(orig)
    print(new)