从 2D 重建相机坐标系中的 3D 坐标 - 具有侧面条件的像素

Question

我正在尝试使用侧面条件（在 MatLab 中）从相机图片中的 2D 像素坐标重建 3D 坐标。我确实有外在和内在的相机参数。

使用齐次变换，我可以将 3D 坐标从初始世界坐标系转换为我的相机坐标系。所以在这里我的变换矩阵 R_world_to_Camera 中有我的外在参数：

R_world_to_Camera = [ r_11, r_12, r_13, t1;
r_21, r_22, r_23, t2;
r_31, r_32, r_33, t3;
0, 0, 0, 1];

对于内在参数，我使用了加州理工学院的“MatLab 相机校准工具箱”并得到了这些参数：

Calibration results (with uncertainties): 

 Focal Length:          fc = [ 1017.21523   1012.54901 ] ± [ NaN   NaN ] 
 Principal point:       cc = [ 319.50000   239.50000 ] ± [ NaN   NaN ] 
 Skew:             alpha_c = [ 0.00000 ] ± [ NaN  ]   => angle of pixel axes = 90.00000 ± NaN degrees 
 Distortion:            kc = [ 0.00000   0.00000   0.00000   0.00000  0.00000 ] ± [ NaN   NaN   NaN   NaN    NaN ] 
 Pixel error:          err = [ 0.11596   0.14469 ] 

 Note: The numerical errors are approximately three times the standard deviations (for reference).

然后我得到 Camera-Calibration-Matrix K (3x3)

K = [1.017215234570303e+03, 0, 3.195000000000000e+02; 
0, 1.012549014668498e+03,2.395000000000000e+02; 
0, 0, 1.0000];

使用它我可以计算 3D -> 2D - Projection-Matrix P (3x4)：

P = K * [eye(3), zeros(3,1)];

在转换世界坐标 [X, Y, Z]_World 中的点时，我首先将其转换为相机坐标，然后将其投影到 2D：

% Transformation
P_world = [X; Y; Z; 1]; % homogenous coordinates in World coordinate System
P_camera = R_world_to_Camera * [X; Y; Z; 1];

% Projection
P_pixels = P * camera;
P_pixels = P_pixels / P_pixels(3); % normalize coordinates

所以我现在的问题是如何扭转这些步骤？作为附带条件，我想将 Z 坐标设置为已知（世界坐标为零）。我尝试了here on Stackoverflow 提出的解决方案，但不知何故我得到了错误的坐标。任何想法？非常感谢每一个帮助！

Answer 1

您通常无法反转该步骤：当 3D 点投影到 2D 图像上时，深度和比例信息会丢失。但是，如果，正如您所指出的，所有 3D 点都在 Z=0 平面上，那么从它们的投影中取回它们是微不足道的：计算相机矩阵的逆 Ki = K^-1，并将其应用于图像点在齐次坐标中。

P_camera = Ki * [u, v, 1]'

其中 [u, v] 是图像坐标，撇号表示转置。 您想要的 3D 点位于从相机中心到 P_camera 的光线上。用世界坐标表达两者：

P_world = [R|t]_camera_to_world * [P_camera, 1]'

C_world = [R|t]_camera_to_world * [0, 0, 0, 1]'

其中 [R|t] 是 4x4 坐标变换。 现在，每条射线上的点集表示为

P = C_world + lambda * P_world;

其中 lambda 是一个标量（沿射线的坐标）。您现在可以强加 P(3) = 0 的条件来找到将您的点放在 Z = 0 平面上的 lambda 值。

Answer 2

感谢一些红酒和彻底的阅读，我在一篇（德国）硕士论文中找到了答案：） 我的“测试代码”如下：

% Clean-Up First 
clear all; 
close all; 
clc; 

% Caamera-Calibration-Matrix
 K = [1.017215234570303e+03, 0, 3.195000000000000e+02, 0; 
0, 1.012549014668498e+03,2.395000000000000e+02, 0; 
0, 0, 1.0000, 0; 
0, 0, 0, 0]; 

% Transforma Matrix from 3D-World-Coordinate System to 3D-Camera-Coordinate System (Origin on CCD-Chip) 
% 
% The Camera is oriented "looking" into positive X-Direction of the World-Coordinate-System. On the picture,
% positive Y-Direction will be to the left, positive Z-Direction to the top. (right hand coordinate system!)
 R_World_to_Cam = [-0.0113242625465167   -0.999822053685344   0.0151163536128891   141.173585444427; 
0.00842007509644635   -0.0152123858102325   -0.999848810645587   1611.96528372161; 
0.999900032304804   -0.0111955728474261   0.00859117128537919   847.090629282911; 
0   0   0   1]; 

% Projection- and Transforma Matrix P 
 P = K * R_World_to_Cam; 

% arbitrary Points X_World in World-Coordinate-System [mm] (homogenous Coordinates) 
% forming a square of size 10m x 4m 
 X_World_1 = [20000; 2000; 0; 1]; 
 X_World_2 = [20000; -2000; 0; 1]; 
 X_World_3 = [10000; 2000; 0; 1]; 
 X_World_4 = [10000; -2000; 0; 1]; 

% Transform and Project from 3D-World -> 2D-Picture 
 X_Pic_1 = P * X_World_1; 
 X_Pic_2 = P * X_World_2; 
 X_Pic_3 = P * X_World_3; 
 X_Pic_4 = P * X_World_4; 

% normalize homogenous Coordinates (3rd Element has to be 1!) 
 X_Pic_1 = X_Pic_1 / X_Pic_1(3); 
 X_Pic_2 = X_Pic_2 / X_Pic_2(3); 
 X_Pic_3 = X_Pic_3 / X_Pic_3(3); 
 X_Pic_4 = X_Pic_4 / X_Pic_4(3); 

% Now for reverse procedure take arbitrary points in Camera-Picture... 
% (for simplicity, take points from above and "go" 30px to the right and 40px down) 
 X_Pic_backtransform_1 = X_Pic_1(1:3) + [30; 40; 0]; 
 X_Pic_backtransform_2 = X_Pic_2(1:3) + [30; 40; 0]; 
 X_Pic_backtransform_3 = X_Pic_3(1:3) + [30; 40; 0]; 
 X_Pic_backtransform_4 = X_Pic_4(1:3) + [30; 40; 0]; 

% ... and transform back following the formula from the Master Thesis (in German):
% Ilker Savas, "Entwicklung eines Systems zur visuellen Positionsbestimmung von Interaktionspartnern" 
 M_Mat = P(1:3,1:3);                 % Matrix M is the "top-front" 3x3 part 
 p_4 = P(1:3,4);                     % Vector p_4 is the "top-rear" 1x3 part 
 C_tilde = - inv( M_Mat ) * p_4;     % calculate C_tilde 

% Invert Projection with Side-Condition ( Z = 0 ) and Transform back to 
% World-Coordinate-System 
 X_Tilde_1 = inv( M_Mat ) * X_Pic_backtransform_1; 
 X_Tilde_2 = inv( M_Mat ) * X_Pic_backtransform_2; 
 X_Tilde_3 = inv( M_Mat ) * X_Pic_backtransform_3; 
 X_Tilde_4 = inv( M_Mat ) * X_Pic_backtransform_4; 

 mue_N_1 = -C_tilde(3) / X_Tilde_1(3); 
 mue_N_2 = -C_tilde(3) / X_Tilde_2(3); 
 mue_N_3 = -C_tilde(3) / X_Tilde_3(3); 
 mue_N_4 = -C_tilde(3) / X_Tilde_4(3); 

% Do the inversion of above steps... 
 X_World_backtransform_1 = mue_N_1 * inv( M_Mat ) * X_Pic_backtransform_1 + C_tilde; 
 X_World_backtransform_2 = mue_N_2 * inv( M_Mat ) * X_Pic_backtransform_2 + C_tilde; 
 X_World_backtransform_3 = mue_N_3 * inv( M_Mat ) * X_Pic_backtransform_3 + C_tilde; 
 X_World_backtransform_4 = mue_N_4 * inv( M_Mat ) * X_Pic_backtransform_4 + C_tilde; 


% Plot everything
figure(1); 
% First Bird Perspective of World-Coordinate System... 
subplot(1,2,1); 
xlabel('Y-World'); 
ylabel('X-World'); 
grid on; 
axis([-3000 3000 0 22000]); 
hold on; 


plot( -X_World_1(2), X_World_1(1), 'bo' ); 
plot( -X_World_2(2), X_World_2(1), 'bo' ); 
plot( -X_World_3(2), X_World_3(1), 'bo' ); 
plot( -X_World_4(2), X_World_4(1), 'bo' ); 
line([-X_World_1(2) -X_World_2(2) -X_World_4(2) -X_World_3(2) -X_World_1(2)], [X_World_1(1) X_World_2(1) X_World_4(1) X_World_3(1) X_World_1(1)], 'Color', 'blue' ); 

plot( -X_World_backtransform_1(2), X_World_backtransform_1(1), 'ro' ); 
plot( -X_World_backtransform_2(2), X_World_backtransform_2(1), 'ro' ); 
plot( -X_World_backtransform_3(2), X_World_backtransform_3(1), 'ro' ); 
plot( -X_World_backtransform_4(2), X_World_backtransform_4(1), 'ro' ); 
line([-X_World_backtransform_1(2) -X_World_backtransform_2(2) -X_World_backtransform_4(2) -X_World_backtransform_3(2) -X_World_backtransform_1(2)], [X_World_backtransform_1(1) X_World_backtransform_2(1) X_World_backtransform_4(1) X_World_backtransform_3(1) X_World_backtransform_1(1)], 'Color', 'red' ); 


hold off; 

% ...then the camera picture (perspective!)
subplot(1,2,2); 
hold on; 
image(ones(480,640).*255); 
colormap(gray(256)); 
axis([0 640 -480 0]); 
line([X_Pic_1(1) X_Pic_2(1) X_Pic_4(1) X_Pic_3(1) X_Pic_1(1)], -1*[X_Pic_1(2) X_Pic_2(2) X_Pic_4(2) X_Pic_3(2) X_Pic_1(2)], 'Color', 'blue' ); 
line([X_Pic_backtransform_1(1) X_Pic_backtransform_2(1) X_Pic_backtransform_4(1) X_Pic_backtransform_3(1) X_Pic_backtransform_1(1)], -1*[X_Pic_backtransform_1(2) X_Pic_backtransform_2(2) X_Pic_backtransform_4(2) X_Pic_backtransform_3(2) X_Pic_backtransform_1(2)], 'Color', 'red' ); 
hold off;