我试图自己在 MATLAB 中实现 IBVS 算法（此处介绍中解释的算法），但我面临以下问题：该算法似乎仅适用于相机不必改变其方向的情况到世界框架。例如，如果我只是尝试使初始（几乎）正方形的一个顶点更靠近其相反顶点，则该算法不起作用，如下图所示
红色 x 是所需的投影，蓝色圆圈是初始的，绿色的是我从算法中得到的。
此外，错误并没有像应有的那样呈指数下降。
我究竟做错了什么？我附上了完全可运行的 MATLAB 代码。如果有人可以看看，我将非常感激。我取出了执行绘图的代码。我希望它现在更具可读性。视觉伺服必须至少有 4 个目标点，否则问题没有唯一的解决方案。如果您愿意提供帮助，我建议您查看该calc_Rotation_matrix()函数以检查是否正确计算了旋转矩阵，然后验证输入的行ds = vc;是否euler_ode正确。根据此约定，相机方向以欧拉角表示。最后，可以检查交互矩阵L是否正确计算。
function VisualServo()
global A3D B3D C3D D3D A B C D Ad Bd Cd Dd
%coordinates of the 4 points wrt camera frame
A3D = [-0.2633;0.27547;0.8956];
B3D = [0.2863;-0.2749;0.8937];
C3D = [-0.2637;-0.2746;0.8977];
D3D = [0.2866;0.2751;0.8916];
%initial projections (computed here only to show their relation with the desired ones)
A=A3D(1:2)/A3D(3);
B=B3D(1:2)/B3D(3);
C=C3D(1:2)/C3D(3);
D=D3D(1:2)/D3D(3);
%initial camera position and orientation
%orientation is expressed in Euler angles (X-Y-Z around the inertial frame
%of reference)
cam=[0;0;0;0;0;0];
%desired projections
Ad=A+[0.1;0];
Bd=B;
Cd=C+[0.1;0];
Dd=D;
t0 = 0;
tf = 50;
s0 = cam;
%time step
dt=0.01;
t = euler_ode(t0, tf, dt, s0);
end
function ts = euler_ode(t0,tf,dt,s0)
global A3D B3D C3D D3D Ad Bd Cd Dd
s = s0;
ts=[];
for t=t0:dt:tf
ts(end+1)=t;
cam = s;
% rotation matrix R_WCS_CCS
R = calc_Rotation_matrix(cam(4),cam(5),cam(6));
r = cam(1:3);
% 3D coordinates of the 4 points wrt the NEW camera frame
A3D_cam = R'*(A3D-r);
B3D_cam = R'*(B3D-r);
C3D_cam = R'*(C3D-r);
D3D_cam = R'*(D3D-r);
% NEW projections
A=A3D_cam(1:2)/A3D_cam(3);
B=B3D_cam(1:2)/B3D_cam(3);
C=C3D_cam(1:2)/C3D_cam(3);
D=D3D_cam(1:2)/D3D_cam(3);
% computing the L matrices
L1 = L_matrix(A(1),A(2),A3D_cam(3));
L2 = L_matrix(B(1),B(2),B3D_cam(3));
L3 = L_matrix(C(1),C(2),C3D_cam(3));
L4 = L_matrix(D(1),D(2),D3D_cam(3));
L = [L1;L2;L3;L4];
%updating the projection errors
e = [A-Ad;B-Bd;C-Cd;D-Dd];
%compute camera velocity
vc = -0.5*pinv(L)*e;
%change of the camera position and orientation
ds = vc;
%update camera position and orientation
s = s + ds*dt;
end
ts(end+1)=tf+dt;
end
function R = calc_Rotation_matrix(theta_x, theta_y, theta_z)
Rx = [1 0 0; 0 cos(theta_x) -sin(theta_x); 0 sin(theta_x) cos(theta_x)];
Ry = [cos(theta_y) 0 sin(theta_y); 0 1 0; -sin(theta_y) 0 cos(theta_y)];
Rz = [cos(theta_z) -sin(theta_z) 0; sin(theta_z) cos(theta_z) 0; 0 0 1];
R = Rx*Ry*Rz;
end
function L = L_matrix(x,y,z)
L = [-1/z,0,x/z,x*y,-(1+x^2),y;
0,-1/z,y/z,1+y^2,-x*y,-x];
end
有效的案例：
Ad=2*A;
Bd=2*B;
Cd=2*C;
Dd=2*D;
Ad=A+1;
Bd=B+1;
Cd=C+1;
Dd=D+1;
Ad=2*A+1;
Bd=2*B+1;
Cd=2*C+1;
Dd=2*D+1;
不起作用的情况：旋转 90 度并缩小（单独缩小有效，但我在这里这样做是为了更好地可视化）
Ad=2*D;
Bd=2*C;
Cd=2*A;
Dd=2*B;
