matlab练习程序（对应点集配准的四元数法）

clear all;
close all;
clc;

%生成原始点集
X=[];Y=[];Z=[];
for i=-180:2:180
    for j=-90:2:90
        x = i * pi / 180.0;
        y = j * pi / 180.0;   
        X =[X,cos(y) * cos(x)];
        Y =[Y,sin(y) * cos(x)];
        Z =[Z,sin(x)]; 
    end
end
P=[X(1:3000)‘ Y(1:3000)‘ Z(1:3000)‘];

%生成变换后点集
i=0.5;j=0.3;k=0.7;
Rx=[1 0 0;0 cos(i) -sin(i); 0 sin(i) cos(i)];
Ry=[cos(j) 0 sin(j);0 1 0;-sin(j) 0 cos(j)];
Rz=[cos(k) -sin(k) 0;sin(k) cos(k) 0;0 0 1];
R=Rx*Ry*Rz;
X=P*R + [0.2,0.3,0.4];

plot3(P(:,1),P(:,2),P(:,3),‘b.‘);
hold on;
plot3(X(:,1),X(:,2),X(:,3),‘r.‘);

%计算点集均值
up = mean(P);
ux = mean(X);

P1=P-up;
X1=X-ux;

%计算点集协方差
sigma=P1‘*X1/(length(X1));
sigma_mi = sigma - sigma‘;
M=sigma+sigma‘-trace(sigma)*[1,0,0;0,1,0;0,0,1];

%由协方差构造4*4对称矩阵
Q=[trace(sigma) sigma_mi(2,3) sigma_mi(3,1) sigma_mi(1,2);
   sigma_mi(2,3) M(1,1) M(1,2) M(1,3);
   sigma_mi(3,1) M(2,1) M(2,2) M(2,3);
   sigma_mi(1,2) M(3,1) M(3,2) M(3,3)];

%计算特征值与特征向量
[x,y] = eig(Q);
e = diag(y);

%计算最大特征值对应的特征向量
lamda=max(e);
for i=1:length(Q)
    if lamda==e(i)
        break;
    end
end
q=x(:,i);

q0=q(1);q1=q(2);q2=q(3);q3=q(4);

%由四元数构造旋转矩阵
RR=[q0^2+q1^2-q2^2-q3^2 ,2*(q1*q2-q0*q3), 2*(q1*q3+q0*q2);
   2*(q1*q2+q0*q3), q0^2-q1^2+q2^2-q3^2, 2*(q2*q3-q0*q1);
   2*(q1*q3-q0*q2), 2*(q2*q3+q0*q1), q0^2-q1^2-q2^2+q3^2];

%计算平移向量
qr=ux-up*RR‘;

%验证旋转矩阵与平移向量正确性
Pre = P*RR‘+qr;

figure;
plot3(P(:,1),P(:,2),P(:,3),‘b.‘);
hold on;
plot3(X(:,1),X(:,2),X(:,3),‘r.‘);

plot3(Pre(:,1),Pre(:,2),Pre(:,3),‘go‘);