這里只羅列出可以用于手眼標(biāo)定（相機(jī)坐標(biāo)系和機(jī)械臂基坐標(biāo)系之間的變換矩陣，3X4的。我的項(xiàng)目是eye to hand,所以得到的結(jié)果對(duì)于eye in hand怎么用還不太清楚）

原文鏈接：http://math.loyola.edu/~mili/Calibration/index.html

分為三種情況：

第一種：

你只想用幾對(duì)同源點(diǎn)的空間三維位置坐標(biāo)作為輸入數(shù)據(jù)，得到變換矩陣。

方法1：論文：K.S. Arun, T.S. Huang, S.D. Blostein, Least-squares fitting of two 3-d point sets, IEEE Trans. Pattern Anal. Mach. Intell. 9 (5) (1987) 698-700

matlab代碼：

function [R,t] = arun(A,B) % Registers two sets of 3DoF data % Assumes A and B are d,n sets of data % where d is the dimension of the system % typically d = 2,3 % and n is the number of points % typically n>3 % % Mili Shah % July 2014[d,n]=size(A);%Mean Center Data Ac = mean(A')'; Bc = mean(B')'; A = A-repmat(Ac,1,n); B = B-repmat(Bc,1,n);%Calculate Optimal Rotation [u,s,v]=svd(A*B'); R = v*u'; if det(R)<0, disp('Warning: R is a reflection'); end%Calculate Optimal Translation t = Bc - R*Ac;

方法2：論文：B.K.P. Horn, H.M. Hilden, S. Negahdaripour, Closed-form solution of absolute orientation using orthonormal matrices, J. Opt. Soc. Am. A 5 (1988) 1127-1135

代碼：

function [R,t] = horn(A,B) % Registers two sets of 3DoF data % Assumes A and B are d,n sets of data % where d is the dimension of the system % typically d = 2,3 % and n is the number of points % typically n>3 % % Mili Shah % July 2014[d,n]=size(A);%Mean Center Data Ac = mean(A')'; Bc = mean(B')'; A = A-repmat(Ac,1,n); B = B-repmat(Bc,1,n);%Calculate Optimal Rotation M = B*A'; R = M*(M'*M)^(-1/2); if det(R)<0, disp('Warning: R is a reflection'); end%Calculate Optimal Translation t = Bc - R*Ac; end

方法3：

• B.K.P. Horn, Closed-form solution of absolute orientation using unit quaternions, J. Opt. Soc. Am. A 4 (1987) 629-642.

function [R,t] = hornQuar(A,B) % Registers two sets of 3DoF data % Assumes A and B are d,n sets of data % where d is the dimension of the system % typically d = 2,3 % and n is the number of points % typically n>3 % % Mili Shah % July 2014[d,n]=size(A);%Mean Center Data Ac = mean(A')'; Bc = mean(B')'; A = A-repmat(Ac,1,n); B = B-repmat(Bc,1,n);%Calculate Optimal Rotation M = A*B'; N = [M(1,1) + M(2,2) + M(3,3) M(2,3)-M(3,2) M(3,1)-M(1,3) M(1,2)-M(2,1);...M(2,3)-M(3,2) M(1,1)-M(2,2)-M(3,3) M(1,2)+M(2,1) M(3,1)+M(1,3);...M(3,1)-M(1,3) M(1,2)+M(2,1) -M(1,1)+M(2,2)-M(3,3) M(2,3)+M(3,2);...M(1,2)-M(2,1) M(3,1)+M(1,3) M(2,3)+M(3,2) -M(1,1)-M(2,2)+M(3,3)]; [u,v]=eig(N); [m,ind]=max(diag(v)); r = u(:,ind); r = [r(2:4);r(1)]; R = (r(4)^2-r(1:3)'*r(1:3))*eye(3) + 2*r(1:3)*r(1:3)'+2*r(4)*[0 -r(3) r(2);r(3) 0 -r(1);-r(2) r(1) 0]; %Calculate Optimal Translation t = Bc - R*Ac; end

方法4：M.W. Walker, L. Shao, R.A. Volz, Estimating 3-d location parameters using dual number quaternions, CVGIP: Image Underst. 54 (3) (1991) 358-367.

function [R,t] = walker(A,B) % Registers two sets of 3DoF data % Assumes A and B are 3,n sets of data % and n is the number of points % typically n>3 % % Mili Shah % July 2014[d,n]=size(A);C1 = zeros(4,4); C2 = n*eye(4); C3 = zeros(4,4); for i = 1:nC1 = C1 + -2*Q([1/2*B(:,i);0])'*W([1/2*A(:,i);0]);C3 = C3 + 2*(W([1/2*A(:,i);0])-Q([1/2*B(:,i);0])); end A = 1/2*(C3'*inv(C2+C2')*C3-C1-C1'); [u,v]=eig(A); [m,ind]=max(diag(v)); r = u(:,ind); s = -inv(C2+C2')*C3*r; R = (r(4)^2-r(1:3)'*r(1:3))*eye(3) + 2*r(1:3)*r(1:3)'+2*r(4)*[0 -r(3) r(2);r(3) 0 -r(1);-r(2) r(1) 0]; t = W(r)'*s*2; t = t(1:3); endfunction Q = Q(r) K = [0 -r(3) r(2);r(3) 0 -r(1);-r(2) r(1) 0]; Q = [r(4)*eye(3)+K r(1:3);-r(1:3)' r(4)]; end function W = W(r) K = [0 -r(3) r(2);r(3) 0 -r(1);-r(2) r(1) 0]; W = [r(4)*eye(3)-K r(1:3);-r(1:3)' r(4)]; end

方法5：S. Umeyama, Least-squares estimation of transformation parameters between two point patterns, IEEE Trans. Pattern Anal. Mach. Intell. 13 (4) (1991) 376- 380.

?

function [R,t] = umeyama(A,B) % Registers two sets of 3DoF data % Assumes A and B are d,n sets of data % where d is the dimension of the system % typically d = 2,3 % and n is the number of points % typically n>3 % % Mili Shah % July 2014[d,n]=size(A);%Mean Center Data Ac = mean(A')'; Bc = mean(B')'; A = A-repmat(Ac,1,n); B = B-repmat(Bc,1,n);%Calculate Optimal Rotation [u,s,v]=svd(A*B'); R = v*diag([1 1 det(v*u')])*u';%Calculate Optimal Translation t = Bc - R*Ac;

第二種情況：輸入的數(shù)據(jù)為位置和姿態(tài)，解方程AX=XB

方法1：

• Y. Shiu, S. Ahmad Calibration of Wrist-Mounted Robotic Sensors by Solving Homogeneous Transform Equations of the Form AX = XB. In IEEE Transactions on Robotics and Automation, 5(1):16-29, 1989.

function X = shiu(A,B,X) % Calculates the least squares solution of % AX = XB % From % Calibration of Wrist-Mounted Robotic Sensors % by Solving Homogeneous Transform Equations of the Form AX = XB % Shiu and Ahmad % % Mili Shah % July 2014[m,n] = size(A); n = n/4; AA = zeros(9*(n-1),2*n); bb = zeros(9*(n-1),1); %Calculate best rotation R for i = 1:nA1 = logm(A(1:3,4*i-3:4*i-1));B1 = logm(B(1:3,4*i-3:4*i-1));a1 = [A1(3,2) A1(1,3) A1(2,1)]'; a1 = a1/norm(a1);b1 = [B1(3,2) B1(1,3) B1(2,1)]'; b1 = b1/norm(b1);v = cross(b1,a1);w = atan2(norm(v),b1'*a1);v = v/norm(v);XP = (eye(3)*cos(w) + sin(w)*skew(v) + (1-cos(w))*v*v');[Ai,bi] = shiuMatrix(a1,XP);if i == 1AA(:,1:2) = repmat(-Ai,n-1,1);bb(:,1) = repmat(-bi,n-1,1);elseAA(9*(i-2)+1:9*(i-1),2*i-1:2*i) = Ai;bb(9*(i-2)+1:9*(i-1),1) = bb(9*(i-2)+1:9*(i-1),1) + bi;end end beta = AA\bb; theta = atan2(beta(2*n),beta(2*n-1)); RA = (eye(3)*cos(theta) + sin(theta)*skew(a1) + (1-cos(theta))*a1*a1'); R = RA*XP; %Calculate best translation t C = zeros(3*n,3); d = zeros(3*n,1); I = eye(3); for i = 1:nC(3*i-2:3*i,:) = I - A(1:3,4*i-3:4*i-1);d(3*i-2:3*i,:) = A(1:3,4*i)-R*B(1:3,4*i); end t = C\d; %Put everything together to form X X = [R t;0 0 0 1]; endfunction [A,b] = shiuMatrix(ka1,X) A = zeros(9,2); b = zeros(9,1);A(1,1) = X(1,1)-ka1(1)*X(:,1)'*ka1; A(2,1) = X(1,2)-ka1(1)*X(:,2)'*ka1; A(3,1) = X(1,3)-ka1(1)*X(:,3)'*ka1;A(4,1) = X(2,1)-ka1(2)*X(:,1)'*ka1; A(5,1) = X(2,2)-ka1(2)*X(:,2)'*ka1; A(6,1) = X(2,3)-ka1(2)*X(:,3)'*ka1;A(7,1) = X(3,1)-ka1(3)*X(:,1)'*ka1; A(8,1) = X(3,2)-ka1(3)*X(:,2)'*ka1; A(9,1) = X(3,3)-ka1(3)*X(:,3)'*ka1;n = cross(X(:,1),ka1); o = cross(X(:,2),ka1); a = cross(X(:,3),ka1);A(1,2) = -n(1); A(2,2) = -o(1); A(3,2) = -a(1);A(4,2) = -n(2); A(5,2) = -o(2); A(6,2) = -a(2);A(7,2) = -n(3); A(8,2) = -o(3); A(9,2) = -a(3);n = X(:,1); o = X(:,2); a = X(:,3);b(1) = -ka1(1)*n'*ka1; b(2) = -ka1(1)*o'*ka1; b(3) = -ka1(1)*a'*ka1;b(4) = -ka1(2)*n'*ka1; b(5) = -ka1(2)*o'*ka1; b(6) = -ka1(2)*a'*ka1;b(7) = -ka1(3)*n'*ka1; b(8) = -ka1(3)*o'*ka1; b(9) = -ka1(3)*a'*ka1; end

方法2：

• R. Tsai, R. Lenz A New Technique for Fully Autonomous and Efficient 3D Robotics Hand/Eye Calibration. In IEEE Transactions on Robotics and Automation, 5(3):345-358, 1989.

?

function X = tsai(A,B) % Calculates the least squares solution of % AX = XB % % A New Technique for Fully Autonomous % and Efficient 3D Robotics Hand/Eye Calibration % Lenz Tsai % % Mili Shah % July 2014[m,n] = size(A); n = n/4; S = zeros(3*n,3); v = zeros(3*n,1); %Calculate best rotation R for i = 1:nA1 = logm(A(1:3,4*i-3:4*i-1)); B1 = logm(B(1:3,4*i-3:4*i-1));a = [A1(3,2) A1(1,3) A1(2,1)]'; a = a/norm(a);b = [B1(3,2) B1(1,3) B1(2,1)]'; b = b/norm(b);S(3*i-2:3*i,:) = skew(a+b);v(3*i-2:3*i,:) = a-b; end x = S\v; theta = 2*atan(norm(x)); x = x/norm(x); R = (eye(3)*cos(theta) + sin(theta)*skew(x) + (1-cos(theta))*x*x')'; %Calculate best translation t C = zeros(3*n,3); d = zeros(3*n,1); I = eye(3); for i = 1:nC(3*i-2:3*i,:) = I - A(1:3,4*i-3:4*i-1);d(3*i-2:3*i,:) = A(1:3,4*i)-R*B(1:3,4*i); end t = C\d; %Put everything together to form X X = [R t;0 0 0 1];

方法3：

• C. Wang Extrinsic Calibration of a Vision Sensor Mounted on a Robot. In IEEE Transactions on Robotics and Automation, 8(2): 161-175, 1992.

function X = wang(A,B) % Calculates the least squares solution of % AX = XB % From % Extrinsic Calibration of a Vision Sensor % Mounted on a Robot. % C. Wang % % Mili Shah % July 2014[m,n] = size(A); n = n/4; K = zeros(3,n-1); theta = zeros(1,n-1);%Calculate best rotation R for i = 1:n-1A1 = logm(A(1:3,4*i-3:4*i-1)); B1 = logm(B(1:3,4*i-3:4*i-1));A2 = logm(A(1:3,4*(i+1)-3:4*(i+1)-1)); B2 = logm(B(1:3,4*(i+1)-3:4*(i+1)-1));a1 = [A1(3,2) A1(1,3) A1(2,1)]'; a1 = a1/norm(a1);b1 = [B1(3,2) B1(1,3) B1(2,1)]'; b1 = b1/norm(b1);a2 = [A2(3,2) A2(1,3) A2(2,1)]'; a2 = a2/norm(a2);b2 = [B2(3,2) B2(1,3) B2(2,1)]'; b2 = b2/norm(b2);if abs(a1'*b1)<.999 && abs(a2'*b2)<.999 && norm(cross((a1-b1),(a2-b2)))~=0r = cross((a1-b1),(a2-b2)); r = r/norm(r);theta(i) = sign((a1-b1)'*cross(r,a1+b1))*atan(norm(a1-b1)/norm(cross(r,a1+b1)))+...sign((a2-b2)'*cross(r,a2+b2))*atan(norm(a2-b2)/norm(cross(r,a2+b2)));K(:,i) = r;elseif abs(a1'*b1)<.999 && abs(a2'*b2)<.999 && norm(cross((a1-b1),(a2-b2)))==0r = -cross(cross(a1,a2),cross(b1,b2)); r = r/norm(r);theta(i) = acos(cross(a1,a2)'*cross(b1,b2));K(:,i) = r;elseif abs(a1'*b1)>.999 && abs(a2'*b2)<.999 r = (a1+b1)/2; r = r/norm(r);theta(i) = 2*sign((a2-b2)'*cross(r,a2+b2))*atan(norm(a2-b2)/norm(cross(r,a2+b2)));K(:,i) = r;elseif abs(a1'*b1)<.999 && abs(a2'*b2)>.999 r = (a2+b2)/2; r = r/norm(r);theta(i) = 2*sign((a1-b1)'*cross(r,a1+b1))*atan(norm(a1-b1)/norm(cross(r,a1+b1)));K(:,i) = r;elser = [1;0;0];theta(i) = 0;K(:,i) = r;end end K = [K;theta]; r = mean(K')'; sc = norm(r(1:3)); r = r/sc; R = eye(3)*cos(r(4)) + (1-cos(r(4)))*r(1:3)*r(1:3)' + skew(r(1:3))*sin(r(4)); %Calculate best translation t C = zeros(3*n,3); d = zeros(3*n,1); I = eye(3); for i = 1:nC(3*i-2:3*i,:) = I - A(1:3,4*i-3:4*i-1);d(3*i-2:3*i,:) = A(1:3,4*i)-R*B(1:3,4*i); end t = C\d; %Put everything together to form X X = [R t;0 0 0 1];

方法4：

• R. Liang, J. Mao Hand-Eye Calibration with a New Linear Decomposition Algorithm. In Journal of Zhejiang University, 9(10):1363-1368, 2008.

?

function [X]=liang(AA,BB) % Solves the problem AX=XB % using the formulation of % % Hand-Eye Calibration with a % New Linear Decomposition Algorithm % R. Liang, J. Mao % % Mili Shah % July 2014[m,n]=size(AA); n = n/4;A = zeros(9*n,9); b = zeros(9*n,1); for i = 1:nRa = AA(1:3,4*i-3:4*i-1);Rb = BB(1:3,4*i-3:4*i-1);A(9*i-8:9*i,:) = [kron(Ra,eye(3))+kron(-eye(3),Rb')]; end [u,s,v]=svd(A); x = v(:,end); R=reshape(x(1:9),3,3)'; R = sign(det(R))/abs(det(R))^(1/3)*R; [u,s,v]=svd(R); R = u*v'; if det(R)<0, R = u*diag([1 1 -1])*v'; end C = zeros(3*n,3); d = zeros(3*n,1); I = eye(3); for i = 1:nC(3*i-2:3*i,:) = I - AA(1:3,4*i-3:4*i-1);d(3*i-2:3*i,:) = AA(1:3,4*i)-R*BB(1:3,4*i); end t = C\d; %Put everything together to form X X = [R t;0 0 0 1];

?第三種情況：輸入數(shù)據(jù)為位置和姿態(tài)，求解方程AX=YB

方法1：

• A. Li, L. Wang, D. Wu Simultaneous robot-world and hand-eye calibration using dual-quaternions and kronecker product. InInternational Journal of the Physical Sciences, 5(10): 1530-1536, 1995.

?

function [X,Y]=li(AA,BB) % Solves the problem AX=YB % using the formulation of % % Simultaneous robot-world and hand-eye calibration % using dual-quaternions and Kronecker product % % Aiguo Li, Lin Wang and Defeng Wu % % Mili Shah % July 2014[m,n]=size(AA); n = n/4;A = zeros(12*n,24); b = zeros(12*n,1); for i = 1:nRa = AA(1:3,4*i-3:4*i-1);Rb = BB(1:3,4*i-3:4*i-1);ta = AA(1:3,4*i);tb = BB(1:3,4*i);A(12*i-11:12*i-3,1:18) = [kron(Ra,eye(3)) kron(-eye(3),Rb')];A(12*i-2:12*i,10:24) = [kron(eye(3),tb') -Ra eye(3)];b(12*i-2:12*i) = ta; end x = A\b;X=reshape(x(1:9),3,3)'; [u,s,v]=svd(X); X = u*v'; if det(X)<0, X = u*diag([1 1 -1])*v'; end X = [X x(19:21);[0 0 0 1]]; Y=reshape(x(10:18),3,3)'; [u,s,v]=svd(Y); Y = u*v'; if det(Y)<0, Y = u*diag([1 1 -1])*v'; end Y = [Y x(22:24);[0 0 0 1]];

方法2：

• M. Shah, Solving the Robot-World/Hand-Eye Calibration Problem Using the Kronecker Product, ASME Journal of Mechanisms and Robotics, Volume 5, Issue 3 (2013).

function [X,Y]=shah(AA,BB) % Solves the problem AX=YB % using the formulation of % % Simultaneous Robot/World and Tool/Flange % Calibration by Solving Homogeneous Transformation % Equations of the form AX=YB % M. Shah % % Mili Shah % July 2014[m,n]=size(AA); n = n/4; A = zeros(9*n,18); T = zeros(9,9); b = zeros(9*n,1); for i = 1:nRa = AA(1:3,4*i-3:4*i-1);Rb = BB(1:3,4*i-3:4*i-1);T = T + kron(Rb,Ra); end [u,s,v]=svd(T); x = v(:,1); y = u(:,1); X = reshape(x,3,3); X = sign(det(X))/abs(det(X))^(1/3)*X; [u,s,v]=svd(X); X = u*v'; Y = reshape(y,3,3); Y = sign(det(Y))/abs(det(Y))^(1/3)*Y; [u,s,v]=svd(Y); Y = u*v'; A = zeros(3*n,6); b = zeros(3*n,1); for i = 1:nA(3*i-2:3*i,:) = [-AA(1:3,4*i-3:4*i-1) eye(3)];b(3*i-2:3*i,:) = AA(1:3,4*i) - kron(BB(1:3,4*i)',eye(3))*reshape(Y,9,1); end t = A\b;X = [X t(1:3);[0 0 0 1]]; Y = [Y t(4:6);[0 0 0 1]];

?

總結(jié)

以上是生活随笔為你收集整理的matlab手眼标定的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。