I'm tryng to calculate some number and matrix with my robotics project and facing this problem when run the code. Can anyone explain for me, i'm just a newbie. Thank you very much. The error is after the code.

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Anh Tong
Anh Tong 2018-12-10
编辑： madhan ravi 2018-12-10
clc
syms q1 q2 q3 dq1 dq2 dq3 a1 a2 m1 m2 m3 t g f
assume(a1, 'real');assume(a1>0);
assume(a2,'real');assume(a2>0);
assume(q3,'real');
assume(dq3,'real');
assume(q1,'real');
assume(q2,'real');
assume(dq1,'real');
assume(dq2,'real');
assume(t,'real');assume(t>0);
assume(f,'real');
q = [q1;q2;q3]; %vecto cac toa do suy rong
dq = [dq1;dq2;dq3]; %vecto cac dao ham toa do suy rong
%Nhap cac ma tran D-H
A_01=[ cos(q1) -sin(q1) 0 a1*cos(q1);sin(q1) cos(q1) 0 a1*sin(q1);0 0 1 0;0 0 0 1;];
A_12=[cos(q2) -sin(q2) 0 a2*cos(q2);sin(q2) cos(q2) 0 a2*sin(q2);0 0 1 0;0 0 0 1;];
A_23=[1 0 0 0;0 -1 0 0;0 0 -1 -q3;0 0 0 1;];
% Tính các ma tran cosin chi phuong
R_01=A_01(1:3,1:3);
A_03=A_01*A_12*A_23;
A_02=simplify(A_01*A_12);disp(A_02); % Sau khi tính toán ta thu gon ket qua ngay bang lenh simplify
R_02=A_02(1:3,1:3);
disp('Ma tran bien doi thuan nhat khau cuoi la');
A_03 = simplify(A_03);
R_03 = A_03(1:3,1:3);
disp(A_03);
%Giai bai toan dong hoc thuan
disp('Giai bai toan dong hoc thuan');
disp('toa do diem cuoi');
rE = A_03(1:3,4); disp(rE);% Vector toa do khâu thao tác
disp('van toc diem cuoi')
v_qE = simplify(jacobian(rE,q)*dq);disp(v_qE); % Tính vector van toc khâu tác dong cuoi
jacobian(rE,q); disp('Jacobi tinh tien'); disp(jacobian(rE,q));
R_0E = A_03(1:3,1:3); disp(R_0E);
diff_R_0E = diff(R_0E,q1)*dq1+diff(R_0E,q2)*dq2; %Tinh dao ham cua R
omega_curveE = diff_R_0E*R_0E.'; %omega song khau cuoi
omega_curveE = simplify(omega_curveE);
disp('Van toc goc khau cuoi:');
omega3 = [omega_curveE(3,2) omega_curveE(1,3) omega_curveE(2,1)]; disp(omega3); %omega
jacobian(omega3,dq); disp('Jacobi quay khau cuoi'); disp(jacobian(omega3,dq));
R_01 = A_01(1:3,1:3); disp(R_01);
diff_R_01 = diff(R_01,q1)*dq1+diff(R_01,q2)*dq2;
omega_curve1 = diff_R_01*R_01.';
omega_curve1 = simplify(omega_curve1);
disp('Van toc goc khau thu nhat');
omega1 = [omega_curve1(3,2) omega_curve1(1,3) omega_curve1(2,1)]; disp(omega1); %omega
jacobian(omega1,dq); disp('Jacobi quay khau thu nhat'); disp(jacobian(omega1,dq));
R_02 = A_02(1:3,1:3); disp(R_02);
diff_R_02 = diff(R_02,q1)*dq1+diff(R_02,q2)*dq2;
omega_curve2 = diff_R_02*R_02.';
omega_curve2 = simplify(omega_curve2);
disp('Van toc goc khau thu hai');
omega2 = [omega_curve2(3,2) omega_curve2(1,3) omega_curve2(2,1)]; disp(omega2); %omega
jacobian(omega2,dq); disp('Jacobi quay khau thu hai'); disp(jacobian(omega2,dq));
%tinh gia toc diem cuoi %chua tinh dc
diff_v_qE = diff(v_qE,q1)*dq1 + diff(v_qE,q2)*dq2;
disp('gia toc diem cuoi la:'); disp(diff_v_qE);
%tinh gia toc goc diem cuoi
%diff_omega = diff(omega,q1)*dq1 + diff(omega,q2)*dq2;
%disp('gia toc goc khau thao tac'); disp(diff_omega);
%khao sat can bang tinh
F_3E = [0 0 f]; % luc gia cong tai diem cuoi so voi he 3
F_0E = F_3E*R_03;
P_01 = [0 0 -m1*g]; %trong luc tac dung len C1
P_02 = [0 0 -m2*g];
P_03 = [0 0 -m3*g];
r_0C1 = [(a1/2)*cos(q1) (a1/2)*sin(q1) 0]; %toa do C1 so voi he 0
r_1C2 = [(a2/2)*cos(q2) (a2/2)*sin(q2) 0];
r_0C2 = r_1C2*R_01;
r_2C3 = [-q3/2 0 0];
r_0C3 = r_2C3*R_02;
J_TE = jacobian(rE,q);
J_TC1= jacobian(r_0C1,q);
J_TC2= jacobian(r_0C2,q);
J_TC3= jacobian(r_0C3,q);
%luc dong co
Tdc = -J_TE.'*F_0E - P1.'*J_TC1 - P2.'*J_TC2 - P3.'*J_TC3;
disp('luc dong co co gia tri la'); disp(Tdc);
F_0_32 = F_0E - P_03; disp('luc tac dung tu khau 2 len khau 3'); disp(F_0_32);
F_0_21 = F_0_32 - P_02; disp('luc tac dung tu khau 1 len khau 2'); disp(F_0_21);
F_0_10 = F_0_21 - P_01; disp('luc tac dung tu khau 0 len khau 1'); disp(F_0_10);
Csym = mupadmex(op,args{1}.s, args{2}.s, varargin{:});
Error in * (line 244)
X = privBinaryOp(A, B, 'symobj::mtimes');
Error in ltmp (line 97)
Tdc = -J_TE.'*F_0E - P1.'*J_TC1 - P2.'*J_TC2 - P3.'*J_TC3;

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采纳的回答

madhan ravi
madhan ravi 2018-12-10
编辑：madhan ravi 2018-12-10
EDITED
Rules of matrix multiplication 3 by 3 times 3 by 1 but what you did was you transposed J_TE which is size 3 by 3 is also 3 by 3 after and you multiply it with 1 by 3(size of F_0E) which contradicts the rule.
Tdc = -J_TE*F_0E.' - P1*J_TC1 - P2*J_TC2 - P3*J_TC3; %change to this line
^---- transposed
Note : P1 ,P2 & P3 is not defined , I suspect they should be P_01 , P_02 & P_03
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madhan ravi
madhan ravi 2018-12-10
编辑：madhan ravi 2018-12-10
You don't have to ask the same question multiple times just use size() to check the variable sizes, run the below code:
clc
syms q1 q2 q3 dq1 dq2 dq3 a1 a2 m1 m2 m3 t g f
assume(a1, 'real');assume(a1>0);
assume(a2,'real');assume(a2>0);
assume(q3,'real');
assume(dq3,'real');
assume(q1,'real');
assume(q2,'real');
assume(dq1,'real');
assume(dq2,'real');
assume(t,'real');assume(t>0);
assume(f,'real');
q = [q1;q2;q3]; %vecto cac toa do suy rong
dq = [dq1;dq2;dq3]; %vecto cac dao ham toa do suy rong
%Nhap cac ma tran D-H
A_01=[ cos(q1) -sin(q1) 0 a1*cos(q1);sin(q1) cos(q1) 0 a1*sin(q1);0 0 1 0;0 0 0 1;];
A_12=[cos(q2) -sin(q2) 0 a2*cos(q2);sin(q2) cos(q2) 0 a2*sin(q2);0 0 1 0;0 0 0 1;];
A_23=[1 0 0 0;0 -1 0 0;0 0 -1 -q3;0 0 0 1;];
% Tính các ma tran cosin chi phuong
R_01=A_01(1:3,1:3);
A_03=A_01*A_12*A_23;
A_02=simplify(A_01*A_12);disp(A_02); % Sau khi tính toán ta thu gon ket qua ngay bang lenh simplify
R_02=A_02(1:3,1:3);
disp('Ma tran bien doi thuan nhat khau cuoi la');
A_03 = simplify(A_03);
R_03 = A_03(1:3,1:3);
disp(A_03);
%Giai bai toan dong hoc thuan
disp('Giai bai toan dong hoc thuan');
disp('toa do diem cuoi');
rE = A_03(1:3,4); disp(rE);% Vector toa do khâu thao tác
disp('van toc diem cuoi')
v_qE = simplify(jacobian(rE,q)*dq);disp(v_qE); % Tính vector van toc khâu tác dong cuoi
jacobian(rE,q); disp('Jacobi tinh tien'); disp(jacobian(rE,q));
R_0E = A_03(1:3,1:3); disp(R_0E);
diff_R_0E = diff(R_0E,q1)*dq1+diff(R_0E,q2)*dq2; %Tinh dao ham cua R
omega_curveE = diff_R_0E*R_0E.'; %omega song khau cuoi
omega_curveE = simplify(omega_curveE);
disp('Van toc goc khau cuoi:');
omega3 = [omega_curveE(3,2) omega_curveE(1,3) omega_curveE(2,1)]; disp(omega3); %omega
jacobian(omega3,dq); disp('Jacobi quay khau cuoi'); disp(jacobian(omega3,dq));
R_01 = A_01(1:3,1:3); disp(R_01);
diff_R_01 = diff(R_01,q1)*dq1+diff(R_01,q2)*dq2;
omega_curve1 = diff_R_01*R_01.';
omega_curve1 = simplify(omega_curve1);
disp('Van toc goc khau thu nhat');
omega1 = [omega_curve1(3,2) omega_curve1(1,3) omega_curve1(2,1)]; disp(omega1); %omega
jacobian(omega1,dq); disp('Jacobi quay khau thu nhat'); disp(jacobian(omega1,dq));
R_02 = A_02(1:3,1:3); disp(R_02);
diff_R_02 = diff(R_02,q1)*dq1+diff(R_02,q2)*dq2;
omega_curve2 = diff_R_02*R_02.';
omega_curve2 = simplify(omega_curve2);
disp('Van toc goc khau thu hai');
omega2 = [omega_curve2(3,2) omega_curve2(1,3) omega_curve2(2,1)]; disp(omega2); %omega
jacobian(omega2,dq); disp('Jacobi quay khau thu hai'); disp(jacobian(omega2,dq));
%tinh gia toc diem cuoi %chua tinh dc
diff_v_qE = diff(v_qE,q1)*dq1 + diff(v_qE,q2)*dq2;
disp('gia toc diem cuoi la:'); disp(diff_v_qE);
%tinh gia toc goc diem cuoi
%diff_omega = diff(omega,q1)*dq1 + diff(omega,q2)*dq2;
%disp('gia toc goc khau thao tac'); disp(diff_omega);
%khao sat can bang tinh
F_3E = [0;0;f]; % luc gia cong tai diem cuoi so voi he 3
F_0E = F_3E.'*R_03;
P_01 = [0 0 -m1*g]; %trong luc tac dung len C1
P_02 = [0 0 -m2*g];
P_03 = [0 0 -m3*g];
r_0C1 = [(a1/2)*cos(q1) (a1/2)*sin(q1) 0]; %toa do C1 so voi he 0
r_1C2 = [(a2/2)*cos(q2) (a2/2)*sin(q2) 0];
r_0C2 = r_1C2*R_01;
r_2C3 = [-q3/2 0 0];
r_0C3 = r_2C3*R_02;
J_TE = jacobian(rE,q);
J_TC1= jacobian(r_0C1,q);
J_TC2= jacobian(r_0C2,q);
J_TC3= jacobian(r_0C3,q);
%luc dong co
Tdc = -J_TE*F_0E.' - P_01*J_TC1 - P_02*J_TC2 - P_03*J_TC3;
disp('luc dong co co gia tri la'); disp(Tdc);
F_0_32 = F_0E - P_03; disp('luc tac dung tu khau 2 len khau 3'); disp(F_0_32);
F_0_21 = F_0_32 - P_02; disp('luc tac dung tu khau 1 len khau 2'); disp(F_0_21);
F_0_10 = F_0_21 - P_01; disp('luc tac dung tu khau 0 len khau 1'); disp(F_0_10);

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