How to use a 'for loop' with inputs from the output of a previous equation.

Question

Obaid 2022-10-14

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1826743-how-to-use-a-for-loop-with-inputs-from-the-output-of-a-previous-equation

评论： Ghazwan 2022-10-15

Hello, Below is the code for inverse kinematics of a robot.

I wish to use the output of Theta [2x1] in for loop where value in first row and first colums of Theta is used in first iteration and value in 1st row and second column is used in second iteration.

I am new to Matlab.

gd=[-0.0256,-0.4496,-0.8929,-4.197;0.9758,0.1829,-0.1200,15.369;0.2173,-0.8743,0.4340,13.931;0,0,0,1];

gst0=[1 0 0 33; 0 1 0 -13; 0 0 1 14.7; 0 0 0 1];

g1=gd*inv(gst0);

p1=[24;-13;14.7];

p2=[0;-13;14.7];

r=[12;-13;0];

Delta=norm((g1*[p1;1])-[p2;1])

w4=[0;0;-1];

w=w4;

p=p1;

q=p2;

Theta=Thirdsubproblem (w,Delta,p,q,r)

for i=Theta(1,1):Theta(1:2)

w5=[0;0;-1];

w6=[1;0;0];

I=[1,0,0;0,1,0;0,0,1];

q4=[12;-13;14.7];

expOmega4hatTheta4=[cos(i),-sin(i),0;sin(i),cos(i),0;0,0,1];

expPsihat4Theta4=[expOmega4hatTheta4,((I-expOmega4hatTheta4)*q4);0,0,0,1];

p2=[0;-13;14.7];

p3=expPsihat4Theta4*[p2;1];

gd=[-0.0256,-0.4496,-0.8929,-4.197;0.9758,0.1829,-0.1200,15.369;0.2173,-0.8743,0.4340,13.931;0,0,0,1];

gst0=[1 0 0 33; 0 1 0 -13; 0 0 1 14.7; 0 0 0 1];

g2p2=inv(gd)*gst0*[p2;1];

q=p3([1,2,3],:);

p=g2p2([1,2,3],:);

w1=w5;

w2=w6;

r=[24;-13;14.7];

[Gamma]=Secondsubproblem (w1,w2,p,q,r)

for X=Gamma(1,1):Gamma(1:2)

u=p-r;

v=q-r;

alpha=((w1'*w2)*w2'*u-w1'*v)/(((w1'*w2)^2)-1);

Beta=((w1'*w2)*w1'*v-w2'*u)/(((w1'*w2)^2)-1);

z=(alpha*w1)+(Beta*w2)+(X)*(cross(w1,w2));

c=z+r;

Theta2SP2=Firstsubproblemsecond1 (w2,p,z,r)

Theta1SP1=Firstsubproblemsecond2 (w1,q,z,r)

clear

end

clear

end

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Answer 1

Ghazwan 2022-10-15

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1826743-how-to-use-a-for-loop-with-inputs-from-the-output-of-a-previous-equation#answer_1075293

在 MATLAB Online 中打开

gd=[-0.0256,-0.4496,-0.8929,-4.197;0.9758,0.1829,-0.1200,15.369;0.2173,-0.8743,0.4340,13.931;0,0,0,1];
gst0=[1 0 0 33; 0 1 0 -13; 0 0 1 14.7; 0 0 0 1];
g1=gd*inv(gst0);
p1=[24;-13;14.7];
p2=[0;-13;14.7];
r=[12;-13;0];
Delta=norm((g1*[p1;1])-[p2;1])
w4=[0;0;-1];
w=w4;
p=p1;
q=p2;
Theta=Thirdsubproblem(w,Delta,p,q,r)
for ii=1:2
    i=Theta(1,ii);
    w5=[0;0;-1];
    w6=[1;0;0];
    I=[1,0,0;0,1,0;0,0,1];
    q4=[12;-13;14.7];
    expOmega4hatTheta4=[cos(i),-sin(i),0;sin(i),cos(i),0;0,0,1];
    expPsihat4Theta4=[expOmega4hatTheta4,((I-expOmega4hatTheta4)*q4);0,0,0,1];
    p2=[0;-13;14.7];
    p3=expPsihat4Theta4*[p2;1];
    gd=[-0.0256,-0.4496,-0.8929,-4.197;0.9758,0.1829,-0.1200,15.369;0.2173,-0.8743,0.4340,13.931;0,0,0,1];
    gst0=[1 0 0 33; 0 1 0 -13; 0 0 1 14.7; 0 0 0 1];
    g2p2=inv(gd)*gst0*[p2;1];
    q=p3([1,2,3],:);
    p=g2p2([1,2,3],:);
    w1=w5;
    w2=w6;
    r=[24;-13;14.7];
    [Gamma]=Secondsubproblem (w1,w2,p,q,r)
    for jj=1:2
        X=Gamma(1,jj);
        u=p-r;
        v=q-r;
        alpha=((w1'*w2)*w2'*u-w1'*v)/(((w1'*w2)^2)-1);
        Beta=((w1'*w2)*w1'*v-w2'*u)/(((w1'*w2)^2)-1);
        z=(alpha*w1)+(Beta*w2)+(X)*(cross(w1,w2));
        c=z+r;
        Theta2SP2=Firstsubproblemsecond1 (w2,p,z,r)
        Theta1SP1=Firstsubproblemsecond2 (w1,q,z,r)
        clear
    end
    clear
end