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

3 Ansichten (letzte 30 Tage)
Obaid am 14 Okt. 2022
Kommentiert: Ghazwan am 15 Okt. 2022
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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### Akzeptierte Antwort

Ghazwan am 15 Okt. 2022
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
##### 2 KommentareKeine anzeigenKeine ausblenden
Obaid am 15 Okt. 2022
Thanks bro
Ghazwan am 15 Okt. 2022
My pleasure. If my solution is correct, please mark "Accept this answer." Thank you.

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