For loop and calculating at each interval of t

Question

Andre Khosrow 2023-4-15

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1947793-for-loop-and-calculating-at-each-interval-of-t

评论： Image Analyst 2023-4-16

I just want to know how can I rewrite the code so that the for loop will include a correct dhparams. Since I have 26 values of theta1,theta2,theta3 how can I have my dhparams have the first value and undergo each calculation with the intervals as follows. I want to avoid cell arrays errors ;

->When t = 0

theta1 = -2, theta2= 2, theta 3= 3

theta4= -2, theta5= 2, theta6= 3

Therefore for t=0;

dhparams = {-2,0,145,0;
            2+pi/2,pi/2,145,0;
            3+pi/2,0,0,260;
            -2,pi/2,40,30;
            2,-pi/2,230,0;
            3,pi/2,90,0};

Then this point would be plotted in the axis.

->Then I want this to happen with the next interval at t=0.2;

theta1 = -2.98, theta2= 1.984, theta 3= 2.88

theta4= -2.98, theta5= 1.984, theta6= 2.88

Therefore for t=0.2;

dhparams = {-2.98,0,145,0;
            1.984+pi/2,pi/2,145,0;
            2.88+pi/2,0,0,260;
            -2.98,pi/2,40,30;
            1.984,-pi/2,230,0;
            2.88,pi/2,90,0};

Then this point would be plotted in the axis.

I want this to contiune until t=5 and the figure to have the complete plot.

This is the full code.

theta1 = zeros(1,26)         ;
theta2 = zeros(1,26)         ;
theta3= zeros(1,26)          ;
t = 0:0.2:5                  ;
for k= 1:26            
    theta1(k) = -2*(1-0.1*(k));  
    theta2(k) = 2*(1-0.04*(k));
    theta3(k) = 3*(1-0.2.*(k));
    theta4=theta1(k);
    theta5=theta2(k);
    theta6=theta3(k);
end  
dhparams = {theta1,0,145,0;
            theta2+pi/2,pi/2,145,0;
            theta3+pi/2,0,0,260;
            theta4,pi/2,40,30;
            theta5,-pi/2,230,0;
            theta6,pi/2,90,0};
robot = rigidBodyTree;
body1 = rigidBody('body1');
jnt1 = rigidBodyJoint('jnt1','revolute');
setFixedTransform(jnt1,dhparams(1,:),'dh');
body1.Joint = jnt1;
addBody(robot,body1,'base')
body2 = rigidBody('body2');
jnt2 = rigidBodyJoint('jnt2','revolute');
body3 = rigidBody('body3');
jnt3 = rigidBodyJoint('jnt3','revolute');
body4 = rigidBody('body4');
jnt4 = rigidBodyJoint('jnt4','revolute');
body5 = rigidBody('body5');
jnt5 = rigidBodyJoint('jnt5','revolute');
body6 = rigidBody('body6');
jnt6 = rigidBodyJoint('jnt6','revolute');
setFixedTransform(jnt2,dhparams(2,:),'dh');
setFixedTransform(jnt3,dhparams(3,:),'dh');
setFixedTransform(jnt4,dhparams(4,:),'dh');
setFixedTransform(jnt5,dhparams(5,:),'dh');
setFixedTransform(jnt6,dhparams(6,:),'dh');
body2.Joint = jnt2;
body3.Joint = jnt3;
body4.Joint = jnt4;
body5.Joint = jnt5;
body6.Joint = jnt6;
addBody(robot,body2,'body1')
addBody(robot,body3,'body2')
addBody(robot,body4,'body3')
addBody(robot,body5,'body4')
addBody(robot,body6,'body5')
showdetails(robot)
show(robot);
axis([-10,10,-10,10,-10,10])

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Answer 1

Image Analyst 2023-4-15

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https://ww2.mathworks.cn/matlabcentral/answers/1947793-for-loop-and-calculating-at-each-interval-of-t#answer_1216428

在 MATLAB Online 中打开

Put the theta into a vector and then use them in the loop to build dhparams:

theta = -3 + 6 * rand(1, 26); % Whatever they are....
for k = 1 : numel(theta) - 6
    dhparams{k} = [-theta(k),0,145,0;
        theta(k+1)+pi/2,pi/2,145,0;
        theta(k+2)+pi/2,0,0,260;
        theta(k+3),pi/2,40,30;
        theta(k+4),-pi/2,230,0;
        theta(k+5),pi/2,90,0];
end
celldisp(dhparams)

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Andre Khosrow 2023-4-16

在 MATLAB Online 中打开

The theta's have a formula, (i had k in there for the loop) . t = 0:0.2:5. So for the dhparams the matrix would have 26 values in each row of the first column.

theta1 = -2*(1-0.1*(t));  
theta2 = 2*(1-0.04*(t));
theta3 = 3*(1-0.2.*(t));
theta4 = theta1;
theta5= theta2;
theta6= theta3;

Image Analyst 2023-4-16

在 MATLAB Online 中打开

But you're overwriting theta 4, 5, and 6 in your for loop. Wouldn't you want an index for them too?

t = 0:0.2:5                  ;
for k= 1:26            
    theta1(k) = -2*(1-0.1*(k));  
    theta2(k) = 2*(1-0.04*(k));
    theta3(k) = 3*(1-0.2.*(k));
    theta4(k)=theta1(k);
    theta5(k)=theta2(k);
    theta6(k)=theta3(k);
end  