Forward and Inverse Kinematics for robot

Hello,

Hope you are doing well.

I am verifying the output of my forward kinematics through inverse kinematics and the results are not as desired. As the output of my inverse kinematics is not coming out to be the same as the input of forward kinematics.

The D-H parameters of manipulator is given as:

alpha a theta d

Link 1 -90 0 theta1* d1

Link 2 0 a2 theta2* 0

Link 3 0 a3 theta3* 0

Functions used are:

Forward kinematics:

function ph  = forward_kinematics(q)

%input: [q1 q2 q3]

l2 = 0.28;
l3 = 0.2;
d1 = 0.03;

q1 = q(1);
q2 = q(2);
q3 = q(3);

ph = zeros(3,1);
ph1 = l2*cos(q1)*cos(q2)+l3*cos(q1)*cos(q2+q3);
ph2 = l2*sin(q1)*cos(q2)+l3*sin(q1)*cos(q2+q3);
ph3 = d1-l2*sin(q2)-l3*sin(q2+q3);

ph=[ph1;ph2;ph3];

end

Inverse kinematics

function q = inv_kinematics(ph)

%input: [ph1 ph2 ph3]

l1 = 0.05;
l2 = 0.28;
l3 = 0.2;
d1 = 0.03;

ph1 = ph(1);
ph2 = ph(2);
ph3 = ph(3);

r=sqrt(ph1^2+(ph3-d1)^2);
alpha=acos((r^2+l2^2-l3^2)/(2*r*l2))

q = zeros(3,1);
q1=atan2(ph2,ph1);
q2=atan2(ph1,ph3-d1)-alpha;
q3=atan2(ph1-l2*sin(q2),ph3-d1-l2*cos(q2))-q2;

q=[q1;q2;q3];

end

[Mohsina Zafar]

Simulink model of the same is attached here.

[Mohd Musharaf Hussain Sarwari]

Thank you jazak Allah

[Mohd Musharaf Hussain Sarwari]

comming error

[Mohd Musharaf Hussain Sarwari]

Answer 1

[Mohd Musharaf Hussain Sarwari]

I WANT SCARA ROBOT forward-and-inverse-kinematics MATLAB CODE

[Abdullah]

Hi Bro

can i have scara robot Forward and inverse code ?

[Sam Chak]

Hi @Abdullah

Could you please copy and paste the mathematical algorithms of the Forward and Inverse Kinematics in code form? Otherwise, we cannot verify their correctness.

Answer 2

[Krishna Akella]

Hi Mohsina,

I don't know the answer to your question but I looked at your model and I have a few observations. In your model I assume the inputs to your forward kinematics function are the joint angle values and the output is the end effector location. And I assume the inputs to your inverse kinematics function is the end effector location and the outputs are the joint values.

If this is correct, then why would you not pass the output from forward kinematics back to your inverse kinematics function to validate that you are getting back the same joint values?

I made this change and it seems like the first output value matches.

The reason the other joint angles might still not be matching is because you could have multiple solutions to worry about. For example, in the computation of your inverse kinematics function, you have

r = sqrt(ph1^2+(ph3-d1)^2);

There could be two solutions to the sqrt function. A positive and a negative value. And MATLAB returns the positive value. Similar thing is true for other functions like acos, where multiple angles can give the same result. All these will result in multiple solutions to result in the same end effector location.

Regards,

Krishna

[Mohsina Zafar]

Actually, I am giving output of inverse kinematics to the input of forward kinematics.

Thank you for your reply but unfortunately I am using MATLab 2016a version and am unable to open your code. Can you kindly share it in the previous version?

Answer 3

[Ram Bodhe]

function q = inv_kinematics(ph)

%input: [ph1 ph2 ph3]

l1 = 0.05; l2 = 0.28; l3 = 0.2; d1 = 0.03;

ph1 = ph(1); ph2 = ph(2); ph3 = ph(3);

r=sqrt(ph1^2+(ph3-d1)^2); alpha=acos((r^2+l2^2-l3^2)/(2*r*l2))

q = zeros(3,1); q1=atan2(ph2,ph1); q2=atan2(ph1,ph3-d1)-alpha; q3=atan2(ph1-l2*sin(q2),ph3-d1-l2*cos(q2))-q2;

q=[q1;q2;q3];

end

[Ram Bodhe]

iam getting error anyone can help with this