# Forward and Inverse Kinematics for robot

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Mohsina Zafar on 28 Jun 2019
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 I WANT SCARA ROBOT forward-and-inverse-kinematics MATLAB CODE

Krishna Akella on 28 Jun 2019
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
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Mohsina Zafar on 1 Jul 2019
Edited: Mohsina Zafar on 1 Jul 2019
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?

Ram Bodhe on 12 May 2020
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
##### 1 CommentShowHide None
Ram Bodhe on 12 May 2020
iam getting error anyone can help with this