November 21, 2024

Solve Inverse Kinematics Problem in MATLAB


In this post, we learn how to solve an inverse kinematics problem in MATLAB. In our previous post, which can be found here, we provided an introduction to the inverse kinematics problem. We explained how to analytically solve the inverse kinematics problem. However, in many cases it is either not possible to find the analytical solution or the analytical solution is too complex. In such cases, it is preferable to numerically approximate the solution. In this post and in the accompanying video we will explain how to compute the analytical solution. The YouTube video accompanying this post is given here

Consider a manipulator with two rotational degrees of freedom \theta_{1} and \theta_{2} shown in the figure below.

Figure 1: Robotic manipulator with two degrees of freedom.

The inverse kinematics problem can be formulated as follows:

Given the coordinates x and y of the end-effector point C (world coordinates), compute the angles \theta_{1} and \theta_{2} (local joint coordinates).

In the statement of the inverse kinematics problem, it is assumed that the lengths of the robotic segments l_{1} and l_{2} are given.

From Figure 1, we have

(1)   \begin{align*}x=l_{1}cos(\theta_{1})+l_{2}cos(\theta_{1}+\theta_{2}) \\y=l_{1}sin(\theta_{1})+l_{2}sin(\theta_{1}+\theta_{2})\end{align*}

Now, for given x and y, and for given l_{1} and l_{2}, the equations (1) represent a system of nonlinear equations where unknown variables are \theta_{1} and \theta_{2}.

In the post given here, we explained how to solve a system of nonlinear equations in MATLAB. In this post, we follow the explained procedure. First, we transform the system (1) as follows

(2)   \begin{align*}x-l_{1}cos(\theta_{1})-l_{2}cos(\theta_{1}+\theta_{2}) =0\\y-l_{1}sin(\theta_{1})-l_{2}sin(\theta_{1}+\theta_{2}) =0\end{align*}

First, we define a MATLAB function that describes the system (2)

function F=kinematics(theta,x,y,l1,l2)
F=[x-l1*cos(theta(1))-l2*cos(theta(1)+theta(2));
   y-l1*sin(theta(1))-l2*sin(theta(1)+theta(2))];
end

We save this function in the file called “kinematics.m”, and add the folder in which this function is saved to the MATLAB path.

This MATLAB function is called from the following MATLAB code

clear, pack, clc
% Possible algorithms 'trust-region-dogleg', 'trust-region', or 'levenberg-marquardt'
options = optimoptions(@fsolve,'Algorithm','trust-region','Display','iter','UseParallel',false,'OptimalityTolerance',1.0000e-8,'FunctionTolerance',1.0000e-12)

%% test cases
%x=0.8; y=0; l1=0.3;l2=0.5; % horizonal position, theta1=0 [deg], theta2=0 [deg]
x=0.4; y=0.4; l1=0.4;l2=0.4; % horizonal position, theta1=0 [deg], theta2=90 [deg]

% initial guess in degrees
% initialGuessDegrees=[10; 10]  % for 0-0
initialGuessDegrees=[20; 60]

% initial guess in radians
initialGuessRadians=initialGuessDegrees*(pi/180)

 
[solutionThetaRadians] = fsolve(@(theta)kinematics(theta,x,y,l1,l2),initialGuessRadians,options)
 
kinematics(solutionThetaRadians,x,y,l1,l2)

solutionThetaDegrees=solutionThetaRadians*(180/pi)

To solve the system of nonlinear equations, we need to specify x,y,l_{1} and l_{2}. This is done on code lines 6-7. We consider two test cases. In the first case, \theta_{1}=0 and \theta_{2}=0. The second case is \theta_{1}= 0 and \theta_{2}= 0. Then for these cases, we define initial guesses of \theta_{1} and \theta_{2}. To solve the system of nonlinear equations we need to specify the initial conditions. On code line 14, we transform initial guesses to radians. Code line 17 is used to solve the system of nonlinear equations. Code line 19 is used to verify the solution, by evaluating the right-hand side of (2). Code line 21 is used to transform the solution to degrees.

The test cases used in this tutorial, are explained in the figure given below.

Test cases for solving the inverse kinematics problem.

For an explanation of these test cases, see the youtube video.