Lagrange Equation by MATLAB with Examples

In this post, I will explain how to derive a dynamic equation with Lagrange Equation by MATLAB with Examples. As an example, I will derive a dynamic model of a three-DOF arm manipulator (or triple pendulum). Of course you may get a dynamic model for a two-DOF arm manipulator by simply removing several lines. I am attaching demo codes for both two and three DOF arm manipulators.

If you know all theories and necessary skills and if you just want source code, you can just download from here.

<Download>

In the attached file, “Symb_Development_3DOF.m” generates a dynamic model. “main_sim_three_dof_arm.m” runs a simulation. You may get a result like this.

1. Example system

Let’s suppose a three DOF arm manipulator shown in the below figure. I am assuming all of masses (M1, M2, M3) exist at the end of links for simplicity. Three actuators exist at each joint, and directly actuate the torques (u1,u2,u3). The manipulator kinematics is governed by three joint angles (q1, q2, q3).

three DOF arm manipulator

2. Theoretical Background

We will get a dynamic equation of this system by using Lagrangian mechanics. If you do not have a background knowledge of Lagrangian mechanics, please refer here.

The general dynamic equation is obtained by

$\frac{d}{dt}\left( \frac{\partial T}{\partial \dot{q}_j}\right ) - \frac{\partial T}{\partial q_j} + \frac{\partial V}{\partial q_j} = P_j$

Where, T is the total kinetic energy, V is the total potential energy of the system. where $P_j$ is the generalized force, t is time, $q_j$ is the generalized coordinates, $\dot{q}_j$ is the generalized velocity. For the example of three-DOF arm manipulator problem, $P_j$ is the torque at the j-th joint, $q_j$ is the angle of the j-th joint, $\dot{q}_j$ is the angular velocity of the j-th joint.

3. Using Matlab symbolic toolbox

First, let’s define the symbols.

I am using x to represent q, xd for $\dot{q}$ , xdd for $\ddot{q}$ L is the length of each link. u is the torque of each joint. g is the gravity constant.

syms M1 M2 M3;
syms x1 x1d x1dd x2 x2d x2dd x3 x3d x3dd;
syms L1 L2 L3;
syms u1 u2 u3;
syms g

Second, find the position and velocities of the point masses.

p1x = L1cos(x1);
p1y = L1sin(x1);
p2x = p1x+L2cos(x1+x2);
p2y = p1y+L2sin(x1+x2);
p3x = p2x+L3cos(x1+x2+x3);
p3y = p2y+L3sin(x1+x2+x3);

v1x = -L1sin(x1)x1d;
v1y = +L1cos(x1)x1d;
v2x = v1x-L2sin(x1+x2)(x1d+x2d);
v2y = v1y+L2cos(x1+x2)(x1d+x2d);
v3x = v2x – L3sin(x1+x2+x3)(x1d+x2d+x3d);
v3y = v2y + L3cos(x1+x2+x3)(x1d+x2d+x3d);

Third, define the kinetic energy, and the potential energy.

KE = 0.5M1( v1x^2 + v1y^2) + 0.5M2( v2x^2 + v2y^2) + 0.5M3( v3x^2 + v3y^2);
KE = simplify(KE);

PE = M1gp1y + M2gp2y + M3gp3y;
PE = simplify(PE);

Fourth, define the generalized forces, here torques.

Px1 = u1;
Px2 = u2;
Px3 = u3;

Fifth, solve the Lagrangian equation

We have to get the Lagrangian eqn. $\frac{d}{dt}\left( \frac{\partial T}{\partial \dot{q}_j}\right ) - \frac{\partial T}{\partial q_j} + \frac{\partial V}{\partial q_j} = P_j$

Let’s obtain step by step.

$\left( \frac{\partial T}{\partial \dot{q}}\right )$ is obtained by

pKEpx1d = diff(KE,x1d);

To calculate, $\frac{d}{dt}\left( \frac{\partial T}{\partial \dot{q}}\right )$ , we need the chain rule, that is $\frac{d}{dt}\left( \frac{\partial T}{\partial \dot{q}_j}\right )= \sum^{3}_i \left\{ \frac{\partial }{\partial {q}_i}\left( \frac{\partial T}{\partial \dot{q}_j}\right )\frac{dq_i}{dt} + \frac{\partial }{\partial \dot{q}_i}\left( \frac{\partial T}{\partial {q}_j}\right )\frac{d \dot{q}_i}{dt} \right\}$

ddtpKEpx1d = diff(pKEpx1d,x1)x1d+ …
diff(pKEpx1d,x1d)x1dd+ …
diff(pKEpx1d,x2)x2d + …
diff(pKEpx1d,x2d)x2dd + …
diff(pKEpx1d,x3)x3d + …
diff(pKEpx1d,x3d)x3dd;

$\frac{\partial T}{\partial q_j}, \frac{\partial V}{\partial q_j}$ are easily obtained by

pKEpx1 = diff(KE,x1);
pPEpx1 = diff(PE,x1);

By summing all equations,

eqx1 = simplify( ddtpKEpx1d – pKEpx1 + pPEpx1 – Px1);

By repeating these procedures, we can get all governing equations.

Sixth, rearrange the equations.

We love more simplified forms like

$\frac{d}{dt} \begin{bmatrix} x_1\\ \dot{x}_1\\ x_2\\ \dot{x}_2\\ x_3\\ \dot{x}_3 \end{bmatrix}= f(x_1,\dot{x}_1,x_2,\dot{x}_2,x_3,\dot{x}_3)$

For this form, we need to rearrange the equations by

Sol = solve(eqx1,eqx2,eqx3,’x1dd,x2dd,x3dd’);
Sol.x1dd = simplify(Sol.x1dd);
Sol.x2dd = simplify(Sol.x2dd);
Sol.x3dd = simplify(Sol.x3dd);

Seventh, substitute with y1,y2… variables.

Just for easier implementation of symbolic codes, let’s substitute x1, x1d, x2, … with y1,y2….

syms y1 y2 y3 y4 y5 y6
fx1=subs(Sol.x1dd,{x1,x1d,x2,x2d,x3,x3d},{y1,y2,y3,y4,y5,y6})
fx2=subs(Sol.x2dd,{x1,x1d,x2,x2d,x3,x3d},{y1,y2,y3,y4,y5,y6})
fx3=subs(Sol.x3dd,{x1,x1d,x2,x2d,x3,x3d},{y1,y2,y3,y4,y5,y6})

Eighth, this is the result…. as you can see, it is almost impossible to solve by hand

fx1 =

(L2L3M2u1 – L2L3M2u2 + L2L3M3u1 – L2L3M3u2 – L1L3M2u2cos(y3) + L1L3M2u3cos(y3) – L1L3M3u2cos(y3) + L1L3M3u3cos(y3) – L2L3M3u1cos(y5)^2 + L2L3M3u2cos(y5)^2 + L1L2^2L3M2^2y2^2sin(y3) + L1L2^2L3M2^2y4^2sin(y3) – L1L2L3M2^2gcos(y1) + (L1^2L2L3M2^2y2^2sin(2y3))/2 + L1L2M2u3sin(y3)sin(y5) + L1L2M3u3sin(y3)sin(y5) + L1L3M3u2cos(y3)cos(y5)^2 – L1L3M3u3cos(y3)cos(y5)^2 + L1L2^2L3M2M3y2^2sin(y3) + L1L2^2L3M2M3y4^2sin(y3) – L1L2L3M1M2gcos(y1) – L1L2L3M1M3gcos(y1) – L1L2L3M2M3gcos(y1) + 2L1L2^2L3M2^2y2y4sin(y3) + (L1^2L2L3M2M3y2^2sin(2y3))/2 – L1L3M3u2cos(y5)sin(y3)sin(y5) + L1L3M3u3cos(y5)sin(y3)sin(y5) + L1L2L3M2^2gcos(y1)cos(y3)^2 – L1L2L3M2^2gcos(y3)sin(y1)sin(y3) + 2L1L2^2L3M2M3y2y4sin(y3) + L1L2L3M2M3gcos(y1)cos(y3)^2 + L1L2L3M1M3gcos(y1)cos(y5)^2 + L1L2L3^2M2M3y2^2cos(y5)sin(y3) + L1L2L3^2M2M3y4^2cos(y5)sin(y3) + L1L2L3^2M2M3y6^2cos(y5)sin(y3) + 2L1L2L3^2M2M3y2y4cos(y5)sin(y3) + 2L1L2L3^2M2M3y2y6cos(y5)sin(y3) + 2L1L2L3^2M2M3y4y6cos(y5)sin(y3) – L1L2L3M2M3gcos(y3)sin(y1)sin(y3))/(L1^2L2L3(M2^2 – M2^2cos(y3)^2 + M1M2 + M1M3 + M2M3 – M2M3cos(y3)^2 – M1M3*cos(y5)^2))

fx2 =

-(2L1^2L3M1u3 – 2L1^2L3M1u2 – 2L1^2L3M2u2 + 2L2^2L3M2u1 + 2L1^2L3M2u3 – L1^2L3M3u2 – 2L2^2L3M2u2 + L2^2L3M3u1 + L1^2L3M3u3 – L2^2L3M3u2 + L1L2^2M2u3cos(y3 – y5) + L1L2^2M3u3cos(y3 – y5) – L1^2L2M2u3cos(2y3 + y5) – L1^2L2M3u3cos(2y3 + y5) – L2^2L3M3u1cos(2y5) + L2^2L3M3u2cos(2y5) + L1^2L3M3u2cos(2y3 + 2y5) – L1^2L3M3u3cos(2y3 + 2y5) – L1L2^2M2u3cos(y3 + y5) – L1L2^2M3u3cos(y3 + y5) + 2L1^2L2M1u3cos(y5) + L1^2L2M2u3cos(y5) + L1^2L2M3u3cos(y5) + 2L1L2^3L3M2^2y2^2sin(y3) + 2L1^3L2L3M2^2y2^2sin(y3) + 2L1L2^3L3M2^2y4^2sin(y3) – L1L2L3M3u1cos(y3 + 2y5) + 2L1L2L3M3u2cos(y3 + 2y5) – L1L2L3M3u3cos(y3 + 2y5) + L1^2L2L3M2^2gcos(y1 + y3) – L1L2^2L3M2^2gcos(y1) – L1^2L2L3M2^2gcos(y1 – y3) + L1L2^2L3M2^2gcos(y1 + 2y3) + 2L1^2L2^2L3M2^2y2^2sin(2y3) + L1^2L2^2L3M2^2y4^2sin(2y3) + 2L1L2L3M2u1cos(y3) – 4L1L2L3M2u2cos(y3) + L1L2L3M3u1cos(y3) + 2L1L2L3M2u3cos(y3) – 2L1L2L3M3u2cos(y3) + L1L2L3M3u3cos(y3) + 2L1^3L2L3M1M2y2^2sin(y3) + L1^3L2L3M1M3y2^2sin(y3) + 2L1L2^3L3M2M3y2^2sin(y3) + 2L1^3L2L3M2M3y2^2sin(y3) + 2L1L2^3L3M2M3y4^2sin(y3) + 4L1L2^3L3M2^2y2y4sin(y3) – (L1^2L2L3M1M3gcos(y1 + y3 + 2y5))/2 – L1^3L2L3M1M3y2^2sin(y3 + 2y5) + L1L2^2L3^2M2M3y2^2sin(y3 + y5) + L1L2^2L3^2M2M3y4^2sin(y3 + y5) + L1L2^2L3^2M2M3y6^2sin(y3 + y5) + L1^2L2L3M1M2gcos(y1 + y3) + (L1^2L2L3M1M3gcos(y1 + y3))/2 + L1^2L2L3M2M3gcos(y1 + y3) – 2L1^2L2L3^2M1M3y2^2sin(y5) – L1^2L2L3^2M2M3y2^2sin(y5) – 2L1^2L2L3^2M1M3y4^2sin(y5) – L1^2L2L3^2M2M3y4^2sin(y5) – 2L1^2L2L3^2M1M3y6^2sin(y5) – L1^2L2L3^2M2M3y6^2sin(y5) – 2L1L2^2L3M1M2gcos(y1) – L1L2^2L3M1M3gcos(y1) – L1L2^2L3M2M3gcos(y1) + (L1^2L2L3M1M3gcos(y1 – y3 – 2y5))/2 + L1L2^2L3^2M2M3y2^2sin(y3 – y5) + L1^2L2L3^2M2M3y2^2sin(2y3 + y5) + L1L2^2L3^2M2M3y4^2sin(y3 – y5) + L1^2L2L3^2M2M3y4^2sin(2y3 + y5) + L1L2^2L3^2M2M3y6^2sin(y3 – y5) + L1^2L2L3^2M2M3y6^2sin(2y3 + y5) – L1^2L2L3M1M2gcos(y1 – y3) – (L1^2L2L3M1M3gcos(y1 – y3))/2 – L1^2L2L3M2M3gcos(y1 – y3) + L1L2^2L3M2M3gcos(y1 + 2y3) + (L1L2^2L3M1M3gcos(y1 – 2y5))/2 + (L1L2^2L3M1M3gcos(y1 + 2y5))/2 + 2L1^2L2^2L3M2M3y2^2sin(2y3) – L1^2L2^2L3M1M3y2^2sin(2y5) + L1^2L2^2L3M2M3y4^2sin(2y3) – L1^2L2^2L3M1M3y4^2sin(2y5) + 2L1^2L2^2L3M2^2y2y4sin(2y3) + 4L1L2^3L3M2M3y2y4sin(y3) + 2L1L2^2L3^2M2M3y2y4sin(y3 + y5) + 2L1L2^2L3^2M2M3y2y6sin(y3 + y5) + 2L1L2^2L3^2M2M3y4y6sin(y3 + y5) – 4L1^2L2L3^2M1M3y2y4sin(y5) – 2L1^2L2L3^2M2M3y2y4sin(y5) – 4L1^2L2L3^2M1M3y2y6sin(y5) – 2L1^2L2L3^2M2M3y2y6sin(y5) – 4L1^2L2L3^2M1M3y4y6sin(y5) – 2L1^2L2L3^2M2M3y4y6sin(y5) + 2L1L2^2L3^2M2M3y2y4sin(y3 – y5) + 2L1^2L2L3^2M2M3y2y4sin(2y3 + y5) + 2L1L2^2L3^2M2M3y2y6sin(y3 – y5) + 2L1^2L2L3^2M2M3y2y6sin(2y3 + y5) + 2L1L2^2L3^2M2M3y4y6sin(y3 – y5) + 2L1^2L2L3^2M2M3y4y6sin(2y3 + y5) + 2L1^2L2^2L3M2M3y2y4sin(2y3) – 2L1^2L2^2L3M1M3y2y4sin(2y5))/(L1^2L2^2L3(M2^2 – M2^2cos(2y3) + 2M1M2 + M1M3 + M2M3 – M2M3cos(2y3) – M1M3cos(2*y5)))

fx3 =

-(L1L3^2M3^2u2 – L1L2^2M3^2u3 – L1L2^2M2^2u3 – L1L3^2M3^2u3 – L2L3^2M3^2u1cos(y3) + L2L3^2M3^2u2cos(y3) – 2L1L2^2M1M2u3 – 2L1L2^2M1M3u3 + 2L1L3^2M1M3u2 – 2L1L2^2M2M3u3 – 2L1L3^2M1M3u3 + 2L1L3^2M2M3u2 – 2L1L3^2M2M3u3 + L1L2^2M2^2u3cos(2y3) + L1L2^2M3^2u3cos(2y3) + L1L3^2M3^2u2sin(2y3)sin(2y5) – L1L3^2M3^2u3sin(2y3)sin(2y5) + L1L2L3M3^2u2cos(y5) – 2L1L2L3M3^2u3cos(y5) – 2L2L3^2M2M3u1cos(y3) + 2L2L3^2M2M3u2cos(y3) – 2L2^2L3M3^2u1sin(y3)sin(y5) + 2L2^2L3M3^2u2sin(y3)sin(y5) + L2L3^2M3^2u1cos(2y5)cos(y3) – L2L3^2M3^2u2cos(2y5)cos(y3) + 2L1L2^2M2M3u3cos(2y3) – L2L3^2M3^2u1sin(2y5)sin(y3) + L2L3^2M3^2u2sin(2y5)sin(y3) – L1L3^2M3^2u2cos(2y3)cos(2y5) + L1L3^2M3^2u3cos(2y3)cos(2y5) – L1L2^2L3^2M2M3^2y2^2sin(2y3) – L1L2^2L3^2M2^2M3y2^2sin(2y3) + 2L1L2^2L3^2M1M3^2y2^2sin(2y5) – L1L2^2L3^2M2M3^2y4^2sin(2y3) – L1L2^2L3^2M2^2M3y4^2sin(2y3) + 2L1L2^2L3^2M1M3^2y4^2sin(2y5) + L1L2^2L3^2M1M3^2y6^2sin(2y5) + 2L1L2L3M1M3u2cos(y5) – 4L1L2L3M1M3u3cos(y5) + L1L2L3M2M3u2cos(y5) – 2L1L2L3M2M3u3cos(y5) – 2L2^2L3M2M3u1sin(y3)sin(y5) + 2L2^2L3M2M3u2sin(y3)sin(y5) + 2L1L2L3^3M1M3^2y2^2sin(y5) + 2L1L2^3L3M1M3^2y2^2sin(y5) + L1L2L3^3M2M3^2y2^2sin(y5) + 2L1L2L3^3M1M3^2y4^2sin(y5) + 2L1L2^3L3M1M3^2y4^2sin(y5) + L1L2L3^3M2M3^2y4^2sin(y5) + 2L1L2L3^3M1M3^2y6^2sin(y5) + L1L2L3^3M2M3^2y6^2sin(y5) – L1L2L3M3^2u2cos(2y3)cos(y5) + 2L1L2L3M3^2u3cos(2y3)cos(y5) + L1L2L3M3^2u2sin(2y3)sin(y5) – 2L1L2L3M3^2u3sin(2y3)sin(y5) – L1^2L2L3^2M1M3^2y2^2sin(y3) – 2L1^2L2L3^2M2M3^2y2^2sin(y3) – 2L1^2L2L3^2M2^2M3y2^2sin(y3) – 2L1L2^2L3^2M2M3^2y2y4sin(2y3) – 2L1L2^2L3^2M2^2M3y2y4sin(2y3) + 4L1L2^2L3^2M1M3^2y2y4sin(2y5) + 2L1L2^2L3^2M1M3^2y2y6sin(2y5) + 2L1L2^2L3^2M1M3^2y4y6sin(2y5) + 2L1L2^3L3M1M2M3y2^2sin(y5) + 2L1L2^3L3M1M2M3y4^2sin(y5) – L1L2L3M2M3u2cos(2y3)cos(y5) + 2L1L2L3M2M3u3cos(2y3)cos(y5) + 4L1L2L3^3M1M3^2y2y4sin(y5) + 4L1L2^3L3M1M3^2y2y4sin(y5) + 2L1L2L3^3M2M3^2y2y4sin(y5) + 4L1L2L3^3M1M3^2y2y6sin(y5) + 2L1L2L3^3M2M3^2y2y6sin(y5) + 4L1L2L3^3M1M3^2y4y6sin(y5) + 2L1L2L3^3M2M3^2y4y6sin(y5) + L1L2L3M2M3u2sin(2y3)sin(y5) – 2L1L2L3M2M3u3sin(2y3)sin(y5) – L1L2L3^3M2M3^2y2^2cos(2y3)sin(y5) – L1L2L3^3M2M3^2y2^2sin(2y3)cos(y5) – L1L2L3^3M2M3^2y4^2cos(2y3)sin(y5) – L1L2L3^3M2M3^2y4^2sin(2y3)cos(y5) – L1L2L3^3M2M3^2y6^2cos(2y3)sin(y5) – L1L2L3^3M2M3^2y6^2sin(2y3)cos(y5) + 2L1^2L2^2L3M1M3^2y2^2cos(y3)sin(y5) – 2L1^2L2L3^2M1M2M3y2^2sin(y3) + L1L2L3^2M1M3^2gsin(y1)sin(y3) + 2L1L2L3^2M2M3^2gsin(y1)sin(y3) + 2L1L2L3^2M2^2M3gsin(y1)sin(y3) + L1^2L2L3^2M1M3^2y2^2cos(2y5)sin(y3) + L1^2L2L3^2M1M3^2y2^2sin(2y5)cos(y3) – L1L2L3^2M1M3^2gcos(2y5)sin(y1)sin(y3) – L1L2L3^2M1M3^2gsin(2y5)cos(y3)sin(y1) + 4L1L2^3L3M1M2M3y2y4sin(y5) + 2L1^2L2^2L3M1M2M3y2^2cos(y3)sin(y5) – 2L1L2L3^3M2M3^2y2y4cos(2y3)sin(y5) – 2L1L2L3^3M2M3^2y2y4sin(2y3)cos(y5) – 2L1L2L3^3M2M3^2y2y6cos(2y3)sin(y5) – 2L1L2L3^3M2M3^2y2y6sin(2y3)cos(y5) – 2L1L2L3^3M2M3^2y4y6cos(2y3)sin(y5) – 2L1L2L3^3M2M3^2y4y6sin(2y3)cos(y5) – 2L1L2^2L3M1M3^2gcos(y3)sin(y1)sin(y5) + 2L1L2L3^2M1M2M3gsin(y1)sin(y3) – 2L1L2^2L3M1M2M3gcos(y3)sin(y1)sin(y5))/(L1L2^2L3^2M3(M2^2 – M2^2cos(2y3) + 2M1M2 + M1M3 + M2M3 – M2M3cos(2y3) – M1M3cos(2y5)))

4. Now it is ready!!! let’s run a simulation!

copy the result and paste the symbols in a Matlab function. Then, run an ODE with the function.

A sample code is here.

<Download>

In the attached file, “Symb_Development_3DOF.m” generates a dynamic model. “main_sim_three_dof_arm.m” runs a simulation. Then, you can get this result.

So far, I have explained how to derive a Lagrange Equation by MATLAB with Examples. I hope that this post helps your project and save your time. Please leave a message if you have any question.

Update on 02/21/2016

I updated some code and posting about typo. “simple” -> “simplify” There is no function “simple” Now all programs are running well.

-Mok-

Reference

[1] http://en.wikipedia.org/wiki/Lagrangian_mechanics

[2] https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CB8QFjAA&url=http%3A%2F%2Fsstl.cee.illinois.edu%2Fapss%2Ffiles%2FLagrange_Reference.doc&ei=iF9SVYT0N8TJtQXCwoDgCQ&usg=AFQjCNEyv1Ut3cGu6ylAjtZyG6ZPCYQkIg&sig2=pJ-84gdkjCqH_AH9pB6tjA&bvm=bv.92885102,d.b2w&cad=rja

[3] http://www.mathworks.com/matlabcentral/fileexchange/23037-lagrange-s-equations

[4] https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=4&cad=rja&uact=8&ved=0CDIQFjAD&url=http%3A%2F%2Fweb.stanford.edu%2Fclass%2Fme328%2Flectures%2Flecture6-dynamics.pdf&ei=oJ5kVa7CG4GKsQWDloFQ&usg=AFQjCNGWFZZ4uBP_ObHHnirkEzAIc5OONw&sig2=ISnRQj-kBc0y_eoT_0y7Hw&bvm=bv.93990622,d.b2w

12 thoughts on “Lagrange Equation by MATLAB with Examples”

vivek December 16, 2015 at 10:48 PM

Hello ,

I am trying to use this logic to solve my project and wanted to know how to move the equations into ODE45 ?
could you please upload a sample of that too ?

Reply ↓
1. admin Post authorFebruary 21, 2016 at 12:58 PM
  
  Aha… I should have explained that point.
  Just copy the result, and paste on your m-file. Then, run ODE program.
  Good luck!!
  
  Reply ↓
mohammad March 26, 2016 at 10:46 AM

Hi, thanks for this great help! Could you also say how to decouple the equations in the form of state space? I want to design a PID controller for 4DOF arm .

Reply ↓
1. admin Post authorMarch 28, 2016 at 4:36 PM
  
  Hi, Mohammad. y1,y2…y6 are the state. So at the seventh step, you can get the equation for the state space. For the design of the four DoF arm, you can get it simply by adding several variables and several lines. I believe that you can do it by reading this article carefully. If you make it, can you please share it?
  
  Reply ↓
mohammad April 9, 2016 at 12:03 PM

Hi I’m Mohammad from the last comment . How can I share the code ?!

Reply ↓
1. admin Post authorApril 9, 2016 at 12:21 PM
  
  Great job!! you made it!! If you can send me the file to my email (yunyoungmok@gmail.com), I will update it. Thanks.
  
  Reply ↓
mohammad April 24, 2016 at 8:15 AM

Hi did you receive my email ? could you please answer me?

Reply ↓
Motaz December 30, 2016 at 7:41 PM

Hello,
This is great code!
Can you give brief explanation of the ode function handle (three_dof_arm_dyn_for_ODE).

Reply ↓
Evren January 7, 2017 at 8:12 PM

Hi,

Comments or explanation about three_dof_arm_dyn_for_ODE will help a lot.

Thank you.

Reply ↓
ASMITA SINGH BISEN March 1, 2017 at 5:43 AM

Respected Sir,

Greetings of the day!

I am ASMITA SINGH BISEN, Post Graduate Research Scholar.
i am working on 7 dof manipulator. I have written the code for the same, can you verify the code which i have written. i have taken the help from you code.
is it the right way to go for 7 dof manipulator( the manipulator i am using is not planar).
please find the attached m file.

Reply ↓
Fawad Ahmed April 26, 2017 at 2:05 AM

In this function “three_dof_arm_dyn_for_ODE(t,y)” how you give fx1,fx2,fx3

Reply ↓
L December 6, 2017 at 10:31 AM

The Picture from the chain rule is not right. a point missed in the last (dT / dq). It should be dT/dq_p… am i right?

Reply ↓

Youngmok Yun: Roboticist in Austin

Lagrange Equation by MATLAB with Examples

Lagrange Equation by MATLAB with Examples

1. Example system

2. Theoretical Background