Project on Robotic arm lying on the centre of a square where my end effector lying on one corner and it should move from one corner to other corner and pause for 10 milli seconds until it reaches its initial position.

Hello Sir/Madam,

This is Sumanth reddy i was facing a problem regarding an algorithm of square with 20cm sides and my robotic arm is in centre of square and need to move from each corner to corner and achieve its initial position.

But my end effector should start at point (0.1,0.1,0.233) and at every corner it needs to pause for 100 milli sec and restart. could you please help me.with 6 DOF

DH is a=[ 0 0.21 0 0.03 0 0], alpha = [ 90 0 90 -90 90 0] d= [ 0.183 0 0 0.2215 0 0.0237] theta=[30; 30; 45; 0; 90; 10]

Thank you.

[darova]

What are size of robotic arms?

[Sumanth Kumar Reddy Maddula]

Assuming the Niryo robot whose DH table is provided in the document NiryoOneinfo.pdf available in Classroom and in the lab website (http://webuser.unicas.it/lai/robotica), by resorting to the ttt file named niryo pen.ttt available with this file, starting from the home configuration and assuming the base frame denoted as Σ − x, y, z, the candidate is required to design an Inverse Kinematic controller such that:

• The control objective is position-only. The required path is an horizontal square lying on the xy axis with side of 20 cm length. The end-effector occupies the lower-right corner of the square looking from the z axis with coordinates .1 .1 .233 meters. Each side needs to be executed with a trajectory of duration 2 s and a trapezoidal velocity profile characterized by a proper cruise velocity. The robots stops for 100 ms at each corner and 2 s once back in the initial end effector position.

• In a second run, the control objective is given by both the position and the orientation. While the position needs to move according to the indications above, the orientation needs to be controlled such that it is kept constant at the initial value.

• In a third run the end-effector orientation needs to be changed such that, when running the first side, it is rotated of 15◦ around the zee axis according to a trapezoidal velocity profile of proper angular cruise velocity. During the second side the inverse rotation is commanded such that, in the second corner, the orientation is equal to the initial one. During the third and fourth sides the orientation movement is repeated and, in the end, the final orientation is equal to the initial one

The above is the actual question and with respect to DH table values we get length of arms.

[darova]

Do you have a picture or something?

[Sumanth Kumar Reddy Maddula]

[Sumanth Kumar Reddy Maddula]

Hello DAROVA ([e-mail]) is my mail id please contact me for more accurate information to share with you remotely.

Thank you for such a fast response gesture.

[darova]

What are those lengths?

Please provide necessary information here

[Sumanth Kumar Reddy Maddula]

D1 =0.21 D2=0.03 (metres)

Answer 1

[darova]

Here is an example

clc,clear

D1 = 0.2;
D2 = 0.15;

t = pi/4+(0:pi/2:2*pi);
[x,y] = pol2cart(t,sqrt(2)*0.2);
    % create more points on the square side (interpolation)
tt = linspace(t(1),t(end),100);
x2 = interp1(t,x,tt);
y2 = interp1(t,y,tt);
z2 = 0*x2;
D = hypot(x2,y2);
a1 = acos( (D2^2-D1^2-D.^2)./(2*D*D1) );

x1 = x2./D*D1;
y1 = y2./D*D1;
z1 = D1*sin(a1);
plot3(x2,y2,z2,'.r')
hold on
for i = 1:length(x1)
    plot3([0 x1(i) x2(i)],...
    [0 y1(i) y2(i)],...
    [0 z1(i) z2(i)])
    
    pause(0.1)
end
hold off

Read about law_cosinus

[darova]