Dynamic Obstacle in matlab environment

if i have a obstacle (represented as (x,y)) in the mobile robot environment by using matlab , how can i make that obstacle move randomly with dynamic speed .

[Mischa Kim]

What do you mean by dynamic speed? Changing velocity (that is, direction and speed)?

[Maria]

yes , changing velocity , that is what i meant.

[Auday]

Dear all I have two objects with original and destination positions as in below; how I can make them move using rand function, please

%// Define rectangle values origin_x1 = [9.5 9.5 11.5 11.5 ]; origin_y1 = [12.6 14.6 14.6 12.6]; destination_x1 = origin_x1 + 3; destination_y1 = origin_y1 + 2;

%// Define circle values r = 1; v = linspace(0,2*pi); origin_x2 = 15+r*cos(v); origin_y2 = 10+r*sin(v); destination_x2 = origin_x2 - 1; destination_y2 = origin_y2 + 3;

Answer 1

[Walter Roberson]

x = x + rand() * max_x;
y = y + rand() * max_y;

or

theta = rand() * 2 * pi;
r = rand() * max_velecity;
[deltax, deltay] = pol2cart(r, theta);
x = x + deltax;
y = y + deltay;

[Maria]

Many thanks to you Walter, I will check that code .

[passioncoding]

did this code work?

I am doing same thing I need information

Answer 2

[Mischa Kim]

I strongly recommend not to "randomize" the position vector since this results in a non-differentiable function. Start with acceleration and work your way back to position by integration. This way you'll at least end up with a solid kinematics system. If you even need to go a step further you have to make sure that the kinetics is compatible, as well.

function rand_mov()
tspan = 0:0.01:5;
X0    = [0 0 0 0];
[T, Xsol] = ode45(@EOM, tspan, X0);

figure
subplot(2,1,1)
plot(T, Xsol(:,1),T, Xsol(:,2))
title('x and vx vs. t')
subplot(2,1,2)
plot(Xsol(:,1), Xsol(:,3))
title('x vs. y')
end

function dX = EOM(t, X)
dX = zeros(4,1);
da = 1;
x  = X(1);
vx = X(2);
y  = X(3);
vy = X(4);

dX = [vx; da*(rand()-0.5); vy; da*(rand()-0.5)];
end

[Maria]

Many thanks to you Mischa , Dear the idea you wrote is really interesting although it is most clear to me, so if you give me more clearly explanation i will be grateful to you .

[Mischa Kim]

From your dynamics course you might remember that the first and second derivatives of position are velocity and acceleration. So far so good. When adding random numbers to a position (e.g. x), you will, by design, see jumps in x at every time step. These types of "jumpy" functions, do not have continuous, bounded derivatives. In other words, there is no velocity vx(t) and no acceleration ax(t) that would result in the x(t) you generated.

This might be acceptable in a simulation, where your obstacle just pops up and disappears at random locations. It will be a problem when you need to implement this kind of scenario with real hardware (say, robots). Obstacles don't just pop up and disappear instantaneously but need to satisfy kinematic constraints (such as twice differentiable position vectors).

[Maria]

Dear Mischa kindly , any answer ? I am waiting your answer .

[Mudasser Wahab]

Hey can you help me with a problem. How can I simulate a static obstacle?

I have the kinematic model of a differential drive robot. I have been able to navigate it from one point to another in an obstacle free environment. Now I want to insert a static obstacle and ovoid them using obstacle avoidance techniques.

The problem is I don't know How to simulate an obstacle in simulink.

[Ruqaiya Attaullah]

Mudasser Wahab can you please help me in navigating the robot from one point to another. I am having trouble with that.

Answer 3

[Maria]

Many thanks to you Mischa , now i got it , if i am not wrong that without differentiable function formula i would not could be enabled to sense the velocity and acceleration of the obstacle since my algorithm required that i should compute the the velocity of moving obstacles to compute the relative speed between robot and obstacle , so to here good , but i hope to not be greedy if asked you to give more explanation about the code above because it is not so clear , it need some comments.

thanks