Index exceeds the number of array elements (1).

I want to create a movie with a for loop, but Index exceeds the number of array elements (1) is shown. Someone can tell me were is the error? The for loop without the condition works fine, but where i assign a condition, the error message appears. :( I just want that when y <= 0, the angle theta0 and v0 change to theta0 = atan(y(j-2)/(x(j)-x(j-2))) and v0 = 0.8v0.

clear, clc
g=9.81; theta0=45*pi/180; v0=5;
t(1)=0;x=0;y=0;
plot(x,y,'o','MarkerFaceColor','b','MarkerSize',8)
axis([0 8 0 0.8])
M(1)=getframe;
dt=1/128;

for j = 2:1000
    t(j)=t(j-1)+dt;
    x=v0*cos(theta0)*t(j);
    y=v0*sin(theta0)*t(j)-0.5*g*t(j)^2;
    plot(x,y,'o','MarkerFaceColor','b','MarkerSize',8);
    axis([0 8 0 0.8]);
    M(j)=getframe;
    if y<=0
        theta0 = atan(y(j-2)/(x(j)-x(j-2)));
        v0 = 0.8*v0;
        plot(x(j),0,'o','MarkerFaceColor','b','MarkerSize',8)
        M(j) = getframe;
    end
end
pause
movie(M,1)

[Chris]

I'm not entirely sure what value theta0 is meant to take when within that loop. y first turns negative when j=94, but don't forget that y is just a scalar the way you have it currently, so there's only one entry in position (1). In

theta0 = atan(y(j-2)/(x(j)-x(j-2)));

the first issue it encounters is y(j-2). At that point in the execution it becomes y(94-2) i.e. y(92) - the value of y in the 92nd column but y only has one entry/value, hence why you get that error.

[Marcos Granados Flores]

Oh, the model must be able to predict exactly when the ball hits the ground. At this point, the direction changes (the new angle will equal the negative of the angle at impact), that's why i put a conditional if y<=0. This is an excercise from Chapra´s Applied Methods, Chapter 3, excercise 3.21. I found the exercise made with a while loop, but I want to make it with a for loop if is possible. Here is the code

g = 9.81; 
theta0 = 50*pi/180; 
v0 = 5;
t(1) = 0;
x = 0;
y = 0;
plot(x,y,'o','MarkerFaceColor','b','MarkerSize',8)
axis([0 3 0 0.8])
M(1) = getframe;
dt = 1/128;
j=2;
x0=0;
t_last = 0;
h_stop = 0.01;
h_max = Inf;
while h_max>h_stop % para los saltos
    flag = 0;
    h_max = -1;
    while(flag == 0)
        t(j) = t(j - 1) + dt;
        x(j) = v0*cos(theta0)*(t(j)-t_last)+x0;
        y(j) = v0*sin(theta0)*(t(j)-t_last) - 0.5*g*(t(j)-t_last)^2;
        if y(j)>h_max
            h_max = y(j);
        end
        
        if y(j)<= 0
            % Calcular el nuevo ángulo
            theta0 = atan(y(j-2)/(x(j)-x(j-2)))
            flag = 1;
            v0 = 0.8*v0;
            x0 = x(j);
            t_last = t(j-1);
            plot(x(j),0,'o','MarkerFaceColor','b','MarkerSize',8)
            axis([0 8 0 0.8])
            M(j) = getframe;
        else
            plot(x(j),y(j),'o','MarkerFaceColor','b','MarkerSize',8)
            axis([0 8 0 0.8])
            M(j) = getframe;
        end
        j=j+1;
    end
    
end
movie(M,1)

Answer 1

[Chris]

Right, I see what you're trying to do but the point stands that your modified code doesn't save the different values of y as time increases, it only stores the current value, so going into y(j-2) finds nothing unless j=3. Keeping them as follows would address that:

x(j) = v0*cos(theta0)*(t(j)-t_last)+x0;
y(j) = v0*sin(theta0)*(t(j)-t_last) - 0.5*g*(t(j)-t_last)^2;

I would also say the use of while in this example is well suited to the conditions set, but if you want to replace them with for, that's also possible. As you've done, if you don't require watetight conditioning (i.e. just to run a couple of examples) you can set a high number of for loops (i.e. 1000/2000/etc), but then it's also wise to have a condition that checks if it's not needed to keep on looping and exit (i.e. when the ball no longer bounches).

g = 9.81; 
theta0 = 50*pi/180; 
v0 = 5;
t(1) = 0;
x = 0;
y = 0;
plot(x,y,'o','MarkerFaceColor','b','MarkerSize',8)
axis([0 3 0 0.8])
M(1) = getframe;
dt = 1/128;
x0=0;
t_last = 0;
h_max = -1;
for j=2:2000
        t(j) = t(j - 1) + dt;
        x(j) = v0*cos(theta0)*(t(j)-t_last)+x0;
        y(j) = v0*sin(theta0)*(t(j)-t_last) - 0.5*g*(t(j)-t_last)^2;
        if y(j)>h_max
            h_max = y(j);
        end
        if y(j)<= 0 
            theta0 = atan(y(j-2)/(x(j)-x(j-2)));
            flag = 1;
            v0 = 0.8*v0;
            x0 = x(j);
            t_last = t(j-1);
            plot(x(j),0,'o','MarkerFaceColor','b','MarkerSize',8)
            axis([0 8 0 0.8])
            M(j) = getframe;
        else
            plot(x(j),y(j),'o','MarkerFaceColor','b','MarkerSize',8)
            axis([0 8 0 0.8])
            M(j) = getframe;
        end
        if x(j)==x(j-1)
            break
        end
end
movie(M,1)

This is one way you can do it all with one for loop. The condition:

if x(j)==x(j-1)
    break
end

checks if the object is no longer moving in the x direction and exits the loop if that's the case. So for the default parameters that's 332 steps. If you double the gravitational acceleration, it only needs 170 steps and exits on its own (no point looping for the remaining 1830.)

[Marcos Granados Flores]

Thank you so much!

[Chris]

👍