In an assignment A(I) = B, the number of elements in B and I must be the same. Error in odearguments (line 88) f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0. Error in ode45 (line 114) [neq, tspan, ntspan...

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Miguel Ángel 2013-12-12

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评论： Miguel Ángel 2013-12-13

Alright. First of all I must say I'm quite new in the world of Matlab. I've done a few things but they were not a very big deal. At university we've been asked to represent the animation of a bouy depending on its density and size. Here's the two pieces of code I've written:

Here I define the differential equation that describes the general position of the object

    function dy=fboya(t,y)
    global g l rho0 rho_a radio;
    v_boya=pi*radio^2*l;
    m_boya= rho0*v_boya;
    P=m_boya*g;
    E=(rho_a*v_boya*g).*(z(1)<-l/2)+(rho_a*pi*radio^2.*(l/2-z(1))*g).*((-l/2<z(1))&(z(1)<l/2))+(0).*(l/2<z(1));
    dz=zeros(2,1);
    dz(1)=z(2);
    dz(2)=(E-P)/m_boya;

And here is the rest of the code:

    function [t,y]=mboya2(L,y_0,v_0,rho_agua,rho_0,tf,R)
    global g l rho0 rho_a radio;
    g=9.81; % gravedad
    l=L; 
    rho0=rho_0; % densidad del objeto
    rho_a=rho_agua; % densidad del agua
    radio=R;
    [t,y]=ode45(@fboya2,[0,tf],[y_0,v_0]);
    y=y(:,1);
    x=0;
    close all;
    figure;
    set(gcf(),'Position',[300 250 800 300]);
    subplot(1,2,1)
    plot(x(1),y(1),'or',[0 x(1)],[0 y(1)],'b')
    axis([-l,l,-l,l]);
    axis square;

The thing is that I keep getting this error messages and I don't really undersand why is this happening.

In an assignment A(I) = B, the number of elements in B and I must be the same.

Error in fboya2 (line 14) dy(2)=(E-P)/m_boya; Error in odearguments (line 88) f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.

Error in ode45 (line 114) [neq, tspan, ntspan, next, t0, tfinal, tdir, y0, f0, odeArgs, odeFcn, ...

Error in mboya2 (line 20) [t,y]=ode45(@fboya2,[0,tf],[y_0,v_0]);

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Answer 1

Walter Roberson 2013-12-12

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在 MATLAB Online 中打开

Have another look at

 E=(rho_a*m_boya).*(y<-l/2)+(rho_a*pi*radio^2.*(l/2-y)).*((-l/2<y)&(y<l/2))+(0).*(l/2<y);

Your "y" has multiple elements, so (y<-1/2) has multiple elements, so E is going to have multiple elements.

Then in

dy(2)=(E-P)/m_boya;

since your E has multiple elements (and your m_boya is scalar), the expression on the right hand side has multiple elements. But you are trying to store those multiple elements into the single location dy(2)

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Miguel Ángel 2013-12-13

编辑：Miguel Ángel 2013-12-13

在 MATLAB Online 中打开

Alright, I've changed both codes a bit in order to avoid using y as a vector.This is the first piece of code:

function dz=fboya(t,z)
global g l rho0 rho_a radio;
v_boya=pi*radio^2*l;
m_boya= rho0*v_boya;
P=m_boya*g;
E=(rho_a*v_boya*g).*(z(1)<-l/2)+(rho_a*pi*radio^2.*(l/2-z(1))*g).*((-l/2<z(1))&(z(1)<l/2))+(0).*(l/2<z(1));
dz=zeros(2,1);
dz(1)=z(2);
dz(2)=(E-P)/m_boya;

And this is the other. Here I've also changed the definition of x making it a vector of zeros as the buoy is only supposed to be moving in the 'y' direction. The only change you may notice is the axis' limits. Since this code now executes, I've modified that bit in order to have a better looking animation:

function [t,x,y]=mboya(L,y_0,v_0,rho_agua,rho_0,tf,R)
global g l rho0 rho_a radio;
g=9.81; % gravedad
l=L; 
rho0=rho_0; % densidad del objeto
rho_a=rho_agua; % densidad del agua
radio=R;
[t,z]=ode45(@fboya,[0,tf],[y_0,v_0]);
py=z(:,1);
y=py;
x=zeros(1,size(z));
close all;
figure;
set(gcf(),'Position',[300 250 800 300]);
subplot(1,2,1)
plot(x(1),y(1),'or',[0 x(1)],[0 y(1)],'b')
axis([-l,l,min(min(py))-2,max(max(py))+2]);
axis square;
subplot(1,2,2);
plot(t,py,t(1),py(1),'or');
ylabel('y(t)');
xlabel('t');
for i=2:length(t)
    subplot(1,2,1)
    plot(x(i),y(i),'or',[0 x(i)],[0 y(i)],'b')
    axis([-l,l,min(min(py))-2,max(max(py))+2]);
    axis square;
    subplot(1,2,2);
    plot(t,py,t(i),py(i),'or');
    ylabel('y(t)');
    xlabel('t');
    pause(0.1);
end

The only problem is that I get the following warning:

Warning: Input arguments must be scalar. 
> In mboya at 18

I've tried changing x to

x=zeros

or

x=0

but if I leave it that way, the animation doesn't work.

Thanks for the help. I just need that little thing solved.

Walter Roberson 2013-12-13

在 MATLAB Online 中打开

You define py=z(:,1); so py is a column vector whose length is the number of rows in z. The number of rows is the same as size(z,1) so zeros(1,size(z,1)) is a row vector with the same number of elements as the column vector py.

If you wanted x to be a column vector instead of a row vector, you could instead use

x = zeros(size(py));

and Yes for a row vector you could use

x = zeros(size(py)).' ;

if you wanted.

Miguel Ángel 2013-12-13

Alright, I think I got it. Thank you very much for your help.

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In an assignment A(I) = B, the number of elements in B and I must be the same. Error in odearguments (line 88) f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0. Error in ode45 (line 114) [neq, tspan, ntspan...

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In an assignment A(I) = B, the number of elements in B and I must be the same. Error in odearguments (line 88) f0 = feval(ode,​t0,y0,args​{:}); % ODE15I sets args{1} to yp0. Error in ode45 (line 114) [neq, tspan, ntspan...

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In an assignment A(I) = B, the number of elements in B and I must be the same. Error in odearguments (line 88) f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0. Error in ode45 (line 114) [neq, tspan, ntspan...

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