I encountered error while running program(runga kutta solving using function handle) for solving a set of differential equations and applying a controller 'u'. How to tackle it out?

matrix dimension doesn't match in line 86 while solving the attached file 'abc'

[SHIJINA P P]

I want out(4,i) to be an array, but when that is used in u (via sdot), error occurs, Is it due due the presence of 'u' in function handle?

[Geoff Hayes]

u appears to be used on a line following the one that generates the error. If you use the MATLAB debugger, what do you see?

Answer 1

[Geoff Hayes]

Shijina - use the MATLAB debugger to step through the code and see what is happening. Consider these lines which initializes a variable named out as

 x=[21900;301.4*2;0;0;2*pi/180;0]; 
 out( :,1)=x;

and so out is a 6x1 array. Now, in your for loop, you do some work and then update out as

 out(:,i+1)=x;

and so out is now a 6x2 array. The problem is with line 86

 sdot=(1/d)*(-out(4,:)-(rho*A*out(2,:)*Cl/(2*m)-(g/out(2,:)-...   
      out(2,:)/(R+out(1,:)))*cos(out(3,:)))-rho*A*out(2,:)*Cl/(2*m)-(g/out(2,:)-...
      out(2,:)/(R+out(1,:)))*cos(out(3,:))-lambda*(rho*(out(2,:))^2*A*D/(2*I)*(-...
      Cm+Cmq*D*out(4,:)/out(2,:))))

Look how out is used. On the first iteration, when out is a 6x1 array, out(4,:) is a scalar as are all other instances where we access this variable. But on the subsequent iteration of the loop, out(4,:) (and the others) will be 1x2 arrays. Is this what you want? Or do you just want the scalar from the previous iteration i.e.

 out(4,i)

were i is the indexing variable in your for loop.