Why does one equation become a 1x2 matrix with one NaN, when all variables within the matrix equation are 1x1?

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Hello,
Basically as the title mentions, I am having a problem that an equation I'm using becomes a 1x2 double matrix, with a second member inside being NaN. Whilst debugging I checked the members within the equation, all of them seem to be 1x1, hence I am wondering what could be the reason for this, as at the moment I can't use the member found from the calculations as it causes further errors because it's a larger matrix than it should be.
Here is the code in which the equation goes wrong (the equation itself is x(k-1)=(sqrt(l3^2-d(k-1).^2)) ):
Angle.m
function [t, z] = Angle(tInc)
% Set init;ial conditions
global x;
t(1)=0;
z(1,1)=0; %1.3814; angle in radians
z(2,1) = 0; %angular velocity
mast=6;
l=1.15; %half the widht of the catamaran
l3=sqrt(mast^2+l^2); %resultant of half the width and mast
k=1;
while z(1,k)<=1.381426277 %0.18937005; %d=-mast-r:0;
k=k+1;
d=-mast*cos(z(1,k-1))+l*sin(z(1,k-1));
x(k-1)=(sqrt(l3^2-d(k-1).^2)); %shortest distance between the force of the air bag and hull
t(k)=t(k-1)+tInc;
[z(:,k)] = RKF(t(k-1), z(:,k-1), tInc);
end
end
This function also uses two more functions, I am unsure whether those are necessary, as the NaN occurs before calling the functions out, but just in case here they are:
RKF.m
function [z] = RKF(t, z, tInc)
global x;
%tolh=0.000002; %chosen tolerance for time increment. Used for RK5
%RK4
A=tInc*Dz(t,z);
B=tInc*Dz(tInc/2+t,z+A/2);
C=tInc*Dz(tInc/2+t,z+B/2);
D=tInc*Dz(tInc+t,z+C);
z=z + (A+2*B+2*C+D)/6;
end
Dz.m
function [dz] = Dz(t, z)
global x;
g=9.81; %/sqrt((6400+z(2)/1000)/6400);
m = 44.6; % Mass (kg)
a=1030; %water density
%a = (p*M)/(R*T); % Air density (kg/m^3)
l=1.15; %half the width of main body of catamaran
r=0.3; %diameter of missile (m)
V=(4*pi*r^3)/3; %volume
I=391.5; %moment of inertia for the catamaran
mast=6; %length of mast
l3=sqrt(mast^2+l^2); %distance between air bag and hull
Fb=a*g*V*x; %buoyancy force of the air bag
Fg=m*g*2*l*cos(z(1));
dz(1)=z(2); %dtheta
dz(2)=(Fb-Fg)/I %ddtheta
%Transpose the dz vector
dz=dz';
I hope someone may be able to help with this.
Thank you for your time.
Regards,
Edvardas
  1 Comment
Walter Roberson
Walter Roberson on 28 Apr 2013
At the command line, please command
dbstop if error
and then run your program. When it stops, please show which line it is on, and show size(mast), size(d), size(l3), size(k), size(z), and also show the value of k.

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Accepted Answer

Matt J
Matt J on 28 Apr 2013
At the command line do
>>dbstop if naninf
Then rerun your code. When it stops at the K>> prompt, show us the output of
K>> whos
Also paste the error messages for us.
  2 Comments
Edvardas
Edvardas on 28 Apr 2013
Edited: Edvardas on 28 Apr 2013
Hello,
dbstop didn't seem to stop the code, so I stopped it at the point where the NaN starts occurring manually:
From the whom command it's then:
K>> whos Name Size Bytes Class Attributes
Fb 1x2 16 double
Fg 1x1 8 double
I 1x1 8 double
V 1x1 8 double
a 1x1 8 double
dz 1x1 8 double
g 1x1 8 double
l 1x1 8 double
l3 1x1 8 double
m 1x1 8 double
mast 1x1 8 double
r 1x1 8 double
t 1x1 8 double
x 1x2 16 double global
z 2x1 16 double
If the function is not stopped the error which appears is this:
In an assignment A(I) = B, the number of elements in B and I must be the same.
Error in Dz (line 23)
dz(2)=(Fb-Fg)/I %ddtheta
Error in RKF (line 5)
A=tInc*Dz(t,z);
Error in Angle (line 30)
[z(:,k)] = RKF(t(k-1), z(:,k-1), tInc);
Fb being an equation with the multiplier x, which has the unexpected 1x2 with NaN becomes 1x2 as well, hence I get the A(I)=B error (I think).
Thank you for your time.
Regards,
Edvardas
Matt J
Matt J on 28 Apr 2013
Edited: Matt J on 28 Apr 2013
Do you have a copy of x in your base workspace (where it is also global)? If so, you should clear it before you run. Better yet, don't use global variables at all, since they are generally regarded as dangerous. Pass x explicitly to RFK and Dz or else make them Nested Functions within Angle.m, so that they can share its workspace.
Otherwise, I cannot reproduce the error you are seeing. When I run Angle(1) with tInc=1, I encounter another error
Attempted to access d(2); index out of bounds because numel(d)=1.
Error in test (line 15)
x(k-1)=(sqrt(l3^2-d(k-1).^2));
This makes sense because "d" is 1x1, yet you are accessing d(k-1) as if it were a vector. You should review that.

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More Answers (1)

Walter Roberson
Walter Roberson on 28 Apr 2013
You have
dz(2)=(Fb-Fg)/I
and before that
Fg=m*g*2*l*cos(z(1))
with m, g, l, and z(1) all being scalars, so Fg is a scalar, so look at Fb
Fb=a*g*V*x
a, g are constants, V you can trace back to see must be a scalar. But x? According to whos, x is 1 x 2... which would make Fb 1x2, which would make (Fb-Fb)/I into 1x2, which is not going to fit into dz(2).
You do not have any assignments to x, so the reason it is 1x2 lies in whatever created it outside of this. You will probably find that x(2) is NaN that that that is why the second element of the expression becomes NaN.
  4 Comments
Walter Roberson
Walter Roberson on 28 Apr 2013
Suppose the problem is not occurring on the first call to RKF, and suppose that the while condition is still true after that call leading there to be another round, k becomes 3, x(2) becomes assigned to ?
When it stops because of the error, use
dbup;dbup
to move the stack up to the angle function, and check out the k value at that level.
Edvardas
Edvardas on 28 Apr 2013
Hmm, I am uncertain of whether I understood the "k becomes 3, x(2) becomes assigned to ?". After looking through Matt's analysis it did come to me that using x as a vector might have been the mistake. So I changed x into a regular variable, so that it would keep overwriting itself with new values all the time and it seems to have solved the problem, and the ODE seems like it's being solved now.
In terms of clearing, it wasn't the case. It simply created the 1x2 matrix from scratch. I am still uncertain as to why though. And in terms of the global being dangerous, I have been warned before as well, will try using nested functions.
Thank you both for your time and help, I really appreciate it :)

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