storing script outputs as an array

I wanted to try identifying convex polygons by their projected widths. How can I get the outputs, D, stored in a compact array that I can save and work with?

%loop that will have 1440 iterations (each iteration
%representing a rotation of the plane by 0.5 degrees)
for t= 1:2000

      format('long')

      %imagine points in 2-dimensional euclidean space.  These three vectors
      %represent the three vertices of an equilateral triangle cetered at the
      %origin
      A=[0;1];
      B=[-0.86603; -0.50000];
      C=[0.86603; 0.50000];

      %T is designed to be a 0.5 degree counter-clockwise rotation matrix, sending each 
      %of the previous vectors to their 0.5 degree rotated counterparts
      T= [ 0.99996 -0.00873;0.00873 0.99996];

      %P extracts the x values of each vector above, discarding the rest
      P= [1 0];

      %represent the rotated images of A,B,C, for some iteration 't'
      tA=(T^t)*A;
      tB=(T^t)*B;
      tC=(T^t)*C;

      %represent the projection of tA,tB,tC onto the x-axis. 
      imA=P*tA;
      imB=P*tB;
      imC=P*tC;

      %distances between the respective projections
      distAB=abs(imA-imB);
      distBC=abs(imB-imC);
      distCA=abs(imC-imA);

      %making a vector V with elements 'distAB' 'distBC' 'distCA'
      X=[1 0 0];
      Y=[0 1 0];
      Z=[0 0 1];

      dX=distAB*X;
      dY=distBC*Y;
      dZ=distCA*Z;

      V= dX+dY+dZ;

      % D = max element in V is the width of the 'shadow' of the triangle on the
      %x-axis
      D=max(V)

end

[Jan]

What is your question?

[Erik Anderson]

my question is

"How can I get the outputs, D, stored in a compact array that I can save and work with?"

right now I am using

Output=zeros(1,10000); Output(t)=D; xlswrite('triangleout.xls',Output);

seems to work well

Answer 1

[per isakson]

I guess: this is what you try to do.

D = nan( 1, 2000 );
format('long')

%imagine points in 2-dimensional euclidean space.  These three vectors
%represent the three vertices of an equilateral triangle cetered at the
%origin
A=[0;1];
B=[-0.86603; -0.50000];
C=[0.86603; 0.50000];

%T is designed to be a 0.5 degree counter-clockwise rotation matrix, sending each 
%of the previous vectors to their 0.5 degree rotated counterparts
T= [ 0.99996 -0.00873;0.00873 0.99996];

%P extracts the x values of each vector above, discarding the rest
P= [1 0];

for t= 1:2000
      %represent the rotated images of A,B,C, for some iteration 't'
      tA=(T^t)*A;
      tB=(T^t)*B;
      tC=(T^t)*C;
      %
      %represent the projection of tA,tB,tC onto the x-axis. 
      imA=P*tA;
      imB=P*tB;
      imC=P*tC;
      %
      %distances between the respective projections
      distAB=abs(imA-imB);
      distBC=abs(imB-imC);
      distCA=abs(imC-imA);
      %
      %making a vector V with elements 'distAB' 'distBC' 'distCA'
      X=[1 0 0];
      Y=[0 1 0];
      Z=[0 0 1];
      %
      dX=distAB*X;
      dY=distBC*Y;
      dZ=distCA*Z;
      V= dX+dY+dZ;
      %
      % D = max element in V is the width of the 'shadow' of the triangle on the
      %x-axis
      D(t)=max(V);
end

especially note

D = nan( 1, 2000 );
...
D(t) = max(V);

in the original code D was overwritten two thousand times

[Erik Anderson]

ah, yes. Thank you very much for your help! Much appreciated