Finite Temperature Variation - Heat Transfer

I am writing a code that displays temperature variation. There is a triangle section cut off of a square(nxn) that is exposed to convection heat transfer. The part that is cut off is a 90 degree triangle between i and n-.6*i. I need help defining this section for every row because the slope is not in one row. The cut out part is after a certain meter distance.

[Youssef Khmou]

you mean the triangle has angles (45°,45°,90°)? can you explain where is the difficulty exactly ?

[Kaelyn]

Yes, the triangle is at the top right corner of the square(nxn). I am having trouble identifying the cut off if the triangle on each row. The area is 1meter by 1meter and the missing area cut off is 1/2*.4^2. Giving a 45 degree angle.

[Image Analyst]

If the triangle is formed by two sides and one side is "i" elements long, and the other side is "n-0.6*i" elements long, how is that a 45 degree angle for the hypoteneuse? To get 45 degrees, both sides would have to be the same length.

Answer 1

[Image Analyst]

Do you mean like this:

clc;
n = 100
% Create sample random data.
myArray = randi(9, [n, n], 'uint8') + 100;
subplot(2, 2, 1);
imshow(myArray);
title('Original Image', 'FontSize', 20);

% Chop off i by n-6*i triangle
i = 4
xvertices = [0, n-6*i, 0];
yvertices = [0,0, i];
mask = poly2mask(xvertices, yvertices, n, n);
subplot(2,2,2);
imshow(mask);
title('Mask Image', 'FontSize', 20);

out = myArray .* uint8(~mask);
subplot(2,2,3);
imshow(out);
title('Output Image', 'FontSize', 20);

% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);

Note: requires Image Processing Toolbox because of poly2mask.

[Image Analyst]

Kaelyn's comment moved here, since it's not an answer to her original question, but a comment on mine.

Yes but n and i will change according to the size of the mesh. The cut off of the triangle will be identified and used to calculate other temperatures. So I have been trying to use the mod function.

[Image Analyst]

n and i are variables in my code. You can change them. You can do everything with a for loop, instead of poly2mask(), if you want.

Answer 2

[Youssef Khmou]

Kaelyn,

OK the problem now is clear , here is fast way

   N=500;  % more resolution better 2D heat conduction resolution .

   H=ones(N);

   M=flipud(triu(H));      % TOP RIGHT TRIANGLE
   surface(M);
   shading interp

[Kaelyn]

How can I attach this a bigger matrix. Would I just add on conditions.

[Youssef Khmou]

what do you mean attaching? high number of dimensions? increase N

[Kaelyn]

The triangle part of square matrix starts at .6 meters. The matrix is 1 by 1 meter. So the function you provided starts at n=1 (half the matrix equals zero). So I want to know can I start this function at .6 meters?

[Youssef Khmou]

then its not (45,45,90), as you confirmed,

[Youssef Khmou]

N=100; 
M=rot90(triu(ones(N),-60));
surface(M), shading interp

[Youssef Khmou]

is the problem solved by this last instruction?

Answer 3

[Youssef Khmou]

 N=1000;  % 1 meter sampled with 1000 Hz
 p=0.6*N;       %  your 0.6 starting point .

 M=zeros(N);   % initial matrix 

 A=(N-p)/N;    % the Coefficient for constructing the triangle

 for n=1:N
M(n,1:A*n)=1;
end