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Finite Temperature Variation - Heat Transfer

Asked by Kaelyn on 12 Oct 2013
Latest activity Commented on by Youssef KHMOU on 13 Oct 2013

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 on 12 Oct 2013

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

Kaelyn on 13 Oct 2013

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 on 13 Oct 2013

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.



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3 Answers

Answer by Image Analyst on 13 Oct 2013
Edited by Image Analyst on 13 Oct 2013

Do you mean like this:

n = 100
% Create sample random data.
myArray = randi(9, [n, n], 'uint8') + 100;
subplot(2, 2, 1);
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);
title('Mask Image', 'FontSize', 20);
out = myArray .* uint8(~mask);
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 on 13 Oct 2013

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 on 13 Oct 2013

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.

Image Analyst
Answer by Youssef KHMOU on 13 Oct 2013
Edited by Youssef KHMOU on 13 Oct 2013


OK the problem now is clear , here is fast way

   N=500;  % more resolution better 2D heat conduction resolution .
   M=flipud(triu(H));      % TOP RIGHT TRIANGLE
   shading interp


Youssef KHMOU on 13 Oct 2013

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

Youssef KHMOU on 13 Oct 2013
surface(M), shading interp
Youssef KHMOU on 13 Oct 2013

is the problem solved by this last instruction?

Youssef  KHMOU
Answer by Youssef KHMOU on 13 Oct 2013
 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


Youssef  KHMOU

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