Asked by Kaelyn
on 12 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.

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Answer by Image Analyst
on 13 Oct 2013

Edited by Image Analyst
on 13 Oct 2013

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
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.

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Answer by Youssef Khmou
on 13 Oct 2013

Edited by Youssef Khmou
on 13 Oct 2013

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

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Youssef Khmou
on 13 Oct 2013

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

Youssef Khmou
on 13 Oct 2013

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

Youssef Khmou
on 13 Oct 2013

is the problem solved by this last instruction?

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

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## 3 Comments

## Youssef Khmou (view profile)

Direct link to this comment:https://www.mathworks.com/matlabcentral/answers/90035-finite-temperature-variation-heat-transfer#comment_173857

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

## Kaelyn (view profile)

Direct link to this comment:https://www.mathworks.com/matlabcentral/answers/90035-finite-temperature-variation-heat-transfer#comment_173873

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 (view profile)

Direct link to this comment:https://www.mathworks.com/matlabcentral/answers/90035-finite-temperature-variation-heat-transfer#comment_173875

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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