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Temperature distribution contour plot

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Bavi John
Bavi John on 30 Oct 2020
Edited: Cris LaPierre on 30 Oct 2020
I want to create a contour plot,like above, of a plate with temperatures at specific points on the plate. The data should be interpolated bicubical. Now i have just a few points on my plate but wanted the contour and so the interpolated data over the full size of the plate is this possible?
Below you can find some example data with x- and y-coordinates and also some temperatures corresponding to the coordinates.
X_Plate=200; % length mm
Y_Plate=400; % heigth mm
x=[-80 -50 0 50 60]; % x-coordinates
y=[-150 -100 0 100 130]; % y-coordinates
T=[15 20 30 10 5]; % Temperature °C
% the coordinate system has its origin in the center of the plate
I would be very pleased if someone could help me.

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Accepted Answer

Cris LaPierre
Cris LaPierre on 30 Oct 2020
Edited: Cris LaPierre on 30 Oct 2020
In order to create a contour plot, you will need to have a temperature measurement for each (x,y) permutation. This means T needs to be a matrix with the same number of rows as there are values in y, and the same number of columns as there are values in x. The column and row indices of T are the x and y coordinates in the plane, respectively.
x=[-80 -50 0 50 60]; % x-coordinates
y=[-150 -100 0 100 130]; % y-coordinates
T=[15 20 30 10 5]; % Temperature °C
z=ones(5,1)*T
z = 5×5
15 20 30 10 5 15 20 30 10 5 15 20 30 10 5 15 20 30 10 5 15 20 30 10 5
contourf(x,y,z)

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Cris LaPierre
Cris LaPierre on 30 Oct 2020
Sure. I'm taking advantage of linear indexing. You might find the sub2ind function helpful for this.
Bavi John
Bavi John on 30 Oct 2020
Is it also possible with coordinates like below?
The contour plot can't deal with vectors, if they don't strictly increasing.
x=[0 0 -80 30 60]; % x-coordinates
y=[0 140 0 50 -10]; % y-coordinates
Error using contourf (line 57)
Vector X must be strictly increasing or strictly decreasing with no repeated values.
Cris LaPierre
Cris LaPierre on 30 Oct 2020
You would apply the principle, rather than the implementation. Here, you need to define x and y as matrices I think.Then, using the same T values as before, create z using the linear indexing of the (x,y) coordinates.
x=[-80 0 30 60]; % x-coordinates
y=[-10 0 50 140];
[X,Y] = meshgrid(x,y)
X = 4×4
-80 0 30 60 -80 0 30 60 -80 0 30 60 -80 0 30 60
Y = 4×4
-10 -10 -10 -10 0 0 0 0 50 50 50 50 140 140 140 140
T=[15 20 30 10 5]; % Temperature °C
z=zeros(length(y),length(x));
% Create linear index of the desired x and y value pairs
lind = sub2ind([4,4],[2 4 2 3 1],[2 2 1 3 4]);
z(lind) = T
z = 4×4
0 0 0 5 30 15 0 0 0 0 10 0 0 20 0 0
contourf(X,Y,z)
colorbar

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More Answers (1)

Ameer Hamza
Ameer Hamza on 30 Oct 2020
Edited: Ameer Hamza on 30 Oct 2020
You first need to use scatteredInterpolant to convert the data to a grid format and then call contourf(). Since you have very few data points, so the variation will also be small
x=[-80 -50 0 50 60]; % x-coordinates
y=[-150 -100 0 100 130]; % y-coordinates
T=[15 20 30 10 5]; % Temperature °C
% the coordinate system has its origin in the center of the plate
mdl = scatteredInterpolant(x(:), y(:), T(:), 'natural');
xg = linspace(min(x), max(x), 20);
yg = linspace(min(x), max(x), 20);
[Xg, Yg] = meshgrid(xg, yg);
Zg = mdl(Xg, Yg);
contourf(Xg, Yg, Zg, 10);

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Bavi John
Bavi John on 30 Oct 2020
Thanks for your suggestion. Didn't know this scatteredInterpolant, It will be very useful I think.

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