# Cannot calculate gradient of contour plot

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KostasK 2020 年 9 月 24 日
コメント済み: KostasK 2020 年 9 月 24 日
Hi all,
I am currently attempting to draw a quiver plot on top of a contour plot, such that I can see the gradient of the contour function. However for some reason the gradient calculation does not take place correctly as all the quivers are far from being perpendicuar to the contours. This suprises me as the task itself is pretty simple and I have managed to accomplish it with mock data, see here:
clear ; clc
% Grid matrices
x = -2:0.25:2;
y = x;
[X,Y] = meshgrid(x);
% Function Matrix
F = X.*exp(-X.^2-Y.^2);
[FU, FV] = gradient(F, diff(x(1:2))) ;
% Plot
figure
hold on
contour(X, Y, F)
quiver(X, Y, FU, FV)
However, using my own data (you can find them attached its a simple 30x30 matrix), the same process does not seem to accomplish the task:
clear ; clc
% Grid matrices
x = linspace(-0.1, 0.1, 30) ;
y = linspace(8, 12, 30) ;
[X, Y] = meshgrid(x, y) ;
% Function Matrix
[FU, FV] = gradient(F, diff(x(1:2)), diff(y(1:2))) ;
% Plot
figure
hold on
contour(X, Y, F)
quiver(X, Y, FU, FV)
The weird thing is that the countour plot gets generated correctly, however something is off with the quiver. Even when I scale down the quiver vectors to say 0.2, they don't have the correct orientation relative to the contours.
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KostasK 2020 年 9 月 24 日

Its a fairly complex process of generating F, that's why I have attached it as a .mat file. I assumed that it shouldn't matter how F is generated, but rather that it forms a good contour plot.

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### 採用された回答

Bjorn Gustavsson 2020 年 9 月 24 日
If you modify your call to gradient to:
[FU, FV] = gradient(F, diff(y(1:2)), diff(x(1:2)));
You get a more reasonable set of quivers:
contour(X, Y, F)
hold on
quiver(X(1:2:end,1:2:end), Y(1:2:end,1:2:end), FU(1:2:end,1:2:end)/2, FV(1:2:end,1:2:end)/2,0)
The zero is to turn off the automatic scaling in quiver.
HTH
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KostasK 2020 年 9 月 24 日
Yes, it appears that scaling is the issue here. Because when I use axis equal even though the contour is greately distorted and inneligible, the gradient appears to be correct. So your answer works for me.

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