Extract row from Surface

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Detox
Detox 2016 年 7 月 5 日
コメント済み: José-Luis 2016 年 7 月 6 日
I created a surface with the following commands:
x2 = xCoord2;
y2 = yCoord2;
z2 = zCoord2;
tri2 = delaunay(x2,y2);
figure
trisurf(tri2,x2,y2,z2)
The problem I have now is the following: How can I extract a "row" from the surface in order to generate a x-z-Plot for an arbitrary row? Is this even possible for a triangularized surface? Thanks in advance!

回答 (2 件)

KSSV
KSSV 2016 年 7 月 6 日
編集済み: KSSV 2016 年 7 月 6 日
It is not that easy in Triangular Surface plot to extract a row of (x,z). Because, if you fix to some value of x, there is no fast rule that there exists a straight line with given nodal coordinates for chosen x. The line will deviate in a range of x+dx or x-dx. Three ways to achieve your task.
1) You have an option of plotting contours. If you want to plot a certain value z, you can use contour algorithms. These algorithms give you location (x,y) which have given value 'z'. From these locations you can search your x. Again not exactly x but you need to give a range x+dx to x-dx. You may find the contouring algorithm on triagular unstructured meshes from the below link:
https://in.mathworks.com/matlabcentral/fileexchange/38925-linearly-interpolate-triangulation/content/interptri.m
2) Fix your x,z and extract all the elements which have the given x. You will end up with the elements having the given x and z.
3) Convert your unstructured grid to structured, which makes extracting specific 'x' very easy.
  2 件のコメント
José-Luis
José-Luis 2016 年 7 月 6 日
I don't understand what you mean.
Detox
Detox 2016 年 7 月 6 日
Thanks for the detailed answer mate! I will give it a try later on and tell you how it worked.

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José-Luis
José-Luis 2016 年 7 月 6 日
編集済み: José-Luis 2016 年 7 月 6 日
%Say you want the line at x = 0.2
val = 0.2;
% Define the input grid
[x, y] = meshgrid(linspace(-1, 1));
% Calculate the two surfaces
z1 = y.^2 + 2*x;
z2 = x;
% Visualize the two surfaces
surface(x, y, z1, 'FaceColor', [0.5 1.0 0.5], 'EdgeColor', 'none');
surface(val.*ones(size(x)), y, z2, 'FaceColor', [1.0 0.5 0.0], 'EdgeColor', 'none');
view(3); camlight; axis vis3d
% Take the difference between the two surface heights and find the contour
% where that surface is zero.
zdiff = z1 - z2;
C = contours(val.*ones(size(x)), y, zdiff, [0 0]);
% Extract the x- and y-locations from the contour matrix C.
xL = C(1, 2:end);
yL = C(2, 2:end);
% Interpolate on the first surface to find z-locations for the intersection
% line.
zL = interp2(x, y, z1, xL, yL);
% Visualize the line.
line(xL, yL, zL, 'Color', 'k', 'LineWidth', 3);
xL and yL should be the coordinates of the intersection of the two planes. It hopefully should be evident from the plot. You'd need to replace x, y and z with your actual values.
  2 件のコメント
Detox
Detox 2016 年 7 月 6 日
Thanks for your answer. I do not think I can apply the meshgrid function on my x and y arrays. I will see if it works! :)
José-Luis
José-Luis 2016 年 7 月 6 日
You wouldn't need to apply meshgrid, just replace with the values you actually have.

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