How do I view "hidden" contour lines on a surface?
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The following Matlab code produces a 3d surface plot with a superimposed contour line. The resulting contour line has several missing segments. If I manually rotate the surface, I can see the missing segments behind the surface. Is there some way to make the entire contour line visible?
x=linspace(0,3,10); y=linspace(0,2,10);
z=[ 2.08, 2.04, 1.91, 1.64, 1.21, 0.74, 0.66, 0.55, 0.33, 0.00; 2.01, 1.87, 1.66, 1.39, 1.03, 0.73, 0.65, 0.47, 0.17, 0.00; 1.79, 1.68, 1.29, 1.11, 0.85, 0.73, 0.56, 0.20, 0.00, 0.00; 1.75, 1.62, 1.20, 0.99, 0.66, 0.59, 0.47, 0.00, 0.00, 0.00; 1.51, 1.35, 1.08, 0.67, 0.53, 0.57, 0.22, 0.00, 0.00, 0.00; 1.06, 0.81, 0.69, 0.38, 0.46, 0.40, 0.00, 0.00, 0.00, 0.00; 0.45, 0.38, 0.21, 0.12, 0.31, 0.04, 0.00, 0.00, 0.00, 0.00; 0.06, 0.04, 0.00, 0.01, 0.09, 0.00, 0.00, 0.00, 0.00, 0.00; 0.00, 0.00, 0.00, 0.00, 0.00, 0.00, 0.00, 0.00, 0.00, 0.00; 0.00, 0.00, 0.00, 0.00, 0.00, 0.00, 0.00, 0.00, 0.00, 0.00];
figure; surf(x, y, z); hold on
view(120,40); colormap(winter)
[C,h]=contour3(x, y, z, [1.0 1.0], 'r');
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Respuestas (1)
John D'Errico
el 27 de Feb. de 2014
I'd suggest turning the surface mildly transparent. In the call to surf, try this instead:
surf(x, y, z,'facealpha',.5)
The rest stays the same.
2 comentarios
Walter Roberson
el 27 de Feb. de 2014
One effect of using transparency would be to force OpenGL renderer. But if that was not what was being used then just the switch of renderer might be the trick rather than the transparency itself. Try
set(gcf, 'Renderer', 'opengl')
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