Hai,
I am working on optics. I have plotted a lens in matlab. The rays of light emerging out of the lens, after undergoing refraction in the lens, needs to be projected on a square plane in front of the lens. How is it possible to draw a plane in matlab? Looking forward to hearing from you soon.
BSD

 Respuesta aceptada

Walter Roberson
Walter Roberson el 21 de Sept. de 2011

0 votos

You can use patch() to draw a shape embedded in 3D.

7 comentarios

bsd
bsd el 23 de Sept. de 2011
I gave the command patch([0,0] [0,1] [1,1] [1,0]), it is showing an error "unbalanced or unexpected parenthesis or bracket". The above points are the coordinates of a square plane in 2D.
BSD
Fangjun Jiang
Fangjun Jiang el 23 de Sept. de 2011
You forgot the comma.
patch([0,0], [0,1], [1,1], [1,0])
bsd
bsd el 23 de Sept. de 2011
I added the comma, it is not showing error. But only a line from the point (0,0) to (0,1) is displayed, other lines of the square is not displayed.
BSD
Fangjun Jiang
Fangjun Jiang el 23 de Sept. de 2011
See doc patch for examples.
Walter Roberson
Walter Roberson el 23 de Sept. de 2011
All of the x data must be together, and all of the y data must be together.
patch([0 0 1 1 0], [0 1 1 0 0], [0 0 0 0 0])
Notice I have provided z coordinates here, to make it a 3D patch in order to make it a plane as was required by the original question. If all that is required is a 2D square then you could use rectangle() instead.
bsd
bsd el 24 de Sept. de 2011
http://www.opticasoftware.com/documentation/Rayica/PrinciplesOfRayica/4Customization/index.html#.1.2
go to the above link, and see the figure just immediately above the section 4.3. This is what I want to do using matlab, after the rays emerge out of the lens, and they hit an aperture in front of it, as seen in the above figure.
Walter Roberson
Walter Roberson el 24 de Sept. de 2011
Yup, you will be wanting non-zero z for that. The basic call will be as I gave above, patch([0 0 1 1 0], [0 1 1 0 0], [0 0 0 0 0]), but you will want different coordinates. The first parameter is a list of the x coordinates, the second parameter is a list of the corresponding y coordinates, and the third parameter is a list of the corresponding z coordinates.
You will also want to create a patch() for each of the apertures. You can use the usual sin() vs cos() coordinates if your square is perpendicular to one of the axes. For example,
sides = 17;
radius = 3;
angles = linspace(0,2*pi,sides+1) .';
coords = [radius * sin(angles), zeros(size(angles)), radius * cos(angles)];
xoff = 5; yoff = 2.8; zoff=3.3;
patch(xoff + coords(:,1), yoff + coords(:,2), zoff + coords(:,3));
and to draw the next circle, just change the xoff, yoff, zoff, and repeat the patch() command.

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Más respuestas (2)

Sean de Wolski
Sean de Wolski el 21 de Sept. de 2011

0 votos

You could also use a surf or a slice

1 comentario

Walter Roberson
Walter Roberson el 21 de Sept. de 2011
If your plane was along the z axis, then surf would not be able to handle it, as surf requires unique (x,y) for every z value.

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Adam Danz
Adam Danz el 4 de Oct. de 2024

0 votos

Starting in MATLAB R2024b, you can use constantplane to create a 2D plane in 3D axes. However, unlike surfaces and patches, the ConstantPlane will extend infinitely in all directions within the axis limits.
For example, to create a flat plane along the z axis at z=0,
constantplane('z',0)
grid on

2 comentarios

DGM
DGM el 4 de Oct. de 2024
Finally! This is the logical extension of xline() and yline().
Looks like they also fixed the plane-plotting bug in fimplicit3 mentioned here.
load data E C ax
planefun=@(x,y,z) E(1)*x+E(2)*y+E(3)*z+E(4);
plot3(C(1), C(2),C(3),'rx'); axis(ax); hold on
fimplicit3(planefun,ax,EdgeColor='none',FaceAlpha=0.5); hold off

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bsd
el 21 de Sept. de 2011

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el 5 de Oct. de 2024

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