I'm trying to create 3D plots in Matlab, and I have practically no experience. I'd really like to draw the figure described by these equations:
x = cos(u) sinh(t) / (cosh(t) - cos(u));
y = cos(u) sin(u) / (cosh(t) - cos(u));
z = sin(u);
where both u and t vary from -pi to pi. This is what Paul Bourke calls Ghost Plane.
clc
clear
a = -pi:.01:pi;
b = -pi:.01:pi;
[U,T] = meshgrid(a,b);
X = (cos(U).*sinh(T))./(cosh(T)-cos(U));
Y = (cos(U).*sin(U))./(cosh(T)-cos(U));
Z = sin(U);
figure
surf(X,Y,Z)
With the code, I get something... indescribable. How do I plot the figure?
Your code is correct. You just need to zoom in.
Some tips to make this more viewable:
Use less fine grids, by changing a = ... and b = ... to: linspace(-pi,pi,40);
Add this ,'FaceColor','none','EdgeColor','interp'); to the surf command to only plot the lines and those in color.
Add this axis equal vis3d; after the surf command, so the axis will have correct scaling and behave well while rotating.
Add whitebg('black'); and grid off; to make the background black.
Change to surf(X,Y,Z,-U,...); and add colormap('HSV'); if you want the same colors as in the original.
set(gca,'xtick',[]); set(gca,'xticklabel',[]); set(gca,'yticklabel',[]); set(gca,'ytick',[]); to remove the axis ticks
You might want to use the cameratoolbar to: Change the projection to perspective projection, zoom all the way out, move the camera in very close, to get nice perspective distortions.
Voilà:
Related
Consider the simple code below that generates a straight downward sloping line in MATLAB.
clear, clc, close all
t = 0:0.1:1;
y = -t+1;
plot(t,y)
ax = gca
This is a line with slope -1, so the (acute) angle between the horizontal axis and the line is 45 degrees. Except it isn't when you measure with a protractor on your monitor.
Without changing the range of values displayed on the x and y axes or the height of the figure window, how could I ensure I would measure 45 degrees from the horizontal axis to the line if I held a protractor up to the screen?
My current approach is to change the width of the figure window. In the limit as the figure window is infinitely thin, the line x is a vertical line. Conversely, if the figure window is stretched to the edges of a monitor, it flattens out. Somewhere in the middle, the line has the angle I want. I just can't find a good way to mathematically find this point and instantiate it in code.
Edit: A slightly more generic solution for any acute angle. (I didn't test obtuse angles.)
clear, clc, close all
ang_deg = 70;
slope = tand(ang_deg);
t = 0:0.1:1;
y = -slope*t+1;
f = figure;
f.Position(3) = f.Position(3)*1.5;
plot(t,y)
% For a given height, change the width
ax = gca;
ax.Units = 'pixels';
ax.Position(3) = ax.Position(4)/slope;
You can achieve that with
axis equal
which, according to the documentation,
uses the same length for the data units along each axis.
You may want to also use
axis tight
which
fits the axes box tightly around the data by setting the axis limits equal to the range of the data
Follow up your commands with a declaration that you'll be working in pixels, and then set the width to the height:
ax.Units = 'pixels';
ax.Position(3) = ax.Position(4);
I have this 3D image generated from the simple code below.
% Input Image size
imageSizeY = 200;
imageSizeX = 120;
imageSizeZ = 100;
%# create coordinates
[rowsInImage, columnsInImage, pagesInImage] = meshgrid(1:imageSizeY, 1:imageSizeX, 1:imageSizeZ);
%# get coordinate array of vertices
vertexCoords = [rowsInImage(:), columnsInImage(:), pagesInImage(:)];
centerY = imageSizeY/2;
centerX = imageSizeX/2;
centerZ = imageSizeZ/2;
radius = 28;
%# calculate distance from center of the cube
sphereVoxels = (rowsInImage - centerY).^2 + (columnsInImage - centerX).^2 + (pagesInImage - centerZ).^2 <= radius.^2;
%# Now, display it using an isosurface and a patch
fv = isosurface(sphereVoxels,0);
patch(fv,'FaceColor',[0 0 .7],'EdgeColor',[0 0 1]); title('Binary volume of a sphere');
view(45,45);
axis equal;
grid on;
xlabel('x-axis [pixels]'); ylabel('y-axis [pixels]'); zlabel('z-axis [pixels]')
I have tried plotting the image with isosurface and some other volume visualization tools, but there remains quite a few surprises for me from the plots.
The code has been written to conform to the image coordinate system (eg. see: vertexCoords) which is a left-handed coordinate system I presume. Nonetheless, the image is displayed in the Cartesian (right-handed) coordinate system. I have tried to see this displayed as the figure below, but that’s simply not happening.
I am wondering if the visualization functions have been written to display the image the way they do.
Image coordinate system:
Going forward, there are other aspects of the code I am to write for example if I have an input image sphereVoxels as in above, in addition to visualizing it, I would want to find north, south east, west, top and bottom locations in the image, as well as number and count the coordinates of the vertices, plus more.
I foresee this would likely become confusing for me if I don’t stick to one coordinate system, and considering that the visualization tools predominantly use the right-hand coordinate system, I would want to stick with that from the onset. However, I really do not know how to go about this.
Right-hand coordinate system:
Any suggestions to get through this?
When you call meshgrid, the dimensions x and y axes are switched (contrary to ndgrid). For example, in your case, it means that rowsInImage is a [120x100x200] = [x,y,z] array and not a [100x120x200] = [y,x,z] array even if meshgrid was called with arguments in the y,x,z order. I would change those two lines to be in the classical x,y,z order :
[columnsInImage, rowsInImage, pagesInImage] = meshgrid(1:imageSizeX, 1:imageSizeY, 1:imageSizeZ);
vertexCoords = [columnsInImage(:), rowsInImage(:), pagesInImage(:)];
The background of this problem relates to my attempt to combine output from a ray tracer with Matlab's 3d plotters. When doing ray tracing, there is no need to apply a perspective transformation to the rendered image. You see this in the image below. Basically, the intersections of the rays with the viewport will automatically adjust for the perspective scaling.
Suppose I've gone and created a ray-traced image (so I am given my camera, my focal length, viewport dimensions, etc.). How do I create exactly the same view in Matlab's 3d plotting environment?
Here is an example:
clear
close all
evec = [0 200 300]; % Camera position
recw = 200; % cm width of box
recl = 200; % cm length of box
h = 150; % cm height of box
% Create the front face rectangle
front = zeros(3,5);
front(:,1) = [-recw/2; 0; -recl/2];
front(:,2) = [recw/2; 0; -recl/2];
front(:,3) = [recw/2; h; -recl/2];
front(:,4) = [-recw/2; h; -recl/2];
front(:,5) = front(:,1);
% Back face rectangle
back = zeros(3,5);
back(:,1) = [-recw/2; 0; recl/2];
back(:,2) = [recw/2; 0; recl/2];
back(:,3) = [recw/2; h; recl/2];
back(:,4) = [-recw/2; h; recl/2];
back(:,5) = back(:,1);
% Plot the world view
figure(1);
patch(front(1,:), front(2,:), front(3,:), 'r'); hold all
patch(back(1,:), back(2,:), back(3,:), 'b');
plot3(evec(1), evec(2), evec(3), 'bo');
xlabel('x'); ylabel('y'); zlabel('z');
title('world view'); view([-30 40]);
% Plot the camera view
figure(2);
patch(front(1,:), front(2,:), front(3,:), 'r'); hold all
patch(back(1,:), back(2,:), back(3,:), 'b');
xlabel('x'); ylabel('y'); zlabel('z');
title('Camera view');
campos(evec);
camup([0 1 0]); % Up vector is y+
camproj('perspective');
camtarget([evec(1), evec(2), 0]);
title('camera view');
Now you see the world view
and the camera view
I know how to adjust the camera position, the camera view angle, and orientation to match the output from my ray tracer. However, I do not know how to adjust Matlab's built-in perspective command
camproj('perspective')
for different distortions.
Note: within the documentation, there is the viewmtx command, which allows you to output a transformation matrix corresponding to a perspective distortion of a certain angle. This is not quite what I want. I want to do things in 3D and through Matlab's OpenGL viewer. In essence, I want a command like
camproj('perspective', distortionamount)
so I can match up the amount of distortion in Matlab's viewer with the distortion from the ray tracer. If you use the viewmtx command to create the 2D projections, you will not be able to use patch' orsurf' and keep colours and faces intact.
The MATLAB perspective projection works just like your raytracer. You don't need any transformation matrices to it use it. Perspective distortion is determined entirely by the camera position and direction of projection.
In the terminology of the raytracer diagram above, if the CameraPosition matches your raytracer's pinhole coordinates and the vector between CameraPosition and CameraTarget is perpendicular to your raytracer's viewport, the perspective distortion will also match. The rest is just scaling and alignment.
I am trying to draw the lines or the edges of a cone using plot3 in matlab. Any help please? I do not need the surface. I need the edges only. SO that I can patch something on it. A useful link. But i need the circle at the bottom:
https://patentimages.storage.googleapis.com/US8514658B2/US08514658-20130820-D00021.png
Few horizontal lines are fine. But no tilted line as i need to patch something inside.
cylinder is your friend here...
You just need to pass it a vector of radii* and transpose the output*...
* negative radii tending to zero will flip the order so the apex is on top...
* so it draws rings not lines from the base to the apex
numRings = 10;
numPointsAround = 100;
[x,y,z] = cylinder(linspace(-1,0,nlines),numPointsAround);
plot3(y.',x.',z.','-k')
I think this is what you want. Most of the answer is directly taken from the above answer by #RTL.
numRings = 2;
numPointsAround = 100;
[x,y,z] = cylinder(linspace(-1,0,numRings),numPointsAround);
plot3(y.',x.',z.','-k')
hold on;line([-0.5878;0], [0.809;0],[0;1]);
hold on;line([0.9511;0], [-0.309;0],[0;1]);
axis square
Below is an arbitrary hand-drawn Intensity profile of a line in an image:
The task is to draw the line. The profile can be approximated to an arc of a circle or ellipse.
This I am doing for camera calibration. Since I do not have the actual industrial camera, I am trying to simulate the correction needed for calibration.
The question can be rephrased as I want pixel values which will follow a plot similar to the above. I want to do this using program (Preferably using opencv) and not manually enter these values because I have thousands of pixels in the line.
An algorithm/pseudo code will suffice. Also please note that I do not have any actual Intensity profile, otherwise I would have read those values.
When will you encounter such situation ?
Suppose you take a picture (assuming complete white) from a Camera, your object being placed on table, and camera just above it in vertical direction. The light coming on the center of the picture vertically downward from the camera will be stronger in intensity as compared to the light reflecting at the edges. You measure pixel values across any line in the Image, you will find intensity curve like shown above. Since I dont have camera for the time being, I want to emulate this situation. How to achieve this?
This is not exactly image processing, rather image generation... but anyways.
Since you want an arc, we still need three points on that arc, lets take the first, middle and last point (key characteristics in my opinion):
N = 100; % number of pixels
x1 = 1;
x2 = floor(N/2);
x3 = N;
y1 = 242;
y2 = 255;
y3 = 242;
and now draw a circle arc that contains these points.
This problem is already discussed here for matlab: http://www.mathworks.nl/matlabcentral/newsreader/view_thread/297070
x21 = x2-x1; y21 = y2-y1;
x31 = x3-x1; y31 = y3-y1;
h21 = x21^2+y21^2; h31 = x31^2+y31^2;
d = 2*(x21*y31-x31*y21);
a = x1+(h21*y31-h31*y21)/d; % circle center x
b = y1-(h21*x31-h31*x21)/d; % circle center y
r = sqrt(h21*h31*((x3-x2)^2+(y3-y2)^2))/abs(d); % circle radius
If you assume the middle value is always larger (and thus it's the upper part of the circle you'll have to plot), you can draw this with:
x = x1:x3;
y = b+sqrt(r^2-(x-a).^ 2);
plot(x,y);
you can adjust the visible window with
xlim([1 N]);
ylim([200 260]);
which gives me the following result: