I need to plot a 2D smooth curve in Matlab. I have found this
npts = 10;
xy = [randn(1,npts); randn(1,npts)];
plot(xy(1,:),xy(2,:),'ro','LineWidth',2);
text(xy(1,:), xy(2,:),[repmat(' ',npts,1), num2str((1:npts)')])
ax = gca;
ax.XTick = [];
ax.YTick = [];
hold on
fnplt(cscvn(xy),'r',2)
hold off
This is the kind of curve I am looking for. But I need the exact coordinates of the curve (as many as possible x,y points) and to show the curve in a animated way,controlling the speed. Also, I need to plot on the line a circle or any other object following the trajectory (I need to control the speed the circle follows the line).
thanks
Related
Having this coordinate system:
And this dominant vertical vanishing point:
I would like to rotate the image around x axis so the vanishing point is at infinity. That means that all vertical lines are parallel.
I am using matlab. I find the line segmentes using LSD and the vanishing point using homogeneous coordinates. I would like to use angle-axis representation, then convert it to a rotation matrix and pass this to imwarp and get the rotated image. Also would be good to know how to rotate the segments. The segments are as (x1,y1,x2,y2).
Image above example:
Vanishin point in homogenous coordinates:
(x,y,z) = 1.0e+05 * [0.4992 -2.2012 0.0026]
Vanishin point in cartesian coordinates (what you see in the image):
(x,y) = [190.1335 -838.3577]
Question: With this vanishing point how do I compute the rotation matrix in the world x axis as explained above?
If all you're doing is rotating the image so that the vector from the origin to the vanishing point, is instead pointing directly vertical, here's an example.
I = imread('cameraman.tif');
figure;imagesc(I);set(gcf,'colormap',gray);
vp=-[190.1335 -838.3577,0]; %3d version,just for cross-product use,-ve ?
y=[0,1,0]; %The vertical axis on the plot
u = cross(vp,y); %you know it's going to be the z-axis
theta = -acos(dot(vp/norm(vp),y)); %-ve ?
rotMat = vrrotvec2mat([u, theta]);
J=imwarp(I,affine2d (rotMat));
figure;imagesc(J);set(gcf,'colormap',gray); %tilted image
You can play with the negatives, and plotting, since I'm not sure about those parts applying to your situation. The negatives may come from plotting upside down, or from rotation of the world vs. camera coordinate system, but I don't have time to think about it right now.
EDIT
If you want to rotation about the X-axis, this might work (adapted from https://www.mathworks.com/matlabcentral/answers/113074-how-to-rotate-an-image-along-y-axis), or check out: Rotate image over X, Y and Z axis in Matlab
[rows, columns, numberOfColorChannels] = size(I);
newRows = rows * cos(theta);
rotatedImage = imresize(I, [newRows, columns]);
In MATLAB, say I have the parameters for an ellipse:
(x,y) center
Minor axis radius
Major axis radius
Angle of rotation
Now, I want to generate random points that lie within that ellipse, approximated from a 2D gaussian.
My attempt thus far is this:
num_samps = 100;
data = [randn(num_samps, 1)+x_center randn(num_samps, 1)+y_center];
This gives me a cluster of data that's approximately centered at the center, however if I draw the ellipse over the top some of the points might still be outside.
How do I enforce the axis rules and the rotation?
Thanks.
my assumptions
x_center = h
y_center = k
Minor Axis Radius = b
Major Axis Raduis = a
rotation angle = alpha
h=0;
k=0;
b=5;
a=10;
alpha=30;
num_samps = 100;
data = [randn(num_samps, 1)+h randn(num_samps, 1)+k];
chk=(((((data(:,1)-h).*cos(alpha)+(data(:,2)-k).*sin(alpha))./a).^2) +...
(((data(:,1)-h).*sin(alpha)+(data(:,2)-k).*cos(alpha))./b).^2)<=1;
idx=find(chk==0);
if ~isempty(idx)
data(idx,:)=data(idx,:)-.5*ones(length(idx),2);
end
I use camera calibration in matlab to detect some checkerboard patterns, after
figure; showExtrinsics(cameraParams, 'CameraCentric');
Now, I want to rotate the checkerboard patterns around the x-axis such that all of them have nearly the same y coordinates in the camera frame.
Method:
I get the positions of all patterns in the camera's frame. Then I do optimization,where the objective function is to minimize variance in y and the variable is rotation about x ranging from o to 360.
Problem:
But when I plot the transformed y-coordinates, they are even nearly in a line.
Code:
Get the checkerboad points:
%% Get rotation and translation matrices for each image;
T_cw=cell(num_imgs,1); % stores camera to world rotation and translation for each image
pixel_coordinates=zeros(num_imgs,2); % stores the pixel coordinates of each checkerboard origin
for ii=1:num_imgs,
% Calibrate the camera
im=imread(list_imgs_path{ii});
[imagePoints, boardSize] = detectCheckerboardPoints(im);
[r_wc, t_wc] = extrinsics(imagePoints, worldPoints, cameraParams);
T_wc=[r_wc,t_wc';0 0 0 1];
% World to camera matrix
T_cw{ii} = inv(T_wc);
t_cw{ii}=T_cw{ii}(1:3,4); % x,y,z coordinates in camera's frame
end
Data(num_imgs=10):
t_cw
[-1072.01388542262;1312.20387622761;-1853.34408157349]
[-1052.07856598756;1269.03455126794;-1826.73576892251]
[-1091.85978641218;1351.08261414473;-1668.88197803184]
[-1337.56358084648;1373.78548638383;-1396.87603554914]
[-1555.19509876309;1261.60428874489;-1174.63047408086]
[-1592.39596647158;1066.82210015055;-1165.34417772659]
[-1523.84307918660;963.781819272748;-1207.27444716506]
[-1614.00792252030;893.962075837621;-1114.73528985018]
[-1781.83112607964;708.973204727939;-797.185326205240]
[-1781.83112607964;708.973204727939;-797.185326205240]
Main code (Optimization and transformation):
%% Get theta for rotation
f_obj = #(x)var_ycors(x,t_cw);
opt_theta = fminbnd(f_obj,0,360);
%% Plotting (rotate ycor and check to fix theta)
y_rotated=zeros(1,num_imgs);
for ii=1:num_imgs,
y_rotated(ii)=rotate_cor(opt_theta,t_cw{ii});
end
plot(1:numel(y_rotated),y_rotated);
function var_computed=var_ycors(theta,t_cw)
ycor=zeros(1,numel(t_cw));
for ii =1:numel(t_cw),
ycor(ii)=rotate_cor(theta,t_cw{ii});
end
var_computed=var(ycor);
end
function ycor=rotate_cor(theta,mat)
r_x=[1 0 0; 0 cosd(theta) -sind(theta); 0 sind(theta) cosd(theta)];
rotate_mat=mat'*r_x;
ycor=rotate_mat(2);
end
This is a clear eigenvector problem!
Take your centroids:
t_cw=[-1072.01388542262;1312.20387622761;-1853.34408157349
-1052.07856598756;1269.03455126794;-1826.73576892251
-1091.85978641218;1351.08261414473;-1668.88197803184
-1337.56358084648;1373.78548638383;-1396.87603554914
-1555.19509876309;1261.60428874489;-1174.63047408086
-1592.39596647158;1066.82210015055;-1165.34417772659
-1523.84307918660;963.781819272748;-1207.27444716506
-1614.00792252030;893.962075837621;-1114.73528985018
-1781.83112607964;708.973204727939;-797.185326205240
-1781.83112607964;708.973204727939;-797.185326205240];
t_cw=reshape(t_cw,[3,10])';
compute PCA on them, so we know the principal conponents:
[R]=pca(t_cw);
And.... thats it! R is now the transformation matrix between your original points and the rotated coordinate system. As an example, I will draw in red the old points and in blue the new ones:
hold on
plot3(t_cw(:,1),t_cw(:,2),t_cw(:,3),'ro')
trans=t_cw*R;
plot3(trans(:,1),trans(:,2),trans(:,3),'bo')
You can see that now the blue ones are in a plane, with the best possible fit to the X direction. If you want them in Y direction, just rotate 90 degrees in Z (I am sure you can figure out how to do this with 2 minutes of Google ;) ).
Note: This is mathematically the best possible fit. I know they are not as "in a row" as one would like, but this is because of the data, this is honestly the best possible fit, as that is what the eigenvectors are!
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'm trying to revolve a 2D curve to generate a 3D surface plot.
I've tried using
[X,Z,Y] = cylinder(u);
surf(X,Y,Z), axis square
this, however, revolves my curve around the wrong axis. How do I go about changing the axis?
Thanks alot.
To rotate the axis of the cylinder, you can simply change the order of X, Y, and Z.
[X,Y,Z] = cylinder(u);
surf(X,Y,Z) %# rotation around Z
surf(Z,X,Y) %# rotation around X
surf(Y,Z,X) %# rotation around Y
EDIT
To change the axis of rotation of your curve, you have to calculate the surface. For example, to rotate y = sin(alpha) with alpha = 0:0.1:pi around the y-axis, you can write
r = 0:0.1:pi;
z = sin(r);
theta = 0:pi/20:2*pi;
xx = bsxfun(#times,r',cos(theta));
yy = bsxfun(#times,r',sin(theta));
zz = repmat(z',1,length(theta));
dfig,surf(xx,yy,zz)
axis equal