Imagine a dome with its centre in the +z direction. What I want to do is to move that dome's centre to a different axis (e.g. 20 degrees x axis, 20 degrees y axis, 20 degrees z axis). How can I do that ? Any hint/tip helps.
Add more info:
I've been dabbling with rotation matrices in wiki for a while. The problem is, it is not a commutative operation. RxRyRz is not same as RzRyRx. So based on the way I multiple it I get a different final results. For example, I want my final projection to have 20 degrees from the original X axis, 20 degrees from original Y axis and 20 degrees from original Z axis. Based on the matrix, giving alpha, beta, gamma values 20 (or its corresponding radian) does NOT result the intended rotation. Am I missing something? Is there a matrix that I can just put the intended angles and get it at the end ?
Using a rotation matrix is an easy way to rotate a collection of (x,y,z) points. You can calculate a rotation matrix for your case using the equations in the general rotation section. Note that figuring out the angle values to plug into those equations can be tricky. Think of it as rotating about one axis at a time and remember that the order of your rotations (order of multiplications) does matter.
An alternative to the general rotation equations is to calculate a rotation matrix from axis and angle. It may be easier for you to define correct parameters with this method.
Update: After perusing Wikipedia, I found a simple way to calculate rotation axis and angle between two vectors. Just fill in your starting and ending vectors for a and b here:
a = [0.0 0.0 1.0];
b = [0.5 0.5 0.0];
vectorMag = #(x) sqrt(sum(x.^2));
rotAngle = acos(dot(a,b) / (vectorMag(a) * vectorMag(b)))
rotAxis = cross(a,b)
rotAxis =
-0.5 0.5 0
rotAngle =
1.5708
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]);
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!
We have these logarithmic spirals which are circling around the centre of the coordinate system:
x = ebθ cos(θ)
y = ebθ sin(θ)
where the ebθ is the distance between the point (which is on the spiral) and the centre; and the θ is the angle between the line connecting the point and the origin and the axis x.
Consider a spiral where the angle is θ ϵ <0,10π> and the parameter is b=0.1. By thickening points on the spirals (and the angle θ) calculate the circumference with the relative precision better than 1%. Draw the spiral!
I'm preparing for a (MATLAB) test and I'm stuck with this exercise. Please help, any hint is appreciated.
Start by computing a list of x,y for your range of theta and value of b. For more accurate results, have your theta increment in smaller steps (I chose 5000 arbitrarily). Then, its simply computing the distance for each pair of consecutive points and summing them up.
t = linspace(0,10*pi,5000);
b = 0.1;
x = exp(b*t).*cos(t);
y = exp(b*t).*sin(t);
result = sum(sqrt((x(2:end) - x(1:end-1)).^2 + (y(2:end)-y(1:end-1)).^2))
I want to simulate depth in a 2D space, If I have a point P1 I suppose that I need to project that given point P1 into a plane x axis rotated "theta" rads clockwise, to get P1'
It seems that P1'.x coord has to be the same as the P1.x and the P1'.y has to b shorter than P1.y. In a 3D world:
cosa = cos(theta)
sina = sin(theta)
P1'.x = P1.x
P1'.y = P1.y * cosa - P1.z * sina
P1'.z = P1.y * sina + P1.z * cosa
Is my P1.z = 0? I tried it and P1'.y = P1.y * cosa doesn't result as expected
Any response would be appreciated, Thanks!
EDIT: What I want, now I rotate camera and translate matrix
EDIT 2: an example of a single line with a start1 point and a end1 point (it's an horizontal line, result expected is a falling line to the "floor" as long as tilt angle increases)
I think it's a sign error or an offset needed (java canvas drawing (0,0) is at top-left), because my new line with a tilt of 0 is the one below of all and with a value of 90º the new line and the original one match
The calculation you are performing is correct if you would like to perform a rotation around the x axis clockwise. If you think of your line as a sheet of paper, a rotation of 0 degrees is you looking directly at the line.
For the example you have given the line is horizontal to the x axis. This will not change on rotation around the x axis (the line and the axis around which it is rotating are parallel to one another). As you rotate between 0 and 90 degrees the y co-ordinates of the line will decrease with P1.y*cos(theta) down to 0 at 90 degrees (think about the piece of paper we have been rotating around it's bottom edge, the x axis, at 90 degrees the paper is flat, and the y axis is perpendicular to the page, thus both edges of the page have the same y co-ordinate, both the side that is the "x-axis" and the opposite parallel side will have y=0).
Thus as you can see for your example this has worked correctly.
EDIT: The reason that multiplying by 90 degrees does not give an exactly zero answer is simply floating point rounding
i am working on some camera data. I have some points which consist of azimuth, angle, distance, and of course coordinate field attributes. In postgresql postgis I want to draw shapes like this with functions.
how can i draw this pink range shape?
at first should i draw 360 degree circle then extracting out of my shape... i dont know how?
I would create a circle around the point(x,y) with your radius distance, then use the info below to create a triangle that has a larger height than the radius.
Then using those two polygons do an ST_Intersection between the two geometries.
NOTE: This method only works if the angle is less than 180 degrees.
Note, that if you extend the outer edges and meet it with a 90 degree angle from the midpoint of your arc, you have a an angle, and an adjacent side. Now you can SOH CAH TOA!
Get Points B and C
Let point A = (x,y)
To get the top point:
point B = (x + radius, y + (r * tan(angle)))
to get the bottom point:
point C = (x + radius, y - (r * tan(angle)))
Rotate your triangle to you azimouth
Now that you have the triangle, you need to rotate it to your azimuth, with a pivot point of A. This means you need point A at the origin when you do the rotation. The rotation is the trickiest part. Its used in computer graphics all the time. (Actually, if you know OpenGL you could get it to do the rotation for you.)
NOTE: This method rotates counter-clockwise through an angle (theta) around the origin. You might have to adjust your azimuth accordingly.
First step: translate your triangle so that A (your original x,y) is at 0,0. Whatever you added/subtracted to x and y, do the same for the other two points.
(You need to translate it because you need point A to be at the origin)
Second step: Then rotate points B and C using a rotation matrix. More info here, but I'll give you the formula:
Your new point is (x', y')
Do this for points B and C.
Third step: Translate them back to the original place by adding or subtracting. If you subtracted x last time, add it this time.
Finally, use points {A,B,C} to create a triangle.
And then do a ST_Intersection(geom_circle,geom_triangle);
Because this takes a lot of calculations, it would be best to write a program that does all these calculations and then populates a table.
PostGIS supports curves, so one way to achieve this that might require less math on your behalf would be to do something like:
SELECT ST_GeomFromText('COMPOUNDCURVE((0 0, 0 10), CIRCULARSTRING(0 10, 7.071 7.071, 10 0), (10 0, 0 0))')
This describes a sector with an origin at 0,0, a radius of 10 degrees (geographic coordinates), and an opening angle of 45°.
Wrapping that with additional functions to convert it from a true curve into a LINESTRING, reduce the coordinate precision, and to transform it into WKT:
SELECT ST_AsText(ST_SnapToGrid(ST_CurveToLine(ST_GeomFromText('COMPOUNDCURVE((0 0, 0 10), CIRCULARSTRING(0 10, 7.071 7.071, 10 0), (10 0, 0 0))')), 0.01))
Gives:
This requires a few pieces of pre-computed information (the position of the centre, and the two adjacent vertices, and one other point on the edge of the segment) but it has the distinct advantage of actually producing a truly curved geometry. It also works with segments with opening angles greater than 180°.
A tip: the 7.071 x and y positions used in the example can be computed like this:
x = {radius} cos {angle} = 10 cos 45 ≈ 7.0171
y = {radius} sin {angle} = 10 sin 45 ≈ 7.0171
Corner cases: at the antimeridian, and at the poles.