For a project I want to do a very simple Pythagoras calculation in C++. An object is equiped with an IMU sensor that gives either a Quaternion rotation or Euler angles. What I want to know is the opposite sides of the triangle underneath the object.:
I want to know these sides of the triangle for both the X and Y axis (black arrows):
This is pretty much very simple, except for the fact that the object can rotate. When the object is rotated I still want to use the X and Y axis in world space (black arrows), but when yawing the Euler angles of the IMU provide me with pitch and roll, which are in local space (red arrows):
In what way can I still get the world space angles (black arrows) while yawing, to be able to calculate my simple Pythagoras calculation? If I can't get them, is there a way to calculate the opposite sides I want using Quaternions?
We can do the calculation by taking into account the Euler angles in the following order -
Pitch
First of all, as you change the roll of the sensor, the sensor "ray" sweeps out a plane inclined to the horizon at angle pitch. We need to first calculate the closest distance between (i) the line of intersection between the plane and the ground, and (ii) the point directly below the sensor on the ground. This is given by d = h * tan(pitch).
Roll
Next we need to do another trigonometric step. As before the roll sweeps through the plane. The offset distance along the axis perpendicular to the line joining (i) and (ii) is given by f = h / cos(pitch) * tan(roll). This gives the intersection point on the ground to be (d, f)
Yaw
Previously, we considered a frame in which the yaw was zero. We now need to rotate this intersection point around the Z-axis by yaw. Thus the final intersection point is given by (x, y) = (d * cos(yaw) - f * sin(yaw), d * sin(yaw) + f * cos(yaw)). You can calculate the "space angle" you want by taking atan2(y, x).
I have a given 3x3 rotation matrix and I want to calculate the rotation angle around z axis. How do I get there?
For example, in this case below, how did they calculated the "-30deg rotation around the x axis"? Or how did they get to the "-74deg" value around that axis?
This is my original matrix:
Thank you!
It is simple if the rotation matrix is just a rotation matrix and there is no scaling. Here is a site that explains in more pretty terms then I am willing to diagram here. Basically the rotation matrix is composed of sinf(x) and cosf(x) of euler angles (well you can think of it like that at least). You can therefore use values within it to back calculate the euler angles.
http://nghiaho.com/?page_id=846
If you have scaling involved you will need to normalize each row of the matrix first. Then apply the above method.
I have a 3d object which is free to rotate along x,y and z axis and it is then saved as a transform matrix. In a case where the sequence of rotation is not known and the object is rotated for more than 3 times (eg :-if i rotate the object x-60degress, y-30 degrees, z-45 degrees then again x->30 degrees), is it possible to extract the angles rotated from the transform matrix?.I know that it is possible to get angles if the sequence of rotation is known, but if I have only the final transform matrix with me and nothing else, is it possible to get the angles rotated(x,y,and z) from the transform matrix ?
Euler angle conversion is a pretty well known topic. Just normalize the matrix orientation vectors and then use something like this c source code.
The matrix is the current state of things it has no knowledge of what the transformation has been in the past. It does not know how the matrix was built. You can just take the matrix into and decompose it into any pieces you like, as long as:
The data do not overlap. For example:Two X turns after each other is indistinguishable form each other (no way to know if its 1 2 or three different rotations summed).
The sequence order is known
A decomposition can be built out of the data (for example scale can be measured)
I have a stack of images (about 180 of them) and there are 2 stars (just basic annotations) on every single image. Hence, the position (x,y) of the two stars are provided initially. The dimensions of all these images are fixed and constant.
The 'distance' between the image is about 1o with the origin to be the center (width/2, height/2) of every single 2D image. Note that, if this is plotted out and interpolated nicely, the stars would actually form a ring of an irregular shape.
The dotted red circle and dotted purple circle are there to give a stronger scent of a 3D space and the arrangement of the 2D images (like a fan). It also indicates that each slice is about 1o apart.
With the provided (x,y) that appeared in the 2D image, how do you get the corresponding (x,y,z) in the 3d space knowing that each image is about 1o apart?
I know that MATLAB had 3D plotting capabilities, how should I go about implementing the solution to the above scenario? (Unfortunately, I have very little experience plotting 3D with MATLAB)
SOLUTION
Based on the accepted answer, I looked up a bit further: spherical coordinate system. Based on the computation of phi, rho and theta, I could reconstruct the ring without problems. Hopefully this helps anyone with similar problems.
I have also documented the solution here. I hope it helps someone out there, too:
http://gray-suit.blogspot.com/2011/07/spherical-coordinate-system.html
I believe the y coordinate stays as is for 3D, so we can treat this as converting 2D x and image angle to an x and z when viewed top down.
The 2D x coordinate is the distance from the origin in 3D space (viewed top down). The image angle is the angle the point makes with respect to the x axis in 3D space (viewed top down). So the x coordinate (distance from orign) and the image angle (angle viewed top down) makes up the x and z coordinates in 3D space (or x and y if viewed top down).
That is a polar coordinate.
Read how to convert from polar to cartesian coordinates to get your 3D x and z coordinates.
I'm not great at maths either, here's my go:
3D coords = (2Dx * cos(imageangle), 2Dy, 2Dx * sin(imageangle))
Given the 2D coordinates (x,y) just add the angle A as a third coordinate: (x,y,A). Then you have 3D.
If you want to have the Anotations move on a circle of radius r in 3D you can just calculate:
you can use (r*cos(phi),r*sin(phi),0) which draws a circle in the XY-plane and rotate it with a 3x3 rotation matrix into the orientation you need.
It is not clear from you question around which axis your rotation is taking place. However, my answer holds for a general rotation axis.
First, put your points in a 3D space, lying on the X-Y plane. This means the points have a 0 z-coordinate. Then, apply a 3D rotation of the desired angle around the desired axis - in your example, it is a one degree rotation. You could calculate the transformation matrix yourself (should not be too hard, google "3D rotation matrix" or similar keywords). However, MATLAB makes it easier, using the viewmtx function, which gives you a 4x4 rotational matrix. The extra (fourth) dimension is dependent on the projection you specify (it acts like a scaling coefficient), but in order to make things simple, I will let MATLAB use its default projection - you can read about it in MATLAB documentation.
So, to make the plot clearer, I assume four points which are the vertices of a square lying on the x-y plane (A(1,1), B(1,-1), C(-1,-1), D(1,-1)).
az = 0; % Angle (degrees) of rotation around the z axis, measured from -y axis.
el = 90; % Angle (degrees) of rotation around the y' axis (the ' indicates axes after the first rotation).
x = [1,-1, -1, 1,1]; y = [1, 1, -1, -1,1]; z = [0,0, 0, 0,0]; % A square lying on the X-Y plane.
[m,n] = size(x);
x4d = [x(:),y(:),z(:),ones(m*n,1)]'; % The 4D version of the points.
figure
for el = 90 : -1 :0 % Start from 90 for viewing directly above the X-Y plane.
T = viewmtx(az, el);
x2d = T * x4d; % Rotated version of points.
plot3 (x2d(1,:), x2d(2,:),x2d(3,:),'-*'); % Plot the rotated points in 3D space.
grid
xlim ([-2,2]);
ylim ([-2,2]);
zlim([-2,2]);
pause(0.1)
end
If you can describe your observation of a real physical system (like a binary star system) with a model, you can use particle filters.
Those filters were developed to locate a ship on the sea, when only one observation direction was available. One tracks the ship and estimates where it is and how fast it moves, the longer one follows, the better the estimates become.
I have one problem related to rotation of point in 3D-space.
Suppose I have one point with X, Y and Z coordinates.
And now I want to rotate it, by specifying the rotation in one of these three ways:
By user-defined degree
By user-defined axis of rotation
Around (relative to) user-defined point
I found good link over here, but it doesn't address point 3. Can anyone help me solve that?
All rotations will go around the origin. So you translate to the origin, rotate, then translate back.
T = translate from global coordinates to user-coordinates
R = rotate around the origin (like in your link)
(T^-1) = translate back
point X
X_rotated = (T^-1)*R*T*X
If you have multiple points to rotate then multiply the matrices together:
A = (T^-1)*R*T
X_rotated = A*X