I want to compute the Euler angles from a rotation matrix in order to find out the orientation associated to that rotation. For that purpose, I am using MATLAB and the function rotm2eul that gives me the rotation first about x-axis, then about y-axis and finally the z-axis.
I am using a signal with 1000 frames and for each frame a rotation matrix is computed, as well, the three Euler angles. However, when I am going to see the Euler angles' curve, there is some "jumps" as I shown on the figures below.
On Technique 1, I think it jumps from -180º to 180º which should be the same. In fact, the upper portion of the plot seems a continuation of the lower portion. So in this case I thought I could subtract 360º to the upper portion to get the plot. But I am not sure if I do this I am falsifying the results.
On Technique 2, it makes a jump with a different reason of the previous one. I think it must be because the angle associated with the y-axis reaches 90º which should be a boundary case. But in this case I don't know how should I correct the data or , like previously, if I want to correct the plot is falsifying the euler angle result.
Technique 2: This is a Gimbal lock, known feature of Euler angles. You can't avoid it completely. You can change the rotation order, but it will appear in another position.
Related
Most of the times, I determine contour orientation generating 2D points and computing the closed polygon area. Depending on the area value sign I can understand if the contour is oriented clockwise or not (see How to determine if a list of polygon points are in clockwise order?).
Would it be possible to do the same computations without generating 2D points? I mean, relying only on geometric curve properties?
We are interested in determining the orientation of contours like these ones without sampling them with 2D points.
EDIT: Some interesting solutions can be found here:
https://math.stackexchange.com/questions/423718/general-way-to-find-out-whether-a-curve-is-positively-oriented
Scientific paper: Determining the orientation of closed planar curves, DJ Filip (1990)
How are those geometric curves defined?
Do you have an angle for them? The radius doesn't matter, only the difference between entry-angle and exit-angle of each curve.
In that case, a trivial idea crossing my mind is to just sum up all the angles. If the result is positive, you know you had more curves towards the right meaning it's a clockwise contour. If it was negative, then more curves were leftwards -> anti-clockwise contour. (assuming that positive angels determine a right-curve and vica versa)
After thinking about this for awhile, for polygons that contain arcs I think there are three ways to do this.
One, is to break the arcs into line segments and then use the area formula as described above. The success of this approach seems to be tied to how close the interpolation of the arcs is as this could cause the polygon to intersect itself.
A quicker way than the above would be to do the interpolation of the arcs and then find a vertex in the corner (minimal Y, if tie minimal X) and use the sign of the cross product for that vertex. Positive CCW, negative CW. Again, this is still tied to the accuracy of the interpolation.
I think a better approach would be to find the midpoint of the arc and create two line segments, one from the beginning of the arc to the midpoint and another from the midpoint to the end of the arc and replace the arc with these line segments. Now you have a polygon with only line segments. Then you can add up all the normalized cross products of all the vertices. The sign will tell you the direction. Positive is counter-clockwise, negative is clockwise. In this case it doesn't matter if the polygon self-intersects.
Still need the math: I am trying to calculate the yxy rotation sequence given a quaternion transformation. I can easily do this using Matlab's quat2angle function. However, I need to calculate this by hand using a python script.
This part solved: Please look at this awesome presentation which helped me resolve these issues below:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0CCoQFjAC&url=http%3A%2F%2Fwww.udel.edu%2Fbiology%2Frosewc%2Fkaap686%2Freserve%2Fshoulder%2Fshoulder%2FBluePresentation.ppt&ei=jgRAVLHfOsSrogTJiYHABQ&usg=AFQjCNGFmwh11jEZen80jc3tM4f7HUQcNw&sig2=Dlr8_7TIFPLyUfJy6-pSJA&bvm=bv.77648437,d.cGU
Also, with Matlab, I am seeing strange results with the way they calculate yxy. I have a quaternion transformation of [1.0000 -0.0002 -0.0011 -0.0006] and I get y = 112.4291 x = -0.0719 y1 = -112.5506 (in degrees).
I don't expect to see any rotations here (my sensors aren't rotating). Why is Matlab showing me rotation? And when I try to just move in the x rotation, I see y and y1 also rotate, however, I don't expect y or y1 to be rotating. Any thoughts?
UPDATE:
When I add y + y1 I seem to get the value for the first y (when doing simple rotation around the first y), and this smooths out the data. However, when I combine the three rotations of the shoulder, the data doesn't make sense. I am trying to define shoulder movement based on plane of elevation, elevation and rotation (yxy) in a way that's easy to interpret. When I rotate around x, then the second y, I get "clipping" (data goes to 180 then -180 following positive trend for y1 and opposite happens for y), even though I start my sensors at the zero position. Also, If I try to rotate only around the second y, I see rotation in the x. That doesn't make any sense either. Any additional thoughts?
Note:
I am using 2 IMU sensors, taring them in the same orientation, holding one constant and rotating the other, calculating the relative rotation between them using quaternions, and then calculating the yxy rotation sequence angles.
In case anyone is interested in quaternion calculations and transformations. I solved it using this transformations library:
http://www.lfd.uci.edu/~gohlke/code/transformations.py.html
There are several functions in here using matrices, quaternions, and Euler rotations. And you can convert quaternions to several different Euler rotation sequences. Give thanks to the person who created this script.
I have two ellipsoids in R3 described in terms of their centre points (P), their axes lengths (a,b,c), and their rotation vector (R). I wish to interpolate a tubular structure between these two ellipsoids along a given centre line. This is done by creating an ellipsoid centred at each point along the centre line. Its axes lengths are interpolated linearly between those at the two endpoints, and the rotation is obtained as a quaternion using spherical linear interpolation, or SLERP.
I previously asked a similar question on this problem here. I have since isolated the issue a little further, and thought it warranted a new post. The difference here is that before doing SLERP, I first rotate the two reference ellipsoids by the inverse of the rotation matrix that describes one of them, such that one of them is now axis-aligned (i.e. has no rotation). Previously this appeared to solve the problem, but I have encountered an example where this fix does not work.
The source code to reproduce this issue is available here. The relevant function is ellipsoidSLERP and the functions it calls. Here is a screenshot of the output:
What you are seeing is an interpolation of ellipsoid volumes (blue) between two reference ellipsoid volumes at either end (green) along a centreline (cyan).
Problem Statement
The interpolation on the left works correctly, resulting in a smooth tubular structure. The interpolation on the right does not work correctly, and results in a twist.
What is causing this behaviour, and how can I correct it?
Please let me know if there's anything I can do to clarify.
I have a convex polygon in 3D. For simplicity, let it be a square with vertices, (0,0,0),(1,1,0),(1,1,1),(0,0,1).. I need to arrange these vertices in counter clockwise order. I found a solution here. It is suggested to determine the angle at the center of the polygon and sort them. I am not clear how is that going to work. Does anyone have a solution? I need a solution which is robust and even works when the vertices get very close.
A sample MATLAB code would be much appreciated!
This is actually quite a tedious problem so instead of actually doing it I am just going to explain how I would do it. First find the equation of the plane (you only need to use 3 points for this) and then find your rotation matrix. Then find your vectors in your new rotated space. After that is all said and done find which quadrant your point is in and if n > 1 in a particular quadrant then you must find the angle of each point (theta = arctan(y/x)). Then simply sort each quadrant by their angle (arguably you can just do separation by pi instead of quadrants (sort the points into when the y-component (post-rotation) is greater than zero).
Sorry I don't have time to actually test this but give it a go and feel free to post your code and I can help debug it if you like.
Luckily you have a convex polygon, so you can use the angle trick: find a point in the interior (e.g., find the midpoint of two non-adjacent points), and draw vectors to all the vertices. Choose one vector as a base, calculate the angles to the other vectors and order them. You can calculate the angles using the dot product: A · B = A B cos θ = |A||B| cos θ.
Below are the steps I followed.
The 3D planar polygon can be rotated to 2D plane using the known formulas. Use the one under the section Rotation matrix from axis and angle.
Then as indicated by #Glenn, an internal points needs to be calculated to find the angles. I take that internal point as the mean of the vertex locations.
Using the x-axis as the reference axis, the angle, on a 0 to 2pi scale, for each vertex can be calculated using atan2 function as explained here.
The non-negative angle measured counterclockwise from vector a to vector b, in the range [0,2pi], if a = [x1,y1] and b = [x2,y2], is given by:
angle = mod(atan2(y2-y1,x2-x1),2*pi);
Finally, sort the angles, [~,XI] = sort(angle);.
It's a long time since I used this, so I might be wrong, but I believe the command convhull does what you need - it returns the convex hull of a set of points (which, since you say your points are a convex set, should be the set of points themselves), arranged in counter-clockwise order.
Note that MathWorks recently delivered a new class DelaunayTri which is intended to superseded the functionality of convhull and other older computational geometry stuff. I believe it's more accurate, especially when the points get very close together. However I haven't tried it.
Hope that helps!
So here's another answer if you want to use convhull. Easily project your polygon into an axes plane by setting one coordinate zero. For example, in (0,0,0),(1,1,0),(1,1,1),(0,0,1) set y=0 to get (0,0),(1,0),(1,1),(0,1). Now your problem is 2D.
You might have to do some work to pick the right coordinate if your polygon's plane is orthogonal to some axis, if it is, pick that axis. The criterion is to make sure that your projected points don't end up on a line.
I am simulating a system where I need Direction Cosine Matrix to quaternion conversion. I use the default DCM to Quaternion conversion block available in simulink. However at some points of the simulation, the output quaternion components reverse sign.
Unfortunately I cannot attach the plot image.
Though this is mathematically correct I desire a smooth change. Any idea on how to avoid this and have a smooth curve for the quaternion?
Update 1:
http://tinypic.com/view.php?pic=33dayap&s=6
Above is the simulated plot. The first plot is of the output quaternion. Second plot is of the Direction Cosine Matrix. As you see that even though the dcm components change smoothly, the quaternion changes sign abruptly.
The problem arises because of the double covering property of quaternions: Two unit quaternions correspond to every rotation. At some point, according to some rule, the Matlab implementation switched from one quaternion to the other. There is not much you can do about it.
A messy workaround would be to write your own rotation matrix to quaternion conversion, and pick that representation of the two possibilities that is closer to the previous one, hence avoiding the sudden jumps. It's messy.
Plotting the quaternions is typically not needed in practical applications. Most likely you are rotating an object / vector. If you plot that object / vector (or some projections of it) you won't get any sudden jumps even if there are jumps in the representation of the rotation. Another benefit of plotting the projections of the rotated object is that it is usually much easier to interpret these plots than the quaternions. I don't know whether it makes sense in your application; it worked beautifully in mine.