What units does Matlab's function angvel produce?
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Matlab introduced a function angvel since 2020. According to the documentation, the function is supposed to compute angular velocity tensor from a sequence of quaternions. The details seem puzzling. No units are mentioned in the documentation; explicit passing of timestep dt is required; check-runs on elementary rotations produce mixed results.
Ex 1. Specifying 0, 90, and 180 degree rotations about z-axis
R1 = [1 0 0; 0 1 0; 0 0 1];
R2 = [0 -1 0; 1 0 0; 0 0 1];
R3 = [-1 0 0; 0 -1 0; 0 0 1];
with the constant time step of 1 second, I would expect numeric angular velocities (deg/s) 0, 90, and 90, or (rad/s) 0, 1.58, 1.58, or (Pi-s/s) 0, 0.5 0.5, or (Rev/s) 0, 0.25 0.25. Yet, the following expression
angvel(quaternion(rotm2quat(cat(3,R1,R2,R3))),1,"point")
produce
0 0 0
0 0 1.4142
0 0 1.4142
Why is the computed velocity sqrt(2)?..
But even further, let us specify 0, 180, and 360 degree rotations instead. I would expect the velocity to be double of the previous.
R1 = [1 0 0; 0 1 0; 0 0 1];
R2 = [-1 0 0; 0 -1 0; 0 0 1];
R3 = [1 0 0; 0 1 0; 0 0 1];
But the output is instead
0 0 0
0 0 2
0 0 -2
I understand that there might be a jump caused by the sign-convention in case of 180-degree rotation. But why is there such a weird scaling from sqrt(2) to 2?
P.S.1 changing "frame" to "point" convention does not affect the result. P.S.2 specifying angles smaller than 90 results in "correctly" scaled velocities, e.g. 45-degree rotation produces velocity 0.707
I would like to figure out how to numerically estimate angular speed (not velocities, but presumably the square root of sum of their squares) of a rotation specified as a Nx3x3 sequence of rotation matrices.
4 commentaires
James Tursa
le 18 Sep 2024
Modifié(e) : James Tursa
le 18 Sep 2024
@Paul So a different small angle approximation, I guess (possibly mathematically equivalent?). I don't have a MATLAB version with angvel( ) so couldn't look at the implementation, but based on my testing it seemed pretty obvious something like that was going on. Nevertheless, I don't know when I would ever want to use such a function when exact rotational methods are available and not difficult to implement.
Paul
le 19 Sep 2024
Modifié(e) : James Tursa
le 19 Sep 2024
Fair enough on the small angle approximation, but I don't think exactly mathematically equivalent. I would never want to use such a function and am seriously wondering why the developers did what they did and in what context they thought it would be more useful than an exact method, which I don't think is that much harder to implement than what they did anyway.
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Paul
le 18 Sep 2024
angvel is fundamentally flawed.
Here's the example (truncated) from the doc
eulerAngles = [(0:10:40).',zeros(numel(0:10:40),2)];
q = quaternion(eulerAngles,'eulerd','ZYX','frame');
dt = 1;
format long e
av = angvel(q,dt,'frame') % units in rad/s
That third component should be
10*pi/180/dt %?
Inspecting the code for angvel shows that it intends for the direction of AV(k,:) to be along the axis of rotation for the rotation from qi(k-1,:) to qi(k,:) (resolved in the k-1 frame) and the magnitude of AV(k,:) is equal to the angle of that rotation divided by dt.
However the code is incorrect. In terms of the angle (phi) and axis of rotation (u, a unit vector), the quaternion (expressed as a vector) is:
q = [cos(phi/2), sin(phi/2)*u]
What angvel does is effectively (to within a sign): av = 2/dt*q(2:4).
In the example above, that would be
2/dt*sind(10/2)*[0 0 1]
which is exactly what was produced by angvel (to rounding in the last digit).
So the code is relying on the small angle approximation:
2*sin(phi/2) almost= phi.
Don't know why it needs any approximation, which of course gets worse as the angle gets bigger as in the code in the question
eulerAngles = [(0:90:90).',zeros(numel(0:90:90),2)];
q = quaternion(eulerAngles,'eulerd','ZYX','frame');
dt = 1;
av = angvel(q,dt,'frame') % units in rad/s
We see the function returned
2/dt*sind(90/2)
instead of
90*pi/180/dt % pi/2 rad/sec
I suggest you file a bug report. If you do, please comment back here with a summary of the response from Tech Support.
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