Euler angles and angular velocity

Click For Summary

Discussion Overview

The discussion revolves around the relationship between angular velocity and Euler angles, specifically how to derive the expression that relates the time derivative of each Euler angle to the angular velocity vector in its respective frame. Participants explore the mathematical foundations and implications of this relationship, addressing both conceptual and technical aspects.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asks for a derivation of the relationship between angular velocity and the time derivative of Euler angles, indicating a lack of clarity on the topic.
  • Another participant describes Euler angles as behaving like 2D polar coordinates, suggesting a connection between angular displacement and angular velocity through a specific mathematical formulation.
  • A concern is raised about the addition of angular velocity vectors from different frames, noting that the order of rotations affects the outcome, which adds complexity to the discussion.
  • It is pointed out that the time derivatives of Euler angles do not directly equate to angular velocity, with a distinction made regarding the terminology used (e.g., Tait-Bryan angles vs. Euler angles).
  • A participant suggests that the expression for angular velocity should involve a cross product rather than a simple multiplication, indicating a potential misunderstanding in the original formulation.
  • Another participant provides a detailed mathematical derivation involving the time derivative of a vector in a Cartesian frame, introducing the concept of angular velocity in relation to the vector's motion.
  • A follow-up post corrects a previous omission in the mathematical expression, emphasizing the importance of including all relevant terms to accurately describe the relationship.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between Euler angles and angular velocity, with some asserting that the time derivatives do not directly correspond to angular velocity, while others attempt to clarify the mathematical connections. The discussion remains unresolved with multiple competing perspectives presented.

Contextual Notes

Participants highlight the complexity of rotations and the importance of order in adding angular velocities, as well as the need for careful mathematical formulation when discussing these concepts. There are indications of missing assumptions and unresolved mathematical steps in the derivations presented.

Curl
Messages
756
Reaction score
0
How do you prove that angular velocity is just the time derivative of each Euler angle times the basis vector of its respective frame?
I remember it used to be perfectly clear to me a while back, but now I don't remember how the result was derived, and I couldn't find it in any of my books I looked so far.
Does anyone remember how the result was derived?

Thanks
 
Physics news on Phys.org
Hi Curl! :smile:

Each euler angle is measured in a plane through the origin.
In other words, one specific Euler angle behaves exactly as if it was a 2 dimensional polar coordinate.
If an angle changes by an amount ##d\phi## in an time interval ##dt##, the position changes by ##r d\phi##.
In other words:
$$\frac{ds}{dt}=\frac{r d\phi}{dt}=r \frac{d\phi}{dt} = r \omega$$
 
the problem I have is that in order to add the 3 angular velocity vectors (one in each frame) implies that the rotation can be represented as the sum of the three individual rotations. But rotation order matters, which makes things confusing.
 
Exactly. The time derivatives of a set of Euler angles (better said: Tait-Bryan angles, Bryan angles, or Cardan angles; Euler angles are a z-x-z rotation) are not angular velocity.
 
How do you prove that angular velocity is just the time derivative of each Euler angle times the basis vector of its respective frame?

I think you mean the 'cross product', not times.
 
Well that sentence is flawed actually.
Say you have a vector [itex]A=A_x\hat{i}+A_y\hat{j}+A_z\hat{z}[/itex] in a Cartesian frame. Knowledge of its velocity in this frame is obtained through its time derivative. This yields
[tex] \frac{d\hat{A}}{dt}=\frac{dA_x}{dt}\hat{i}+\frac{dA_y}{dt}\hat{j}+\frac{dA_z}{dt}\hat{z}[/tex]
which can then be deduced to
[tex] \frac{d\hat{A}}{dt}=\dot{A} \vec{A}+\vec{\omega} \times \vec{A}[/tex]
In the above, [itex]\vec{\omega}[/itex] is what you call angular velocity.

Cheers,
 
Oh jeeze. I came to check this just now and I realized I forgot one line. :)

Here it is:
[tex] \frac{d\hat{A}}{dt}=\frac{dA_x}{dt}\hat{i}+\frac{d A_y}{dt}\hat{j}+\frac{dA_z}{dt}\hat{z}+\left( A_x \frac{d\hat{i}}{dt}+A_y\frac{d\hat{j}}{dt}+A_z \frac{d \hat{z}}{dt}\right)[/tex]

Change this line in my above post. This can then deduce the last equation after realizing that the unit vectors have a fixed length.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K