Euler, Tait, Gyroscope: Rotations that cover it all

[Bullwinckle]

I understand that a gyroscope undergoes precession, nutation, spin.
And that the order of the rotations are such that the precession and spin share a common "local axis."

I also understand there are, for totally different purposes, Euler angles to model rotations.
In this case, the order of the rotations are ALSO such that the first and third share a common local axis.

For the Tait-Bryan angles, it so happens that each rotation is about a different local axis (corresponding to pitch, Yaw, roll).

I understand that I can cover all possible rotational configurations of the body with each of these. I can see it by example.

But is there a mathematical statement that "ensures" that I have indeed "covered" all possible rotational configurations with these selections?

Maybe it is so obvious, it escapes me.

But I have this very naive (ignorant?) understanding that just like in translation space, where one must cover all three orthogonal directions, that here, too, one must cover each local angle separately. I know this is silly. But could someone explain how they KNOW that the choice of rotation angles for the three I mentioned -- gyroscope, Euler, Tait -- do, indeed, cover all configurations.

For in my ignorant understanding, the Tait makes the most sense because the three local axes are different -- and yes, I know that is silly, but I am using it as a springboard to solicit guidance.

(I do "see" how they all work. But how can I "know" it?)

(And I am aware of Gimbal Lock and this is not about that.)

 

[vanhees71]

These are just different parametrizations of the rotation group $\mathrm{SO}(3)$. Of course you cannot cover a compact space with one map, and thus there are coordinate singularities, but as you said you are aware of these. You can proof the completeness (up to the coordinate singularities, which have to be covered by another map) either by referring to the geometric meaning of the rotations, as in the Wikipedia article

https://en.wikipedia.org/wiki/Euler_angles

or algebraically using the properties of the SO(3) matrices, i.e., $\hat{R} \hat{R}^{\mathrm{T}}=\mathbb{1}$.

 

[Bullwinckle]

These are just different parametrizations of the rotation group $\mathrm{SO}(3)$. Of course you cannot cover a compact space with one map, and thus there are coordinate singularities, but as you said you are aware of these. You can proof the completeness (up to the coordinate singularities, which have to be covered by another map) either by referring to the geometric meaning of the rotations, as in the Wikipedia article

https://en.wikipedia.org/wiki/Euler_angles

or algebraically using the properties of the SO(3) matrices, i.e., $\hat{R} \hat{R}^{\mathrm{T}}=\mathbb{1}$.

Thank you Vanhees,

But may I elaborate in the context of your response?

How do I KNOW that the Tait or Euler or Gyro are parameterizations of the rotation group?

Surely I can rotate about the local 3-axis three times. And I know, obviously, that it will not cover all the rotations.

The Tait has a rotation about each of the three.
The Euler and Gyro about only two (one is repeated).

How do I KNOW that that will be a parameterization of SO(3)

(Yes, as you acknowledged, I know that you need two maps due to singularties.)

 

[vanhees71]

To get a first idea, look at the figure about the Euler coordinates in the cited Wikipedia article.

You can consider the rotations SO(3) as exactly those linear transformations of vectors in Euclidean $\mathbb{R}^3$ that map a fixed Cartesian right-handed basis to any other Cartesian right-handed basis. So just draw two such bases and look at the figure: Given the blue and the red Cartesian bases you define the line of nodes (green) as the intersection of the $xy$ plane with the $XY$ plane. Further the angle $\beta \in (0,\pi)$ is defined as the angle between the $z$ and the $Z$ axis Now if you carefully think about the drawing, it's not too difficult to see, how the Euler angles describe successively the complete rotation of the blue basis to the red one:

(1) rotate the blue frame around the $z$-axis such that the $x$-axis points along the line of nodes. This leads to a new basis, which we can denote as $\vec{e}_{x'}$, $\vec{e}_{y'}$, $\vec{e}_{z'}$. Since we rotate around the $z$-axis we have $\vec{e}_{z'}=\vec{e}_z$. Its clear that $\alpha \in [0,2 \pi)$ (or any other semi-open interval of length of $2 \pi$).

(2) now we rotate around the new $x'$-axis (i.e. the line of nodes) by the angle $\beta$ such that $\vec{e}_{z'}$ coincides with $\vec{e}_{z''}=\vec{e}_Z$.

(3) finally we just have to rotate by the angle $\gamma$ around the $z''$-axis, which is by constrcution the same as the $Z$-axis, such that the vector $\vec{e}_{x''}=\vec{e}_{x'}$ is mapped to $\vec{e}_{X}$. It's clear that also $\gamma$ must be taken out of a semiopen interval of length $2 \pi$. Then also $\vec{e}_{y''}$ must coincide with $\vec{e}_{Y}$, because by assumption both Cartesian bases are right-handed.

This is of course only an intuitive proof, but you can as well use algebra to show that any SO(3) matrix can be written in terms of the sequence of rotations around coordinate axes. These matrices are defined to represent the corresponding operations on the column vectors of real numbers, representing the components of vectors with respect to the various bases in the above construction. Thus the order is the opposite of what we have explained above, i.e., the given SO(3) matrix $\hat{R}$ reads
$$\hat{R}=\hat{R}_3(\alpha) \hat{R}_1(\beta) \hat{R}_3(\gamma).$$
The proof is simple but a bit lengthy.

 

[Bullwinckle]

To get a first idea, look at the figure about the Euler coordinates in the cited Wikipedia article.

You can consider the rotations SO(3) as exactly those linear transformations of vectors in Euclidean $\mathbb{R}^3$ that map a fixed Cartesian right-handed basis to any other Cartesian right-handed basis. So just draw two such bases and look at the figure: Given the blue and the red Cartesian bases you define the line of nodes (green) as the intersection of the $xy$ plane with the $XY$ plane. Further the angle $\beta \in (0,\pi)$ is defined as the angle between the $z$ and the $Z$ axis Now if you carefully think about the drawing, it's not too difficult to see, how the Euler angles describe successively the complete rotation of the blue basis to the red one:

(1) rotate the blue frame around the $z$-axis such that the $x$-axis points along the line of nodes. This leads to a new basis, which we can denote as $\vec{e}_{x'}$, $\vec{e}_{y'}$, $\vec{e}_{z'}$. Since we rotate around the $z$-axis we have $\vec{e}_{z'}=\vec{e}_z$. Its clear that $\alpha \in [0,2 \pi)$ (or any other semi-open interval of length of $2 \pi$).

(2) now we rotate around the new $x'$-axis (i.e. the line of nodes) by the angle $\beta$ such that $\vec{e}_{z'}$ coincides with $\vec{e}_{z''}=\vec{e}_Z$.

(3) finally we just have to rotate by the angle $\gamma$ around the $z''$-axis, which is by constrcution the same as the $Z$-axis, such that the vector $\vec{e}_{x''}=\vec{e}_{x'}$ is mapped to $\vec{e}_{X}$. It's clear that also $\gamma$ must be taken out of a semiopen interval of length $2 \pi$. Then also $\vec{e}_{y''}$ must coincide with $\vec{e}_{Y}$, because by assumption both Cartesian bases are right-handed.

This is of course only an intuitive proof, but you can as well use algebra to show that any SO(3) matrix can be written in terms of the sequence of rotations around coordinate axes. These matrices are defined to represent the corresponding operations on the column vectors of real numbers, representing the components of vectors with respect to the various bases in the above construction. Thus the order is the opposite of what we have explained above, i.e., the given SO(3) matrix $\hat{R}$ reads
$$\hat{R}=\hat{R}_3(\alpha) \hat{R}_1(\beta) \hat{R}_3(\gamma).$$
The proof is simple but a bit lengthy.

 

[Bullwinckle]

This is great!
This I understand.

I follow your "proof" that...
$$\hat{R}=\hat{R}_3(\alpha) \hat{R}_1(\beta) \hat{R}_3(\gamma).$$
is a parameterization of the rotation group SO(3)

And I can easily intuit similar proof for the Tait angles (and ditto for the gyroscope)

So is that the issue then? One must PROVE this for any other potential set of three rotations?

It is not possible to make some general proof that all such possibilities must adhere to?

It is OK to say "yes, but it is very complicated." For then I would content myself with knowing I do not know it immediately, but could.

Also, (and this only seems redundant because your continue re-expression is enabling me to focus)

You wrote:

But you can as well use algebra to show that any SO(3) matrix can be written in terms of the sequence of rotations around coordinate axes.

That, then, is the key to me: are you saying that:

It is possible to prove, algebraically, that certain candidates -- Euler, Tait, Gyroscope -- can "cover" the space of SO(3)

No need to prove the above. Just let me know if that is the chrysalis for me.

IN fact, on closer inspection of the wikipedia page, I read this

• Proper Euler angles (z-x-z, x-y-x, y-z-y, z-y-z, x-z-x, y-x-y)
• Tait–Bryan angles (x-y-z, y-z-x, z-x-y, x-z-y, z-y-x, y-x-z).

Is this basically saying that SO(3) can be parameterized in ONLY these TWELVE possible ways?