Euler Angles: Proper Euler Angles Examples, Exercises, and Solutions

This entry is the exercise companion to Euler Angles: proper Euler angles.

The problems focus on the six proper Euler sequences, repeated outer-axis geometry, the universal singularity $\sin\beta=0$ , and the flagship intrinsic $3$ - $1$ - $3$ sequence.

All exercises are stated first. Complete solutions follow afterward.

Convention summary

PhysicsLibrary uses

$\displaystyle {}^B\mathbf v = {}^BC_A\,{}^A\mathbf v.$

(1)

For intrinsic $i$ - $j$ - $k$ ,

$\displaystyle {}^BC_A = C_k(\gamma)C_j(\beta)C_i(\alpha).$

(2)

For proper Euler angles,

$\displaystyle k=i, \qquad i\neq j,$

(3)

$\displaystyle {}^BC_A = C_i(\gamma)C_j(\beta)C_i(\alpha).$

(4)

The six proper Euler sequences are

$\displaystyle 121,\quad131,\quad212,\quad232,\quad313,\quad323.$

(5)

Their generic singularity is

$\displaystyle \sin\beta=0.$

(6)

Visual reference

Figure. The six intrinsic proper Euler sequences.

Image EA06_proper_euler_middle_angle_singularity_tikz

Figure. The proper Euler singularity illustrated with intrinsic $3$

Exercises

Recognize proper Euler sequences.
Which are proper Euler?

$\displaystyle 121,\quad123,\quad313,\quad312,\quad232,\quad231.$
Generate all six sequences.
Derive all six proper Euler axis orders from the rule that the first and third labels match while the middle label differs.
Write all six passive products.
Write the passive intrinsic DCM product for every proper Euler sequence.
Recover a sequence from a product.
Identify the sequence represented by

$\displaystyle {}^BC_A=C_2(\gamma)C_1(\beta)C_2(\alpha).$
Recover another sequence.
Identify the sequence represented by

$\displaystyle {}^BC_A=C_1(\gamma)C_3(\beta)C_1(\alpha).$
Repeated label versus physical axis.
Explain why the two axis- rotations in intrinsic -- are not generally rotations about the same physical direction.
Single-angle reductions for 3-1-3.
For

$\displaystyle {}^BC_A=C_3(\gamma)C_1(\beta)C_3(\alpha),$

find the result when:
1. $\beta=\gamma=0$ ;
2. $\alpha=\gamma=0$ ;
3. $\alpha=\beta=0$ .
Universal singularity.
State the generic proper Euler singularity and its values on

$\displaystyle 0\leq\beta\leq\pi.$
Outer-angle coupling at $\beta=0$ .
Show algebraically that only $\alpha+\gamma$ remains observable when $\beta=0$ .
Outer-angle coupling at $\beta=\pi$ .
Explain geometrically why the outer angles are again coupled at $\beta=\pi$ .
Derive the third row of the 3-1-3 DCM.
Starting from

$\displaystyle {}^BC_A=C_3(\gamma)C_1(\beta)C_3(\alpha),$

derive the third row.
Extract 3-1-3 angles.
For a nonsingular passive DCM

$\displaystyle C= \begin{bmatrix} C_{11}&C_{12}&C_{13}\ C_{21}&C_{22}&C_{23}\ C_{31}&C_{32}&C_{33} \end{bmatrix},$

write the principal -- extraction formulas.
Numerical 3-1-3 DCM.
Compute the passive DCM for

$\displaystyle \alpha=30^\circ,\qquad \beta=60^\circ,\qquad \gamma=-20^\circ.$
Numerical round trip.
Use the matrix from Exercise 13 to recover the original principal $(\alpha,\beta,\gamma)$ .
Equivalent extrinsic 3-1-3.
Find the extrinsic description equivalent to intrinsic

$\displaystyle 3$ - $\displaystyle 1$ - $\displaystyle 3(\alpha,\beta,\gamma).$

Why is the unchanged digit string potentially misleading?
Another proper Euler conversion.
Convert intrinsic

$\displaystyle 2$ - $\displaystyle 3$ - $\displaystyle 2(\alpha,\beta,\gamma)$

to its equivalent extrinsic description.
Alternate branch.
State an alternate proper Euler triple equivalent to $(\alpha,\beta,\gamma)$ away from singularity and explain why a principal branch is needed.
Passive quaternion counterpart.
Write the passive quaternion product for intrinsic -- and state the quaternion/DCM consistency relation.
Distinguish Euler families from singularity.
Which family has singularity $\cos\beta=0$ ?

Which has singularity $\sin\beta=0$ ?
Convention audit.
A mechanics text says only “use a -- Euler rotation.” List at least five additional convention questions that must be answered before its formulas can be copied into PhysicsLibrary.

Solutions

Solution 1: recognize proper Euler sequences

Proper Euler sequences have equal first and third labels.

Thus

$\displaystyle 121,\qquad313,\qquad232$

are proper Euler.

The others shown are Tait Bryan.

Solution 2: generate all six sequences

Choose the repeated outer axis in three ways.

For each choice, select either of the other two axes as the middle axis.

Hence

$\displaystyle 3\times2=6.$

The sequences are

$\displaystyle 121,\quad131,\quad212,\quad232,\quad313,\quad323.$

(7)

Solution 3: write all six passive products

Using

$\displaystyle {}^BC_A=C_i(\gamma)C_j(\beta)C_i(\alpha),$

we obtain

$\displaystyle C_{121}=C_1(\gamma)C_2(\beta)C_1(\alpha),$

(8)

$\displaystyle C_{131}=C_1(\gamma)C_3(\beta)C_1(\alpha),$

(9)

$\displaystyle C_{212}=C_2(\gamma)C_1(\beta)C_2(\alpha),$

(10)

$\displaystyle C_{232}=C_2(\gamma)C_3(\beta)C_2(\alpha),$

(11)

$\displaystyle C_{313}=C_3(\gamma)C_1(\beta)C_3(\alpha),$

(12)

and

$\displaystyle C_{323}=C_3(\gamma)C_2(\beta)C_3(\alpha).$

(13)

Solution 4: recover a sequence from a product

The rightmost factor contains the first angle and identifies the first axis.

Thus

$\displaystyle C_2(\gamma)C_1(\beta)C_2(\alpha)$

is intrinsic

$\displaystyle 2$ - $\displaystyle 1$ - $\displaystyle 2.$

(14)

Solution 5: recover another sequence

Likewise,

$\displaystyle C_1(\gamma)C_3(\beta)C_1(\alpha)$

is intrinsic

$\displaystyle 1$ - $\displaystyle 3$ - $\displaystyle 1.$

(15)

Solution 6: repeated label versus physical axis

The first axis $3$ belongs to the initial frame.

After the first rotation, the middle rotation about the new axis $1$ changes the current frame orientation.

The final axis $3$ belongs to that second intermediate frame.

Therefore the first and third axis- $3$ directions are generally not parallel. The repeated digit means the same coordinate-axis label in different intermediate frames, not necessarily the same physical line.

Solution 7: single-angle reductions for 3-1-3

From

$\displaystyle {}^BC_A=C_3(\gamma)C_1(\beta)C_3(\alpha),$

we obtain

$\displaystyle \beta=\gamma=0 \quad\Longrightarrow\quad {}^BC_A=C_3(\alpha),$

(16)

$\displaystyle \alpha=\gamma=0 \quad\Longrightarrow\quad {}^BC_A=C_1(\beta),$

(17)

and

$\displaystyle \alpha=\beta=0 \quad\Longrightarrow\quad {}^BC_A=C_3(\gamma).$

(18)

Solution 8: universal singularity

Every proper Euler sequence is singular when

$\displaystyle \sin\beta=0.$

(19)

$\displaystyle 0\leq\beta\leq\pi,$

the singular values are

$\displaystyle \beta=0 \qquad\hbox{or}\qquad \beta=\pi.$

(20)

Solution 9: outer-angle coupling at beta equals zero

$\displaystyle \beta=0,$

the middle rotation becomes the identity:

$\displaystyle C_j(0)=I.$

Thus

$\displaystyle {}^BC_A = C_i(\gamma)C_i(\alpha).$

(21)

Same-axis rotations add, so

$\displaystyle C_i(\gamma)C_i(\alpha) = C_i(\alpha+\gamma).$

(22)

Only

$\displaystyle \alpha+\gamma$

(23)

can therefore be determined from the final orientation.

Solution 10: outer-angle coupling at beta equals pi

$\displaystyle \beta=\pi,$

the middle rotation reverses the repeated outer axis.

The first and third physical axes therefore lie on the same line but point in opposite directions.

Two rotations about that same physical line cannot be recovered independently. Only one signed difference combination of the outer angles remains observable.

Solution 11: derive the third row of the 3-1-3 DCM

Multiplication gives the third row

$\displaystyle \begin{bmatrix} s_\alpha s_\beta & -c_\alpha s_\beta & c_\beta \end{bmatrix}.$

(24)

Hence

$\displaystyle C_{31}=s_\alpha s_\beta,$

(25)

$\displaystyle C_{32}=-c_\alpha s_\beta,$

(26)

and

$\displaystyle C_{33}=c_\beta.$

(27)

These entries directly motivate the inverse formulas.

Solution 12: extract 3-1-3 angles

On the nonsingular principal branch,

$\displaystyle \beta=\arccos(C_{33}),$

(28)

$\displaystyle \alpha=\operatorname{atan2}(C_{31},-C_{32}),$

(29)

and

$\displaystyle \gamma=\operatorname{atan2}(C_{13},C_{23}).$

(30)

These formulas require

$\displaystyle \sin\beta\neq0.$

(31)

Solution 13: numerical 3-1-3 DCM

For

$\displaystyle \alpha=30^\circ,\qquad \beta=60^\circ,\qquad \gamma=-20^\circ,$

$\displaystyle {}^BC_A = C_3(-20^\circ)C_1(60^\circ)C_3(30^\circ)$

(32)

gives

$\displaystyle {}^BC_A \approx \begin{bmatrix} 0.89930&0.32175&-0.29620\ 0.06127&0.57791&0.81380\ 0.43301&-0.75000&0.50000 \end{bmatrix}.$

(33)

Solution 14: numerical round trip

The middle angle is

$\displaystyle \beta = \arccos(C_{33}) = \arccos(0.50000) = 60^\circ.$

The first angle is

$\displaystyle \alpha = \operatorname{atan2}(0.43301,0.75000) = 30^\circ.$

The third angle is

$\displaystyle \gamma = \operatorname{atan2}(-0.29620,0.81380) = -20^\circ.$

Thus

$\displaystyle (\alpha,\beta,\gamma) = (30^\circ,60^\circ,-20^\circ).$

(34)

Solution 15: equivalent extrinsic 3-1-3

Reverse the axis order and reverse the angle association.

Because $3$ - $1$ - $3$ is a palindrome, the digit string remains unchanged:

intrinsic $\displaystyle 3$ - $\displaystyle 1$ - $\displaystyle 3 (\alpha,\beta,\gamma) \equiv$ extrinsic $\displaystyle 3$ - $\displaystyle 1$ - $\displaystyle 3 (\gamma,\beta,\alpha).$

(35)

The unchanged digits can hide the fact that the axis construction and chronological angle assignment have changed.

Solution 16: another proper Euler conversion

Similarly,

intrinsic $\displaystyle 2$ - $\displaystyle 3$ - $\displaystyle 2 (\alpha,\beta,\gamma) \equiv$ extrinsic $\displaystyle 2$ - $\displaystyle 3$ - $\displaystyle 2 (\gamma,\beta,\alpha).$

(36)

Solution 17: alternate branch

Away from singularity, an equivalent proper Euler triple is

$\displaystyle (\alpha',\beta',\gamma') = (\alpha+\pi,-\beta,\gamma+\pi),$

(37)

with the outer angles wrapped by multiples of $2\pi$ if needed.

Therefore Euler coordinates are not globally unique.

The principal choice

$\displaystyle 0\leq\beta\leq\pi$

selects a standard representative.

Solution 18: passive quaternion counterpart

For intrinsic $3$ - $1$ - $3$ ,

$\displaystyle {}^Bq_A = q_3^P(\gamma)q_1^P(\beta)q_3^P(\alpha).$

(38)

The quaternion and DCM describe the same passive map:

$\displaystyle {}^BC_A = C({}^Bq_A).$

(39)

Solution 19: distinguish Euler families from singularity

The condition

$\displaystyle \cos\beta=0$

belongs to the Tait Bryan family.

The condition

$\displaystyle \sin\beta=0$

belongs to the proper Euler family.

Solution 20: convention audit

The sequence label alone is incomplete.

At minimum, determine:

whether the sequence is intrinsic or extrinsic;
whether the transformation is active or passive;
the coordinate-map direction;
whether row or column vectors are used;
the positive-angle convention;
whether the listed angles are chronological first, second, and third angles;
whether the written matrix order is operator order or chronological prose order;
the inverse-map principal ranges.

Only after those conventions agree should formulas be copied directly.

Compact review

The six proper Euler sequences are

$\displaystyle 121,\quad131,\quad212,\quad232,\quad313,\quad323.$

(40)

For intrinsic $i$ - $j$ - $i$ ,

$\displaystyle {}^BC_A = C_i(\gamma)C_j(\beta)C_i(\alpha).$

(41)

Their generic singularity is

$\displaystyle \sin\beta=0.$

(42)

For intrinsic $3$ - $1$ - $3$ ,

$\displaystyle {}^BC_A = C_3(\gamma)C_1(\beta)C_3(\alpha).$

(43)

Away from singularity,

$\displaystyle \beta=\arccos(C_{33}),$

(44)

$\displaystyle \alpha=\operatorname{atan2}(C_{31},-C_{32}),$

(45)

and

$\displaystyle \gamma=\operatorname{atan2}(C_{13},C_{23}).$

(46)

Sources and exercise provenance

The exercises and solutions in this companion are newly written for PhysicsLibrary to reinforce the framework developed in Euler angles: proper Euler angles.

Bibliography

1: H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison Wesley, 2002. Publisher search
2: D. M. Henderson, Euler Angles, Quaternions, and Transformation Matrices: Working Relationships, JSC-12960, NASA Johnson Space Center, 1977. NASA Technical Reports Server
3: J. Diebel, “Representing Attitude: Euler Angles, Unit Quaternions, and Rotation Vectors,” Stanford University, 2006. Online PDF

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