example of Euler angles: 3-1-3 proper Euler sequence

Euler Angles: 3-1-3 Proper Euler Sequence Examples, Exercises, and Solutions

This entry is the self study companion to Euler Angles: 3-1-3 proper Euler sequence.

The problems develop intrinsic 3-1-3 proper Euler angles from their moving-axis geometry through the passive direction cosine matrix, inverse extraction, singularity structure, quaternion representation, angular velocity relation, classical precession-nutation-spin language, and numerical implementation.

All exercises are stated first. Complete worked solutions follow afterward.

1 Convention summary

PhysicsLibrary uses the passive coordinate map

Bv = BC Av. A

(1)

For intrinsic 3-1-3,

α : 3A, β : 11, γ : 32,

(2)

and therefore

BCA = C3(γ)C1 (β )C3(α ).

(3)

A common principal branch is

− π < α ≤ π, 0 ≤ β ≤ π, − π < γ ≤ π.

(4)

The coordinate singularities occur at

β = 0 and β = π.

(5)

2 Visual reference

Figure. Intrinsic 3-1-3 frame chain: first rotate about z_A, then about the moved axis x₁, then about the moved axis z₂.

Figure. Moving-axis interpretation. Although the first and third axis labels are both 3, the third rotation is generally about z₂, not the original z_A line.

Figure. At β = 0 the first and third rotation axes are aligned; at β = π they are anti-aligned.

3 Exercises

Map direction and chronological order.
State the meaning of

$Bv = BCA Av.$
Then state the chronological intrinsic 3-1-3 frame rotations associated with (α,β,γ).
Intermediate-frame construction.
Introduce A₀,A₁,A₂,A₃ for intrinsic 3-1-3. Write the three frame-labeled passive coordinate maps and show how their frame labels compose to give ^BC_A.
Repeated axis label versus repeated physical axis.
Explain why the first and third rotations both carry the digit 3 but are generally not rotations about the same physical line. Under what conditions do their physical axes become collinear?
Elementary passive matrices.
Write C₁(λ) and C₃(λ) in the PhysicsLibrary passive convention. Then write the three factors used in the 3-1-3 product.
Derive the full passive 3-1-3 DCM.
Starting from

$B CA = C3(γ)C1 (β )C3(α ),$
multiply the matrices and obtain all nine elements of ^BC_A.
Single-angle reductions.
Show that the complete DCM reduces to C₃(α), C₁(β), and C₃(γ) when the other two angles vanish in each case.
Rows, columns, and the middle angle.
For C = ^BC_A, state the geometric meaning of C_ij, of row i, and of column j. Use the 3-1-3 matrix to show that

$C33 = cosβ.$
What geometric angle does β measure?
The line of nodes.
Show that the middle rotation leaves x₁ unchanged, so that

$x2 = x1.$
Explain why this axis is naturally interpreted as the line of nodes between the planes perpendicular to z_A and z₂.
Principal ranges.
State a common principal branch for α,β,γ. What sign property of sin β follows on this branch, and why is it useful in inverse extraction?
Inverse DCM extraction.
For a nonsingular passive DCM

$C = BC , A$
derive principal-branch formulas for α, β, and γ using atan2 and arccos.
Numerically robust middle-angle extraction.
Starting from the DCM elements, show that

$2 2 2 C 31 + C 32 = sin β.$
Use this fact to derive an atan2 formula for β that uses both sine-like and cosine-like information.
Numerical DCM round trip.
Consider

$⌊ ⌋ 0.95193 0.20876 − 0.22414 C = ⌈ 0.07803 0.54236 0.83652 ⌉ . 0.29620 − 0.81380 0.50000$
Assuming the principal branch, recover (α,β,γ).
Locate the singularities.
Show from the inverse extraction formulas that intrinsic 3-1-3 becomes singular when sin β = 0. Identify the singular values on the principal branch and explain the geometry of the outer rotation axes at each one.
Exact coupling at β = 0.
Set β = 0 in the matrix product. Show that the orientation depends only on

$α + γ.$
Explain why α and γ cannot then be recovered independently.
Exact coupling at β = π.
Set β = π in the full 3-1-3 matrix and show that the orientation depends only on

$α − γ.$
Contrast this result with the β = 0 case.
A singular-case software policy.
Suppose a software implementation detects | sin β| below a selected threshold and chooses γ = 0 at singularity. Derive formulas for the reported α at β = 0 and at β = π in terms of C₁₁ and C₁₂.
Alternate Euler branch.
Away from singularity, verify that

$′ ′ ′ (α ,β ,γ ) = (α + π,− β, γ + π)$
represents the same orientation, up to wrapping the outer angles by integer multiples of 2π. Why does imposing 0 ≤ β ≤ π help select a principal representative?
Intrinsic-extrinsic reverse-order equivalence.
Write the extrinsic sequence equivalent to intrinsic 3-1-3 with chronological angles (α,β,γ). Explain why the digit string still reads 3-1-3 even though the chronology reverses.
Active versus passive matrices.
If C_passive is the PhysicsLibrary passive 3-1-3 matrix, state the active matrix for the same geometric orientation relationship. Explain why confusing the two produces a transpose or sign error.
Passive quaternion equivalent.
Using

$qP(λ) = cos λ-− eisin λ, i 2 2$
form the Hamilton product for intrinsic 3-1-3 and reduce it to a scalar-first quaternion in terms of the sum and difference of the outer angles.
Quaternion view of the singularities.
Evaluate the passive quaternion from the previous exercise at β = 0 and at β = π. Show directly that only α + γ survives at the first singularity and only α − γ survives at the second.
Angular velocity from Euler rates.
Let

$⌊ ⌋ p B ωB ∕A = ⌈q⌉ . r$
Resolve the three instantaneous rotation axes in frame B and derive the matrix that maps (α,β,γ) to (p,q,r).
Euler rates from body angular velocity.
Invert the rate matrix away from singularity. Derive formulas for α, β, and γ in terms of p,q,r,β,γ. What is the determinant of the forward rate matrix?
Numerical rate example and classical interpretation.
At

$β = 60∘, γ = − 15 ∘,$
suppose

$p = 0.10 rad∕s, q = − 0.05 rad∕s, r = 0.20 rad ∕s.$
Compute (α,β,γ). Then state the common classical mechanics names often associated with the first, middle, and third 3-1-3 angles.
relative attitude and implementation audit.
Two frames B₁ and B₂ are each represented by 3-1-3 Euler triples relative to the same reference frame A. Explain why componentwise subtraction of the two triples is not, in general, the relative attitude. Write the correct DCM expression for ^B₂C_B₁.
Finally, list at least five convention or numerical checks that should be applied to a 3-1-3 implementation.

4 Solutions

Map direction and chronological order.
The equation

$Bv = BC Av A$ (6)

means that the same physical vector is being re-expressed from coordinates in frame A to coordinates in frame B. The vector itself is not physically rotated by this coordinate map.
For intrinsic 3-1-3, the chronological frame rotations are

$+ α about zA,$
then

$+ β about x1,$
then

$+ γ about z2.$
Positive frame rotation follows the right hand rule.
Intermediate-frame construction.
Let

$A0 = A, A3 = B.$
The three passive coordinate maps are

$A1CA = C3 (α),$ (7)

$A2CA1 = C1(β ),$ (8)

and

$B CA2 = C3(γ).$ (9)

Matching adjacent frame labels gives

$BC = BC A2C A1C . A A2 A1 A$ (10)

Therefore

$B CA = C3(γ)C1 (β )C3(α ).$ (11)

The chronological rotations occur from left to right through the frames, but the rightmost matrix acts first on a coordinate column.
Repeated axis label versus repeated physical axis.
The first axis-3 rotation is about z_A. Since that rotation does not change its own axis,

$z = z . 1 A$
The middle rotation is about x₁. A nonzero rotation about x₁ changes the directions of y₁ and z₁, producing a new z₂. The third rotation is therefore about z₂, not generally about z_A.
The two axis-3 directions become collinear when

$β = 0$
or

$β = π.$
They are aligned at β = 0 and anti-aligned at β = π.
Elementary passive matrices.
The required elementary passive matrices are

$⌊ ⌋ 1 0 0 C (λ) = ⌈0 cos λ sin λ⌉ , 1 0 − sin λ cosλ$ (12)

and

$⌊ ⌋ cos λ sin λ 0 C (λ) = ⌈− sin λ cosλ 0⌉ . 3 0 0 1$ (13)

Thus the 3-1-3 factors are

$C (α), C (β), C (γ), 3 1 3$
with the final product ordered as

$BC = C (γ)C (β )C (α ). A 3 1 3$ (14)
Derive the full passive 3-1-3 DCM.
Write

$c = cosα, s = sinα, α α$
with analogous definitions for β and γ.
First,

$⌊ ⌋ cα sα 0 C1 (β)C3 (α ) = ⌈− cβsα cβcα sβ⌉ . sβsα − sβcα cβ$ (15)

Premultiplication by C₃(γ) gives

$⌊ ⌋ cαcγ − cβs αsγ cαcβsγ + cγsα sβsγ BCA = ⌈ − cαs γ − cβcγs α cαcβcγ − s αsγ cγsβ⌉ . sαsβ − cαsβ cβ$ (16)
Single-angle reductions.
If β = γ = 0, then

$BCA = C3 (α ).$ (17)

If α = γ = 0, then

$BCA = C1 (β).$ (18)

If α = β = 0, then

$B CA = C3 (γ).$ (19)

These checks verify both sign convention and multiplication order.
Rows, columns, and the middle angle.
For orthonormal frames,

$C = eB ⋅ eA . ij i j$ (20)

Thus row i is the ith basis vector of B expressed in A, while column j is the jth basis vector of A expressed in B.
From the full matrix,

$C = c = cosβ. 33 β$ (21)

Therefore

$eA ⋅ eB = cosβ. 3 3$ (22)

Hence β is the angle between the initial and final axis-3 directions, often interpreted as a nutation or tilt angle.
The line of nodes.
The middle transformation is a rotation about x₁. Any rotation leaves its own rotation axis unchanged, so

$x = x . 2 1$ (23)

This axis lies in the plane perpendicular to z_A after the first rotation and is also the axis about which z_A is tilted into z₂. It therefore forms the common line of intersection of the two planes perpendicular to z_A and z₂, which motivates the classical name line of nodes.
Principal ranges.
A common principal branch is

$− π < α ≤ π, 0 ≤ β ≤ π, − π < γ ≤ π.$ (24)

On this branch,

$sin β ≥ 0.$ (25)

That sign information allows the positive square root of sin ²β to be used in the robust atan2 extraction of the middle angle.
Inverse DCM extraction.
From

$C33 = cosβ,$
the principal middle angle is

$β = arccos(C ). 33$ (26)

Away from singularity,

$C31 = sαs β, C32 = − cαsβ,$
so

$α = atan2 (C31,− C32).$ (27)

Likewise,

$C13 = sβs γ, C23 = sβc γ,$
so

$γ = atan2(C13, C23).$ (28)

These formulas require sin β≠0.
Numerically robust middle-angle extraction.
From the third row,

$C31 = sαs β, C32 = − cαsβ.$
Hence

$( ) C231 + C232 = s2β s2α + c2α = sin2β.$ (29)

On the principal branch sin β ≥ 0, so

$∘ --2-----2- C 31 + C 32 = sin β.$
Therefore

$(∘ ---------- ) β = atan2 C231 + C232,C33 .$ (30)

This uses both the sine-like and cosine-like information for the middle angle.
Numerical DCM round trip.
First,

$β = arccos(0.50000 ) = 60∘.$ (31)

Next,

$α = atan2(0.29620, 0.81380 ) ≈ 20∘.$ (32)

Finally,

$γ = atan2 (− 0.22414, 0.83652 ) ≈ − 15∘.$ (33)

Thus

$(α, β,γ) = (20∘,60 ∘,− 15∘).$ (34)

This reproduces the principal triple used to construct the matrix.
Locate the singularities.
The outer-angle extraction formulas divide the orientation information by the common factor sin β. Therefore the chart is singular when

$sin β = 0.$ (35)

On 0 ≤ β ≤ π, this occurs at

$β = 0 and β = π.$ (36)

At β = 0, z₂ is aligned with z_A. At β = π, z₂ is anti-aligned with z_A. In either case the first and third physical rotation axes are collinear, so the two outer Euler coordinates lose independent observability.
Exact coupling at β = 0.
Setting β = 0 gives

$BCA = C3 (γ)C3(α ).$ (37)

Rotations about the same axis add:

$BC = C (α + γ). A 3$ (38)

Therefore

$⌊ ⌋ cos(α + γ) sin(α + γ) 0 BCA = ⌈− sin(α + γ) cos(α + γ ) 0⌉ . 0 0 1$ (39)

Only

$α + γ$ (40)

is observable. Infinitely many pairs (α,γ) with the same sum describe the same orientation.
Exact coupling at β = π.
Substituting β = π into the full matrix gives

$⌊cos(α − γ) sin(α − γ ) 0 ⌋ B ⌈ ⌉ CA = sin(α − γ) − cos(α − γ ) 0 . 0 0 − 1$ (41)

Only

$α − γ$ (42)

is observable.
Thus the β = 0 singularity preserves the sum of the outer angles, while the β = π singularity preserves their difference.
A singular-case software policy.
At β = 0,

$C11 = cos(α + γ), C12 = sin(α + γ ).$
Define

$δ0 = atan2 (C12, C11) = α + γ.$ (43)

If the chosen policy sets γ = 0, report

$α = δ. 0$ (44)

At β = π,

$C11 = cos(α − γ), C12 = sin(α − γ ).$
Define

$δπ = atan2 (C12, C11) = α − γ.$ (45)

With the same γ = 0 policy, report

$α = δ . π$ (46)

The exact policy is conventional; consistency is the essential requirement.
Alternate Euler branch.
The nonprincipal transformation

$(α ′,β ′,γ ′) = (α + π,− β, γ + π)$ (47)

produces the same orientation away from singularity. This follows because changing the sign of the middle rotation while shifting each outer rotation by π compensates for the reversed orientation of the intermediate axes.
Since Euler coordinates are not unique globally, restricting

$0 ≤ β ≤ π$ (48)

selects one standard branch for the middle angle, while the outer angles are wrapped into their chosen 2π intervals.
Intrinsic-extrinsic reverse-order equivalence.
Intrinsic 3-1-3 with chronological angles

$(α,β,γ )$
is equivalent to extrinsic 3-1-3 with chronological angles

$(γ, β,α).$
Symbolically,

$intrinsic 3-1-3(α,β,γ ) ≡ extrinsic 3-1- 3 (γ, β,α ).$ (49)

Because the axis sequence 3-1-3 is palindromic, reversing the digit order leaves the visible digit string unchanged. The angle association and chronology still reverse.
Active versus passive matrices.
For the same orientation relationship,

$T −1 Ractive = Cpassive = C passive.$ (50)

The transpose arises because the passive matrix changes coordinates between frames, whereas the active matrix rotates the physical vector in a fixed coordinate frame.
Using an active formula where a passive map is required typically produces a transpose, reversed product order, or angle-sign error.
Passive quaternion equivalent.
For a positive passive frame rotation,

$qP(λ) = cos λ-− eisin λ. i 2 2$ (51)

The intrinsic 3-1-3 product is

$BqA = qP3 (γ)qP1 (β)qP3 (α).$ (52)

Using Hamilton multiplication with scalar-first storage,

$⌊ ⌋ β- α-+-γ- | cos 2 cos 2 | || β α − γ || B ||− sin 2-cos--2---|| qA = | β α − γ | . ||− sin --sin ------|| ⌈ 2β α 2+ γ ⌉ − cos --sin ------ 2 2$ (53)

The DCM and quaternion represent the same passive map:

$BCA = C (BqA ).$ (54)
Quaternion view of the singularities.
At β = 0,

$β β cos --= 1, sin --= 0. 2 2$
Hence

$⌊ ⌋ cos α-+-γ | 2 | Bq = || 0 || . A |⌈ 0 |⌉ α-+--γ − sin 2$ (55)

Only α + γ appears.
At β = π,

$β- β- cos 2 = 0, sin 2 = 1.$
Thus

$⌊ 0 ⌋ | α − γ| B || − cos---2--|| qA = | α-−--γ| . ⌈ − sin 2 ⌉ 0$ (56)

Only α − γ appears. The quaternion itself remains nonsingular; the loss is in the Euler coordinate chart.
Angular velocity from Euler rates.
The physical angular velocity is the sum of the three instantaneous frame rotations:

$ωB ∕A = α˙eA3 + β˙eA11 + γ˙eA32.$ (57)

Expressed in frame B,

$⌊s s ⌋ B A ⌈ β γ⌉ e 3 = s βcγ , cβ$ (58)

$⌊ ⌋ cγ BeA11 = ⌈ − sγ⌉ , 0$ (59)

and

$⌊ ⌋ 0 BeA23 = ⌈0 ⌉. 1$ (60)

Therefore

$⌊ ⌋ ⌊ ⌋ ⌊ ⌋ p sβsγ cγ 0 α˙ ⌈ q⌉ = ⌈ sβcγ − sγ 0⌉ ⌈β˙⌉ . r cβ 0 1 γ˙$ (61)
Euler rates from body angular velocity.
Away from sin β = 0, inversion gives

$⌊ ⌋ ⌊ ˙α⌋ sγ- -cγ- 0 ⌊p ⌋ ⌈ ˙⌉ | sβ s β | ⌈ ⌉ β = ⌈ cγ − sγ 0⌉ q . ˙γ − sγ cot β − cγ cot β 1 r$ (62)

Equivalently,

$sγp-+-cγq- α˙= s β ,$ (63)

$˙ β = cγp − sγq,$ (64)

and

$˙γ = r − cot β (sγp + cγq) .$ (65)

The determinant of the forward rate matrix is

$− sinβ.$ (66)

Thus the rate mapping becomes singular at exactly the same configurations as the Euler coordinate extraction.
Numerical rate example and classical interpretation.
For

$∘ ∘ β = 60 , γ = − 15 ,$
we have approximately

$s = 0.866025, c = 0.500000, s = − 0.258819, c = 0.965926. β β γ γ$
Using the inverse-rate formulas,

$˙α = (−-0.258819)(0.10)-+-(0.965926-)(−-0.05-)≈ − 0.08565 rad∕s. 0.866025$ (67)

Next,

$˙ β = (0.965926 )(0.10) − (− 0.258819 )(− 0.05 ) ≈ 0.08365 rad ∕s.$ (68)

Finally,

$˙γ = 0.20 − cot(60∘)[(− 0.258819 )(0.10 ) + (0.965926)(− 0.05)] ≈ 0.24283 rad ∕s.$ (69)

Thus

$⌊ ⌋ ⌊ ⌋ ˙α − 0.08565 ⌈ ˙β⌉ ≈ ⌈ 0.08365 ⌉ rad∕s. ˙γ 0.24283$ (70)

In common classical rigid-body language, α is often called precession, β nutation, and γ spin. The names are application dependent; the frame-axis definitions remain primary.
Relative attitude and implementation audit.
Euler angles are ordered coordinates on the rotation group, not ordinary Cartesian vector components. Therefore subtracting two Euler triples does not, in general, produce the relative orientation.
If

$B1CA$

and

$B2 CA$

describe the two frames relative to the same reference frame, then the passive relative map from B₁ coordinates to B₂ coordinates is

$B2 B2 (B1 )T CB1 = CA CA .$ (71)

If a 3-1-3 triple is desired for this relative attitude, extract it from this relative DCM using the same branch and singularity policy.
Useful implementation checks include:
- verify the passive map direction ^Bv = ^BC_A ^Av;
- verify the product order C₃(γ)C₁(β)C₃(α);
- check all three single-angle reductions;
- check CC^T = I and det C = 1;
- verify DCM to Euler to DCM round trips away from singularity;
- explicitly test β = 0 and β = π;
- use atan2 for the outer angles;
- define and test a singular-case branch policy;
- verify the passive quaternion against the DCM;
- distinguish active and passive matrices by transpose;
- test the Euler-rate matrix determinant − sin β;
- avoid componentwise Euler-angle subtraction for finite relative attitudes.

5 Further study

The parent entry Euler angles: 3-1-3 proper Euler sequence develops the same formulas as a continuous derivation. The broader articles Proper Euler angles, Euler sequence composition and the twelve standard sequences, and Euler-angle singularities and gimbal lock place 3-1-3 within the full family of Euler coordinate charts.

The 3-1-3 sequence is especially important in classical mechanics because it makes the geometry of precession, nutation, and spin explicit. For numerical attitude propagation near either singular endpoint, a nonsingular representation such as a quaternion or direction cosine matrix is generally preferable as the propagated state.

6 References and further reading

References

[1] D. M. Henderson, Euler Angles, Quaternions, and Transformation Matrices: Working Relationships, JSC-12960, NASA Johnson Space Center, 1977. NASA Technical Reports Server

[2] J. K. Moore, Learn Multibody Dynamics, chapters “Orientation of Reference Frames” and “Angular Kinematics,” 2026 edition. Licensed CC BY 4.0. Learn Multibody Dynamics

[3] J. Diebel, “Representing Attitude: Euler Angles, Unit Quaternions, and Rotation Vectors,” Stanford University, 2006. Online PDF

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