Review of the 3-2-1 Euler Angles: a yaw–pitch–roll sequence by H. Alemi Ardakani & T. J. Bridges Department of Mathematics, University of Surrey, Guildford GU2 7XH UK — April 15, 2010— 1 Introduction The 3 − 2 − 1 Euler angles are one of the most widely used parameterisations of rotations. O’Reilly gives a history on page 184 of [4]. This review will give an overview of the important feautures of this set of Euler angles, and show that they are the ones used in [2] and [3]. Let {E1, E2, E3} be a basis for a fixed spatial frame, and let {e1, e2, e3} be a basis for a body frame. The 3 − 2 − 1 Euler angles provide an orthogonal matrix Q which maps {E1, E2, E3} to {e1, e2, e3}, Q {E1, E2, E3} −−−−−→ {e1, e2, e3} by breaking the action of Q up into three steps {E1, E2, E3} −→ {a1, a2, a3} −→ {b1, b2, b3} −→ {e1, e2, e3} , where {a1, a2, a3} and {b1, b2, b3} are intermediate frames of reference. 2 Map from Ej to aj : the yaw rotation In this section the details of the map L(ψ,E3) {E1, E2, E3} −−−−−→ {a1, a2, a3} are constructed using the rotation tensor L(ψ, E3)=(I − E3 ⊗ E3) cos ψ + E3 sin ψ + E3 ⊗ E3 , (2.1) b where (a ⊗ b) c := (b · c) a , a, b, c ∈ R3 , 1 ψ , yaw θ , pitch φ , roll Figure 1: Schematic of the yaw-pitch-roll motion in terms of the Euler angles ψ , φ and θ. and 0 −a a 3 2 a := a3 0 −a1 . (2.2) b −a2 a1 0 The hat-matrix has the property that a b = a × b , a, b ∈ R3 . b The rotation (2.1) is a counterclockwise rotation about the Z−axis as shown schemat- ically in Figure 1. The new basis is a1 = L(ψ, E3)E1 = E1 cos ψ + E3 × E1 sin ψ a2 = L(ψ, E3)E2 = E2 cos ψ + E3 × E2 sin ψ a3 = L(ψ, E3)E3 = E3 . Using the identities E1 × E2 = E3 , E3 × E1 = E2 , E2 × E3 = E1 , the matrix representation of the map is a cos ψ sin ψ 0 E 1 1 a2 = − sin ψ cos ψ 0 E2 . a3 0 01 E3 This is the first equation of (6.23) on page 184 of [4]. 2 3 Map from aj to bj : the pitch rotation In this section the details of the map L(θ,a2) {a1, a2, a3} −−−−−→ {b1, b2, b3} are constructed using the rotation tensor L(θ, a2)=(I − a2 ⊗ a2) cos θ + a2 sin θ + a2 ⊗ a2 , (3.3) where b a2 = −E1 sin ψ + E2 cos ψ . The new basis is b1 = L(θ, a2)a1 = a1 cos θ + a2 × a1 sin θ b2 = L(θ, a2)a2 = a2 b3 = L(θ, a2)a3 = a3 cos θ + a2 × a3 sin θ , with matrix representation, b cos θ 0 − sin θ a 1 1 b2 = 0 1 0 a2 . b3 sin θ 0 cos θ a3 This is the second equation of (6.23) on page 184 of [4]. 4 Map from bj to ej : the roll rotation In this section the details of the map L(φ,b1) {b1, b2, b3} −−−−−→ {e1, e2, e3} are constructed using the rotation tensor L(φ, b1)=(I − b1 ⊗ b1) cos φ + b1 sin φ + b1 ⊗ b1 , (4.4) where b b1 = a1 cos θ − a3 sin θ . The new basis is e1 = L(φ, b1)b1 = b1 e2 = L(φ, b1)b2 = b2 cos φ + b1 × b2 sin φ e3 = L(φ, b1)b3 = b3 cos φ + b1 × b3 sin φ , with matrix representation, e 1 0 0 b 1 1 e2 = 0 cos φ sin φ b2 . e3 0 − sin φ cos φ b3 This is the third equation of (6.23) on page 184 of [4]. 3 5 The composite rotation from Ej to ej In this section the details of the map L(φ,b1)L(θ,a2)L(ψ,a3) {E1, E2, E3} −−−−−−−−−−−−−−−→ {e1, e2, e3} are constructed. Using the above results e 1 0 0 cos θ 0 − sin θ cos ψ sin ψ 0 E 1 1 e2 = 0 cos φ sin φ 0 1 0 − sin ψ cos ψ 0 E2 e3 0 − sin φ cos φ sin θ 0 cos θ 0 01 E3 cos θ 0 − sin θ cos ψ sin ψ 0 E 1 = sin φ sin θ cos φ sin φ cos θ − sin ψ cos ψ 0 E2 cos φ sin θ − sin φ cos φ cos θ 0 01 E3 cos θ cos ψ cos θ sin ψ − sin θ E 1 = sin φ sin θ cos ψ − cos φ sin ψ sin φ sin θ sin ψ + cos φ cos ψ sin φ cos θ E2 . cos φ sin θ cos ψ + sin φ sin ψ cos φ sin θ sin ψ − sin φ cos ψ cos φ cos θ E3 Write this as e = QT E . Then cos θ cos ψ sin φ sin θ cos ψ − cos φ sin ψ cos φ sin θ cos ψ + sin φ sin ψ Q = cos θ sin ψ sin φ sin θ sin ψ + cos φ cos ψ cos φ sin θ sin ψ − sin φ cos ψ . − sin θ sin φ cos θ cos φ cos θ (5.5) To see that this is the correct Q, construct Q directly. 6 Direct construction of Q By definition Q = L(φ, b1)L(θ, a2)L(ψ, E3) . In order to construct a matrix representation of the composite rotation, use the property of rotation tensors L(θ, Rb)= R L(θ, b) RT . when R is any proper rotation. Define A3 = L(ψ, E3) , A2 = L(θ, E2) , A1 = L(φ, E1) . Then T L(θ, a2)= L(θ, A3E2)= A3A2A3 , 4 and L(φ, b1) = L(φ, L(θ, a2)a1) T = L(θ, a2)L(φ, a1)L(θ, a2) T T T = A3A2A3 L(φ, a1)A3A2 A3 T T T = A3A2A3 L(φ, A3E1)A3A2 A3 T T T T = A3A2A3 A3A1A3 A3A2 A3 T T = A3A2A1A2 A3 . Hence T Q = A A A AT AT A A AT A = A A A = AT AT AT . 3 2 1 2 3 3 2 3 3 3 2 1 1 2 3 with matrix representation T 1 0 0 cos θ 0 − sin θ cos ψ sin ψ 0 Q = 0 cos φ sin φ 0 1 0 − sin ψ cos ψ 0 , 0 − sin φ cos φ sin θ 0 cos θ 0 01 confirming the form of Q. 7 Angular velocities To compute angular velocities start with the representation cos ψ − sin ψ 0 cos θ 0 sin θ 10 0 Q = A3A2A1 = sin ψ cos ψ 0 0 1 0 0 cos φ − sin φ , 0 0 1 − sin θ 0 cos θ 0 sin φ cos φ and use the properties ˙ T ˙ ˙ T ˙ ˙ T ˙ A1A1 = φ E1 , A2A2 = θ E2 , A3A3 = ψ E3 . b b b Now differentiate Q, Q˙ = A˙ 3A2A1 + A3A˙ 2A1 + A3A2A˙ 1 . The body angular velocity is Ωb := QT Q˙ c T T T ˙ = A1 A2 A3 Q T T T ˙ ˙ ˙ = A1 A2 A3 A3A2A1 + A3A2A1 + A3A2A1 ˙ T T ˙ T ˙ = ψ A1 A2 E3A2A1 + θ A1 E2A1 + φ E1 . b b b 5 But T \T A1 E2A1 = (A1 E2) , and b T A1 E2 = h(I − E1 ⊗ E1) cos φ − sin φE1 + E1 ⊗ E1i E2 b = E2 cos φ − sin φE1 × E2 = E2 cos φ − E3 sin φ . Similarly T \T A2 E3A2 = (A2 E3) , and b T θ θ A2 E3 = h(I − E2 ⊗ E2) cos − sin E2 + E2 ⊗ E2i E3 b = E3 cos θ − sin θE2 × E3 = E3 cos θ − E1 sin θ . Continuing T T T \ A1 A2 E3A2A1 = (A1 (E3 cos θ − E1 sin θ) , and b T T A1 (E3 cos θ − E1 sin θ) = −E1 sin θ + cos θ A1 E3 = −E1 sin θ + cos θ(E3 cos φ − E1 × E3 sin φ) = −E1 sin θ + cos θ(E3 cos φ + E2 sin φ) = −E1 sin θ + E2 cos θ sin φ + E3 cos θ cos φ . Substitute into the expression for the body angular velocity, b ˙ Ω = ψ(−E1 sin θ + E2 cos θ sin φ + E3 cos θ cos φ) +θ˙(E2 cos φ − E3 sin φ)+ φ˙E1 = E1(φ˙ − ψ˙ sin θ)+ E2(ψ˙ cos θ sin φ + θ˙ cos φ) +E3(ψ˙ cos θ cos φ − θ˙ sin φ) . b In [4] the body angular velocity is denoted by ω0R and the above expression for Ω agrees with ω0R on page 188 of [4]. In matrix form b ˙ Ω1 1 0 − sin θ φ b ˙ Ω2 = 0 cos φ cos θ sin φ θ . b Ω3 0 − sin φ cos θ cos φ ψ˙ Write this is Ωb = B−1Θ˙ . 6 Then 1 sin φ tan θ cos φ tan θ B = 0 cos φ − sin φ . 0 sin φ sec θ cos φ sec θ As shown below in §9, this expression agrees with [B] := B in [2] and in [3]. 8 Small angle approximation For small angle approximation rewrite the body representation of the angular ve- locity as b ˙ ˙ Ω1 φ 0 0 − sin θ φ b ˙ ˙ Ω2 = θ + 0 cos φ − 1 cos θ sin φ θ . b Ω3 ψ˙ 0 − sin φ cos θ cos φ − 1 ψ˙ Hence for small angles b ˙ ˙ Ω1 φ 0 0 −θ φ b ˙ 1 2 ˙ Ω2 = θ + 0 − 2 φ φ θ + ··· .