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Desenvolvimento de um Veículo Aéreo Não-Tripulado - LARA ...

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∂q b n<br />

∂ω b ib<br />

=<br />

⎡<br />

⎤<br />

α x q 0 + β x Σ q0 − γq 1 α y q 0 + β y Σ q0 − γq 2 α z q 0 + β z Σ q0 − γq 3<br />

α x q 1 + β x Σ q1 + γq 0 α y q 1 + β y Σ q1 − γq 3 α z q 1 + β z Σ q1 + γq 2<br />

⎢<br />

⎣α x q 2 + β x Σ q2 + γq 3 α y q 2 + β y Σ q2 + γq 0 α z q 2 + β z Σ q2 − γq 1<br />

⎥<br />

⎦<br />

α x q 3 + β x Σ q3 − γq 2 α y q 3 + β y Σ q3 + γq 1 α z q 3 + β z Σ q3 + γq 0<br />

(5.67a)<br />

∂v n<br />

∂q b n<br />

= 2Φ∆T<br />

∂v n<br />

∂b f<br />

= −Θ∆T<br />

∂v n<br />

= Θ∆T<br />

∂f b<br />

(5.67b)<br />

∂r n<br />

∂q b n<br />

∂e pseudo<br />

∂q b n<br />

= 2Φ ∆T 2<br />

2<br />

[<br />

= − q 0<br />

||qn||<br />

b<br />

∂r n<br />

= −Θ ∆T 2<br />

∂b f 2<br />

− q 1<br />

||q b n||<br />

− q 2<br />

||q b n||<br />

∂r n<br />

2<br />

∂f b = Θ∆T 2<br />

]<br />

− q 3<br />

||qn||<br />

b<br />

(5.67c)<br />

(5.67d)<br />

α i = − sen(d/2) δ i ∆T<br />

2d<br />

(5.68a)<br />

β i = [− d 2cos(d/2) + sen(d/2)]<br />

d 3 δ i ∆T (5.68b)<br />

γ = sen(d/2) ∆T<br />

d<br />

(5.68c)<br />

Σ q0 = ( δ x q 1 + δ y q 2 + δ z q 3 ) (5.68d)<br />

Σ q1 = (−δ x q 0 + δ y q 3 − δ z q 2 )<br />

Σ q2 = (−δ x q 3 − δ y q 0 + δ z q 1 )<br />

(5.68e)<br />

(5.68f)<br />

Σ q3 = ( δ x q 2 − δ y q 1 − δ z q 0 ) (5.68g)<br />

(5.68h)<br />

⎡<br />

⎤<br />

(q 0 f x − q 3 f y + q 2 f z ) (q 1 f x + q 2 f y + q 3 f z ) (−q 2 f x + q 1 f y + q 0 f z ) (−q 3 f x − q 0 f y + q 1 f z )<br />

Φ = ⎢<br />

⎣(q 3 f x + q 0 f y − q 1 f z ) (q 2 f x − q 1 f y − q 0 f z ) ( q 1 f x + q 2 f y + q 3 f z ) ( q 0 f x − q 3 f y + q 2 f z ) ⎥<br />

⎦<br />

(q 2 f x − q 1 f y + q 0 f z ) (q 3 f x + q 0 f y − q 1 f z ) (−q 0 f x + q 3 f y − q 2 f z ) ( q 1 f x + q 2 f y + q 3 f z )<br />

⎡<br />

(q0 2 + q2 1 − q2 2 − q2 3 ) (2q ⎤<br />

1q 2 − 2q 0 q 3 ) (2q 1 q 3 + 2q 0 q 2 )<br />

Θ = ⎢<br />

⎣ (2q 1 q 2 + 2q 0 q 3 ) (q0 2 − q2 1 + q2 2 − q2 3 ) (2q 2q 3 − 2q 0 q 1 ) ⎥<br />

⎦<br />

(2q 1 q 3 − 2q 0 q 2 ) (2q 2 q 3 + 2q 0 q 1 ) (q0 2 − q2 1 − q2 2 + q2 3 )<br />

(5.68i)<br />

(5.68j)<br />

110

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