Diretório de artigos
prefácio
O quaternion dual também é chamado de operador espiral, que pode efetivamente descrever a relação de acoplamento entre posição e atitude.
1. O que é um quaternion dual?
O dual quaternion é uma ferramenta para descrever a posição e a atitude ao mesmo tempo.Este artigo explica sua multiplicação.
2. Multiplicação de quatérnio duplo
Dados dois quatérnios duais, sua multiplicação é tratada como multiplicação de quatérnios.
a = [ ai ] ∈ R 4 \boldsymbol{a}=[a_i]\in\mathbb{R}^4a=[ umeu]∈R4，b = [ bi ] ∈ R 4 \boldsymbol{b}=[b_i]\in\mathbb{R}^4b=[ beu]∈R4，c = [ ci ] ∈ R 4 \boldsymbol{c}=[c_i]\in\mathbb{R}^4c=[ ceu]∈R4，d = [ di ] ∈ R 4 \boldsymbol{d}=[d_i]\in\mathbb{R}^4d=[ deu]∈R4 é quatro quaternions, ondea 1 a_1a1，b 1 b_1b1，c 1 c_1c1，d 1 d_1d1Para os quatro escalares correspondentes, ε \varepsilonε é a unidade do quaternion dual.
[ a 1 + ε b 1 a 2 + ε b 2 a 3 + ε b 3 a 4 + ε b 4 ] ∘ [ c 1 + ε d 1 c 2 + ε d 2 c 3 + ε d 3 c 4 + ε d 4 ] \left[\begin{array}{c} a_1+\varepsilon b_1\\ a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+\varepsilon b_4 \end{array}\right] \circ \left[\ begin{array}{c} c_1+\varepsilon d_1\\ c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right] a1+ε b1a2+ε b2a3+ε b3a4+ε b4 ∘ c1+e d1c2+e d2c3+e d3c4+e d4
Referindo-se à multiplicação de quaternions , a fórmula acima pode ser expandida.
[ a 1 + b 1 a 2 + b 2 a 3 + b 3 a 4 + b 4 ] ∘ [ c 1 + b 1 c 2 + b 2 c 3 + b 3 c 4 + b d 4 ] = [ ( a 1 + e b 1 ) ( c 1 + e d 1 ) − ( a 2 + e b 2 ) ( c 2 + e d 2 ) − ( a 3 + e b 3 ) ( c + ε d 3 ) − ( a 4 + ε b 4 ) ( c 4 + ε d 4 ) ( c 1 + ε d 1 ) [ a 2 + ε b 2 a 3 + ε b 3 a 4 + ε b + ( a 1 + b 1 ) [ c 2 + e d 2 c 3 + e d 3 c 4 + e d 4 ] + [ a 2 + e b 2 a 3 + e b 3 a 4 + e b × [ c 2 + ε d 2 c 3 + ε d 3 c 4 + ε d 4 ] ] \left [\begin{array}{c} a_1+\varepsilon b_1\\ a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\; \ a_4+\varepsilon b_4 \end{array}\right] \circ \left[\begin{array} {c} c_1+\varepsilon d_1\\ c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end {array}\right] =\left[\begin{array}{c} (a_1+\varepsilon b_1 )(c_1+\varepsilon d_1)-(a_2+\varepsilon b_2)(c_2+\varepsilon d_2)-(a_3+\varepsilon b_3); (c_3+\varepsilond_3)-(a_4+\varepsilon b_4)(c_4+\varepsilon d_4)\\ (c_1+\varepsilon d_1)\left[\begin{array}{c} a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+\varepsilon b_4 \end{array}\right] +(a_1+\varepsilon b_1)\left[\begin{array}{c} c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right] + \left[\begin{array}{c} a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+\varepsilon b_4 \end{array}\right]^{\times} \left[\begin{array}{ c} c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right] \end{array}\right]\left[\begin{array}{c} a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+\varepsilon b_4 \end{array}\right]^{\times} \left[\begin{array}{c } c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right] \end{array}\right]\left[\begin{array}{c} a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+\varepsilon b_4 \end{array}\right]^{\times} \left[\begin{array}{c } c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right] \end{array}\right]
a1+ε b1a2+ε b2a3+ε b3a4+ε b4
∘
c1+e d1c2+e d2c3+e d3c4+e d4
=
( um1+ε b1) ( c1+e d1)−( um2+ε b2) ( c2+e d2)−( um3+ε b3) ( c3+e d3)−( um4+ε b4) ( c4+e d4)( c1+e d1)
a2+ε b2a3+ε b3a4+ε b4
+( um1+ε b1)
c2+e d2c3+e d3c4+e d4
+
a2+ε b2a3+ε b3a4+ε b4
×
c2+e d2c3+e d3c4+e d4
Então, equação
[ a 2 + x 2 a 3 + x 3 a 4 + x 4 ] × = [ 0 − ( a 4 + x 4 ) a 3 + x 3 a 4 + x 4 0 − ( a 2 + ε b 2 ) − ( a 3 + ε b 3 ) a 2 + ε b 2 0 ] \left[\begin{array}{c} a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+\ varepsilon b_4 \end{array }\right]^\times=\left[\begin{array}{ccc} 0 & -(a_4+\varepsilon b_4) & a_3+\varepsilon b_3 \\ a_4+\varepsilon b_4 & 0 & -( a_2+\varepsilon b_2)\ \ -(a_3+\varepsilon b_3) & a_2+\varepsilon b_2 & 0 \end{array}\right]
a2+ε b2a3+ε b3a4+ε b4
×=
0a4+ε b4− ( um3+ε b3)− ( um4+ε b4)0a2+ε b2a3+ε b3− ( um2+ε b2)0
expandir ainda mais,
[ a 1 + b 1 a 2 + b 2 a 3 + b 3 a 4 + b 4 ] ∘ [ c 1 + b 1 c 2 + b 2 c 3 + b 3 c 4 + b d 4 ] = [ a 1 c 1 − a 2 c 2 − a 3 c 3 − a 4 c 4 + ε ( b 1 c 1 + a 1 d 1 − b 2 c 2 − a 2 d 2 − b 3 c 3 − a 3 d 3 − b 4 c 4 − a 4 d 4 ) [ c 1 a 2 + ε ( c 1 b 2 + d 1 a 2 ) c 1 a 3 + ε ( c 1 b 3 + d 1 a 3 ) c 1 a 4 + ε ( c 1 b 4 + d 1 a 4 ) ] + [ a 1 c 2 + ε ( a 1 d 2 + b 1 c 2 ) a 1 c 3 + ε ( a 1 d + b 1 c 3 ) a 1 c 4 + ε ( a 1 d 4 + b 1 c 4 ) ] + [ 0 − ( a 4 + ε b 4 ) a 3 + ε b 3 a 4 + ε b 4 0 − ( a 2 + b 2 ) − ( a 3 + b 3 ) a 2 + b 2 0 ] [ c 2 + b 2 c 3 + b 3 c 4 + b 4 ] ] \left[\ begin{array}{c} a_1+\varepsilon b_1\\ a_2+ \varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+\varepsilon b_4 \end{array}\right] \circ \left[\begin{array}{c} c_1+\varepsilon d_1\\ c_2+\varepsilon d_2\\ c_3+\ varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right]=\left[\begin{array}{c} a_1c_1-a_2c_2-a_3c_3-a_4c_4+\varepsilon(b_1c_1+a_1d_1-b_2c_2-a_2d_2-b_3c_3-a_3d_3-b_4c_4-a_4d_4) \\ \left[\begin{array}{c } c_1a_2+\varepsilon (c_1b_2+d_1a_2)\\ c_1a_3+\varepsilon (c_1b_3+d_1a_3)\\ c_1a_4+\varepsilon (c_1b_4+d_1a_4) \end{array}\right] + \left[\begin{array}{c} a_1c_2+ \varepsilon (a_1d_2+b_1c_2)\\ a_1c_3+\varepsilon (a_1d_3+b_1c_3)\\ a_1c_4+\varepsilon (a_1d_4+b_1c_4) \end{array}\right] + \left[\begin{array}{ccc} 0 & - (a_4+\varepsilon b_4) & a_3+\varepsilon b_3 \\ a_4+\varepsilon b_4 & 0 & -(a_2+\varepsilon b_2)\\ -(a_3+\varepsilon b_3) & a_2+\varepsilon b_2 & 0 \end{array}\right ] \left[\begin{array}{c} c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right] \end{array}\right]0 \end{array}\right] \left[\begin{array}{c} c_2+\varepsilon d_2\\ c_3+\varepsilon d_3\\ c_4+\varepsilon d_4 \end{array}\right] \end{array}\ certo]
a1+ε b1a2+ε b2a3+ε b3a4+ε b4
∘
c1+e d1c2+e d2c3+e d3c4+e d4
=
a1c1−a2c2−a3c3−a4c4+e ( b1c1+a1d1−b2c2−a2d2−b3c3−a3d3−b4c4−a4d4)
c1a2+e ( c1b2+d1a2)c1a3+e ( c1b3+d1a3)c1a4+e ( c1b4+d1a4)
+
a1c2+e ( um1d2+b1c2)a1c3+e ( um1d3+b1c3)a1c4+e ( um1d4+b1c4)
+
0a4+ε b4− ( um3+ε b3)− ( um4+ε b4)0a2+ε b2a3+ε b3− ( um2+ε b2)0
c2+e d2c3+e d3c4+e d4
Após a ordenação, podemos obter
[ a 1 + b 1 a 2 + b 2 a 3 + b 3 a 4 + b 4 ] ∘ [ c 1 + b 1 c 2 + b 2 c 3 + b 3 c 4 + b d 4 ] = [ a 1 c 1 − a 2 c 2 − a 3 c 3 − a 4 c 4 + ε ( b 1 c 1 + a 1 d 1 − b 2 c 2 − a 2 d 2 − b 3 c 3 − a 3 d 3 − b 4 c 4 − a 4 d 4 ) [ c 1 a 2 + a 1 c 2 + ε ( c 1 b 2 + d 1 a 2 + a 1 d 2 + b 1 c 2 ) c 1 a 3 + a 1 c 3 + ε ( c 1 b 3 + d 1 a 3 + a 1 d 3 + b 1 c 3 ) c 1 a 4 + a 1 c 4 + ε ( c 1 b 4 + d 1 a 4 + a 1 d 4 + b 1 c 4 ) ] + [ − ( a 4 + ε b 4 ) ( c 3 + ε d 3 ) + ( a 3 + ε b 3 ) ( c 4 + ε d ) ( a 4 + e b 4 ) ( c 2 + e d 2 ) − ( a 2 + e b 2 ) ( c 4 + e b 4 ) − ( a 3 + e b 3 ) ( c 2 + e d ) + ( a 2 + ε b 2 ) ( c 3 + ε d 3 ) ] ] \left[\begin{array}{c} a_1+\varepsilon b_1\\ a_2+\varepsilon b_2\\ a_3+\varepsilon b_3\\ a_4+ \varepsilon b_4 \end{array}\right] \circ \left [\begin{array}{c} c_1+\varepsilon d_1\\ c_2+\varepsilon d_2\\ c_3+\varepsilond_3\\ c_4+\varepsilon d_4 \end{array}\right] =\left[\begin{array}{c} a_1c_1-a_2c_2-a_3c_3-a_4c_4+\varepsilon(b_1c_1+a_1d_1-b_2c_2-a_2d_2-b_3c_3-a_3d_3-b_4c_4 -a_4d_4) \\ \left[\begin{array}{c} c_1a_2+a_1c_2+\varepsilon (c_1b_2+d_1a_2+a_1d_2+b_1c_2)\\ c_1a_3+a_1c_3+\varepsilon (c_1b_3+d_1a_3+a_1d_3+b_1c_3)\\ c_1a_4+ a_1c_4+\varepsilon (c_1b_4+d_1a_4+a_1d_4+b_1c_4) \end{array}\right] + \left[\begin{array}{c} -(a_4+\varepsilon b_4)(c_3+\varepsilon d_3)+(a_3+\varepsilon b_3)(c_4+\varepsilon d_4)\\ (a_4+\varepsilon b_4)(c_2+\varepsilon d_2)-(a_2+\varepsilon b_2)(c_4+\varepsilon d_4)\\ -(a_3+\varepsilon b_3)(c_2+\varepsilon d_2) +(a_2+\varepsilon b_2)(c_3+\varepsilon d_3) \end{array}\right] \end{array}\right]a_1c_1-a_2c_2-a_3c_3-a_4c_4+\varepsilon(b_1c_1+a_1d_1-b_2c_2-a_2d_2-b_3c_3-a_3d_3-b_4c_4-a_4d_4) \\ \left[\begin{array}{c} c_1a_2+a_1c_2+\varepsi lon (c_1b_2+d_1a_2+a_1d_2 +b_1c_2)\\ c_1a_3+a_1c_3+\varepsilon (c_1b_3+d_1a_3+a_1d_3+b_1c_3)\\ c_1a_4+a_1c_4+\varepsilon (c_1b_4+d_1a_4+a_1d_4+b_1c_4) \end{array}\right] + \left[\ começar{ array}{c} -(a_4+\varepsilon b_4)(c_3+\varepsilon d_3)+(a_3+\varepsilon b_3)(c_4+\varepsilon d_4)\\ (a_4+\varepsilon b_4)(c_2+\varepsilon d_2)-(a_2+\varepsilon b_2)(c_4+\varepsilon d_4)\\ -(a_3+\varepsilon b_3)(c_2+\varepsilon d_2)+(a_2+\varepsilon b_2)(c_3+\varepsilon d_3) \end{array}\right] \end{array}\ certo]a_1c_1-a_2c_2-a_3c_3-a_4c_4+\varepsilon(b_1c_1+a_1d_1-b_2c_2-a_2d_2-b_3c_3-a_3d_3-b_4c_4-a_4d_4) \\ \left[\begin{array}{c} c_1a_2+a_1c_2+\varepsi lon (c_1b_2+d_1a_2+a_1d_2 +b_1c_2)\\ c_1a_3+a_1c_3+\varepsilon (c_1b_3+d_1a_3+a_1d_3+b_1c_3)\\ c_1a_4+a_1c_4+\varepsilon (c_1b_4+d_1a_4+a_1d_4+b_1c_4) \end{array}\right] + \left[\ começar{ array}{c} -(a_4+\varepsilon b_4)(c_3+\varepsilon d_3)+(a_3+\varepsilon b_3)(c_4+\varepsilon d_4)\\ (a_4+\varepsilon b_4)(c_2+\varepsilon d_2)-(a_2+\varepsilon b_2)(c_4+\varepsilon d_4)\\ -(a_3+\varepsilon b_3)(c_2+\varepsilon d_2)+(a_2+\varepsilon b_2)(c_3+\varepsilon d_3) \end{array}\right] \end{array}\ certo]\end{array}\right] + \left[\begin{array}{c} -(a_4+\varepsilon b_4)(c_3+\varepsilon d_3)+(a_3+\varepsilon b_3)(c_4+\varepsilon d_4)\\ (a_4+ \varepsilon b_4)(c_2+\varepsilon d_2)-(a_2+\varepsilon b_2)(c_4+\varepsilon d_4)\\ -(a_3+\varepsilon b_3)(c_2+\varepsilon d_2)+(a_2+\varepsilon b_2)(c_3+\varepsilon d_3 ) \end{array}\right] \end{array}\right]\end{array}\right] + \left[\begin{array}{c} -(a_4+\varepsilon b_4)(c_3+\varepsilon d_3)+(a_3+\varepsilon b_3)(c_4+\varepsilon d_4)\\ (a_4+ \varepsilon b_4)(c_2+\varepsilon d_2)-(a_2+\varepsilon b_2)(c_4+\varepsilon d_4)\\ -(a_3+\varepsilon b_3)(c_2+\varepsilon d_2)+(a_2+\varepsilon b_2)(c_3+\varepsilon d_3 ) \end{array}\right] \end{array}\right]
a1+ε b1a2+ε b2a3+ε b3a4+ε b4
∘
c1+e d1c2+e d2c3+e d3c4+e d4
=
a1c1−a2c2−a3c3−a4c4+e ( b1c1+a1d1−b2c2−a2d2−b3c3−a3d3−b4c4−a4d4)
c1a2+a1c2+e ( c1b2+d1a2+a1d2+b1c2)c1a3+a1c3+e ( c1b3+d1a3+a1d3+b1c3)c1a4+a1c4+e ( c1b4+d1a4+a1d4+b1c4)
+
− ( um4+ε b4) ( c3+e d3)+( um3+ε b3) ( c4+e d4)( um4+ε b4) ( c2+e d2)−( um2+ε b2) ( c4+e d4)− ( um3+ε b3) ( c2+e d2)+( um2+ε b2) ( c3+e d3)
Resumir
A multiplicação de quatérnios duais tem a mesma forma que a multiplicação de quatérnios.