本文仅供学习使用,总结很多本现有讲述运动学或动力学书籍后的总结,从矢量的角度进行分析,方法比较传统,但更易理解,并且现有的看似抽象方法,两者本质上并无不同。
2024年底本人学位论文发表后方可摘抄
若有帮助请引用
本文参考:
.
食用方法
如何表达刚体在空间中的位置与姿态
姿态参数如何表达?不同表达方式直接的转换关系?
旋转矩阵?转换矩阵?有什么意义和性质?转置代表什么?
如何表示连续变换?——与RPY有关
齐次坐标的意义——简化公式?
务必自己推导全部公式,并理解每个符号的含义
同样基于罗德里格旋转公式,定义四个欧拉参数为:
q
?
=
[
s
v
?
]
=
[
cos
?
θ
2
→
s
c
a
l
e
??
p
a
r
t
v
?
sin
?
θ
2
→
v
e
c
t
o
r
??
p
a
r
t
]
=
[
cos
?
θ
2
v
1
sin
?
θ
2
v
2
sin
?
θ
2
v
3
sin
?
θ
2
]
=
[
q
1
q
2
q
3
q
4
]
\vec{q}=\left[ \begin{array}{c} s\\ \vec{v}\\ \end{array} \right] =\left[ \begin{matrix} \cos \frac{\theta}{2}& \rightarrow scale\,\,part\\ \vec{v}\sin \frac{\theta}{2}& \rightarrow vector\,\,part\\ \end{matrix} \right] =\left[ \begin{array}{c} \cos \frac{\theta}{2}\\ v_1\sin \frac{\theta}{2}\\ v_2\sin \frac{\theta}{2}\\ v_3\sin \frac{\theta}{2}\\ \end{array} \right] =\left[ \begin{array}{c} q_1\\ q_2\\ q_3\\ q_4\\ \end{array} \right]
q?=[sv?]=[cos2θ?vsin2θ??→scalepart→vectorpart?]=
?cos2θ?v1?sin2θ?v2?sin2θ?v3?sin2θ??
?=
?q1?q2?q3?q4??
?
根据上述定义,可将轴角变换的旋转矩阵
[
Q
B
A
]
\left[ Q_{\mathrm{B}}^{A} \right]
[QBA?] 改写为:
[
Q
B
A
]
=
[
1
?
2
q
3
2
?
2
q
4
2
2
(
q
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3
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2
(
q
2
q
4
+
q
1
q
3
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2
(
q
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q
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+
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1
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1
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2
q
2
2
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2
q
4
2
2
(
q
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4
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q
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2
(
q
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q
1
q
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2
(
q
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+
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1
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1
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q
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2
]
=
[
2
q
1
2
+
2
q
2
2
?
1
2
(
q
2
q
3
?
q
1
q
4
)
2
(
q
2
q
4
+
q
1
q
3
)
2
(
q
2
q
3
+
q
1
q
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)
2
q
1
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+
2
q
3
2
?
1
2
(
q
3
q
4
?
q
1
q
2
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2
(
q
2
q
4
?
q
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q
3
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2
(
q
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+
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1
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2
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2
q
1
2
+
2
q
4
2
?
1
]
\left[ Q_{\mathrm{B}}^{A} \right] =\left[ \begin{matrix} 1-2{q_3}^2-2{q_4}^2& 2\left( q_2q_3-q_1q_4 \right)& 2\left( q_2q_4+q_1q_3 \right)\\ 2\left( q_2q_3+q_1q_4 \right)& 1-2{q_2}^2-2{q_4}^2& 2\left( q_3q_4-q_1q_2 \right)\\ 2\left( q_2q_4-q_1q_3 \right)& 2\left( q_3q_4+q_1q_2 \right)& 1-2{q_2}^2-2{q_3}^2\\ \end{matrix} \right] =\left[ \begin{matrix} 2{q_1}^2+2{q_2}^2-1& 2\left( q_2q_3-q_1q_4 \right)& 2\left( q_2q_4+q_1q_3 \right)\\ 2\left( q_2q_3+q_1q_4 \right)& 2{q_1}^2+2{q_3}^2-1& 2\left( q_3q_4-q_1q_2 \right)\\ 2\left( q_2q_4-q_1q_3 \right)& 2\left( q_3q_4+q_1q_2 \right)& 2{q_1}^2+2{q_4}^2-1\\ \end{matrix} \right]
[QBA?]=
?1?2q3?2?2q4?22(q2?q3?+q1?q4?)2(q2?q4??q1?q3?)?2(q2?q3??q1?q4?)1?2q2?2?2q4?22(q3?q4?+q1?q2?)?2(q2?q4?+q1?q3?)2(q3?q4??q1?q2?)1?2q2?2?2q3?2?
?=
?2q1?2+2q2?2?12(q2?q3?+q1?q4?)2(q2?q4??q1?q3?)?2(q2?q3??q1?q4?)2q1?2+2q3?2?12(q3?q4?+q1?q2?)?2(q2?q4?+q1?q3?)2(q3?q4??q1?q2?)2q1?2+2q4?2?1?
?
进而定义矩阵:
B
3
×
4
=
[
?
q
2
q
1
?
q
4
q
3
?
q
3
q
4
q
1
?
q
2
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q
4
?
q
3
q
2
q
1
]
B
ˉ
3
×
4
=
[
?
q
2
q
1
q
4
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q
3
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q
3
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q
4
q
1
q
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q
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3
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1
]
\begin{split} B_{3\times 4}&=\left[ \begin{array}{cccc} -q_2& q_1& -q_4& q_3\\ -q_3& q_4& q_1& -q_2\\ -q_4& -q_3& q_2& q_1\\ \end{array} \right] \\ \bar{B}_{3\times 4}&=\left[ \begin{array}{cccc} -q_2& q_1& q_4& -q_3\\ -q_3& -q_4& q_1& q_2\\ -q_4& q_3& -q_2& q_1\\ \end{array} \right] \end{split}
B3×4?Bˉ3×4??=
??q2??q3??q4??q1?q4??q3???q4?q1?q2??q3??q2?q1??
?=
??q2??q3??q4??q1??q4?q3??q4?q1??q2???q3?q2?q1??
??
则有:
[
Q
B
A
]
=
B
3
×
4
B
ˉ
3
×
4
T
\left[ Q_{\mathrm{B}}^{A} \right] =B_{3\times 4}{\bar{B}_{3\times 4}}^{\mathrm{T}}
[QBA?]=B3×4?Bˉ3×4?T
上述矩阵具有如下性质:
B
3
×
4
B
3
×
4
T
=
B
ˉ
3
×
4
B
ˉ
3
×
4
T
=
E
B
3
×
4
T
B
3
×
4
=
B
ˉ
3
×
4
T
B
ˉ
3
×
4
=
E
4
×
4
?
q
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?
q
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T
B
ˉ
3
×
4
?
q
?
=
0
?
B_{3\times 4}{B_{3\times 4}}^{\mathrm{T}}=\bar{B}_{3\times 4}{\bar{B}_{3\times 4}}^{\mathrm{T}}=E \\ {B_{3\times 4}}^{\mathrm{T}}B_{3\times 4}={\bar{B}_{3\times 4}}^{\mathrm{T}}\bar{B}_{3\times 4}=E_{4\times 4}-\vec{q}\cdot \vec{q}^{\mathrm{T}} \\ \bar{B}_{3\times 4}\cdot \vec{q}=\vec{0}
B3×4?B3×4?T=Bˉ3×4?Bˉ3×4?T=EB3×4?TB3×4?=Bˉ3×4?TBˉ3×4?=E4×4??q??q?TBˉ3×4??q?=0
因此,若已知旋转矩阵:
[
Q
B
A
]
=
[
q
11
q
12
q
13
q
21
q
22
q
23
q
31
q
23
q
33
]
\left[ Q_{\mathrm{B}}^{A} \right] =\left[ \begin{matrix} q_{11}& q_{12}& q_{13}\\ q_{21}& q_{22}& q_{23}\\ q_{31}& q_{23}& q_{33}\\ \end{matrix} \right]
[QBA?]=
?q11?q21?q31??q12?q22?q23??q13?q23?q33??
?,则可求解四元数参数为:
[ q 1 q 2 q 3 q 4 ] = [ 1 2 q 11 + q 22 + q 33 + 1 q 32 ? q 23 4 q 1 q 13 ? q 31 4 q 1 q 21 ? q 12 4 q 1 ] \left[ \begin{array}{c} q_1\\ q_2\\ q_3\\ q_4\\ \end{array} \right] =\left[ \begin{array}{c} \frac{1}{2}\sqrt{q_{11}+q_{22}+q_{33}+1}\\ \frac{q_{32}-q_{23}}{4q_1}\\ \frac{q_{13}-q_{31}}{4q_1}\\ \frac{q_{21}-q_{12}}{4q_1}\\ \end{array} \right] ?q1?q2?q3?q4?? ?= ?21?q11?+q22?+q33?+1?4q1?q32??q23??4q1?q13??q31??4q1?q21??q12??? ?
对于任意矢量
R
?
F
\vec{R}^F
RF,可通过上述四元数矢量进行旋转变化:
[
0
R
?
′
F
]
=
q
?
F
?
[
0
R
?
F
]
?
(
q
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F
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?
1
=
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F
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[
0
R
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?
(
q
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F
)
T
=
L
(
q
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R
(
q
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T
[
0
R
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F
]
=
R
(
q
)
T
L
(
q
)
[
0
R
?
F
]
\left[ \begin{array}{c} 0\\ {\vec{R}^{\prime}}^F\\ \end{array} \right] =\vec{q}^F\cdot \left[ \begin{array}{c} 0\\ \vec{R}^F\\ \end{array} \right] \cdot \left( \vec{q}^F \right) ^{-1}=\vec{q}^F\cdot \left[ \begin{array}{c} 0\\ \vec{R}^F\\ \end{array} \right] \cdot \left( \vec{q}^F \right) ^{\mathrm{T}} \\ =L\left( q \right) R\left( q \right) ^{\mathrm{T}}\left[ \begin{array}{c} 0\\ \vec{R}^F\\ \end{array} \right] =R\left( q \right) ^{\mathrm{T}}L\left( q \right) \left[ \begin{array}{c} 0\\ \vec{R}^F\\ \end{array} \right]
[0R′F?]=q?F?[0RF?]?(q?F)?1=q?F?[0RF?]?(q?F)T=L(q)R(q)T[0RF?]=R(q)TL(q)[0RF?]
同理也可以将上述视为矢量的坐标系变换,其转换矩阵拓展为
4
×
4
4\times 4
4×4,进而有转换矩阵
[
Q
B
A
]
4
×
4
T
\left[ Q_{\mathrm{B}}^{A} \right]^{\mathrm{T}} _{4\times 4}
[QBA?]4×4T?:
[
Q
B
A
]
4
×
4
=
L
(
q
)
R
(
q
)
T
=
R
(
q
)
T
L
(
q
)
\left[ Q_{\mathrm{B}}^{A} \right] _{4\times 4}=L\left( q \right) R\left( q \right) ^{\mathrm{T}}=R\left( q \right) ^{\mathrm{T}}L\left( q \right)
[QBA?]4×4?=L(q)R(q)T=R(q)TL(q)
此时,则有:
[
0
R
?
B
]
=
[
Q
B
A
]
4
×
4
T
[
0
R
?
A
]
\left[ \begin{array}{c} 0\\ \vec{R}^B\\ \end{array} \right] =\left[ Q_{\mathrm{B}}^{A} \right]^{\mathrm{T}} _{4\times 4}\left[ \begin{array}{c} 0\\ \vec{R}^A\\ \end{array} \right]
[0RB?]=[QBA?]4×4T?[0RA?]
欧拉角是一种较为原始的旋转表示方式,在实际的算法运用过程中,除了描述已知姿态的刚体角度外,在实际计算中,效果很差(具有奇异性、高度非线性、计算复杂)。因此,并不推荐用欧拉角来描述转换矩阵。本节仅对部分重点内容进行介绍。
欧拉角(ZYX变换)的旋转变换描述为:绕固定坐标系的基矢量 k ? F \vec{k}^{F} kF回转 γ \gamma γ,得到新标架 { F 1 : ( i ? 1 F , j ? 1 F , k ? 1 F ) } \left\{ F_1:\left( \vec{i}_{1}^{F},\vec{j}_{1}^{F},\vec{k}_{1}^{F} \right) \right\} {F1?:(i1F?,j?1F?,k1F?)};再绕基矢量 j ? 1 F \vec{j}_{1}^{F} j?1F?回转 β \beta β,得到新标架 { F 2 : ( i ? 2 F , j ? 2 F , k ? 2 F ) } \left\{ F_2:\left( \vec{i}_{2}^{F},\vec{j}_{2}^{F},\vec{k}_{2}^{F} \right) \right\} {F2?:(i2F?,j?2F?,k2F?)};最后绕基矢量 i ? 2 F \vec{i}_{2}^{F} i2F?回转 α \alpha α,得到新标架 { F 3 : ( i ? 3 F , j ? 3 F , k ? 3 F ) } \left\{ F_3:\left( \vec{i}_{3}^{F},\vec{j}_{3}^{F},\vec{k}_{3}^{F} \right) \right\} {F3?:(i3F?,j?3F?,k3F?)}为最终的变换姿态(运动坐标系下连续转动,右乘)。因此对于多次连续转动而言有:
[ Q M F ] = [ Q F 1 F ( k ? F , γ ) ] [ Q F 2 F 1 ( j ? 1 F , β ) ] [ Q F 3 ( M ) F 2 ( i ? 2 F , α ) ] ] \left[ Q_{\mathrm{M}}^{F} \right] =\left[ Q_{\mathrm{F}_1}^{F}\left( \vec{k}^F,\gamma \right) \right] \left[ Q_{\mathrm{F}_2}^{F_1}\left( \vec{j}_{1}^{F},\beta \right) \right] \left[ Q_{\mathrm{F}_3\left( M \right)}^{F_2}\left( \vec{i}_{2}^{F},\alpha \right) \right] ] [QMF?]=[QF1?F?(kF,γ)][QF2?F1??(j?1F?,β)][QF3?(M)F2??(i2F?,α)]]
而对于RPY角(滚转角Roll,仰俯角Pitch,偏航角Yaw)而言,其旋转变换描述为:绕固定坐标系的基矢量
i
?
F
\vec{i}^{F}
iF回转
α
\alpha
α,得到新标架
{
F
1
:
(
i
?
1
F
,
j
?
1
F
,
k
?
1
F
)
}
\left\{ F_1:\left( \vec{i}_{1}^{F},\vec{j}_{1}^{F},\vec{k}_{1}^{F} \right) \right\}
{F1?:(i1F?,j?1F?,k1F?)};再绕固定坐标系的基矢量
j
?
F
\vec{j}^{F}
j?F回转
β
\beta
β,得到新标架
{
F
2
:
(
i
?
2
F
,
j
?
2
F
,
k
?
2
F
)
}
\left\{ F_2:\left( \vec{i}_{2}^{F},\vec{j}_{2}^{F},\vec{k}_{2}^{F} \right) \right\}
{F2?:(i2F?,j?2F?,k2F?)};最后绕固定坐标系的基矢量
k
?
F
\vec{k}^{F}
kF回转
γ
\gamma
γ,得到新标架
{
F
3
:
(
i
?
3
F
,
j
?
3
F
,
k
?
3
F
)
}
\left\{ F_3:\left( \vec{i}_{3}^{F},\vec{j}_{3}^{F},\vec{k}_{3}^{F} \right) \right\}
{F3?:(i3F?,j?3F?,k3F?)}为最终的变换姿态(固定坐标系下连续转动,左乘)。因此对于多次连续转动而言有:
[
Q
M
F
]
=
[
Q
F
3
(
M
)
F
2
(
k
?
F
,
γ
)
]
[
Q
F
2
F
1
(
j
?
F
,
β
)
]
[
Q
F
1
F
(
i
?
F
,
α
)
]
\left[ Q_{\mathrm{M}}^{F} \right] =\left[ Q_{\mathrm{F}_3\left( M \right)}^{F_2}\left( \vec{k}^F,\gamma \right) \right] \left[ Q_{\mathrm{F}_2}^{F_1}\left( \vec{j}^F,\beta \right) \right] \left[ Q_{\mathrm{F}_1}^{F}\left( \vec{i}^F,\alpha \right) \right]
[QMF?]=[QF3?(M)F2??(kF,γ)][QF2?F1??(j?F,β)][QF1?F?(iF,α)]
进而求解出其转换矩阵为:
[
Q
M
F
]
=
[
cos
?
β
cos
?
γ
?
cos
?
β
sin
?
γ
sin
?
β
sin
?
α
sin
?
β
cos
?
γ
+
cos
?
α
sin
?
γ
?
sin
?
α
sin
?
β
sin
?
γ
+
cos
?
α
cos
?
γ
?
sin
?
α
cos
?
β
?
cos
?
α
sin
?
β
cos
?
γ
+
sin
?
α
sin
?
γ
cos
?
α
sin
?
β
cos
?
γ
+
sin
?
α
cos
?
γ
cos
?
α
cos
?
β
]
\left[ Q_{\mathrm{M}}^{F} \right] =\left[ \begin{matrix} \cos \beta \cos \gamma& -\cos \beta \sin \gamma& \sin \beta\\ \sin \alpha \sin \beta \cos \gamma +\cos \alpha \sin \gamma& -\sin \alpha \sin \beta \sin \gamma +\cos \alpha \cos \gamma& -\sin \alpha \cos \beta\\ -\cos \alpha \sin \beta \cos \gamma +\sin \alpha \sin \gamma& \cos \alpha \sin \beta \cos \gamma +\sin \alpha \cos \gamma& \cos \alpha \cos \beta\\ \end{matrix} \right]
[QMF?]=
?cosβcosγsinαsinβcosγ+cosαsinγ?cosαsinβcosγ+sinαsinγ??cosβsinγ?sinαsinβsinγ+cosαcosγcosαsinβcosγ+sinαcosγ?sinβ?sinαcosβcosαcosβ?
?
同理,若已知转换矩阵:
[
Q
B
A
]
=
[
q
11
q
12
q
13
q
21
q
22
q
23
q
31
q
32
q
33
]
\left[ Q_{\mathrm{B}}^{A} \right] =\left[ \begin{matrix} q_{11}& q_{12}& q_{13}\\ q_{21}& q_{22}& q_{23}\\ q_{31}& q_{32}& q_{33}\\ \end{matrix} \right]
[QBA?]=
?q11?q21?q31??q12?q22?q32??q13?q23?q33??
?, 则有:
[
α
β
γ
]
=
[
a
r
c
tan
?
(
?
q
23
q
33
)
a
r
c
sin
?
(
q
13
)
a
r
c
tan
?
(
?
q
12
q
11
)
]
\left[ \begin{array}{c} \alpha\\ \beta\\ \gamma\\ \end{array} \right] =\left[ \begin{array}{c} \mathrm{arc}\tan \left( -\frac{q_{23}}{q_{33}} \right)\\ \mathrm{arc}\sin \left( q_{13} \right)\\ \mathrm{arc}\tan \left( -\frac{q_{12}}{q_{11}} \right)\\ \end{array} \right]
?αβγ?
?=
?arctan(?q33?q23??)arcsin(q13?)arctan(?q11?q12??)?
?
绕两条不同轴进行转动的转换矩阵乘积不可交换。
3.1~3.6 只考虑了坐标系姿态的表达,专注于如何求解/表达
[
Q
M
F
]
\left[ Q_{\mathrm{M}}^{F} \right]
[QMF?], 而对于更一般的情况:(忽略原点重合)
R
?
P
F
=
[
Q
M
F
]
R
?
P
M
+
R
?
M
F
\vec{R}_{\mathrm{P}}^{F}=\left[ Q_{\mathrm{M}}^{F} \right] \vec{R}_{\mathrm{P}}^{M}+\vec{R}_{\mathrm{M}}^{F}
RPF?=[QMF?]RPM?+RMF?
引入齐次矩阵Homogeneous Transformation Matrix
:
[
T
M
F
]
\left[ T_{\mathrm{M}}^{F} \right]
[TMF?]
R
?
P
F
=
[
Q
M
F
]
R
?
P
M
+
R
?
M
F
?
[
R
?
P
F
1
]
=
[
[
Q
F
M
]
R
?
M
F
0
1
×
3
1
]
4
×
4
[
R
?
P
M
1
]
\vec{R}_{\mathrm{P}}^{F}=\left[ Q_{\mathrm{M}}^{F} \right] \vec{R}_{\mathrm{P}}^{M}+\vec{R}_{\mathrm{M}}^{F}\Rightarrow \left[ \begin{array}{c} \vec{R}_{\mathrm{P}}^{F}\\ 1\\ \end{array} \right] =\left[ \begin{matrix} \left[ Q_{\mathrm{F}}^{M} \right]& \vec{R}_{\mathrm{M}}^{F}\\ 0_{1\times 3}& 1\\ \end{matrix} \right] _{4\times 4}\left[ \begin{array}{c} \vec{R}_{\mathrm{P}}^{M}\\ 1\\ \end{array} \right]
RPF?=[QMF?]RPM?+RMF??[RPF?1?]=[[QFM?]01×3??RMF?1?]4×4?[RPM?1?]
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\Rightarrow \left[ T_{\mathrm{M}}^{F} \right] =\left[ \begin{matrix} \left[ Q_{\mathrm{M}}^{F} \right]& \vec{R}_{\mathrm{M}}^{F}\\ 0& 1\\ \end{matrix} \right]
?[TMF?]=[[QMF?]0?RMF?1?]
令:
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∈
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\left[ \vec{R}_{\mathrm{P}}^{F} \right] =\left[ \begin{array}{c} \vec{R}_{\mathrm{P}}^{F}\\ 1\\ \end{array} \right] \in \mathbb{R} ^4
[RPF?]=[RPF?1?]∈R4,则有:
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\left[ \vec{R}_{\mathrm{P}}^{F} \right] =\left[ T_{\mathrm{M}}^{F} \right] \left[ \vec{R}_{\mathrm{P}}^{M} \right]
[RPF?]=[TMF?][RPM?]
对于向量 R ? P 1 P 2 F \vec{R}_{\mathrm{P}_1\mathrm{P}_2}^{F} RP1?P2?F? 而言,则有:
[ R ? P 1 P 2 F ] = [ R ? P 2 F ? R ? P 1 F ] = [ R ? P 2 F 1 ] ? [ R ? P 1 F 1 ] = [ R ? P 2 F ? R ? P 1 F 0 ] = [ R ? P 1 P 2 F 0 ] \left[ \vec{R}_{\mathrm{P}_1\mathrm{P}_2}^{F} \right] =\left[ \vec{R}_{\mathrm{P}_2}^{F}-\vec{R}_{\mathrm{P}_1}^{F} \right] =\left[ \begin{array}{c} \vec{R}_{\mathrm{P}_2}^{F}\\ 1\\ \end{array} \right] -\left[ \begin{array}{c} \vec{R}_{\mathrm{P}_1}^{F}\\ 1\\ \end{array} \right] =\left[ \begin{array}{c} \vec{R}_{\mathrm{P}_2}^{F}-\vec{R}_{\mathrm{P}_1}^{F}\\ 0\\ \end{array} \right] =\left[ \begin{array}{c} \vec{R}_{\mathrm{P}_1\mathrm{P}_2}^{F}\\ 0\\ \end{array} \right] [RP1?P2?F?]=[RP2?F??RP1?F?]=[RP2?F?1?]?[RP1?F?1?]=[RP2?F??RP1?F?0?]=[RP1?P2?F?0?]
对于固定坐标系下同一点/向量,在不同坐标系
{
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}
,
{
B
}
\left\{ A \right\} ,\left\{ B \right\}
{A},{B}下进行表达,存在如下转换关系:
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\vec{R}_{\mathrm{Vector}}^{A}=\left[ Q_{\mathrm{B}}^{A} \right] \vec{R}_{\mathrm{Vector}}^{B}
RVectorA?=[QBA?]RVectorB?
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\vec{R}_{\mathrm{P}}^{A}=\left[ Q_{\mathrm{B}}^{A} \right] \vec{R}_{\mathrm{P}}^{B}+\vec{R}_{\mathrm{B}}^{A}
RPA?=[QBA?]RPB?+RBA?
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