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
机构运动学与动力学分析与建模 Ch00-3刚体的位形 Configuration of Rigid BodyPart3
-
- 3.8 点、线、面、向量在坐标系下的表达
-
- 3.8.1 线的特征
- 3.8.2 面的特征
- 3.9 简单的示例与计算
3.8 点、线、面、向量在坐标系下的表达
对于固定坐标系下同一
{
A
}
,
{
B
}
left{ A
ight} ,left{ B
ight}
{A},{B}下进行表达,存在如下转换关系:
R
?
V
e
c
t
o
r
A
=
[
Q
B
A
]
R
?
V
e
c
t
o
r
B
vec{R}_{mathrm{Vector}}^{A}=left[ Q_{mathrm{B}}^{A}
ight] vec{R}_{mathrm{Vector}}^{B}
R
R
?
P
A
=
[
Q
B
A
]
R
?
P
B
+
R
?
B
A
vec{R}_{mathrm{P}}^{A}=left[ Q_{mathrm{B}}^{A}
ight] vec{R}_{mathrm{P}}^{B}+vec{R}_{mathrm{B}}^{A}
R
对于固定坐标系下同一
{
A
}
,
{
B
}
left{ A
ight} ,left{ B
ight}
{A},{B}下进行表达,存在如下转换关系:
R
?
P
A
+
λ
R
?
V
e
c
t
o
r
A
=
[
Q
B
A
]
R
?
P
B
+
R
?
B
A
+
λ
[
Q
B
A
]
R
?
V
e
c
t
o
r
B
=
[
Q
B
A
]
(
R
?
P
B
+
λ
R
?
V
e
c
t
o
r
B
)
+
R
?
B
A
vec{R}_{mathrm{P}}^{A}+lambda vec{R}_{mathrm{Vector}}^{A}=left[ Q_{mathrm{B}}^{A}
ight] vec{R}_{mathrm{P}}^{B}+vec{R}_{mathrm{B}}^{A}+lambda left[ Q_{mathrm{B}}^{A}
ight] vec{R}_{mathrm{Vector}}^{B}=left[ Q_{mathrm{B}}^{A}
ight] left( vec{R}_{mathrm{P}}^{B}+lambda vec{R}_{mathrm{Vector}}^{B}
ight) +vec{R}_{mathrm{B}}^{A}
R
R
?
P
A
+
λ
1
R
?
V
e
c
t
o
r
1
A
+
λ
2
R
?
V
e
c
t
o
r
2
A
=
[
Q
B
A
]
R
?
P
B
+
R
?
B
A
+
λ
1
[
Q
B
A
]
R
?
V
e
c
t
o
r
1
B
+
λ
2
[
Q
B
A
]
R
?
V
e
c
t
o
r
2
B
=
[
Q
B
A
]
(
R
?
P
B
+
λ
1
R
?
V
e
c
t
o
r
1
B
+
λ
2
R
?
V
e
c
t
o
r
2
B
)
+
R
?
B
A
vec{R}_{mathrm{P}}^{A}+lambda _1vec{R}_{mathrm{Vector}_1}^{A}+lambda _2vec{R}_{mathrm{Vector}_2}^{A}=left[ Q_{mathrm{B}}^{A}
ight] vec{R}_{mathrm{P}}^{B}+vec{R}_{mathrm{B}}^{A}+lambda _1left[ Q_{mathrm{B}}^{A}
ight] vec{R}_{mathrm{Vector}_1}^{B}+lambda _2left[ Q_{mathrm{B}}^{A}
ight] vec{R}_{mathrm{Vector}_2}^{B}=left[ Q_{mathrm{B}}^{A}
ight] left( vec{R}_{mathrm{P}}^{B}+lambda _1vec{R}_{mathrm{Vector}_1}^{B}+lambda _2vec{R}_{mathrm{Vector}_2}^{B}
ight) +vec{R}_{mathrm{B}}^{A}
R
3.8.1 线的特征
- 线的单位方向
l
?
vec{l}
l
已知平面上存在点P
1
(
x
1
,
y
1
,
z
1
)
,
P
2
(
x
2
,
y
2
,
z
2
)
P_1left( x_1,y_1,z_1
ight) ,P_2left( x_2,y_2,z_2
ight)P1?(x1?,y1?,z1?),P2?(x2?,y2?,z2?),则其单位方向向量
l
?
vec{l}
l
为: l
?
=
P
1
P
2
?
∣
P
1
P
2
?
∣
=
(
x
2
?
x
1
)
i
?
+
(
y
2
?
y
1
)
j
?
+
(
z
2
?
z
1
)
k
?
(
x
2
?
x
1
)
2
+
(
y
2
?
y
1
)
2
+
(
z
2
?
z
1
)
2
vec{l}=frac{overrightharpoon{P_1P_2}}{left| overrightharpoon{P_1P_2}
ight|}=frac{left( x_2-x_1
ight) vec{i}+left( y_2-y_1
ight) vec{j}+left( z_2-z_1
ight) vec{k}}{sqrt{left( x_2-x_1
ight) ^2+left( y_2-y_1
ight) ^2+left( z_2-z_1
ight) ^2}}l
=
?P1?P2?
?
?P1?P2?
??=(x2??x1?)2+(y2??y1?)2+(z2??z1?)2
?(x2??x1?)i
+(y2??y1?)j
?+(z2??z1?)k
? - 线的姿态参数
(
θ
,
?
)
left( heta ,phi
ight)(θ,?)
(见1.2.3)
已知平面上存在点P
1
(
x
1
,
y
1
,
z
1
)
,
P
2
(
x
2
,
y
2
,
z
2
)
P_1left( x_1,y_1,z_1
ight) ,P_2left( x_2,y_2,z_2
ight)P1?(x1?,y1?,z1?),P2?(x2?,y2?,z2?),则其
球坐标 系姿态角(
θ
,
?
)
left( heta ,phi
ight)(θ,?), 为:
R
?
P
1
P
2
F
=
(
x
2
?
x
1
)
i
?
+
(
y
2
?
y
1
)
j
?
+
(
z
2
?
z
1
)
k
?
=
∣
P
1
P
2
?
∣
(
cos
?
?
sin
?
θ
i
?
+
sin
?
?
sin
?
θ
j
?
+
cos
?
?
k
?
)
?
{
?
=
a
r
c
cos
?
(
z
2
?
z
1
∣
P
1
P
2
?
∣
)
θ
=
a
r
c
sin
?
(
(
x
2
?
x
1
)
2
+
(
y
2
?
y
1
)
2
)
?
π
2
(
y
2
?
y
1
∣
y
2
?
y
1
∣
?
1
)
∣
P
1
P
2
?
∣
vec{R}_{P_1P_2}^{F}=left( x_2-x_1
ight) vec{i}+left( y_2-y_1
ight) vec{j}+left( z_2-z_1
ight) vec{k}=left| overrightharpoon{P_1P_2}
ight|left( cos phi sin heta vec{i}+sin phi sin heta vec{j}+cos phi vec{k}
ight) \ Rightarrow egin{cases} phi =mathrm{arc}cos left( frac{z_2-z_1}{left| overrightharpoon{P_1P_2}
ight|}
ight)\ heta =mathrm{arc}sin frac{left( sqrt{left( x_2-x_1
ight) ^2+left( y_2-y_1
ight) ^2}
ight) -frac{pi}{2}left( frac{y_2-y_1}{left| y_2-y_1
ight|}-1
ight)}{left| overrightharpoon{P_1P_2}
ight|}\ end{cases}R
P1?P2?F?=(x2??x1?)i
+(y2??y1?)j
?+(z2??z1?)k
=
?P1?P2?
?
?(cos?sinθi
+sin?sinθj
?+cos?k
)??
?
???=arccos
?
?P1?P2?
?
?z2??z1??
?θ=arcsin
?P1?P2?
?
?((x2??x1?)2+(y2??y1?)2
?)?2π?(∣y2??y1?∣y2??y1???1)??
3.8.2 面的特征
-
法矢量
n
?
vec{n}
n
已知平面上存在点P
1
(
x
1
,
y
1
,
z
1
)
,
P
2
(
x
2
,
y
2
,
z
2
)
,
P
3
(
x
3
,
y
3
,
z
3
)
P_1left( x_1,y_1,z_1
ight) ,P_2left( x_2,y_2,z_2
ight) ,P_3left( x_3,y_3,z_3
ight)P1?(x1?,y1?,z1?),P2?(x2?,y2?,z2?),P3?(x3?,y3?,z3?), 则其法矢量
n
?
vec{n}
n
为: n
?
=
P
1
P
2
?
×
P
1
P
3
?
=
∣
i
?
j
?
k
?
x
2
?
x
1
y
2
?
y
1
z
2
?
z
1
x
3
?
x
1
y
3
?
y
1
z
3
?
z
1
∣
=
a
i
?
+
b
j
?
+
c
k
?
;
n
?
(
a
,
b
,
c
)
{
a
=
(
y
2
?
y
1
)
(
z
3
?
z
1
)
?
(
y
3
?
y
1
)
(
z
2
?
z
1
)
b
=
(
z
2
?
z
1
)
(
x
3
?
x
1
)
?
(
z
3
?
z
1
)
(
x
2
?
x
1
)
c
=
(
x
2
?
x
1
)
(
y
3
?
y
1
)
?
(
x
3
?
x
1
)
(
y
2
?
y
1
)
??
vec{n}=overrightharpoon{P_1P_2} imes overrightharpoon{P_1P_3}=left| egin{matrix} vec{i}& vec{j}& vec{k}\ x_2-x_1& y_2-y_1& z_2-z_1\ x_3-x_1& y_3-y_1& z_3-z_1\ end{matrix}
ight|=avec{i}+bvec{j}+cvec{k};vec{n}left( a,b,c
ight) \ egin{cases} a=left( y_2-y_1
ight) left( z_3-z_1
ight) -left( y_3-y_1
ight) left( z_2-z_1
ight)\ b=left( z_2-z_1
ight) left( x_3-x_1
ight) -left( z_3-z_1
ight) left( x_2-x_1
ight)\ c=left( x_2-x_1
ight) left( y_3-y_1
ight) -left( x_3-x_1
ight) left( y_2-y_1
ight) ,,\ end{cases}n
=P1?P2?
?×P1?P3?
?=
?i
x2??x1?x3??x1??j
?y2??y1?y3??y1??k
z2??z1?z3??z1??
?=ai
+bj
?+ck
;n
(a,b,c)?
?
??a=(y2??y1?)(z3??z1?)?(y3??y1?)(z2??z1?)b=(z2??z1?)(x3??x1?)?(z3??z1?)(x2??x1?)c=(x2??x1?)(y3??y1?)?(x3??x1?)(y2??y1?)? -
平面的姿态参数
已知平面上存在点P
1
(
x
1
,
y
1
,
z
1
)
,
P
2
(
x
2
,
y
2
,
z
2
)
,
P
3
(
x
3
,
y
3
,
z
3
)
P_1left( x_1,y_1,z_1
ight) ,P_2left( x_2,y_2,z_2
ight) ,P_3left( x_3,y_3,z_3
ight)P1?(x1?,y1?,z1?),P2?(x2?,y2?,z2?),P3?(x3?,y3?,z3?), 令
i
?
M
=
P
1
P
2
?
∣
P
1
P
2
?
∣
=
(
x
2
?
x
1
)
i
?
+
(
y
2
?
y
1
)
j
?
+
(
z
2
?
z
1
)
k
?
(
x
2
?
x
1
)
2
+
(
y
2
?
y
1
)
2
+
(
z
2
?
z
1
)
2
vec{i}^M=frac{overrightharpoon{P_1P_2}}{left| overrightharpoon{P_1P_2}
ight|}=frac{left( x_2-x_1
ight) vec{i}+left( y_2-y_1
ight) vec{j}+left( z_2-z_1
ight) vec{k}}{sqrt{left( x_2-x_1
ight) ^2+left( y_2-y_1
ight) ^2+left( z_2-z_1
ight) ^2}}i
M=
?P1?P2?
?
?P1?P2?
??=(x2??x1?)2+(y2??y1?)2+(z2??z1?)2
?(x2??x1?)i
+(y2??y1?)j
?+(z2??z1?)k
?, k
?
M
=
a
i
?
F
+
b
j
?
F
+
c
k
?
F
a
2
+
b
2
+
c
2
,
{
a
=
(
y
2
?
y
1
)
(
z
3
?
z
1
)
?
(
y
3
?
y
1
)
(
z
2
?
z
1
)
b
=
(
z
2
?
z
1
)
(
x
3
?
x
1
)
?
(
z
3
?
z
1
)
(
x
2
?
x
1
)
c
=
(
x
2
?
x
1
)
(
y
3
?
y
1
)
?
(
x
3
?
x
1
)
(
y
2
?
y
1
)
??
vec{k}^M=frac{avec{i}^F+bvec{j}^F+cvec{k}^F}{sqrt{a^2+b^2+c^2}},egin{cases} a=left( y_2-y_1
ight) left( z_3-z_1
ight) -left( y_3-y_1
ight) left( z_2-z_1
ight)\ b=left( z_2-z_1
ight) left( x_3-x_1
ight) -left( z_3-z_1
ight) left( x_2-x_1
ight)\ c=left( x_2-x_1
ight) left( y_3-y_1
ight) -left( x_3-x_1
ight) left( y_2-y_1
ight) ,,\ end{cases}k
M=a2+b2+c2
?ai
F+bj
?F+ck
F?,?
?
??a=(y2??y1?)(z3??z1?)?(y3??y1?)(z2??z1?)b=(z2??z1?)(x3??x1?)?(z3??z1?)(x2??x1?)c=(x2??x1?)(y3??y1?)?(x3??x1?)(y2??y1?)?, 根据笛卡尔坐标系的基矢量转换关系: j
?
M
=
k
?
M
×
i
?
M
vec{j}^M=vec{k}^M imes vec{i}^M
j
?M=k
M×i
M
可得:[
i
?
M
j
?
M
k
?
M
]
=
[
Q
F
M
]
[
i
?
F
j
?
F
k
?
F
]
;
[
Q
M
F
]
=
[
Q
F
M
]
T
=
[
q
11
q
12
q
13
q
21
q
22
q
23
q
31
q
32
q
33
]
left[ egin{array}{c} vec{i}^M\ vec{j}^M\ vec{k}^M\ end{array}
ight] =left[ Q_{mathrm{F}}^{M}
ight] left[ egin{array}{c} vec{i}^F\ vec{j}^F\ vec{k}^F\ end{array}
ight] ;left[ Q_{mathrm{M}}^{F}
ight] =left[ Q_{mathrm{F}}^{M}
ight] ^{mathrm{T}}=left[ egin{matrix} q_{11}& q_{12}& q_{13}\ q_{21}& q_{22}& q_{23}\ q_{31}& q_{32}& q_{33}\ end{matrix}
ight]?i
Mj
?Mk
M?
?=[QFM?]
?i
Fj
?Fk
F?
?;[QMF?]=[QFM?]T=
?q11?q21?q31??q12?q22?q32??q13?q23?q33??
?
将该矩阵内的元素带入上述小节中对应的转换关系,即可得到对应表达下的姿态参数。
3.9 简单的示例与计算