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点/向量
,在不同坐标系
{
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}
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{
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\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
对于固定坐标系下同一线/面
,在不同坐标系
{
A
}
,
{
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{P}}^{A}+\lambda \vec{R}_{\mathrm{Vector}}^{A}=\left[ Q_{\mathrm{B}}^{A} \right] \vec{R}_{\mathrm{P}}^{B}+\vec{R}_{\mathrm{B}}^{A}+\lambda \left[ Q_{\mathrm{B}}^{A} \right] \vec{R}_{\mathrm{Vector}}^{B}=\left[ Q_{\mathrm{B}}^{A} \right] \left( \vec{R}_{\mathrm{P}}^{B}+\lambda \vec{R}_{\mathrm{Vector}}^{B} \right) +\vec{R}_{\mathrm{B}}^{A}
RPA+λRVectorA=[QBA]RPB+RBA+λ[QBA]RVectorB=[QBA](RPB+λRVectorB)+RBA
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\vec{R}_{\mathrm{P}}^{A}+\lambda _1\vec{R}_{\mathrm{Vector}_1}^{A}+\lambda _2\vec{R}_{\mathrm{Vector}_2}^{A}=\left[ Q_{\mathrm{B}}^{A} \right] \vec{R}_{\mathrm{P}}^{B}+\vec{R}_{\mathrm{B}}^{A}+\lambda _1\left[ Q_{\mathrm{B}}^{A} \right] \vec{R}_{\mathrm{Vector}_1}^{B}+\lambda _2\left[ Q_{\mathrm{B}}^{A} \right] \vec{R}_{\mathrm{Vector}_2}^{B}=\left[ Q_{\mathrm{B}}^{A} \right] \left( \vec{R}_{\mathrm{P}}^{B}+\lambda _1\vec{R}_{\mathrm{Vector}_1}^{B}+\lambda _2\vec{R}_{\mathrm{Vector}_2}^{B} \right) +\vec{R}_{\mathrm{B}}^{A}
RPA+λ1RVector1A+λ2RVector2A=[QBA]RPB+RBA+λ1[QBA]RVector1B+λ2[QBA]RVector2B=[QBA](RPB+λ1RVector1B+λ2RVector2B)+RBA
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_1\left( x_1,y_1,z_1 \right) ,P_2\left( x_2,y_2,z_2 \right) 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} \right|}=\frac{\left( x_2-x_1 \right) \vec{i}+\left( y_2-y_1 \right) \vec{j}+\left( z_2-z_1 \right) \vec{k}}{\sqrt{\left( x_2-x_1 \right) ^2+\left( y_2-y_1 \right) ^2+\left( z_2-z_1 \right) ^2}} l= P1P2 P1P2=(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
,
ϕ
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\left( \theta ,\phi \right)
(θ,ϕ)(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_1\left( x_1,y_1,z_1 \right) ,P_2\left( x_2,y_2,z_2 \right) P1(x1,y1,z1),P2(x2,y2,z2),则其球坐标
系姿态角 ( θ , ϕ ) \left( \theta ,\phi \right) (θ,ϕ), 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 ϕ sin θ i ⃗ + sin ϕ sin θ j ⃗ + cos ϕ k ⃗ ) ⇒ { ϕ = 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 \right) \vec{i}+\left( y_2-y_1 \right) \vec{j}+\left( z_2-z_1 \right) \vec{k}=\left| \overrightharpoon{P_1P_2} \right|\left( \cos \phi \sin \theta \vec{i}+\sin \phi \sin \theta \vec{j}+\cos \phi \vec{k} \right) \\ \Rightarrow \begin{cases} \phi =\mathrm{arc}\cos \left( \frac{z_2-z_1}{\left| \overrightharpoon{P_1P_2} \right|} \right)\\ \theta =\mathrm{arc}\sin \frac{\left( \sqrt{\left( x_2-x_1 \right) ^2+\left( y_2-y_1 \right) ^2} \right) -\frac{\pi}{2}\left( \frac{y_2-y_1}{\left| y_2-y_1 \right|}-1 \right)}{\left| \overrightharpoon{P_1P_2} \right|}\\ \end{cases} RP1P2F=(x2−x1)i+(y2−y1)j+(z2−z1)k= P1P2 (cosϕsinθi+sinϕsinθj+cosϕk)⇒⎩ ⎨ ⎧ϕ=arccos P1P2 z2−z1 θ=arcsin P1P2 ((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_1\left( x_1,y_1,z_1 \right) ,P_2\left( x_2,y_2,z_2 \right) ,P_3\left( x_3,y_3,z_3 \right) 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}\times \overrightharpoon{P_1P_3}=\left| \begin{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} \right|=a\vec{i}+b\vec{j}+c\vec{k};\vec{n}\left( a,b,c \right) \\ \begin{cases} a=\left( y_2-y_1 \right) \left( z_3-z_1 \right) -\left( y_3-y_1 \right) \left( z_2-z_1 \right)\\ b=\left( z_2-z_1 \right) \left( x_3-x_1 \right) -\left( z_3-z_1 \right) \left( x_2-x_1 \right)\\ c=\left( x_2-x_1 \right) \left( y_3-y_1 \right) -\left( x_3-x_1 \right) \left( y_2-y_1 \right) \,\,\\ \end{cases} n=P1P2×P1P3= ix2−x1x3−x1jy2−y1y3−y1kz2−z1z3−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_1\left( x_1,y_1,z_1 \right) ,P_2\left( x_2,y_2,z_2 \right) ,P_3\left( x_3,y_3,z_3 \right) 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} \right|}=\frac{\left( x_2-x_1 \right) \vec{i}+\left( y_2-y_1 \right) \vec{j}+\left( z_2-z_1 \right) \vec{k}}{\sqrt{\left( x_2-x_1 \right) ^2+\left( y_2-y_1 \right) ^2+\left( z_2-z_1 \right) ^2}} iM= P1P2 P1P2=(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{a\vec{i}^F+b\vec{j}^F+c\vec{k}^F}{\sqrt{a^2+b^2+c^2}},\begin{cases} a=\left( y_2-y_1 \right) \left( z_3-z_1 \right) -\left( y_3-y_1 \right) \left( z_2-z_1 \right)\\ b=\left( z_2-z_1 \right) \left( x_3-x_1 \right) -\left( z_3-z_1 \right) \left( x_2-x_1 \right)\\ c=\left( x_2-x_1 \right) \left( y_3-y_1 \right) -\left( x_3-x_1 \right) \left( y_2-y_1 \right) \,\,\\ \end{cases} kM=a2+b2+c2aiF+bjF+ckF,⎩ ⎨ ⎧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\times \vec{i}^M jM=kM×iM
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[ \begin{array}{c} \vec{i}^M\\ \vec{j}^M\\ \vec{k}^M\\ \end{array} \right] =\left[ Q_{\mathrm{F}}^{M} \right] \left[ \begin{array}{c} \vec{i}^F\\ \vec{j}^F\\ \vec{k}^F\\ \end{array} \right] ;\left[ Q_{\mathrm{M}}^{F} \right] =\left[ Q_{\mathrm{F}}^{M} \right] ^{\mathrm{T}}=\left[ \begin{matrix} q_{11}& q_{12}& q_{13}\\ q_{21}& q_{22}& q_{23}\\ q_{31}& q_{32}& q_{33}\\ \end{matrix} \right] iMjMkM =[QFM] iFjFkF ;[QMF]=[QFM]T= q11q21q31q12q22q32q13q23q33
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