Figure shows the schematic representation of the robot( 1 , 1 ) (1,1) ( 1 , 1 )
【考點1】Table of parameters of (1,1) robot
wheel
L
α \alpha α d
β \beta β γ \gamma γ φ \varphi φ Φ = α + β + γ \Phi=\alpha+\beta+\gamma Φ = α + β + γ
1 f 1f 1 f a
− π 2 -\frac{\pi}{2} − 2 π 0
0
π 2 \frac{\pi}{2} 2 π φ 1 f \varphi_{1f} φ 1 f 0
2 f 2f 2 f a
π 2 \frac{\pi}{2} 2 π 0
0
− π 2 -\frac{\pi}{2} − 2 π φ 2 f \varphi_{2f} φ 2 f 0
3 s 3s 3 s b
0
0
β 3 s \beta_{3s} β 3 s 0
φ 3 s \varphi_{3s} φ 3 s β 3 s \beta_{3s} β 3 s
【考點2】configuration vector q q = [ x , y , θ , β 3 s , φ 1 f , φ 2 f , φ 3 s ] T q=\left[x, y, \theta, \beta_{3 s}, \varphi_{1 f}, \varphi_{2 f}, \varphi_{3 s}\right]^{T} q = [ x , y , θ , β 3 s , φ 1 f , φ 2 f , φ 3 s ] T
【考點3】J,C矩陣求解 ( v t = 0 ) (v_t =0) ( v t = 0 ) J 1 = [ 1 0 a 1 0 − a cos β 3 c sin β 3 c b sin β 3 c ] J_{1}=\left[\begin{array}{ccc}
1 & 0 & a \\
1 & 0 & -a \\
\cos \beta_{3 c} & \sin \beta_{3 c} & b \sin \beta_{3 c}
\end{array}\right] J 1 = ⎣ ⎡ 1 1 cos β 3 c 0 0 sin β 3 c a − a b sin β 3 c ⎦ ⎤ J 2 = − r I 3 × 3 J_2 = -rI_{3\times 3} J 2 = − r I 3 × 3
( v n = 0 ) (v_n =0) ( v n = 0 ) C 1 = [ 0 1 0 0 1 0 − sin β 3 c cos β 3 c b cos β 3 c ] C_{1}=\left[\begin{array}{ccc}
0 & 1 & 0 \\
0 & 1 & 0 \\
-\sin \beta_{3 c} & \cos \beta_{3 c} & b \cos \beta_{3 c}
\end{array}\right] C 1 = ⎣ ⎡ 0 0 − sin β 3 c 1 1 cos β 3 c 0 0 b cos β 3 c ⎦ ⎤ C 2 = [ ] C_2 = [\quad] C 2 = [ ] 【考點4】Degree of mobility C 1 ∗ = [ C 1 f C 1 s ] = = [ 0 1 0 0 1 0 − sin β 3 c cos β 3 c b cos β 3 c ] C_{1}^{*}=\left[\begin{array}{c}
C_{1 f} \\
C_{1 s}
\end{array}\right]==\left[\begin{array}{ccc}
0 & 1 & 0 \\
0 & 1 & 0 \\
-\sin \beta_{3 c} & \cos \beta_{3 c} & b \cos \beta_{3 c}
\end{array}\right] C 1 ∗ = [ C 1 f C 1 s ] = = ⎣ ⎡ 0 0 − sin β 3 c 1 1 cos β 3 c 0 0 b cos β 3 c ⎦ ⎤ δ m = d i m ( K e r ( C 1 ∗ ) ) = 3 − r a n k ( C 1 ∗ ) = 3 − 2 = 1 \delta_m = dim(Ker(C_{1}^{*}))=3-rank(C_{1}^{*})=3-2 =1 δ m = d i m ( K e r ( C 1 ∗ ) ) = 3 − r a n k ( C 1 ∗ ) = 3 − 2 = 1 【易錯點】Professor:As I have already pointed out after continuous evaluation , you have to justify that the rank of C1* is 2 whatever β 3 s \beta_{3s} β 3 s
C 1 ⋅ m Ω 0 ( θ ) ⋅ ξ ˙ = C 1 ⋅ m ξ ˙ = 0 C_{1} \cdot^{m} \Omega_{0}(\theta) \cdot \dot{\xi}=C_{1} \cdot m \dot{\xi}=0 C 1 ⋅ m Ω 0 ( θ ) ⋅ ξ ˙ = C 1 ⋅ m ξ ˙ = 0 m y ˙ = 0 ^m\dot y = 0 m y ˙ = 0 − s i n ( β 3 s ) m x ˙ + 0 + b c o s ( β 3 c ) m θ ˙ = 0 -sin(\beta_{3s})^m\dot x + 0 + bcos(\beta_{3c})^m\dot \theta = 0 − s i n ( β 3 s ) m x ˙ + 0 + b c o s ( β 3 c ) m θ ˙ = 0 ∑ = [ 1 s i n ( β 3 s ) , 0 , 1 b c o s ( β 3 s ) ] T \sum = [\frac{1}{sin(\beta_{3s})},0,\frac{1}{bcos(\beta_{3s})}]^T ∑ = [ s i n ( β 3 s ) 1 , 0 , b c o s ( β 3 s ) 1 ] T
u s = β s ˙ u_s =\dot { \beta_s} u s = β s ˙ u m u_m u m ξ \xi ξ 0 Ω m ( θ ) = [ cos θ − sin θ 0 sin θ cos θ 0 0 0 1 ] ^0\Omega_m(\theta) = \left[\begin{array}{ccc}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{array}\right] 0 Ω m ( θ ) = ⎣ ⎡ cos θ sin θ 0 − sin θ cos θ 0 0 0 1 ⎦ ⎤ ∑ = [ 1 s i n ( β 3 s ) , 0 , 1 b c o s ( β 3 s ) ] T \sum = [\frac{1}{sin(\beta_{3s})},0,\frac{1}{bcos(\beta_{3s})}]^T ∑ = [ s i n ( β 3 s ) 1 , 0 , b c o s ( β 3 s ) 1 ] T ∑ \sum ∑ ξ = [ cos θ s i n ( β 3 s ) sin θ s i n ( β 3 s ) 1 b c o s ( β 3 s ) ] u m \xi=\left[\begin{array}{c}
\frac{\cos \theta}{sin(\beta_{3s})} \\
\frac{\sin \theta}{sin(\beta_{3s})} \\
\frac{1}{bcos(\beta_{3s})}
\end{array}\right] u_m ξ = ⎣ ⎢ ⎡ s i n ( β 3 s ) cos θ s i n ( β 3 s ) sin θ b c o s ( β 3 s ) 1 ⎦ ⎥ ⎤ u m
【考點4】D,E矩陣求解 D = [ ] D = [\quad] D = [ ] S ( q ) S(q) S ( q ) E ( β s , β c ) = − J 2 − 1 ⋅ J 1 ( β s , β c ) = − ( − r I 3 × 3 ) − 1 J i = 1 r J 1 ( β s , β c ) \mathbf{E}\left(\beta_{s}, \beta_{c}\right)=-\mathbf{J}_{2}^{-1} \cdot \mathbf{J}_{1}\left(\beta_{s}, \beta_{c}\right)=-(-rI_{3\times 3 })^{-1}J_i=\frac{1}{r}J1(\beta_{s}, \beta_{c}) E ( β s , β c ) = − J 2 − 1 ⋅ J 1 ( β s , β c ) = − ( − r I 3 × 3 ) − 1 J i = r 1 J 1 ( β s , β c ) E = 1 r [ 1 0 a 1 0 − a cos β 3 c sin β 3 c b sin β 3 c ] E =\frac{1}{r}\left[\begin{array}{ccc}
1 & 0 & a \\
1 & 0 & -a \\
\cos \beta_{3 c} & \sin \beta_{3 c} & b \sin \beta_{3 c}
\end{array}\right] E = r 1 ⎣ ⎡ 1 1 cos β 3 c 0 0 sin β 3 c a − a b sin β 3 c ⎦ ⎤
For the work to be complete, you need to study alternative motorization, determine possible singularities and, if any, interpret what happens physically at the singularity.
E ∑ = 1 r [ 1 0 a 1 0 − a cos β 3 c sin β 3 c b sin β 3 c ] [ 1 s i n ( β 3 s ) , 0 , 1 b c o s ( β 3 s ) ] T = 1 r [ b cos β 3 s + a sin β 3 s b cos β 3 s − a sin β 3 s b ] E\sum =\frac{1}{r}\left[\begin{array}{ccc}
1 & 0 & a \\
1 & 0 & -a \\
\cos \beta_{3 c} & \sin \beta_{3 c} & b \sin \beta_{3 c}
\end{array}\right] [\frac{1}{sin(\beta_{3s})},0,\frac{1}{bcos(\beta_{3s})}]^T=\frac{1}{r}\left[\begin{array}{c}
b\cos\beta_{3s}+a\sin\beta_{3s} \\
b\cos\beta_{3s}-a\sin\beta_{3s} \\
b
\end{array}\right] E ∑ = r 1 ⎣ ⎡ 1 1 cos β 3 c 0 0 sin β 3 c a − a b sin β 3 c ⎦ ⎤ [ s i n ( β 3 s ) 1 , 0 , b c o s ( β 3 s ) 1 ] T = r 1 ⎣ ⎡ b cos β 3 s + a sin β 3 s b cos β 3 s − a sin β 3 s b ⎦ ⎤
【考點5】wheel motorization [ φ ˙ 1 f φ ˙ 2 f φ ˙ 3 s ] = 1 r [ b cos β 3 s + a sin β 3 s b cos β 3 s − a sin β 3 s b ] \left[\begin{array}{c}
\dot \varphi_{1f} \\
\dot \varphi_{2f} \\
\dot \varphi_{3s}
\end{array}\right]=\frac{1}{r}\left[\begin{array}{c}
b\cos\beta_{3s}+a\sin\beta_{3s} \\
b\cos\beta_{3s}-a\sin\beta_{3s} \\
b
\end{array}\right] ⎣ ⎡ φ ˙ 1 f φ ˙ 2 f φ ˙ 3 s ⎦ ⎤ = r 1 ⎣ ⎡ b cos β 3 s + a sin β 3 s b cos β 3 s − a sin β 3 s b ⎦ ⎤
motorizing φ 1 f \varphi_{1f} φ 1 f
motorizing φ 2 f \varphi_{2f} φ 2 f
motorizing φ 3 s \varphi_{3s} φ 3 s
[ β ˙ 3 s φ ˙ 1 f ] = [ 0 1 b r cos β 3 s + a r sin β 3 s 0 ] [ u s u m ] \left[\begin{array}{c}
\dot \beta_{3s} \\
\dot{\varphi}_{1f}
\end{array}\right]=\left[\begin{array}{cc}
0 & 1 \\
\frac{b}{r}\cos\beta_{3s}+\frac{a}{r}\sin\beta_{3s} &0
\end{array}\right]\left[\begin{array}{c}
u_{s} \\
u_{m}
\end{array}\right] [ β ˙ 3 s φ ˙ 1 f ] = [ 0 r b cos β 3 s + r a sin β 3 s 1 0 ] [ u s u m ]
[ β ˙ 3 s φ ˙ 2 f ] = [ 0 1 b r cos β 3 s − a r sin β 3 s 0 ] [ u s u m ] \left[\begin{array}{c}
\dot \beta_{3s} \\
\dot{\varphi}_{2f}
\end{array}\right]=\left[\begin{array}{cc}
0 & 1 \\
\frac{b}{r}\cos\beta_{3s}-\frac{a}{r}\sin\beta_{3s} &0
\end{array}\right]\left[\begin{array}{c}
u_{s} \\
u_{m}
\end{array}\right] [ β ˙ 3 s φ ˙ 2 f ] = [ 0 r b cos β 3 s − r a sin β 3 s 1 0 ] [ u s u m ]
[ β ˙ 3 s φ ˙ 3 s ] = [ 0 1 1 r b 0 ] [ u s u m ] \left[\begin{array}{c}
\dot \beta_{3s} \\
\dot{\varphi}_{3s}
\end{array}\right]=\left[\begin{array}{cc}
0 & 1 \\
\frac{1}{r}b&0
\end{array}\right]\left[\begin{array}{c}
u_{s} \\
u_{m}
\end{array}\right] [ β ˙ 3 s φ ˙ 3 s ] = [ 0 r 1 b 1 0 ] [ u s u m ]
Interpret what happens at the singularity using the ICR and blocking joints selectively.
【考點6】Singularity
今日寫文背景音樂《時代號子》畢竟是勞動節
力量攥在手，夢想在前頭
，是已經知道的，我們要看看與之間的關係。
【考點4】D,E矩陣求解
第三種情況由於沒有半徑無窮大的輪子，其實奇異情況是不存在的。之後到底什麼纔是好的驅動方式，交給以後生活中的實踐吧（